信号与系统（第二版）英文版 (Simon Haykin 著) 电子工业出版社 课后答案 635页

'课后答案网：www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！课后答案网您最真诚的朋友www.hackshp.cn网团队竭诚为学生服务，免费提供各门课后答案，不用积分，甚至不用注册，旨在为广大学生提供自主学习的平台！课后答案网：www.hackshp.cn视频教程网：www.efanjy.comPPT课件网：www.ppthouse.com CHAPTER11.1to1.41-partoftext1.42(a)Periodic:Fundamentalperiod=0.5s(b)Nonperiodic(c)PeriodicFundamentalperiod=3s(d)PeriodicFundamentalperiod=2samples(e)Nonperiodic(f)Periodic:Fundamentalperiod=10samples(g)Nonperiodic(h)Nonperiodic(i)Periodic:Fundamentalperiod=1sample课后答案网2æöæöpl.43yt()=3cos200t+---èøèø62www.hackshp.cnp=9cosæö200t+---èø69æöp=---cos400t+---12èø39(a)DCcomponent=---29æöp(b)Sinusoidalcomponent=---cos400t+---2èø39Amplitude=---21 200Fundamentalfrequency=---------Hzp1.44TheRMSvalueofsinusoidalx(t)isA¤2.Hence,theaveragepowerofx(t)ina1-ohm22resistoris()A¤2=A/2.1.45LetNdenotethefundamentalperiodofx[N].whichisdefinedby2pN=------WTheaveragepowerofx[n]isthereforeN-112P=----åx[]nNn=0N-1122æö2pn=----åAcos----------+fNèøNn=02N-1A2æö2pn=------åcos----------+fNèøNn=01.46Theenergyoftheraisedcosinepulseis课后答案网pw¤12E=ò---()cos()wt+1td–pw¤41pw¤2www.hackshp.cn=---ò()cos()wt+2cos()wt+1td201pw¤æö11=---òèø---cos()2wt++---2cos()wt+1td20221---æö3---æö---p-==3p¤4w2èø2èøw1.47Thesignalx(t)iseven;itstotalenergyistherefore52E=2òx()ttd02 4252=21ò()dt+25ò()–ttd041543=2[]t+2–---()5–tt=03t=4226==8+---------331.48(a)Thedifferentiatoroutputisì1for–5<0)=10;x1(t>5)=0;%Generateshiftedfunctiondelay=1.5;x2=zeros(size(t));x2(t>(0+delay))=10;x2(t>(5+delay))=0;42 %Plotdatafigure(1);clf;plot(t,x1,’b’)holdonplot(t,x2,’r:’)xlabel(’Time(sec)’)ylabel(’Amplitude’)title(’RectangularPulse’)axis([-210-111])legend(’ZeroDelay’,’Delay=1.5’);RectangularPulseZeroDelayDelay=1.51086Amplitude42课后答案网0−2www.hackshp.cn0246810Time(sec)43 SolutionstoAdditionalProblems2.32.Adiscrete-timeLTIsystemhastheimpulseresponseh[n]depictedinFig.P2.32(a).Uselinear-ityandtimeinvariancetodeterminethesystemoutputy[n]iftheinputx[n]isUsethefactthat:δ[n−k]∗h[n]=h[n−k](ax1[n]+bx2[n])∗h[n]=ax1[n]∗h[n]+bx2[n]∗h[n](a)x[n]=3δ[n]−2δ[n−1]y[n]=3h[n]−2h[n−1]=3δ[n+1]+7δ[n]−7δ[n−2]+5δ[n−3]−2δ[n−4](b)x[n]=u[n+1]−u[n−3]x[n]=δ[n]+δ[n−1]+δ[n−2]y[n]=h[n]+h[n−1]+h[n−2]=δ[n+1]+4δ[n]+6δ[n−1]+4δ[n−2]+2δ[n−3]+δ[n−5](c)x[n]asgiveninFig.P2.32(b)课后答案网.x[n]=2δ[n−3]+2δwww.hackshp.cn[n]−δ[n+2]y[n]=2h[n−3]+2h[n]−h[n+2]=−δ[n+3]−3δ[n+2]+7δ[n]+3δ[n−1]+8δ[n−3]+4δ[n−4]−2δ[n−5]+2δ[n−6]2.33.Evaluatethediscrete-timeconvolutionsumsgivenbelow.(a)y[n]=u[n+3]∗u[n−3]Letu[n+3]=x[n]andu[n−3]=h[n]1 x[k]h[n−k]......−3−2−1kn−3kFigureP2.33.(a)Graphofx[k]andh[n−k]forn−3<−3n<0y[n]=0forn−3≥−3n≥0n−3y[n]=1=n+1k=−3n+1n≥0y[n]=0n<0(b)y[n]=3nu[−n+3]∗u[n−2]x[k]h[n−k]k3...课后答案网...123kn−2kwww.hackshp.cnFigureP2.33.(b)Graphofx[k]andh[n−k]forn−2≤3n≤5n−2y[n]=3kk=−∞1ny[n]=36forn−2≥4n≥63y[n]=3kk=−∞2 81y[n]=213nn≤5y[n]=681n≥62n(c)y[n]=1u[n]∗u[n+2]4x[k]h[n−k]k()14......12kn+2kFigureP2.33.(c)Graphofx[k]andh[n−k]forn+2<0n<−2y[n]=0forn+2≥0n≥−2n+2k1y[n]=4k=0n411y[n]=−3124n4−11n≥−2y[n]=3124课后答案网0n<−2(d)y[n]=cos(πn)u[n]∗u[n−1]2www.hackshp.cnforn−1<0n<1y[n]=0forn−1≥0n≥1n−1πy[n]=cosk2k=01n=4v+1,4v+2y[n]=0n=4v,4v+3y[n]=u[n−1]f[n]where1n=4v+1,4v+2f[n]=0n=4v,4v+33 (e)y[n]=(−1)n∗2nu[−n+2]∞y[n]=(−1)k2n−kk=n−2∞kn1=2−2k=n−21n−2−=2n21−−128n=(−1)3n(f)y[n]=cos(πn)∗1u[n−2]22n−2n−kπ1y[n]=cosk22k=−∞substitutingp=−k∞n+pπ1y[n]=cosp22p=−(n−2)∞p1n+p(−1)2nevenp=−(n−2)2y[n]=∞p1n+p(−1)2noddp=−(n−3)21(−1)nneveny[n]=5n+11(−1)nodd10(g)y[n]=βnu[n]课后答案网∗u[n−3],|β|<1forn−3<0n<3www.hackshp.cny[n]=0forn−3≥0n≥3n−3y[n]=βkk=01−βn−2y[n]=1−βn−21−βn≥3y[n]=1−β0n<3(h)y[n]=βnu[n]∗αnu[n−10],|β|<1,|α|<1forn−10<0n<104 y[n]=0forn−10≥0n≥10n−10kβny[n]=ααk=0βn−9n1−(α)αβα=βy[n]=1−αα(n−9)α=β(i)y[n]=(u[n+10]−2u[n]+u[n−4])∗u[n−2]forn−2<−10n<−8y[n]=0forn−2<0−8≤n<2n−2y[n]=1=n+9k=−10forn−2≤32≤n≤5−1n−2y[n]=1−1=11−nk=−10k=0forn−2≥4n≥6−13y[n]=1−1=6k=−10k=00n<−8n+9−8≤n<2y[n]=11−n−2≤n≤5课后答案网6n>5(j)y[n]=(u[n+10]−2u[n]+u[n−4])∗βnu[n],|β|<1forn3−1k3kn1n1y[n]=β−βββk=−10k=0βn+11−βn+1βn+1−βn−3y[n]=−β−1β−10n<−10n+11β−1−10≤n<0β−1y[n]=n+11n+1n+1β−β−β−10≤n≤3β−1β−1n+11n+1n+1n−3β−ββ−β−n>3β−1β−1(k)y[n]=(u[n+10]−2u[n+5]+u[n−6])∗cos(πn)2Therearefourdifferentcases:(i)n=4vvisanyintegery[n]=(1)[−1+0+1+0−1]+(−1)[0+1+0−1+0+1+0−1+0+1+0]=−2(ii)n=4v+2y[n]=(1)[1+0−1+0+1]+(−1)[0−1+0+1+0−1+0+1+0−1+0]=2(iii)n=4v+3y[n]=(1)[0−1+0+1+0]+(−1)[−1+0+1+0−1+0+1+0−1+0+1]=0(iv)n=4v+1y[n]=0−2n=4vy[n]=2n=4v+20otherwise∞(l)y[n]=u[n]∗δ[n−4p]课后答案网p=0forn<0www.hackshp.cny[n]=0forn≥0n=0,4,8,...ny[n]=+14forn≥0n=0,4,8,...ny[n]=4wherexisthesmallestintegerlargerthanx.Ex.3.2=4n∞(m)y[n]=βu[n]∗δ[n−4p],|β|<1p=0forn<06 y[n]=0forn≥0n=0,4,8,...n4y[n]=β4kk=04(n+1)1−β4y[n]=1−β4forn≥0n=1,5,9,...n−14y[n]=β4k−1k=0n−14(+1)11−β4y[n]=β1−β4forn≥0n=2,6,10,...n−24y[n]=β4k−2k=0n−24(+1)11−β4y[n]=β21−β4forn≥0n=3,7,11,...n−34y[n]=β4k−3k=0n−34(+1)11−β4y[n]=β31−β4n(n)y[n]=1u[n+2]∗γ|n|2课后答案网forn+2≤0n≤−2n+2n−k1−ky[n]=γwww.hackshp.cn2k=−∞nn+2−k1γy[n]=22k=−∞letl=−kn∞l1γy[n]=22l=−(n+2)nγ−(n+2)1y[n]=2γ21−2n212γy[n]=γγ1−2forn+2≥0n>−27 0n−kn+2n−k1−k1ky[n]=γ+γ22k=−∞k=1n0−knn+21γ1ky[n]=+(2γ)222k=−∞k=1nn+3111−(2γ)y[n]=γ+−121−1−2γ22.34.Considerthediscrete-timesignalsdepictedinFig.P2.34.Evaluatetheconvolutionsumsindi-catedbelow.(a)m[n]=x[n]∗z[n]forn+5<0n<−5m[n]=0forn+5<4−5≤n<−1n+5m[n]=1=n+6k=0forn−1<1−1≤n<23n+5m[n]=1+21=2n+8k=0k=4forn+5<92≤n<43n+5m[n]=1+21=9+nk=n−1k=4forn−1<44≤n<538m[n]=1+21=15−nk=n−1k=4课后答案网forn−1<95≤n<108m[n]=21=20−2nk=n−1forwww.hackshp.cnn−1≥9n≥10m[n]=00n<−5n+6−5≤n<−12n+8−1≤n<2m[n]=9+n2≤n<415−n4≤n<520−2n5≤n<100n≥10(b)m[n]=x[n]∗y[n]forn+5<−3n<−88 m[n]=0forn+5<1−8≤n<−4n+5m[n]=1=n+9k=−3forn−1<−2−4≤n<−10n+5m[n]=1−1=−n−1k=−3k=1forn+5<5−1≤n<00n+5m[n]=1−1=−2n−4k=n−1k=1forn−1<10≤n<204m[n]=1−1=−n−2k=n−1k=1forn−1<52≤n<64m[n]=−1=n−6k=n−1forn−1≥5n≥6m[n]=00n<−8n+3−8≤n<−4−n−1−4≤n<−1m[n]=−2n−4−1≤n<0−n−20≤n<2n−62≤n<60n≥6(c)m[n]=x[n]∗课后答案网f[n]forn+51课后答案网1y(t)=cos(πt)dτ−1y(t)=0www.hackshp.cn1sin(πt)−1≤t<1y(t)=π0otherwise(d)y(t)=(u(t+3)−u(t−1))∗u(−t+4)x[]h[t−]11...−31t−428 FigureP2.39.(d)Graphofx[τ]andh[t−τ]fort−4<−3t<11y(t)=dτ=4−3fort−4<1t<51y(t)=dτ=5−tt−4fort−4≥1t≥5y(t)=04t<1y(t)=5−t1≤t<50t≥5(e)y(t)=(tu(t)+(10−2t)u(t−5)−(10−t)u(t−10))∗u(t)fort<0y(t)=0for0≤t<5t12y(t)=τdτ=t02for5≤t<105t12y(t)=τdτ+(10−τ)dτ=−t+10t−25052fort≥10510y(t)=τdτ+(10−τ)dτ=2505课后答案网0t<01t20≤t<5y(t)=2−1t2+10t−255≤t<10www.hackshp.cn225t≥10(f)y(t)=2t2(u(t+1)−u(t−1))∗2u(t+2)fort+2<−1t<−3y(t)=0fort+2<1−3≤t<−1t+2243y(t)=22τdτ=(t+2)+1−13fort+2≥−1t≥−1128y(t)=22τdτ=−1329 0t<−3y(t)=4(t+2)3+1−3≤t<−138t≥−13(g)y(t)=cos(πt)(u(t+1)−u(t−1))∗(u(t+1)−u(t−1))fort+1<−1t<−2y(t)=0fort+1<1−2≤t<0t+11y(t)=cos(πt)dτ=sin(π(t+1))−1πfort−1<10≤t<211y(t)=cos(πt)dτ=−sin(π(t−1))t−1πfort−1≥1t≥2y(t)=00t<−21sin(π(t+1))−2≤t<0y(t)=π−1sin(π(t−1))0≤t<2π0t≥2(h)y(t)=cos(2πt)(u(t+1)−u(t−1))∗e−tu(t)fort<−1y(t)=0fort<1−1≤t<1t课后答案网y(t)=e−(t−τ)cos(2πt)dτ−1eτty(t)=e−t(cos(2πτ)+2πsin(2πτ))1+4π2www.hackshp.cn−1cos(2πt)+2πsin(2πt)−e−(t+1)y(t)=1+4π2fort≥1t≥11y(t)=e−(t−τ)cos(2πt)dτ−1eτ1y(t)=e−t(cos(2πτ)+2πsin(2πτ))1+4π2−1e−(t−1)−e−(t+1)y(t)=1+4π20t<−1−(t+1)y(t)=cos(2πt)+2πsin(2πt)−e1+4π2−1≤t<−1−(t−1)−(t+1)e−et≥11+4π230 (i)y(t)=(2δ(t+1)+δ(t−5))∗u(t−1)fort−1<−1t<0y(t)=0fort−1<50≤t<6Bythesiftingproperty.t−1y(t)=2δ(t+1)dτ=2−∞fort−1≥5t≥6t−1y(t)=(2δ(t+1)+δ(t−5))dτ=3−∞0t<0y(t)=20≤t<63t≥6(j)y(t)=(δ(t+2)+δ(t−2))∗(tu(t)+(10−2t)u(t−5)−(10−t)u(t−10))fort<−2y(t)=0fort<2−2≤t<2ty(t)=(t−τ)δ(τ+2)dτ=t+2t−10fort−5<−22≤t<3tty(t)=(t−τ)δ(τ+2)dτ+(t−τ)δ(τ−2)dτ=2tt−10t−10fort−5<23≤t<7tty(t)=[10−(t−τ)]δ(τ+2)dτ+(t−τ)δ(τ−2)dτ=6课后答案网t−10t−10fort−10<−27≤t<8tty(t)=[10−(t−τ)]δ(τ+2)dτ+[10−(t−τ)]δ(τ−2)dτ=20−2twww.hackshp.cnt−10t−10fort−10<28≤t<12ty(t)=[10−(t−τ)]δ(τ−2)dτ=12−tt−10fort−10≥2t≥12y(t)=00t<−2t+2−2≤t<22t2≤t<3y(t)=63≤t<720−2t7≤t<812−t8≤t<120t≥1231 (k)y(t)=e−γtu(t)∗(u(t+2)−u(t))fort<−2y(t)=0fort<0−2≤t<0ty(t)=e−γ(t−τ)dτ−21−γ(t+2)y(t)=1−eγfort≥00y(t)=e−γ(t−τ)dτ−21−γt−γ(t+2)y(t)=e−eγ0t<−2y(t)=11−e−γ(t+2)−2≤t<0γ1e−γt−e−γ(t+2)t≥0γ−γt∞1p(l)y(t)=eu(t)∗δ(t−2p)p=04fort<0y(t)=0fort≥0t∞p−γτ1y(t)=eδ(t−2p−τ)dτ04p=0∞pt1−γτy(t)=eδ(t−2p−τ)dτ40p=0课后答案网Usingthesiftingpropertyyields∞p1−γ(t−2p)y(t)=eu(t−2p)4www.hackshp.cnp=0for0≤t<2y(t)=e−γtfor2≤t<4−γt1−γ(t−2)y(t)=e+e4for4≤t<6−γt1−γ(t−2)1−γ(t−4)y(t)=e+e+e416for2l≤t<2l+2l1py(t)=e2pγe−γt4p=0∞1p(m)y(t)=(2δ(t)+δ(t−2))∗δ(t−p)p=0232 ∞p1letx1(t)=δ(t−p)2p=0fort<0y(t)=0fort<2y(t)=2δ(t)∗x1(t)=2x1(t)fort≥2y(t)=2δ(t)∗x1(t)+δ(t−2)∗x1(t)=2x1(t)+x1(t−2)0t<0∞1py(t)=2δ(t−p)0≤t<2p=02pp2∞1δ(t−p)+∞1δ(t−p−2)t≥0p=02p=02(n)y(t)=e−γtu(t)∗eβtu(−t)fort<0∞y(t)=eβte−(β+γ)τdτ0eβty(t)=β+γfort≥0∞y(t)=eβte−(β+γ)τdτteβty(t)=e−(β+γ)tβ+γe−γty(t)=β+γβtet<0y(t)=β+γ−γt课后答案网et≥0β+γe2tt<0(o)y(t)=u(t)∗h(t)whereh(t)=www.hackshp.cne−3tt≥0fort<0ty(t)=e2τdτ−∞12ty(t)=e2fort≥00ty(t)=e2τdτ+e−3τdτ−∞011y(t)=+1−e−3t231e2tt<0y(t)=21+11−e−3tt≥02333 2.40.Considerthecontinuous-timesignalsdepictedinFig.P2.40.Evaluatethefollowingconvolutionintegrals:.(a)m(t)=x(t)∗y(t)fort+1<0t<−1m(t)=0fort+1<2−1≤t<1t+1m(t)=dτ=t+10fort+1<41≤t<32t+1m(t)=dτ+2dτ=t+1t−12fort−1<43≤t<54m(t)=2dτ=10−2tt−1fort−1≥4t≥5m(t)=00t<−1t+1−1≤t<1m(t)=t+11≤t<310−2t3≤t<50t≥5(b)m(t)=x(t)∗z(t)课后答案网fort+1<−1t<−2m(t)=0fort+1<0−2≤t<−1www.hackshp.cnt+1m(t)=−dτ=−t−2−1fort+1<1−1≤t<00t+1m(t)=−dτ+dτ=t−10fort−1<00≤t<101m(t)=−dτ+dτ=tt−10fort−1<11≤t<21m(t)=dτ=2−tt−1fort−1≥1t≥2m(t)=034 0t<−2−t−2−2≤t<−1t−1≤t<0m(t)=t0≤t<12−t1≤t<20t≥2(c)m(t)=x(t)∗f(t)fort<−1m(t)=0fort<0−1≤t<0tm(t)=e−(t−τ)dτ=1−e−(t+1)−1fort<10≤t<1tm(t)=e−(t−τ)dτ=1−e−1t−1fort<21≤t<21m(t)=e−(t−τ)dτ=e1−t−e−1t−1fort≥2m(t)=00t<−11−e−(t+1)−1≤t<0m(t)=1−e−10≤t<1e1−t−e−11≤t<20t≥2(d)m(t)=x(t)∗课后答案网a(t)Byinspection,sincex(τ)hasawidthof2anda(t−τ)hastheperiod2anddutycycle1,theareaunder2theoverlappingsignalsisalways1,thuswww.hackshp.cnm(t)=1forallt.(e)m(t)=y(t)∗z(t)fort<−1m(t)=0fort<0−1≤t<0tm(t)=−dτ=−t−1−1fort<10≤t<10tm(t)=−dτ+dτ=−1+t−10fort<21≤t<235 t−201m(t)=−2dτ−dτ+dτ=−t+1−1t−20fort−2<12≤t<30t−21m(t)=−2dτ+2dτ+dτ=t−3−10t−2fort−4<03≤t<401m(t)=−2dτ+2dτ=2t−6t−40fort−4<14≤t<51m(t)=2dτ=10−2tt−4fort≥5m(t)=00t<−1−t−1−1≤t<0−1+t0≤t<1−t+11≤t<2m(t)=t−32≤t<32t−63≤t<410−2t4≤t<50t≥5(f)m(t)=y(t)∗w(t)fort<0m(t)=0fort<10≤t<1tm(t)=dτ=t课后答案网0fort<21≤t<21tm(t)=dτ−dτ=2−twww.hackshp.cn01fort<32≤t<3t−21tm(t)=2dτ+dτ−dτ=00t−21fort−4<03≤t<41t−23m(t)=2dτ−2dτ−dτ=3−t01t−2fort−4<14≤t<51t−23m(t)=2dτ−2dτ−dτ=11−3tt−41t−2fort−4<35≤t<73m(t)=−2dτ=2t−14t−4fort≥736 m(t)=00t<0t0≤t<12−t1≤t<202≤t<3m(t)=3−t3≤t<411−3t4≤t<52t−145≤t<70t≥7(g)m(t)=y(t)∗g(t)fort<−1m(t)=0fort<1−1≤t<1tm(t)=τdτ=0.5[t2−1]−1fort−2<11≤t<3t−21m(t)=2τdτ+τdτ=0.5t2+0.5(t−2)2−1−1t−2fort−4<13≤t<51m(t)=2τdτ=1−(t−4)2t−4fort≥5m(t)=00t<−10.5[t2−1]−1≤t<1m(t)=0.5t2+0.5(t−2)2−11≤t<3课后答案网1−(t−4)23≤t<50t≥5(h)m(t)=y(t)∗c(t)www.hackshp.cnfort+2<0t<−2m(t)=0fort+2<2−2≤t<0m(t)=1fort−1<00≤t<1tm(t)=dτ+2=t+20fort<21≤t<2tm(t)=dτ+2=3t−137 fort<32≤t<32tm(t)=−1+dτ+2dτ=t−2t−12fort<43≤t<4tm(t)=−1+2dτ=1t−2fort<54≤t<54m(t)=−2+2dτ=8−2tt−1fort−2<45≤t<6m(t)=−2fort≥6m(t)=00t<−21−2≤t<0t+20≤t<131≤t<2m(t)=t−22≤t<313≤t<48−2t5≤t<5−25≤t<60t≥6(i)m(t)=z(t)∗f(t)fort+1<0t<−1m(t)=0课后答案网fort+1<1−1≤t<0t+1m(t)=−e−τdτ=e−(t+1)−10fort0y(l,0)=x(l),00k=−JT12wealsohave|x(t)|dt>0T0J2AsJincreases:,|X[k]|increasesorremainsconstantandminMSEJdecreases.k=−J3.100.GeneralizedFourierSeries.TheconceptoftheFourierSeriesmaybegeneralizedtosumsofsignalsotherthancomplexsinusoids.Thatis,givenasignalx(t)wemayapproximatex(t)onaninterval[t1,t2]asaweightedsumofNfunctionsφ0(t),φ2(t),...,φN−1(t)N−1x(t)≈ckφk(t)k=0WeshallassumethattheseNfunctionsaremutuallyorthogonalon[t1,t2].Thismeansthatt20,k=l,φ(t)φ∗(t)dt=klt1fk,k=lTheMSEinusingthisapproximationis))N))21t2))MSE=)x(t)−ckφk(t))dtt2−t1t1)k=1)t12∗(a)Showthatthe课后答案网MSEisminimizedbychoosingck=fkt1x(t)φk(t)dt.Hint:Generalizestepsout-linedinProblem3.99(a)-(d)tothisproblem.)www.hackshp.cn)2)N)N−1N))2∗∗∗)x(t)−ckφk(t))=|x(t)|−x(t)ckφk(t)−x(t)ckφk(t)))k=1k=0k=0N−1N−1+cc∗φ(t)φ∗(t)kmkmm=0k=0Thus:t2N−1t212∗1∗MSE=|x(t)|dt−ckx(t)φk(t)dtt2−t1t1k=0t2−t1t1N−1t2N−1N−1t21∗∗1∗−ckx(t)φk(t)dt+ckcmφk(t)φm(t)dtt2−t1t1m=0t2−t1t1k=0k=0t21∗LetA[k]=x(t)φk(t)dtanduseorthogonalityfkt173 t2N−112fk∗MSE=|x(t)|dt−ckA[k]t2−t1t1t2−t1k=0fN−1fN−1−kcA∗[k]+k|c|2kkt2−t1t2−t1k=0k=01t2fN−1fN−12k2k2=|x(t)|dt+|ck−A[k]|−|A[k]|t2−t1t1t2−t1t2−t1k=0k=0t12AnalogoustoProblem3.99(d),c=A[k]=x(t)φ∗(t)dtkkfkt1(b)ShowthattheMSEiszeroift2N−1|x(t)|2dt=f|c|2kkt1k=0Ifthisrelationshipholdsforallx(t)inagivenclassoffunctions,thenthebasisfunctionsφ0(t),φ2(t),...,φN−1(t)issaidtobe“complete”forthatclass.t2N−112fk2minMSE=|x(t)|dt−|ck|t2−t1t1t2−t1k=01t2N−1=|x(t)|2dt−f|c|2kkt2−t1t1k=0minMSE=0when:t2N−1|x(t)|2dt=f|c|2kkt1k=0(c)TheWalshfunctionsareonesetoforthogonalfunctionsthatareusedforsignalrepresentationon[0,1].DeterminetheckandMSEobtainedbyapproximatingthefollowingsignalswiththefirstsixWalshfunctionsdepictedinFig.P3.100.SketchthesignalandtheWalshfunctionapproximation.课后答案网2,1T,andthat2thelimitasTapproachesinfinityistakeninasymmetricmanner.Definethefinitedurationnonperiodicsignalx(t)asoneperiodoftheTperiodicsignal˜x(t)x˜(t),−TT2(a)Graphanexampleofx(t)and˜x(t)todemonstratethatasTincreases,theperiodicreplicatesofx(t)in˜x(t)aremovedfartherandfartherawayfromtheorigin.Eventually,asTapproachesinfinity,thesereplicatesareremovedtoinfinity.Thus,wewritex(t)=limx˜(t)T→∞NoticehowasT→∞,x(t)→x˜(t)Plotofxhat(t),withT=15,x(t)=T/3−t543xhat(t)210−20−15−10课后答案网−505101520timePlotofxhat(t),withT=30,x(t)=T/3−t54www.hackshp.cn3xhat(t)210−40−30−20−10010203040timeFigureP3.101.Plot1of1(b)TheFSrepresentationfortheperiodicsignal˜x(t)is∞x˜(t)=X[k]ejkωotk=−∞T12X[k]=x˜(t)e−jkωotdtT−T278 ShowthatX[k]=1X(jkω)whereTo∞X(jω)=x(t)ejωtdt−∞T12X[k]=x˜(t)e−jkωotdtT−T2TTSince˜x(t)=x(t)for−T2ComparewiththeequationoftheFT,andonecansee,1X[k]=X(jkωo)T(c)SubstitutethisdefinitionforX[k]intotheexpressionfor˜x(t)in(b)andshowthat∞1jkωotx˜(t)=X(jkωo)eωo2πk=−∞∞1jkωotx˜(t)=X(jkωo)eTk=−∞2πsinceT=ωo∞1jkωot=X(jkωo)eωo2πk=−∞(d)Usethelimitingexpressionfor课后答案网x(t)in(a)anddefineω≈kωotoexpressthelimitingformofthesumin(c)astheintegral∞1jωtx(t)=X(jω)edωwww.hackshp.cn2π−∞x(t)=limx˜(t)T→∞∞1jkωot=limX(jkωo)eωoT→∞2πk=−∞letω=kωoasT→∞,ωo→0,k→∞,ω→∞implies:∞1jωtx(t)=limX(jω)eωoωo→02πω=−∞∞1jωt=X(jω)edω2π−∞79 SolutionstoComputerExperiments3.102.UseMATLABtorepeatExample3.7forN=50and(a)M=12.(b)M=5.(c)M=20.0.510[n]0.51xB[1]cos(omega*n)−0.50−20−1001020−20−1001020nn0.210.10[n]0.53x−0.1B[3]cos(omega*n)0−0.2−20−1001020−20−1001020nn0.110.050[n]0.55x−0.05B[5]cos(omega*n)−0.10−20−1001020−20−1001020nnFigureP3.102.Plot1of60.0410.80.020.6[n]023x0.4B[23]cos(omega*n)−0.02课后答案网0.2−0.040−20−1001020−20−1001020nwww.hackshp.cnn0.0210.80.010.6[n]025x0.4B[25]cos(omega*n)−0.010.2−0.020−20−1001020−20−1001020nnFigureP3.102.Plot2of680 0.40.60.20.40[n]0.21x−0.20B[1]cos(omega*n)−0.4−20−1001020−20−1001020nn10.10[n]0.53x−0.1B[3]cos(omega*n)0−20−1001020−20−1001020nn0.0410.020[n]0.55x−0.02B[5]cos(omega*n)0−20−1001020−20−1001020nnFigureP3.102.Plot3of6−3x101640.820.6[n]023x0.4−2B[23]cos(omega*n)−40.2−6课后答案网0−20−1001020−20−1001020nn0.02www.hackshp.cn10.80.010.6[n]025x0.4B[25]cos(omega*n)−0.010.2−0.020−20−1001020−20−1001020nnFigureP3.102.Plot4of681 10.20[n]1x0.5−0.2B[1]cos(omega*n)0−20−1001020−20−1001020nn0.210.10[n]3x0.5−0.1B[3]cos(omega*n)−0.20−20−1001020−20−1001020nn10.020[n]0.55x−0.02B[5]cos(omega*n)0−20−1001020−20−1001020nnFigureP3.102.Plot5of60.01510.010.80.0050.6[n]023x0.4−0.005B[23]cos(omega*n)−0.010.2−0.0150−20−1001020−20−1001020nn0.0210.80.01课后答案网0.6[n]025x0.4B[25]cos(omega*n)−0.01www.hackshp.cn0.2−0.020−20−1001020−20−1001020nnFigureP3.102.Plot6of682 3.103.UseMATLAB’sfftcommandtorepeatProblem3.48.0.41000.3[k]|a[k]deg00.2a|X1000)valuesofΩintheinterval−π<Ω≤π.PlotthefrequencyresponsemagnitudeindB(20log|H(ejΩ)|)forthefollowingvaluesofM.10t(a)M=4(b)M=10课后答案网(c)M=25(d)M=50DiscusstheeffectofincreasingMontheaccuracywithwhichH(ejΩ)approximatesH(ejΩ).www.hackshp.cntItcanbeseenthatasMincreases:(1)Therippledecreases(2)ThetransitionrollsofffasterObservethath[n]=1sin(πn)issymmetric,hence:πn2M−1H(ejΩ)=h[n]e−jΩn+h[n]e−jΩn+h[0]tn=1n=−MM=2h[n]cos(Ωn)+h[0]n=1ThisisthesamebasicformasaFourierSeriesapproximationusingMterms.Theapproximationis97 moreaccurateasMincreases.10.80.6(Omega)|M=40.4t|H0.2−3−2−10123Omega10.80.60.4(Omega)|M=10t|H0.2−3−2−10123OmegaFigureP3.112.Plot1of210.80.60.4(Omega)|M=25t|H0.2−3−2−10123Omega10.8课后答案网0.60.4(Omega)|M=50t|H0.2www.hackshp.cn−3−2−10123OmegaFigureP3.112.Plot2of23.113.UsetheMATLABcommandfreqsorfreqztoplotthemagnituderesponseofthefollowingsys-tems.Determinewhetherthesystemhasalow-pass,high-pass,orband-passcharacteristic.(a)H(jω)=8(jω)3+4(jω)2+8jω+8Lowpassfilter3(jω)(b)H(jω)=(jω)3+2(jω)2+2jω+1Highpassfilter98 −jΩ−j2Ω−j3Ω(c)H(ejΩ)=1+3e+3e+e6+2e−j2ΩLowpassfilter−jΩ4(d)H(ejΩ)=0.02426(1−e)(1+1.10416e−jΩ+0.4019e−j2Ω)(1+0.56616e−jΩ+0.7657e−j2Ω)Highpassfilter(a)(b)00−20−20−40−40−60−60−80MagnitudeindBMagnitudeindB−100−80−120−100−140−202−202101010101010omegaomega(c)(d)00−50−50−100−100−150MagnitudeindB−150MagnitudeindB−200−250−200−4−2024−202OmegaOmegaFigureP3.113.Plot1of1课后答案网3.114.UseMATLABtoverifythatthetime-bandwidthproductforadiscrete-timesquarewaveiswww.hackshp.cnapproximatelyindependentofthenumberofnonzerovaluesineachperiodwhendurationisdefinedasthenumberofnonzerovaluesinthesquarewaveandbandwidthisdefinedasthemainlobewidth.Defineoneperiodofthesquarewaveas1,0≤n66.659τ>0.0884s4.20.ConsiderthesystemdepictedinFig.P4.20(a).TheFToftheinputsignalisdepictedinFig.4.20课后答案网FTFT(b).Letz(t)←−−−→Z(jω)andy(t)←−−−→Y(jω).SketchZ(jω)andY(jω)forthefollowingcases.sin(6πt)(a)w(t)=cos(5πt)andh(t)=www.hackshp.cnπt1Z(jω)=X(jω)∗W(jω)2πW(jω)=π(δ(ω−5π)+δ(ω+5π))1Z(jω)=(X(j(ω−5π))+X(j(ω+5π)))21|ω|<6πH(jω)=0otherwiseC(jω)=H(jω)Z(jω)=Z(jω)1Y(jω)=C(jω)∗[π(δ(ω−5π)+δ(ω+5π))]2π1Y(jω)=[Z(j(ω−5π))+Z(j(ω+5π))]212 1=[X(j(ω−10π))+2X(jω)+X(j(ω+10π))]4Z(jw)0.5w5FigureP4.20.(a)SketchofZ(jω)Y(jw)0.50.25w10FigureP4.20.(a)SketchofY(jω)sin(5πt)(b)w(t)=cos(5πt)andh(t)=πt1Z(jω)=[X(j(ω−5π))+X(j(ω+5π))]21|ω|<5πH(jω)=课后答案网0otherwiseC(jω)=H(jω)Z(jω)1Y(jω)=C(jω)∗[π(δ(ω−5π)+δ(ω+5π))]2πwww.hackshp.cn1Y(jω)=[C(j(ω−5π))+C(j(ω+5π))]2Z(jw)0.5w5FigureP4.20.(b)SketchofZ(jω)13 Y(jw)0.25w10FigureP4.20.(b)SketchofY(jω)sin(2πt)(c)w(t)depictedinFig.P4.20(c)andh(t)=cos(5πt)πtT=2,ω=π,T=1oo2∞2sin(kπ)W(jω)=2δ(ω−kπ)kk=−∞1Z(jω)=X(jω)∗W(jω)2π∞sin(kπ)=2X(j(ω−kπ))kπk=−∞C(jω)=H(jω)Z(jω)7sin(kπ)−7sin(kπ)=2X(j(ω−kπ))+2X(j(ω−kπ))kπkπk=3k=−31Y(jω)=[C(j(ω−5π))+C(j(ω+5π))]课后答案网2www.hackshp.cnZ(jw)0.5......−7−337wFigureP4.20.(c)SketchofZ(jω)14 Y(jw)21812w−5−51073−5021FigureP4.20.(c)SketchofY(jω)4.21.ConsiderthesystemdepictedinFig.P4.21.Theimpulseresponseh(t)isgivenbysin(11πt)h(t)=πtandwehave∞1x(t)=cos(k5πt)k2k=110g(t)=cos(k8πt)k=1UsetheFTtodeterminey(t).课后答案网y(t)=[(x(t)∗h(t))(g(t)∗h(t))]∗h(t)=[xh(t)gh(t)]∗h(t)=m(t)∗h(t)www.hackshp.cnsin(11πt)FT1|ω|≤11πh(t)=←−−−→H(jω)=πt0otherwise∞1FT∞1x(t)=cos(k5πt)←−−−→X(jω)=π[δ(ω−5kπ)+δ(ω+5kπ)]k2k2k=1k=11010g(t)=cos(k8πt)=π[δ(ω−8kπ)+δ(ω+8kπ)]k=1k=1Xh(jω)=X(jω)H(jω)21=π[δ(ω−5kπ)+δ(ω+5kπ)]k2k=115 Gh(jω)=G(jω)H(jω)=πδ(ω−8π)+πδ(ω−8π)1M(jω)=Xh(jω)∗Gh(jω)2π1=[Xh(j(ω−8π))+Xh(j(ω+8π))]221=π[(δ(ω−8π−5kπ)+δ(ω−8π+5kπ))+(δ(ω+8π−5kπ)+δ(ω+8π+5kπ))]k2k=1Y(jω)=M(jω)H(jω)ππ=[δ(ω+3π)+δ(ω−3π)]+[δ(ω−2π)+δ(ω+2π)]2811y(t)=cos(3πt)+cos(2πt)284.22.Theinputtoadiscrete-timesystemisgivenbyπ3πx[n]=cos(n)+sin(n)84UsetheDTFTtofindtheoutputofthesystem,y[n],forthefollowingimpulseresponsesh[n],firstnotethatjΩπππ3π3πX(e)=πδ(Ω−)+δ(Ω+)+δ(Ω−)−δ(Ω+)88j44sin(πn)(a)h[n]=4πn课后答案网1|Ω|≤πH(ejΩ)=40π≤|Ω|<π,2πperiodic.4www.hackshp.cnY(ejΩ)=H(ejΩ)X(ejΩ)ππ=πδ(Ω−)+δ(Ω+)88πy[n]=cos(n)8πn)(b)h[n]=(−1)nsin(2πnsin(πn)h[n]=ejπn2πn1|Ω−π|≤πH(ejΩ)=40π≤|Ω−π|<π,2πperiodic.4Y(ejΩ)=H(ejΩ)X(ejΩ)16 π3π3π=δ(Ω−)+δ(Ω+)j443πy[n]=sin(n)4sin(πn)(c)h[n]=cos(πn)52πnsin(πn)πh[n]=cos(n)52πn1|Ω−π|≤π1|Ω+π|≤πH(ejΩ)=225+2250π≤|Ω−π|<π0π≤|Ω+π|<π,2πperiodic.5252Y(ejΩ)=H(ejΩ)X(ejΩ)=0y[n]=0sin(πn)4.23.Considerthediscrete-timesystemdepictedinFig.P4.23.Assumeh[n]=2.UsetheDTFTπntodeterminetheoutput,y[n]forthefollowingcases:AlsosketchG(ejΩ),theDTFTofg[n].y[n]=(x[n]w[n])∗h[n]=g[n]∗h[n]ππsin(n)FTjΩ1|Ω|≤h[n]=2←−−−→H(e)=2πn0π≤|Ω|<π2H(ejΩ)is2πperiodic.sin(πn)(a)x[n]=4,w[n]=(−1)nπn课后答案网ππsin(n)DTFTjΩ1|Ω|≤x[n]=4←−−−−→X(e)=4πn0π≤|Ω|<πwww.hackshp.cn4DTFTw[n]=ejπn←−−−−→W(ejΩ)=2πδ(Ω−π)jΩ1jΩjΩG(e)=X(e)∗W(e)2π1|Ω−π|≤π=40π≤|Ω−π|<π4sin(πn)g[n]=ejπn4πnY(ejΩ)=G(ejΩ)H(ejΩ)=0y[n]=017 jG(e)13544FigureP4.23.(a)TheDTFTofg[n]sin(πn)(b)x[n]=δ[n]−4,w[n]=(−1)nπnππsin(n)DTFTjΩ0|Ω|≤x[n]=δ[n]−4←−−−−→X(e)=4πn1π≤|Ω|<π4DTFTw[n]=ejπn←−−−−→W(ejΩ)=2πδ(Ω−π)jΩ1jΩjΩG(e)=X(e)∗W(e)2π0|Ω−π|≤3π=413π≤|Ω−π|<π4sin(3πn)g[n]=4πnY(ejΩ)=G(ejΩ)H(ejΩ)=H(ejΩ)sin(πn)y[n]=2课后答案网πnjwww.hackshp.cnG(e)134FigureP4.23.(b)TheDTFTofg[n]sin(πn)(c)x[n]=2,w[n]=cos(πn)πn218 jΩππW(e)=πδ(Ω−)+δ(Ω+),2πperiodic22jΩ1jΩjΩG(e)=X(e)∗W(e)2π1|Ω−π|≤π1|Ω+π|≤π=222+2220π≤|Ω−π|<π0π≤|Ω+π|<π2222sin(πn)12jπn−jπng[n]=e2+e22πnsin(πn)π=2cos(n)πn2sin(πn)=2πn1=δ(n)2y[n]=g[n]∗h[n]1=h[n]2sin(πn)=2πnjG(e)0.5......课后答案网FigureP4.23.(c)TheDTFTofg[n](d)x[n]=1+sin(πn)+2cos(www.hackshp.cn3πn),w[n]=cos(3πn)1648jΩπππ3π3πX(e)=2πδ(Ω)+[δ(Ω−)+δ(Ω+)+2πδ(Ω−)+δ(Ω+),2πperiodic.j161644jΩ3π3πW(e)=πδ(Ω−)+δ(Ω+),2πperiodic88jΩ1jΩjΩG(e)=X(e)∗W(e)2π1j(Ω−3π)j(Ω+3π)=X(e8)+X(e8)2π13ππ7π5π9π3π=2πδ(Ω−)+δ(Ω−)−δ(Ω−)+2πδ(Ω−)+δ(Ω+)2π8j16168813ππ5π7π3π9π+2πδ(Ω+)+δ(Ω+)−δ(Ω+)+2πδ(Ω−)+δ(Ω+)2π8j16168819 23π17π15π19πg[n]=cos(n)+sin(n)−sin(n)+cos(n)π82π162π16π8Y(ejΩ)=G(ejΩ)H(ejΩ)23π17π15πy[n]=cos(n)+sin(n)−sin(n)π82π162π16jG(e)210.556791616168jG(e)27557161616162FigureP4.23.(d)TheDTFTof课后答案网g[n]4.24.DetermineandsketchtheFTrepresentation,www.hackshp.cnXδ(jω),forthefollowingdiscrete-timesignalswiththesamplingintervalTsasgiven:∞X(jω)=e−jωTsnδn=−∞=X(ejΩ)Ω=ωTssin(πn)(a)x[n]=3,T=2πns1|Ω|≤πX(ejΩ)=30π≤|Ω|<π320 1|2ω|≤πXδ(jω)=30π≤|2ω|<π31|ω|≤π=60π≤|ω|<π66Xδ(jω)isπperiodicX(jw)1......w6FigureP4.24.(a)FTofXδ(jω)sin(πn)(b)x[n]=3,T=1πns41|Ω|≤πX(ejΩ)=30π≤|Ω|<π31|1ω|≤πXδ(jω)=430π≤|1ω|<π341|ω|≤4π=304π≤|ω|<4π课后答案网3Xδ(jω)is8πperiodicwww.hackshp.cnX(jw)1......4w83FigureP4.24.(b)FTofXδ(jω)21 sin(πn)(c)x[n]=cos(πn)4,T=22πns1|Ω−π|≤π1|Ω+π|≤πX(ejΩ)=224+2240π≤|Ω−π|<π0π≤|Ω+π|<π42421|2ω−π|≤π1|2ω+π|≤πXδ(jω)=223+2230π≤|2ω−π|<π0π≤|2ω+π|<π32321π<ω<3π1−3π<ω<−π=288+2880otherwise0otherwiseXδ(jω)isπperiodicX(jw)0.5......3w88FigureP4.24.(c)FTofXδ(jω)(d)x[n]depictedinFig.P4.17(a)with课后答案网Ts=4.www.hackshp.cnπDTFS:N=8Ωo=4sin(k5π)X[k]=8,k∈[−3,4]8sin(kπ)8∞jΩπDTFT:X(e)=2πX[k]δ(Ω−k)4k=−∞π∞sin(k5π)πFT:X(jω)=8δ(4ω−k)δπ48sin(k)4k=−∞8π∞sin(k5π)π=8δ(ω−k)168sin(kπ)16k=−∞8X(jω)isπperiodicδ222 (d)PlotofFTrepresentationofX(jw)d10.8)|0.6w(jd|X0.40.20−0.8−0.6−0.4−0.200.20.40.60.8w3.532.5))w(j2d1.5arg(X10.50−0.8−0.6−0.4−0.200.20.40.60.8wFigureP4.24.(d)FTofXδ(jω)∞1(e)x[n]=p=−∞δ[n−4p],Ts=8课后答案网www.hackshp.cnπDTFS:N=4Ωo=21X[k]=,k∈[0,3]4∞jΩππDTFT:X(e)=δ(Ω−k)22k=−∞∞π1πFT:Xδ(jω)=δ(ω−k)282k=−∞∞=4πδ(ω−k4π)k=−∞Xδ(jω)is4πperiodic23 (e)PlotofFTrepresentationofX(jw)d141210)|8w(jd|X6420−30−20−100102030w10.5))w(jd0arg(X−0.5−1−30−20−100102030wFigureP4.24.(e)FTofXδ(jω)课后答案网www.hackshp.cn4.25.Considersamplingthesignalx(t)=1sin(2πt).πt(a)SketchtheFTofthesampledsignalforthefollowingsamplingintervals:(i)T=1s8(ii)T=1s3(iii)T=1s2(iv)T=2s3Inpart(iv),aliasingoccurs.Thesignalsoverlapandadd,whichcanbeseeninthefollowingfigure.24 PlotofFTofthesampledsignal10=1/8s5),Tw(jdX0−80−60−40−200204060804=1/3s),T2w(jdX0−40−30−20−1001020304032=1/2s1),Tw(j0dX−1−20−15−10−5051015204=2/3s2),Tw(j0dX−8−6−4−202468wFigureP4.25.(a)SketchofXδ(jω)(b)Letx[n]=x(nT).SketchtheDTFTofx[n],X(ejΩ),foreachofthesamplingintervalsgivensin(a).课后答案网www.hackshp.cn1DTFTx[n]=x(nT)=sin(2πnT)←−−−−→X˜(ejΩ)ssπnTs1|Ω|≤TπX˜(ejΩ)=Tss2πperiodic0Tsπ≤|Ω|<πNoticethatthedifferencebetweenthefigurein(a)and(b)isthatthe’ω’axishasbeenscaledbythesamplingrate.25 PlotofFTofthesampledsignal10=1/8s5),TjWX(e0−8−6−4−2024684=1/3s),T2jWX(e0−15−10−505101532=1/2s1),TjW0X(e−1−10−8−6−4−202468104=2/3s2),TjW0X(e−15−10−5051015WFigureP4.25.(b)SketchofX(ejΩ)4.26.Thecontinuous-timesignalx(t)withFTasdepictedinFig.P4.26issampled.Identifyineachcaseifaliasingoccurs.(a)SketchtheFTofthesampledsignalforthefollowingsamplingintervals:课后答案网(i)T=1s14Noaliasingoccurs.www.hackshp.cnX(jw)14......141038w381014FigureP4.26.(i)FTofthesampledsignal(ii)T=1s7SinceT>1,aliasingoccurs.s1126 X(jw)7......41024w24104FigureP4.26.(ii)FTofthesampledsignal(iii)T=1s5SinceT>1,aliasingoccurs.s11X(jw)......w1010FigureP4.26.(iii)FTofthesampledsignal(b)Letx[n]=x(nT).SketchtheDTFTofx[n],X(ejΩ),foreachofthesamplingintervalsgivensin(a).TheDTFTsimplescalesthe’x’axisbythesamplingrate.(i)T=1s14课后答案网j)X(ewww.hackshp.cn14141038......141414381014141414FigureP4.26.(i)DTFTofx[n](ii)T=1s727 jX(e)471024......77724104777FigureP4.26.(ii)DTFTofx[n](iii)T=1s5jX(e)......101055FigureP4.26.(iii)DTFTofx[n]sin(πn)4.27.Considersubsamplingthesignalx[n]=6sothaty[n]=x[qn].SketchY(ejΩ)fortheπnfollowingchoicesofq:1|ω|≤π课后答案网X(ejΩ)=60π≤|ω|<π2πperiodic6q[n]=x[qn]q−1www.hackshp.cnjΩ1j1(Ω−m2π)Y(e)=Xeqqm=0(a)q=21jΩ1j1(Ω−m2π)Y(e)=Xe22m=0(b)q=43jΩ1j1(Ω−m2π)Y(e)=Xe44m=028 (c)q=87jΩ1j1(Ω−m2π)Y(e)=Xe88m=0Plotofthesubsampledsignalx[n]10.5),q=2jWY(e0−10−8−6−4−202468100.30.2),q=4jW0.1Y(e0−0.1−10−8−6−4−202468100.40.3),q=80.2jWY(e0.10−10−8−6−4−20246810课后答案网WFigureP4.27.Sketchofwww.hackshp.cnY(ejΩ)4.28.Thediscrete-timesignalx[n]withDTFTdepictedinFig.P4.28issubsampledtoobtainy[n]=x[qn].SketchY(ejΩ)forthefollowingchoicesofq:(a)q=3(b)q=4(c)q=829 Plotofthesubsampledsignalx[n]0.80.6),q=30.4jW......Y(e0.20−15−10−50510150.40.3),q=40.2jWY(e0.10−20−15−10−5051015200.20.15),q=80.1jWY(e0.050−20−15−10−505101520WFigureP4.28.SketchofY(ejΩ)4.29.ForeachofthefollowingsignalssampledwithsamplingintervalTs,determinetheboundsonTsthatgauranteetherewillbenoaliasing.(a)x(t)=1sin3πt+cos(2πt)t11FT|ω|≤3π课后答案网sin(3πt)←−−−→πt0otherwiseFTcos(2πt)←−−−→πδ(Ω−2π)+πδ(Ω+2π)www.hackshp.cnωmax=3ππT<ωmax1T<3sin(πt)(b)x(t)=cos(12πt)2t1|ω−12π|≤π1|ω+12π|≤πX(jω)=4π+4π0otherwise0otherwiseωmax=13ππT<ωmax30 1T<13(c)x(t)=e−6tu(t)∗sin(Wt)πt1X(jω)=[u(ω+W)−u(ω−W)]6+jωωmax=WπT<ωmaxπTωm.FindthelargestvalueofTsuchthatx(t)canbereconstructedfromy(t).DetermineasystemthatwillperformthereconstructionforthismaximumvalueofT.Forreconstruction,weneedtohavew>2w,orT<π.Afinitedutycycleresultsindistortion.smaxωmaxsin(πk)2−jπk课后答案网W[k]=e2kπ∞2πW(jω)=2πW[k]δ(ω−k)Twww.hackshp.cnk=−∞Aftermultiplication:∞sin(πk)2−jπk2πY(jω)=e2X(j(ω−k))kπTk=−∞Toreconstruct:2πHr(jω)Y(jω)=X(jω),|ω|<ωmax,>2ωmaxTk=01Hr(jω)X(jω)=X(jω)22|ω|<ωmaxH(jω)=don’tcareω<|ω|<2π−ωrmaxTmax0|ω|>2π−ωTmax31 4.31.Let|X(jω)|=0for|ω|>ωm.Formthesignaly(t)=x(t)[cos(3πt)+sin(10πt)].Determinethemaximumvalueofωmforwhichx(t)canbereconstructedfromy(t)andspecifyasystemthatthatwillperformthereconstruction.1Y(jω)=[X(j(ω−2π))+X(j(ω+2π))−jX(j(ω−10π))+jX(j(ω+10π))]2x(t)canbereconstructedfromy(t)ifthereisnooverlapamoungthefourshiftedversionsofX(jω),yetx(t)canstillbereconstructedwhenoverlapoccurs,providedthatthereisatleastoneshiftedX(jω)thatisnotcontaminated.10π−4πωmmax==4π2Y(jw)1......3−wm3+wmw10FigureP4.31.Y(jω)Werequire10π−ω>3π+ω,thusω<7π.Torecoverthesignal,bandpassfilterwithpassbandmmm26.5π≤ω≤13.5πandmultiplywith2sin(10πt)toretrievex(t).4.32.Areconstructionsystemconsistsofazero-orderholdfollowedbyacontinuous-timeanti-imagingfilterwithfrequencyresponseHc(jω).Theoriginalsignalx(t)isbandlimitedtoωm,thatis,X(jω)=0for|ω|>ωmandissampledwithasamplingintervalofTs.Determinetheconstraintsonthemagnituderesponseoftheanti-imagingfiltersothattheoverallmagnituderesponseofthisreconstructionsystemisbetween0.99and1.01inthesignalpassbandandlessthan10课后答案网−4totheimagesofthesignalspectrumforthefollowingvalues:H(jw)owww.hackshp.cnx[n]zerox(t)srorderH(jw)choldx[n]=x(nT)ssFigureP4.32.Reconstructionsystem.(1)0.99<|Ho(jω)||Hc(jω)|<1.01,−ωm≤ω≤ωmThus:0.99ω(i)|Hc(jω)|>2sin(ωTs)232 1.01ω(ii)|Hc(jω)|<2sin(ωTs)2Thepassbandconstraintforeachcaseis:0.99ω<|H(jω)|<1.01ωTscTs2sin(ω)2sin(ω)22Thestopbandconstraintis|H(jω)||H(jω)|<10−4,attheworstcaseω=2π−ω.ocTsm−410ω|Hc(jω)|<T2sin(ωs)2(a)ωm=10π,Ts=0.12πω=−ωmTs2π=−10π=10π0.110−410π|Hc(j10π)|<0.12sin(10π())2=5π(10−4)=0.001571(b)ωm=10π,Ts=0.052πω=−10π=30π0.0530π(10−4)|Hc(j30π)|<0.052sin(30π())230π课后答案网=√(10−4)=0.0066642(c)ωm=10π,Ts=0.02www.hackshp.cn2πω=−10π=90π0.0210−490π|Hc(j90π)|<0.022sin(90π())290π(10−4)==0.045752sin(0.9π)(d)ωm=2π,Ts=0.052πω=−10π=30π0.0533 30π(10−4)|Hc(j30π)|<0.052sin(30π())210−430π==0.0066642sin(3π)44.33.Thezero-orderholdproducesastairstepapproximationtothesampledsignalx(t)fromsamplesx[n]=x(nTs).Adevicetermedafirst-orderholdlinearlyinterpolatesbetweenthesamplesx[n]andthusproducesasmootherapproximationtox(t).Theoutputofthefirst-orderholdmaybedescribedas∞x1(t)=x[n]h1(t−nTs)n=−∞whereh1(t)isthetriangularpulseshowninFig.P4.33(a).Therelationshipbetweenx[n]andx1(t)isdepictedinFig.P4.33(b).(a)Identifythedistortionsintroducedbythefirst-orderholdandcomparethemtothoseintroducedbythezero-orderhold.Hint:h1(t)=ho(t)∗ho(t).∞x1(t)=x[n]h1(t−nTs)n=−∞∞=ho(t)∗ho(t)∗x[n]δ(t−nTs)n=−∞Thus:X1(jω)=Ho(jω)Ho(jω)X∆(jω)Whichimplies:4sin2(ωTs)H(jω)=e−jωTs21ω2Distortions:课后答案网(1)AlinearphaseshiftcorrespondingtoatimedelayofTsseconds(aunitofsamplingtime).2(2)sin(.)termintroducesmoredistortiontotheportionofXδ(jω),especiallythehigherfrequencypartisseverelyattenuatedcomparedtothelowfrequencypartwhichfallswithinthemainlobe,betweenwww.hackshp.cn−ωmandωm.(3)DistortedandattenuatedversionsofX(jω)stillremainatthenonzeromultiplesofωm,yetitislowerthanthecaseofthezeroorderhold.(b)Considerareconstructionsystemconsistingofafirst-orderholdfollowedbyananti-imagingfil-terwithfrequencyresponseHc(jω).FindHc(jω)sothatperfectreconstructionisobtained.X∆(jω)H1(jω)Hc(jω)=X(jω)ejωTsω2Hc(jω)=2TsTsHLPF(jω)4sin(ω)2whereHLPF(jω)isanideallowpassfilter.34 jωTs2eωT|ω|≤ω4sin2(ωTs)sm2Hc(jω)=don’tcareω≤|ω|<2π−ωmTsm0|ω|>2π−ωTsmAssumingX(jω)=0for|ω|>ωm(c)Determinetheconstraintson|Hc(jω)|sothattheoverallmagnituderesponseofthisreconstruc-tionsystemisbetween0.99and1.01inthesignalpassbandandlessthan10−4totheimagesofthesignalspectrumforthefollowingvalues.Assumex(t)isbandlimitedto12π,thatis,X(jω)=0for|ω|>12π.Constraints:(1)Inthepassband:0.99<|H1(jω)||Hc(jω)|<1.010.99ω21.01ω2<|Hc(jω)|<4sin2(ωTs)4sin2(ωTs)22(2)Intheimageregion:ω=2π−ωTsm|H(jω)||H(jω)|<10−41c10−4ω2|Hc(jω)|<4sin2(ωTs)课后答案网2(i)Ts=.05www.hackshp.cn2πω=−ωmTs2π=−12π=28π0.05−4210(28π)|Hc(jω)|<20.054sin(28π())2≈0.29560.99ω21.01ω2<|Hc(jω)|<4sin2(ωTs)4sin2(ωTs)222926.01<|Hc(jω)|<2985.1235 (ii)Ts=.022πω=−12π=88π0.02−4210(88π)|Hc(jω)|<20.024sin(88π())2≈14.10.99ω21.01ω2<|Hc(jω)|<4sin2(ωTs)4sin2(ωTs)22139589<|Hc(jω)|<1424094.34.Determinethemaximumfactorqbywhichx[n]withDTFTX(ejΩ)depictedinFig.P4.34canbedecimatedwithoutaliasing.SketchtheDTFTofthesequencethatresultswhenx[n]isdecimatedbythisamount.Lookingatthefollowingequation:q−1jΩ1j1(Ω−m2π)Y(e)=Xeqqm=0Forthebandlimitedsignal,overlapstartswhen:课后答案网2qW>2πThus:πqmax==3www.hackshp.cnWAfterdecimation:2jΩ1j1(Ω−m2π)Y(e)=Xe33m=0jY(e)1...3...2436 FigureP4.34.SketchoftheDTFT4.35.Adiscrete-timesystemforprocessingcontinuous-timesignalsisshowninFig.P4.35.Sketchthemagnitudeofthefrequencyresponseofanequivalentcontinuous-timesystemforthefollowingcases:Ts1jωTs2sin(ω2)|HT(jω)|=|Ha(jω)||H(e)||Hc(jω)|Tsω(a)Ω=π,W=20π14cπωmax=min(10π,(20),20π)=5π4H(jw)T1w5FigureP4.35.(a)Magnitudeofthefrequencyresponse(b)Ω=3π,W=20π14课后答案网c3πωmax=min(10π,(20),20π)=10πwww.hackshp.cn4H(jw)T110wFigureP4.35.(b)Magnitudeofthefrequencyresponse(c)Ω=π,W=2π14c37 πωmax=min(10π,(20),2π)=2π4H(jw)T12wFigureP4.35.(c)MagnitudeofthefrequencyresponsejΩsin(11Ω)jkΩDTFS;Ωo4.36.LetX(e)=2anddefineX˜[k]=X(eo).Findandsketch˜x[n]where˜x[n]←−−−−→sin(Ω2X˜[k]forthefollowingvaluesofΩo:X˜[k]=X(ejkΩo)sin(11kΩo)DTFS;ΩoN|n|≤5X˜[k]=2←−−−−→x˜[n]=sin(kΩo)05<|n|120πrad/s.Determinethefrequencyrange−ωa<ω<ωaoverwhichtheDTFSoffersareasonableapproximationto40 X(jω),theeffectiveresolutionofthisapproximation,ωr,andthefrequencyintervalbetweeneachDTFScoefficient,∆ω.x(t)Ts=0.01100samplesM=100UseN=200DTFStoapproximateX(jω),|X(jω)|≈0,|ω|>120π,ωm=120π2πTs<ωm+ωa2πωa<−ωmTsTherefore:ωa<80π2πMTs>ωrTherefore:ωr>2πωsN>∆ωωs∆ω>N2π∆ω=NTsTherefore:∆ω>π4.39.Asignalx(t)issampledatintervalsofTs=0.1s.Assume|X(jω)|≈0for|ω|>12πrad/s.Determinethefrequencyrange课后答案网−ωa<ω<ωaoverwhichtheDTFSoffersareasonableapproximationtoX(jω),theminimumnumberofsamplesrequiredtoobtainaneffectiveresolutionωr=0.01πrad/s,andthelengthoftheDTFSrequiredsothefrequencyintervalbetweenDTFScoefficientsis∆ω=0.001πrad/s.www.hackshp.cn2πTs<ωm+ωaωm=12πTs=0.1ωa<8πThefrequencyrange|ω|<8πprovidesareasonableapproximationtotheFT.ωsM≥ωr41 2πωs==20πTsωr=0.01πM≥2000M=2000samplesissufficientforthegivenresolution.ωsN≥∆ω∆ω=0.001πN≥20,000TherequiredlengthoftheDTFSisN=20,000.4.40.Letx(t)=asin(ωot)besampledatintervalsofTs=0.1s.Assume100samplesofx(t),x[n]=x(nTs),n=0,1,...99,areavailable.WeusetheDTFSofx[n]toapproximatetheFTofx(t)andwishtodetermineafromtheDTFScoefficientoflargestmagnitude.Thesamplesx[n]arezero-paddedtolengthNbeforetakingtheDTFS.DeterminetheminimumvalueofNforthefollowingvaluesofωo:DeterminewhichDTFScoefficienthasthelargestmagnitudeineachcase.Choose∆ωsothatωoisanintegermultipleof∆ω,(ωo=p∆ω),wherepisaninteger,andsetN=M=100.UsingthesetwoconditionsresultsintheDTFSsamplingWδ(j(ω−ωo))atthepeakofthemainlobeandatallofthezerocrossings.Consequently,ak=pY[k]=课后答案网0otherwiseon0≤k≤N−1(a)ωo=3.2πwww.hackshp.cnωsN=∆ω20πp=ωo20πp=3.2π=25p=4SincepandNhavetobeintegersak=4Y[k]=0otherwiseon0≤k≤2442 (b)ωo=3.1πωsN=∆ω20πp=ωo20πp=3.1π=200p=31SincepandNhavetobeintegersak=31Y[k]=0otherwiseon0≤k≤199(c)ωo=3.15πωsN=∆ω20πp=ωo20πp=3.15π=400p=63SincepandNhavetobeintegers课后答案网ak=63Y[k]=0otherwiseon0≤k≤399SolutionstoAdvancedProblemswww.hackshp.cn4.41.Acontinuous-timesignalliesinthefrequencyband|ω|<5π.Thissignaliscontaminatedbyalargesinusoidalsignaloffrequency120π.Thecontaminatedsignalissampledatasamplingrateofωs=13π.(a)Aftersampling,atwhatfrequencydoesthesinusoidalinteferingsignalappear?X(jω)isbandlimitedto5πs(t)=x(t)+Asin(120πt)240πs[n]=s(nTs)=x[n]+Asin(n)1343 6π=x[n]+Asin(9(2π)n+n)136π=x[n]+Asin(n)136πΩsin=13Ω6π13ω===3πTs132Thesinusoidappearsatω=3πrads/secinSδ(jω).(b)Thecontaminatedsignalispassedthroughananti-aliasingfilterconsistingoftheRCcircuitde-pictedinFig.P4.41.FindthevalueofthetimeconstantRCrequiredsothatthecontaminatingsinusoidisattenuatedbyafactorof1000priortosampling.Beforethesampling,s(t)ispassedthroughaLPF.1jωCT(jω)=R+1jωC1=1+jωRC1=1+jωτ1|T(jω)|=√1+ω2τ2ω=120π1=课后答案网1000τ=2.65swww.hackshp.cn(c)SketchthemagnituderesponseindBthattheanti-aliasingfilterpresentstothesignalofinterestforthevalueofRCidentifiedin(b).1T(jω)=1+jω2.651|T(jω)|=1+ω2(2.65)244 P4.41Magnituderesponseoftheanti−aliasingfilterindB0−5−10)indBwT(j−15−20−25−5−4−3−2−1012345w,−3dBpointatw=±1/2.65FigureP4.41.Sketchofthemagnituderesponse4.42.Thisproblemderivesthefrequency-domainrelationshipforsubsamplinggiveninEq.(4.27).UseEq.(4.17)torepresentx[n]astheimpulse-sampledcontinuous-timesignalwithsamplingintervalTs,andthuswrite∞xδ(t)=x[n]δ(t−nTs)n=−∞Supposex[n]arethesamplesofacontinuous-timesignal课后答案网x(t),obtainedatintegermultiplesofTs.Thatis,FTx[n]=x(nTs).Letx(t)←−−−→X(jω).Definethesubsampledsignaly[n]=x[qn]sothaty[n]=x(nqTs)isalsoexpressedassamplesofx(t).(a)ApplyEq.(4.23)toexpresswww.hackshp.cnXδ(jω)asafunctionofX(jω).Showthat∞1kYδ(jω)=X(j(ω−ωs))qTsqk=−∞Sincey[n]isformedbyusingeveryqthsampleofx[n],theeffectivesamplingrateisT=qTss∞FT1∞y(t)=x(t)δ(t−nT)←−−−→Y(jω)=X(j(ω−kω))δsδTsn=−∞sk=−∞ωsSubstitutingTs=qTs,andωs=yields:q∞1kYδ(jω)=X(j(ω−ωs))qTsqk=−∞45 (b)ThegoalistoexpressY(jω)asafunctionofX(jω)sothatY(ejΩ)canbeexpressedintermsofδδX(ejΩ).Tothisend,writekinY(jω)astheproperfractionqδkm=l+qqwherelistheintegerportionofkandmistheremainder.ShowthatwemaythusrewriteY(jω)asqδ1q−11∞mYδ(jω)=X(j(ω−lωs−ωs))qTsqm=0l=−∞Lettingktorangefrom−∞to∞correspondstohavinglrangefrom−∞to∞andmfrom0toq−1,whichpermitsustorewriteYδ(jω)as:1q−11∞mYδ(jω)=X(j(ω−lωs−ωs))qTsqm=0l=−∞Nextshowthatq−11mYδ(jω)=Xδ(j(ω−ωs))qqm=0RecognizingthattheterminbracescorrespondstoX(j(ω−mω),allowsustorewritetheequationasδqsthefollowingdoublesum:q−11mYδ(jω)=Xδ(j(ω−ωs))qqm=0(c)NowweconvertfromtheFTrepresentationbacktotheDTFTinordertoexpressY(ejΩ)asafunctionofX(ejΩ).ThesamplingintervalassociatedwithY(jω)isqT.UsingtherelationshipΩ=ωqTinδssY(ejΩ)=Y(jω)|δω=ΩqTsshowthatq−1jΩ1jΩmY(e)=Xδ−2πqTsqq课后答案网m=0Substitutingω=ΩyieldsqTsq−1www.hackshp.cnjΩ1jΩmY(e)=Xδ−2πqTsqqm=0(d)Lastly,useX(ejΩ)=X(jΩ)toobtainδTsq−1jΩ1j1(Ω−m2π)Y(e)=Xeqqm=0ThesamplingintervalassociatedwithX(jω)isT,soX(ejΩ)=X(jΩ).HencewemaysubstituteδsδTsj(Ω−m2πjXeqq)forX(Ω−m2π)andobtainδTsqqq−1jΩ1j1(Ω−m2π)Y(e)=Xeqqm=046 4.43.Abandlimitedsignalx(t)satisfies|X(jω)|=0for|ω|<ω1and|ω|>ω2.Assumeω1>ω2−ω1.Inthiscasewecansamplex(t)ataratelessthanthatindicatedbythesamplingintervalandstillperformperfectreconstructionbyusingabandpassreconstructionfilterHr(jω).Letx[n]=x(nTs).DeterminethemaximumsamplingintervalTssuchthatx(t)canbeperfectlyreconstructedfromx[n].Sketchthefrequencyresponseofthereconstructionfilterrequiredforthiscase.Wecantoleratealiasingaslongasthereisnooverlaponω1≤|ω|≤ω2Werequire:ωs−ω2≥−ω1ωs≥ω2−ω1Implies:2πTs≤ω2−ω1Tsω1≤|ω|≤ω2Hr(jω)=0otherwise4.44.Supposeaperiodicsignalx(t)hasFScoefficients(3)k,|k|≤4X[k]=40,otherwiseTheperiodofthissignalisT=1.(a)Determinetheminimumsamplingintervalforthissignalthatwillpreventaliasing.43kX(jω)=2π()δ(ω−k2π)4k=−4ωm=8π2π>2(8π)课后答案网Ts1minTs=www.hackshp.cn8(b)Theconstraintsofthesamplingtheoremcanberelaxedsomewhatinthecaseofperiodicsignalsifweallowthereconstructedsignaltobeatime-scaledversionoftheoriginal.SupposewechooseasamplingintervalT=20anduseareconstructionfilters191,|ω|<πHr(jω)=0,otherwiseShowthatthereconstructedsignalisatime-scaledversionofx(t)andidentifythescalingfactor.20Ts=19∞19Xδ(jω)=X(j(ω−l1.9π))20l=−∞47 Aliasingproducesa“frequencyscaled”replicaofX(jω)centeredatzero.Thescalingisbyafactorof20fromω=2πtoω=0.1π.ApplyingtheLPF,|ω|<πgivesx(t),andx(t)=19x(t)oo20reconstructed2020(c)FindtheconstraintsonthesamplingintervalTssothatuseofHr(jω)in(b)resultsinthere-constructionfilterbeingatime-scaledversionofx(t)anddeterminetherelationshipbetweenthescalingfactorandTs.ThechoiceofTsissothatnoaliasingoccurs.2π(1)<2πTsTs>1periodoftheoriginalsignal.2π12π(2)2π−410π:Xp(jω)=Hp(jω)X∆(jω)To2sin(ω)To2−jωwww.hackshp.cnHp(jω)=e2ToωConstraints:(1)Passband0.99<|Hp(jω)||Hc(jω)|<1.01,usingωmax=10π0.99Toω1.01Toω<|Hc(jω)|<2sin(ωTo)2sin(ωTo)22(2)Intheimagelocation10−4Tωo|Hp(jω)|<2sin(ωTo)248 2πwhereω=−10πTs(a)Ts=0.08,To=0.04(1)Passband1.1533<|Hc(jω)|<1.1766(2)Intheimagelocationω=15π|H(j15π)|<1.165×10−4c(b)Ts=0.08,To=0.02(1)Passband1.0276<|Hc(jω)|<1.0484(2)Intheimagelocationω=15π|H(j15π)|<1.038×10−4c(c)Ts=0.04,To=0.02(1)Passband1.3081<|Hc(jω)|<1.3345(2)Intheimagelocation课后答案网ω=40π|H(j40π)|<1.321×10−4c(d)Ts=0.04,To=0.01www.hackshp.cn(1)Passband1.0583<|Hc(jω)|<1.0796(2)Intheimagelocationω=40π|H(j40π)|<1.0690×10−4c4.46.Anon-idealsamplingoperationobtainsx[n]fromx(t)asnTsx[n]=x(t)dt(n−1)Ts49 (a)Showthatthiscanbewrittenasidealsamplingofafilteredsignaly(t)=x(t)∗h(t),thatis,x[n]=y(nTs),andfindh(t).Tsy(t)=x(τ)dτbyinspectiont−Ts∞=x(τ)h(t−τ)dτ−∞=x(t)∗h(t)10≤t≤Tschooseh(t)=0otherwise1t−Ts≤τ≤Tsh(t−τ)=0otherwiseh(t)=u(t)−u(t−Ts)(b)ExpresstheFTofx[n]intermsofX(jω),H(jω),andTs.Y(jω)=X(jω)H(jω)∞FT12πy(nTs)←−−−→Y(j(ω−k))TsTsk=−∞∞12π2πFT{x[n]}=X(j(ω−k))H(j(ω−k))TsTsTsk=−∞(c)Assumethatx(t)isbandlimitedtothefrequencyrange|ω|<3π.Determinethefrequencyresponse4Tsofadiscrete-timesystemthatwillcorrectthedistortioninx[n]introducedbynon-idealsampling.x(t)isbandlimitedto:课后答案网3π2π|ω|<0.9200πTsimplies:Ts(100π)<0.785课后答案网maxTs=0.0025(5)minW3=200π2πmaxW4=−200π=600πwww.hackshp.cnTs(3)Ωa=0.25πΩb=0.5π(1)and(2)minW1=200π12πmaxW2==400π,Nooverlap.2Ts4.52.Atime-domaininterpretationoftheinterpolationproceduredescribedinFig.4.50(a)isderivedDTFTinthisproblem.Leth[n]←−−−−→H(ejΩ)beanideallow-passfilterwithtransitionbandofzerowidth.ii56 Thatisq,|Ω|<πH(ejΩ)=qiπ0,<|Ω|<πq(a)Substituteforhi[n]intheconvolutionsum∞xi[n]=xz[k]∗hi[n−k]k=−∞sinπnqhi[n]=qπn∞qsinπ(n−k)qxi[n]=xz[k]π(n−k)k=−∞(b)Thezero-insertionprocedureimpliesxz[k]=0unlessk=qmwheremisinteger.Rewritexi[n]usingonlythenon-zerotermsinthesumasasumovermandsubstitutex[m]=xz[qm]toobtainthefollowingexpressionforidealdiscrete-timeinterpolation:∞qsinπ(n−qm)qxi[n]=x[m]π(n−qm)m=−∞Substitutingk=qmyields:∞qsinπ(n−qm)qxi[n]=xz[qm]π(n−qm)m=−∞Nowusexz[qm]=x[m].课后答案网DTFS;2πN4.53.Thecontinuous-timerepresentationforaperiodicdiscrete-timesignalx[n]←−−−−→X[k]ispe-riodicandthushasaFSrepresentation.ThisFSrepresentationisafunctionoftheDTFScoefficientswww.hackshp.cnX[k],asweshowinthisproblem.TheresultestablishestherelationshipbetweentheFSandDTFS∞representations.Letx[n]haveperiodNandletxδ(t)=n=−∞x[n]δ(t−nTs).x[n]=x[n+N]∞xδ(t)=x[n]δ(t−nTs)n=−∞(a)Showxδ(t)isperiodicandfindtheperiod,T.∞xδ(t+T)=x[n]δ(t+T−nTs)n=−∞57 Nowusex[n]=x[n−N]torewrite∞xδ(t+T)=x[n−N]δ(t+T−nTs)n=−∞letk=n−N∞xδ(t+T)=x[k]δ(t−kTs+(T−NTs))k=−∞ItisclearthatifT=NTs,thenxδ(t+T)=xδ(t).Therefore,xδ(t)isperiodicwithT=NTs.(b)BeginwiththedefinitionoftheFScoefficientsT1−jkωotXδ[k]=xδ(t)edt.T0SubstituteforT,ωo,andoneperiodofxδ(t)toshow1Xδ[k]=X[k].TsT1−jkωotXδ[k]=xδ(t)edtT0TN−11−jkωot=x[n]δ(t−nTs)edtT0n=0TN−11−jkωot=x[n]δ(t−nTs)edtNTs0n=0N−1T1−jkωot=x[n]δ(t−nTs)edtNTs0n=0N−1=1x[n]e−jkωonTsNTs课后答案网n=0N−111−jk2πn=x[n]eNTsNwww.hackshp.cnn=01=X[k]Ts4.54.ThefastalgorithmforevaluatingtheDTFS(FFT)maybeusedtodevelopacomputationallyefficientalgorithmfordeterminingtheoutputofadiscrete-timesystemwithafinite-lengthimpulsereponse.Insteadofdirectlycomputingtheconvolutionsum,theDTFSisusedtocomputetheoutputbyperformingmultiplicationinthefrequencydomain.ThisrequiresthatwedevelopacorrespondencebetweentheperiodicconvolutionimplementedbytheDTFSandthelinearconvolutionassociatedwiththesystemoutput,thegoalofthisproblem.Leth[n]beanimpulseresponseoflengthMsothath[n]=0forn<0,n≥M.Thesystemoutputy[n]isrelatedtotheinputviatheconvolutionsumM−1y[n]=h[k]x[n−k]k=058 (a)ConsidertheN-pointperiodicconvolutionofh[n]withNconsecutivevaluesoftheinputsequencex[n]andassumeN>M.Let˜x[n]andh˜[n]beNperiodicversionsofx[n]andh[n],respectivelyx˜[n]=x[n],for0≤n≤N−1x˜[n+mN]=˜x[n],forallintegerm,0≤n≤N−1h˜[n]=h[n],for0≤n≤N−1h˜[n+mN]=h˜[n],forallintegerm,0≤n≤N−1Theperiodicconvolutionbetweenh˜[n]and˜x[n]isN−1y˜[n]=h˜[k]˜x[n−k]k=0Usetherelationshipbetweenh[n],x[n]andh˜[n],x˜[n]toprovethat˜y[n]=y[n],M−1≤n≤N−1.Thatis,theperiodicconvolutionisequaltothelinearconvolutionatL=N−M+1valuesofn.N−1y˜[n]=h[k]˜x[n−k](1)k=0Nowsince˜x[n]=x[n],for0≤n≤N−1,weknowthatx˜[n−k]=x[n−k],for0≤n−k≤N−1In(1),thesumoverkvariesfrom0toM−1,andsothecondition0≤n−k≤N−1isalwayssatisfiedprovidedM−1≤n≤N−1.Substitutingx[n−k]=˜x[n−k],M−1≤n≤N−1into(1)yieldsM−1y˜[n]=h[k]x[n−k]M−1≤n≤N−1k=0课后答案网=y[n](b)Showthatwemayobtainvaluesofwww.hackshp.cny[n]otherthanthoseontheintervalM−1≤n≤N−1byshiftingx[n]priortodefining˜x[n].Thatis,ifx˜p[n]=x[n+pL],0≤n≤N−1x˜p[n+mN]=˜xp[n],forallintegerm,0≤n≤N−1andy˜p[n]=h˜[n]∗x˜p[n]thenshowy˜p[n]=y[n+pL],M−1≤n≤N−1ThisimpliesthatthelastLvaluesinoneperiodof˜yp[n]correspondtoy[n]forM−1+pL≤n≤N−1+pL.EachtimeweincrementptheNpointperiodicconvolutiongivesusLnewvaluesofthelinearconvolu-tion.Thisresultisthebasisfortheso-calledoverlapandsavemethodforevaluatingalinearconvolution59 withtheDTFS.OverlapandSaveMethodofImplementingConvolution¯DTFS;2π/N1.ComputetheNDTFScoefficientsH[k]:h[n]←−−−−→H[k]2.Setp=0andL=N−M+13.Define˜xp[n]=x[n−(M−1)+pL],0≤n≤N−1DTFS;2π/N4.ComputetheNDTFScoefficientsX˜p[k]:˜xp[n]←−−−−→X˜p[k]5.ComputetheproductY˜p[k]=NH[k]X˜p[k]DTFS;2π/N6.Computethetimesignal˜yp[n]fromtheDTFScoefficients,Y˜p[k]:˜yp[n]←−−−−→Y˜p[k].7.SavetheLoutputpoints:y[n+pL]=˜yp[n+M−1],0≤n≤L−18.Setp=p+1andreturntostep3.SolutionstoComputerExperiments4.55.RepeatExample4.7usingzero-paddingandtheMATLABcommandsfftandfftshifttosampleandplotY(ejΩ)at512pointson−π<Ω≤πforeachcase.P4.5580604020|Y(Omega)|:M=80−3−2−10123Omega108642|Y(Omega)|:M=12课后答案网−3−2−10123Omega108www.hackshp.cn64|Y(Omega)|:M=82−3−2−10123OmegaFigureP4.55.PlotofY(ejΩ)4.56.Therectangularwindowisdefinedas1,0≤n≤Mwr[n]=0,otherwise60 Wemaytruncateasignaltotheinterval0≤n≤Mbymultiplyingthesignalwithw[n].InthefrequencydomainweconvolvetheDTFTofthesignalwithΩ(M+1)sinjΩ−jMΩ2Wr(e)=e2sinΩ2Theeffectofthisconvolutionistosmeardetailandintroduceripplesinthevicinityofdiscontinuities.Thesmearingisproportionaltothemainlobewidth,whiletherippleisproportionaltothesizeofthesidelobes.Avarietyofalternativewindowsareusedinpracticetoreducesidelobeheightinreturnforincreasedmainlobewidth.Inthisproblemweevaluatetheeffectwindowingtime-domainsignalsontheirDTFT.TheroleofwindowinginfilterdesignisexploredinChapter8.TheHanningwindowisdefinedas0.5−0.5cos(2πn),0≤n≤Mwh[n]=M0,otherwise(a)AssumeM=50andusetheMATLABcommandffttoevaluatemagnitudespectrumoftherectan-gularwindowindBatintervalsofπ,π,andπ.50100200(b)AssumeM=50andusetheMATLABcommandffttoevaluatethemagnitudespectrumoftheHanningwindowindBatintervalsofπ,π,andπ.50100200(c)Usetheresultsfrom(a)and(b)toevaluatethemainlobewidthandpeaksidelobeheightindBforeachwindow.Usinganintervalofπ,themainlobewidthandpeaksidelobeforeachwindowcanbeestimatedfrom200thefigure,orfindingthelocalminimaandlocalnullsinthevicinityofthemainlobe.Ω:rad(dB)MainlobeSidelobewidthheightRectangular0.25-13.48Hanning0.5-31.48Note:sidelobeheightisrelativetothemainlobe.课后答案网TheHanningwindowhaslowersidelobes,butatthecostofawidermainlobewhencomparedtotherectangularwindowwww.hackshp.cn(d)Lety[n]=x[n]w[n]andy[n]=x[n]w[n]wherex[n]=cos(26πn)+cos(29πn)andM=50.rrhh100100UsethetheMATLABcommandffttoevaluate|Y(ejΩ)|indBand|Y(ejΩ)|indBatintervalsofπ.rh200Doesthewindowchoiceaffectwhetheryoucanidentifythepresenceoftwosinusoids?Why?Yes,sincethetwosinusoidsareveryclosetooneanotherinfrequency,26πand29π.100100SincetheHanningwindowhasawidermainlobe,itscapabilitytoresolvethesetwosinusoidisinferiortotherectangularwindow.Noticefromtheplotthattheexistenceoftwosinusoidsarevisiblefortherectangularwindow,butnotfortheHanning.(e)Lety[n]=x[n]w[n]andy[n]=x[n]w[n]wherex[n]=cos(26πn)+0.02cos(51πn)andM=50.rrhh100100UsethetheMATLABcommandffttoevaluate|Y(ejΩ)|indBand|Y(ejΩ)|indBatintervalsofπ.rh200Doesthewindowchoiceaffectwhetheryoucanidentifythepresenceoftwosinusoids?Why?61 Yes,herethetwosinusoidfrequenciesaresignificantlydifferentfromoneanother.Theseparationofπfromeachotherissignificantlylargerthanthemainlobewidthofeitherwindow.Hence,resolutionis4notaproblemfortheHanningwindow.Sincethesidelobemagnitudeisgreaterthan0.02ofthemainlobeintherectangularwindow,thesinusoidat51πisnotdistinguishable.Incontrast,thesidelobesofthe100Hanningwindowaremuchlower,whichallowsthetwosinusoidstoberesolved.P4.56(a)0−20−40(Omega)|dB:pi/50r|W−60−3−2−10123Omega0−20−40(Omega)|dB:pi/100r|W−60−3−2−10123Omega0−20−40(Omega)|dB:pi/200r|W−60−3−2−10123Omega课后答案网www.hackshp.cnFigureP4.56.Magnitudespectrumoftherectangularwindow62 P4.56(b)0−50(Omega)|dB:pi/50−100h|W−3−2−10123Omega0−50−100(Omega)|dB:pi/100h|W−3−2−10123Omega0−50−100(Omega)|dB:pi/200h|W−3−2−10123OmegaFigureP4.56.MagnitudespectrumoftheHanningwindowP4.56(c)0−5−10−15|zoomed−r20|W−25课后答案网−30−35−0.4−0.3−0.2−0.100.10.20.30.4www.hackshp.cnOmega0−10−20−30|zoomedh|W−40−50−60−0.4−0.3−0.2−0.100.10.20.30.4OmegaFigureP4.56.ZoomedinplotsofW(ejΩ)andW(ejΩ)rh63 P4.56(d)0−5−10−15−20(Omega)|dB−25r1|Y−30−35−3−2−10123Omega0−20−40(Omega)|dB−h160|Y−80−3−2−10123OmegaFigureP4.56.(d)Plotsof|Y(ejΩ)|and|Y(ejΩ)|rhP4.56(e)0−20−40(Omega)|dB−60r2|Y−80课后答案网−3−2−10123www.hackshp.cnOmega0−20−40−60(Omega)|dBh2−80|Y−100−3−2−10123OmegaFigureP4.56.(e)Plotsof|Y(ejΩ)|and|Y(ejΩ)|rh64 4.57.Letadiscrete-timesignalx[n]bedefinedas2(0.1n)−e2,|n|≤50x[n]=0,otherwiseUsetheMATLABcommandsfftandfftshifttonumericallyevaluateandplottheDTFTofx[n]andthefollowingsubsampledsignalsat500valuesofΩontheinterval−π<Ω≤π:(a)y[n]=x[2n](b)z[n]=x[4n]P4.5725201510DTFTx[n]5−3−2−10123Omega121086DTFTx[2n]42−3−2−10123Omega64DTFTx[4n]2课后答案网0−3−2−10123www.hackshp.cnOmegaFigureP4.57.PlotoftheDTFTofx[n]4.58.RepeatProblem4.57assuming2(0.1n)π−cos(n)e2,|n|≤50x[n]=20,otherwise65 P4.58121086DTFTx[n]42−3−2−10123Omega121086DTFTx[2n]42−3−2−10123Omega64DTFTx[4n]20−3−2−10123OmegaFigureP4.58.PlotoftheDTFTofx[n]4.59.Asignalx(t)isdefinedas23π−tx(t)=cos(t)e22(a)EvaluatetheFT课后答案网X(jω)andshowthat|X(jω)|≈0for|ω|>3π.ω2−3π3πe2X(jω)=πδ(ω−)+δ(ω+)∗√222πwww.hackshp.cn√−(ω−3π)2−(ω+3π)222=2π3e2+e2for|ω|>3π√−(ω−3π)2−(ω+3π)222|X(jω)|=2π3e2+e23π29π2√−(2)−(2)|X(jω)|≤2π3e2+e2≈1.2×10−4Whichissmallenoughtobeapproximatedaszero.66 P4.59(a)765)w4X(j321−8−6−4−202468wFigureP4.59.TheFourierTransformofx(t)Inparts(b)-(d),wecompareX(jω)totheFTofthesampledsignal,x[n]=x(nTs),forseveralsam-FTplingintervals.Letx[n]←−−−→Xδ(jω)betheFTofthesampledversionofx(t).UseMATLABtonumericallydetermineXδ(jω)byevaluating25X(jω)=x[n]e−jωTsδn=−25at500valuesofω课后答案网ontheinterval−3π<ω<3π.Ineachcase,compareX(jω)andXδ(jω)andexplainanydifferences.Noticethatx[n]issymmetricwithrespecttowww.hackshp.cnn,so25Xδ(jω)=x[0]+2x[n]cos(ωnTs)n=125=1+2x[n]cos(ωnTs)n=1Asidefromthemagnitudedifferenceof1,foreachT,X(jω)andX(jω)aredifferentwithin[−3π,3π]Tssδonlywhenthesamplingperiodislargeenoughtocausenoticeablealiasing.ItisobviousthattheworstaliasingoccurswhenT=1.s2(b)T=1s3(c)T=2s567 (d)T=1s2P4.59(b)3)w(j21Xd1−8−6−4−202468P4.59(c)w3)2w(j2Xd1−8−6−4−202468P4.59(d)w2.52)w1.5(j3Xd10.5−8−6−4−202468wFigureP4.59.ComparisonoftheFTtothesampledFT课后答案网www.hackshp.cn4.60.UsetheMATLABcommandffttorepeatExample4.14.68 P4.601510N=3250−3−2−10123Omega1510N=6050−3−2−10123Omega1510N=12050−3−2−10123OmegaFigureP4.60.Plotof|X(ejΩ)|andN|X[k]|4.61.UsetheMATLABcommandffttorepeatExample4.15.P4.61(a)654课后答案网32N=4000,M=1001www.hackshp.cn002468101214161820w:rad/sP4.61(b)65432N=4000,M=5001002468101214161820w:rad/s69 FigureP4.61.DTFTapproximationtotheFT,graphsof|X(ejΩ)|andNT|Y[k]|sP4.61(c)65432N=4000,M=25001002468101214161820w:rad/sP4.61(d)65432N=16000,M=25001099.51010.51111.51212.513w:rad/sFigureP4.61.DTFTapproximationtotheFT,graphsof|X(ejΩ)|andNT|Y[k]|课后答案网swww.hackshp.cn4.62.UsetheMATLABcommandffttorepeatExample4.16.AlsodepicttheDTFSapproximationandtheunderlyingDTFTforM=2001andM=2005.70 P4.62−110.80.60.4|Y[k]:M=400.2000.511.522.533.544.55Frequency(Hz)0.50.40.3|Y[k]:M=20000.20.100.340.360.380.40.420.440.460.480.5Frequency(Hz)FigureP4.62.DTFSApproximationP4.62−20.50.40.3|Y[k]:M=20010.20.1课后答案网00.340.360.380.40.420.440.460.480.5www.hackshp.cnFrequency(Hz)0.50.40.3|Y[k]:M=20050.20.100.340.360.380.40.420.440.460.480.5Frequency(Hz)FigureP4.62.DTFSApproximation71 P4.62−30.50.40.3|Y[k]:M=20100.20.100.340.360.380.40.420.440.460.480.5Frequency(Hz)FigureP4.62.DTFSApproximation4.63.Considerthesumofsinusoids1x(t)=cos(2πt)+2cos(2π(0.8)t)+cos(2π(1.1)t)2Assumethefrequencybandofinterestis−5π<ω<5π.(a)DeterminethesamplingintervalTssothattheDTFSapproximationtotheFTofx(t)spansthedesiredfrequencyband.ωa=5π2πTs<=0.1333ωachoose:Ts=0.1(b)Determinetheminimumnumberofsamples课后答案网MosothattheDTFSapproximationconsistsofdiscrete-valuedimpulseslocatedatthefrequencycorrespondingtoeachsinusoid.ForagivenM,thefrequencyintervalforsamplingtheDTFSis2πowww.hackshp.cnMo2πk1=2πTsMok1Mo=Ts2πk2=2π(0.8)TsMok2Mo=0.8Ts2πk3=2π(1.1)TsMok3Mo=1.1Ts72 wherek1,k2,k3areintegers.BychoosingTs=0.1,theminimumMo=100withk1=10,k2=8,k3=11.(c)UseMATLABtoplot1|Y(jω)|and|Y[k]|forthevalueofTchoseninparts(a)andM=M.Mδso(d)Repeatpart(c)usingM=Mo+5andM=Mo+8.P4.631.510.5|Y[k]:M=100000.511.522.53Frequency(Hz)1.510.5|Y[k]:M=105000.511.522.53Frequency(Hz)1.510.5|Y[k]:M=108000.511.522.53Frequency(Hz)FigureP4.63.Plotsof(1/M)|Yδ(jω)|and|Y[k]|fortheappropriatevaluesofTsandM课后答案网4.64.WedesiretousetheDTFStoapproximatetheFTofacontinuoustimesignalx(t)ontheband−ωa<ω<ωawithresolutionwww.hackshp.cnωrandamaximumsamplingintervalinfrequencyof∆ω.FindthesamplingintervalTs,numberofsamplesM,andDTFSlengthN.Youmayassumethatthesignaliseffectivelybandlimitedtoafrequencyωmforwhich|X(jωa)|≥10|X(jω)|,ω>ωm.PlottheFTandtheDTFSapproximationforeachofthefollowingcasesusingtheMATLABcommandfft.WeuseT<2π;M≥ωs;N≥ωstofindtherequiredT,M,Nforpart(a)and(b).sωm+ωaωr∆ωs1,|t|<1(a)x(t)=,ω=3π,ω=3π,and∆ω=πa2r480,otherwise2sin(ω)X(jω)=ωWewant:sin(ω)1≤for|ω|≥ωmω15π73 whichgivesωm=15πHence:2πTs<3π≈0.1215π+2chooseTs=0.1M≥26.67chooseM=28N≥160chooseN=16021−t11(b)x(t)=e2,ωa=3,ωr=,and∆ω=2π2821−ωX(jω)=√e22πWewant:2−ω1−9e2≤e210whichgivesωm=3.69Hence:Ts<0.939chooseTs=0.9M≥3.49chooseM=4N≥55.85课后答案网chooseN=56(c)x(t)=cos(20πt)+cos(21πt),ω=40π,ω=π,and∆ω=πwww.hackshp.cnar310ωs=2ωa=80πX(jω)=π(δ(ω−20π)+δ(ω+20π))Hence:2πTs<=0.01673ωachooseTs=0.01M≥600chooseM=600chooseN=M=60074 (d)Repeatcase(c)usingω=π.r10Hint:Besuretosamplethepulsesin(a)and(b)symmetricallyaboutt=0.chooseTs=0.01M≥2000chooseM=2000chooseN=M=2000For(c)and(d),oneneedstoconsiderX(jω)∗Wδ(jω),whereωTs−jωTM−1sin(M)W(jω)=es(2)2δωTsin(s)2P4.64(a)2.52)|w1.51Part(a):|Y(j0.50课后答案网−1−0.8−0.6−0.4−0.200.20.40.60.81Freq(Hz)www.hackshp.cnP4.64(b)0.4)|0.3w0.2Part(b):|Y(j0.10−0.5−0.4−0.3−0.2−0.100.10.20.30.40.5Freq(Hz)FigureP4.64.FTandDTFSapproximation75 P4.64(c)0.50.40.3Part(c):|Y[k]|0.20.1088.599.51010.51111.51212.513Freq(Hz)P4.64(d)0.50.40.3Part(d):|Y[k]|0.20.109.51010.511Freq(Hz)FigureP4.64.FTandDTFSapproximation4.65.TheoverlapandsavemethodforlinearfilteringisdiscussedinProblem4.54.WriteaMATLABm-filethatimplementstheoverlapandsavemethodusingffttoevaluatetheconvolutiony[n]=h[n]∗x[n]on0≤nM.Bysettingt=nTs,dt≈∆t=Ts,so∞Mt2|x(t)|2dt≈(nT)2|x(nT)|2Tsss−∞n=−MM=T3n2|x[n]|2sn=−M78 similarly∞M|x(t)|2dt≈T|x[n]|2s−∞n=−MTherefore1Mn2|x[n]|22n=−MTd≈TsM|x[n]|2n=−M(b)UsetheDTFSapproximationtotheFTandtheRiemannsumapproximationtoanintegraltoshowthat1∞ω2|X(jω)|2dω2−∞Bw=∞|X(jω)|2dω−∞1M|k|2|X[k]|22ωsk=−M≈2M+1M|X[k]|2k=−MDTFS;2π2M+1wherex[n]←−−−−→X[k],ω=2πisthesamplingfrequency,andX(jkωs)≈0for|k|>M.sTs2M+1Usingthe(2M+1)-pointDTFSapproximation,wehave:ωsωk=k2M+1henceωsdω≈∆ω=2M+11ωsX[k]=X(jk)(2M+1)Ts2M+1∞3M22ωs22ω|X(jω)|dω=k|X[k]|−∞2M+1k=−M∞M课后答案网2ωs2|X(jω)|dω=|X[k]|−∞2M+1k=−Mthereforewww.hackshp.cnM2212ωsk=−M|k||X[k]|Bw≈M2M+1|X[k]|2k=−M(c)Usetheresultfrom(a)and(b)andEq.(3.65)toshowthatthetime-bandwidthproductcomputedusingtheDTFSapproximationsatisfies11Mn2|x[n]|22M|k|2|X[k]|22n=−Mk=−M2M+1MM≥|x[n]|2|X[k]|24πn=−Mk=−M11Mn2|x[n]|22M|k|2|X[k]|22Tsωsn=−Mk=−MNoteTdBw≈M2M22M+1|x[n]||X[k]|n=−Mk=−M79 SinceTB≥1,andTsωs=2π,wehavedw22M+12M+111Mn2|x[n]|22M|k|2|X[k]|22n=−Mk=−M2M+1MM≥|x[n]|2|X[k]|24πn=−Mk=−M(d)RepeatComputerExperiment3.115todemonstratethattheboundin(c)issatisfiedandthatGaussianpulsessatisfytheboundwithequality.P4.67(d):o=(2M+1)/(4p)6000bound4000Product200000100200300400500600700800M:rectangularpulse800600400Product20000100200300400500600700800M:raisedcosine150100Product5000100200300400500600700800M:GaussianpulseFigureP4.67.PlotofComputerExperiment3.115课后答案网www.hackshp.cn80 CHAPTER5AdditionalProblems5.17(a)TheAMsignalisdefinedbyst()=A()1+kmt()cos()wtcacæöka=Acç÷1+-------------2cos()wctèø1+tToobtain50%modulation,wechooseka=1,whichresultsinthemodulatedwaveformofFig.1fors(t)withAc=1volt:−−(1+ka)Ac−−−−−−Figure1−−−(b)TheDSB-SCmodulatedsignalisdefinedbyst()=Amt()cos()wtccAc=-------------cos()wt2c课后答案网1+twhichhasthemodulatedwaveformofFig.2forAc=1volt:www.hackshp.cnAcFigure2Typically,aDSB-SCsignalexhibitsphasereversals.NosuchreversalsareexhibitedinFig.2becausethemodulatingsignalm(t)isnonnegativeforalltimet.1 5.18TheAMsignalisdefinedbyst()=A()1+kmt()cos()wtcacwhereAcos()wtisthecarrierandkisaconstant.Wearegivenw=2px105rad/sec;ccacthatis,fc=100kHz.3(a)mt()=Acos()2p´10t03w=2p´10rad/sec0f=1kHzcThefrequencycomponentsofs(t)forpositivefrequenciesare:f=100kHz0f+f==100+1101kHzc0f–f==100–199kHzc033(b)mt()=Acos()2p´10t+Acos()4p´10t00whichconsistsoftwosinusoidalcomponentswithfrequenciesf0=1kHzandf1=2kHz.Hence,thefrequencycomponentsofs(t)forpositivefrequenciesare:f=100kHzcf+f==100+1101kHzc0f–f==100–199kHzc0f+f==100+2102kHzc1f–f==100–298kHzc1(c)First,wenotethat课后答案网11cos()Xsin()Y=---sin()YX++---sin()YX–www.hackshp.cn22Hence,33mt()=Acos()2p´10tsin()4p´10t0A03A03=------sin()6p´10t+------sin()2p´10t22whichconsistsoftwosinusoidalcomponentswithfrequenciesf0=3kHzandf1=1kHz.Correspondingly,thefrequencycomponentsofs(t)forpositivefrequenciesare:f=100kHzcf+f==100+3103kHzc0f–f==100–397kHzc0f+f==100+1101kHzc12 f–f==100–199kHzc1(d)Hereweusetheformula:21cosq=---[]12+cos()q2Hence,23mt()=Acos()2p´10t0A03=------[]14+cos()p´10t2whichconsistsofdccomponentandsinusoidalcomponentoffrequencyf0=2kHz.Thefrequencycomponentsofs(t)forpositivefrequenciesaretherefore:f=100kHzcf+f==100+2102kHzc0f–f==100–298kHzc0(e)Forthispartoftheproblem,wefinduseoftheformulas:21cosq=---[]12+cos()q221sinq=---[]12–cos()q2Hence,1313mt()=---[]14+cos()p´10t+---[]18–cos()p´10t221313=1+---cos()4p´10t–---cos()8p´10t课后答案网22whichconsistsofdccomponent,andtwosinusoidalcomponentswithf0=2kHzandf1=4kHz.Correspondingly,thefrequencycomponentsofs(t)forpositivefrequenciesare:www.hackshp.cnf=100kHzcf+f==100+2102kHzc0f–f==100–298kHzc0f+f==100+4104kHzc1f–f==100–496kHzc1(f)Wefirstusetheformula31cosq=cosq×---[]12+cos()q23 11=---cosq+---cosqcos()2q221111=---cosq+------cos()q+2q+---cos()q–2q222213=---cos()3q+---cosq44Hence,23mt()=Acos()2p´10t0A033A03=------cos()6p´10t+---------cos()2p´10t44whichconsistsoftwosinusoidalcomponentswithfrequenciesf0=1kHzandf1=3kHz.Thefrequencycomponentsofs(t)aretherefore:f=100kHzcf+f==100+1101kHzc0f–f==100–199kHzc0f+f==100+3103kHzc1f–f==100–397kHzc1Note:Fornegativefrequencies,thefrequencycomponentsofs(t)arethenegativeofthoseforpositivefrequencies.5.19(a)Forasquarewavewithequalmark-to-spaceratioandfrequencyf0=500Hz,alternatingbetween0and1,hasthefrequencyexpansion(seeExample3.13ofthe课后答案网text):¥www.hackshp.cnjkw0t2pmt()=åmk[]e,w0=------Tk=-¥2sin()kw0T02p2ppwheremk[]=--------------------------------,wT==------×T------=---Tkw00T04201æökp=------sin------kpèø2Thatis,4 ¥11æökpjkw0tmt()=---å---sin------epkèø2k=-¥1222=---+---cos()wt+------cos3()wt+------cos5()¼wt+2p03p05p0Withw==2pf2p´500rad/sec,m(t)consistsofadccomponent,and00sinusoidalcomponentsoffrequencies0.5kHz,1.5kHz,2.5kHz(i.e.,oddharmonics).Hence,thefollowingcomponentsoftheAMsignals(t)forpositivefrequencies(withprogressivelydecreasingamplitude)areasfollows:f=100kHz0f+0.5=100.5kHz0f–0.5=99.5kHz0f+1.5=101.5kHz0f–1.5=98.5kHz0f+2.5=102.5kHz0f–2.5=97.5kHz0andsoon.(b)Whenthesquaremodulatingwavem(t)alternatesbetween-1and+1,thedccomponentiszero.Nevertheless,thefrequencycomponentsoftheAMsignalm(t)remainthesameasinpart(a);theonlydifferenceisinthecarrieramplitude.Note:Thefrequencycomponentsfornegativefrequenciesarethenegativeofthosefor课后答案网positivefrequencies.5.20(a)Forpositivefrequencies,thespectralcontentoftheAMsignalconsistsofthewww.hackshp.cnfollowing:Carrier:fc=100kHzUppersideband,occupyingthebandfrom100.3to103.1kHzLowersideband,occupyingthebandfrom99.7to96.9kHz(b)Forpositivefrequencies,thespectralcontentoftheAMsignalconsistsofthefollowing:Carrier:fc=100kHzUppersideband,occupyingthebandfrom100.05to115kHzLowersideband,occupyingthebandfrom99.95to85kHz5 Note:Fornegativefrequencies,thespectralcontentsoftheAMsignals(t)arethenegativeofthoseforpositivefrequencies.5.21ThepercentagemodulationisdefinedbyAmax–Amin9.75–0.25-----------------------------=---------------------------A+A9.75+0.25maxmin9.5==-------95%105.22BuildingonthesolutiontoProblem5.17,thefrequencycomponentsoftheDSB-Scsignalforpositivefrequenciesareasfollows:(a)f+f=101kHzc0f–f=99kHzc0(b)f+f=101kHzc0f–f=99kHzc0f+f=102kHzc1f–f=98kHzc1(c)f+f=103kHzc0f–f=97kHzc课后答案网0f+f=101kHzc1f–f=99kHzc1www.hackshp.cn(d)f+f=102kHzc0f–f=98kHzc0(e)f+f=102kHzc0f–f=98kHzc0f+f=104kHzc1f–f=96kHzc1(f)f+f=101kHzc06 f–f=99kHzc0f+f=103kHzc1f–f=97kHzc1Note:Fornegativefrequencies,thefrequencycomponentsoftheDSB-SCsignalarethenegativeofthoseforpositivefrequencies.5.23BuildingonthesolutionforProblem5.19,wefindthattheDSB-SCsignalhasthesamefrequencycontentforbothformsofthesquarewavedescribedinparts(a)and(b),asshownby:•Forpositivefrequencies,wehavethefrequencycomponents100.5,99.5,101.5,98.5,102.5,97.5kHz,andsoon,withprogressivelydecreasingamplitude.•Fornegativefrequencies,thefrequencycomponentsarethenegativeofthoseforpositivefrequencies.Note:Bydefinition,thecarrierissuppressedfromthemodulatedsignal.5.24(a)Forpositivefrequencies,wehaveUppersideband,extendingfrom100.3to103.1kHzLowersideband,extendingfrom99.7to96.9kHz(b)Forpositivefrequencies,wehaveUppersideband,extendingfrom100.05to115kHzLowersideband,extendingfrom99.95to85kHzNote:Fornegativefrequencies,thespectralcontentsoftheDSB-SCsignalarethe课后答案网negativesofthoseforpositivefrequencies.www.hackshp.cn5.25BuildingonthesolutiontoProblem5.18,wemaydescribethefrequencycomponentsofSSBmodulatorforpositivefrequenciesasfollows:(i)Uppersidebandtransmission:(a)f+f=101kHzc0(b)f+f=101kHzc0f–f=102kHzc17 (c)f+f=103kHzc0f–f=101kHzc1(d)f+f=102kHzc0(e)f+f=102kHzc0f–f=104kHzc1(f)f+f=101kHzc0f+f=103kHzc1(ii)Lowersidebandtransmission:(a)f–f=99kHzc0(b)f–f=99kHzc0f–f=98kHzc1(c)f–f=97kHzc0f–f=99kHzc1(d)f–f=98kHzc0(e)f–f=98kHz课后答案网c0f–f=96kHzc1www.hackshp.cn(f)f–f=99kHzc0f–f=97kHzc1Note:Forboth(i)and(ii),thefrequencycontentsofSSBmodulatedsignalfornegativefrequenciesarethenegativeofthefrequencycontentsforpositivefrequencies.5.26Withthecarriersuppressed,thefrequencycontentsoftheSSBsignalarethesameforbothsquarewavesdescribedunder(a)and(b),asshownby(forpositivefrequencies):(i)Uppersidebandtransmission:100.5,101.5,102.5kHz,andsoon.(ii)Lowersidebandtransmission:99.5,98.5,97.5kHz,andsoon.8 Note:Fornegativefrequencies,thefrequencycomponentsoftheSSBsignalarethenegativeofthoseforpositivefrequencies.5.27(i)Uppersidebandtransmission:(a)Uppersideband,occupyingthebandfrom100.3to103.1kHz(b)Uppersideband,occupyingthebandfrom100.05to115kHz(ii)Lowersidebandtransmission:(a)Lowersideband,occupyingthebandfrom99.7to96.9kHz(b)Lowersideband,occupyingthebandfrom99.5to85kHzNote:Fornegativefrequencies,thespectrumcontentsoftheSSBsignalarethenegativeofthoseforpositivefrequencies.5.28Thespectraofthepertinentsignalsareasfollows:MagnitudeCarrierf=w/(2p),kHz-202MagnitudeModulatingf=w/(2p),kHzsignal-4-2024课后答案网MagnitudeModulatingsignalLowersidefrequencyLowersidefrequencyfornegativefrequenciesforpositivefrequencieswww.hackshp.cnUppersidefrequencyCarrierCarrierUppersidefrequencyforfornegativefrequenciespositivefrequenciesf=(w/2p),kHz-6-4-20246Weclearlyseethatthisspectrumsuffersfromfrequencyoverlap,withthelowersidefrequencyfornegativefrequenciescoincidingwiththecarrierforpositivefrequencies;similarly,fornegativefrequencies.ThisspectrumisthereforeradicallydifferentfromthespectrumofaregularAMsignal;hence,itisnotpossibletorecoverthemodulatingsignalusinganenvelopedetector.9 5.29(a)MagnitudeModulatingsignalLowersidefrequencyLowersidefrequencyfornegativefrequenciesforpositivefrequenciesUppersidefrequencyUppersidefrequencyforfornegativefrequenciespositivefrequenciesf=(w/2p),kHz-6-4-20246Themodulatedsignalthereforeconsistsoftwosinusoidalcomponents,oneatfrequency2kHzandtheotheratfrequency6kHz.(b)Aconventionalcoherentdetectorconsistsofaproductmodulatorfollowedbyalow-passfilter.Theproductmodulatorissuppliedwithacarrieroffrequency2kHz.Thespectrumofthismodulatorconsistsofthefollowingcomponents:dccomponent(atzerofrequency)sinusoidalcomponentat4kHzanothersinusoidalcomponentat8kHzToextractthesinusoidalmodulatingsignalof4kHz,thelow-passfilterhasacutofffrequencyslightlyinexcessof4kHz.Theresultingoutputthereforeconsistsofthedesiredsinusoidalmodulatingsignal,plusadccomponentthatisundesired.Tosuppressthedccomponent,wehavetomodifythecoherentdetectorbypassingthedetectoroutputthroughacapacitor.5.30Foramessagebandwidthw=2.5px103rad/sandcarrierfrequencyw=2px103rad/s,课后答案网mcthespectraofthemessagesignalanddoublesideband-suppressedcarrier(DSB-SC)modulatedsignalmaybedepictedasfollows:www.hackshp.cn|M(jw)|M(0)w(103rad/s)-2p02p|S(jw)|w(103rad/s)-4.5p-2.5p-0.5p00.5p2.5p4.5p10 Theimportantpointtonotefromthepicturedepictedhereisthatthereisaclearseparationbetweenthesidebandslyinginthenegativefrequencyregionandthoseinthepositiveregion.Consequently,whentheDSB-SCmodulatedsignalisappliedtoacoherentdetector,theresultingdemodulatedsignalisareplicaoftheoriginalmessagesignalexceptforascalechange.When,however,thecarrierfrequencyisreducedtow=1.5px103rad/s,thelowercsidebandfornegativefrequenciesoverlapsthelowersidebandforpositivefrequencies,andthesituationchangesdramaticallyasdepictedbelow:|M(jw)|M(0)w(103rad/s)-2p-p0p2pS(jw)w(103rad/s)-3.5p-1.5p01.5p3.5p-0.5p0.5p|V(jw)|课后答案网w(103rad/s)-2p02pThespectrumlabeledV(jw)referstothedemodulatedsignalappearingattheoutputofthecoherentdetector.Comparingthisspectrumwiththeoriginalmessagespectrumwww.hackshp.cnM(jw),wenowseethatthemodulation-demodulationprocessresultsinamessagedistortion.Theconclusionstobedrawnfromtheresultspresentedaboveare:1.Toavoidmessagedistortionondemodulationduetosidebandoverlap,wemustchoosethecarrierfrequencyinaccordancewiththeconditionw³w.Theminimumcmacceptablevalueofwcisthereforewm.2.Forwc2w,thesidefrequenciesofthemodulatedsignalareasfollows(forpositivesmfrequencies)k=0:wmk=1:ws-wm,ws+wmk=2:2ws-wm,2ws+wmk=3:3ws-wm,3ws+wmandsoon.Fornegativefrequencies,thesidefrequenciesarethenegativeofthoseforpositivefrequencies.5.33(a)Theradiofrequency(RF)pulseisdefinedbyìTTïAcos()wt,–---££t---st()=ícc22ïî0,otherwiseThemodulatedsignals(t)isobtainedbymultiplyingthecarrierAccos(wct)byarectangularpulseofunitamplitudeanddurationT(centeredabouttheorigin).TheFouriertransformofAccos(wct)isAAcc------dww()–+------dww()+.2c2cTheFouriertransformoftherectangularpulseisequaltothesincfunctionTsinc(wT).Sincemultiplicationinthetimedomainistransformedintoconvolutioninthefrequencydomain,wemayexpresstheFouriertransformofs(t)asAT课后答案网cSj()w=----------[]sinc()T()ww–+sinc()T()ww+(1)2ccWhenwcTismuchgreaterthan2p,theoverlapbetweenthetwosincfunctionssinc(T(w-wcwww.hackshp.cn))andsinc(T(w+wc))iscorrespondinglysmall.WemaythenapproximateEq.(1)asfollows:ìïAcTï----------sinc()T()ww–c,w>02ïSj()w»í0,w=0(2)ïïAcTï----------sinc()T()ww+c,w<02î(b)ForwcT=20p,useoftheapproximateformulaofEq.(2)isjustified.Thewidthofthemainlobeofthesincfunctionis4p/T=wc/5.Thewidthofeachsidelobeiswc/10.WemaythusplotthemagnitudespectrumofS(jw)asshownonthenextpage.13 5.34(a)ThespectrumofeachtransmittedradarpulseiscloselydefinedbyEq.(2)giveninthesolutiontoProblem5.33.Theperiodictransmissionofeachsuchpulsehastheequivalenteffectofsamplingthisspectrumatarateequaltothepulserepetitionfrequencyoftheradar.Accordingly,wemayexpressthespectrumofthetransmittedradarsignals(t)asfollows:ìïåadww()––nwforw>0ïnc0ïnSj()w»í0forw=0(1)ïïåandww()++cnw0forw<0ïnîwhere课后答案网w=fundamentalfrequency02p=------T0www.hackshp.cnandthecoefficientanisdefinedbyTAnTa=------------1csincæö---------1n2TèøT00whereT1isthepulseduration.(b)ThespectrumS(jw)definedinEq.(1)isdiscreteinnature,consistingofasetofimpulsefunctionslocatedatw=+(wc+nw0),wheren=0,+1,+2,...Theenvelopeofthemagnitudespectrum|S(jw)|isthereforeasshownonthenextpage.14 T1Ac2T0Themainlobeofthespectrumhasawidthof4p6------=4p´10rad/sT1andthesidelobeshaveawidthof2px106rad/s.Theimpulsefunctionsareseparatedby3w=2p10rad/s0andthecarrierfrequencyw=2px09rad/s.Thus,thereare2000impulsefunctionscenvelopedbythemainlobeand1000impulsefunctionsenvelopedbyeachsidelobe.5.35TheDSB-SCmodulatedsignalisdefinedby课后答案网st()=Amt()cos()wtccLetthelocaloscillatoroutputinthecoherentdetectorbedenotedbycos(wct+Dwt),whereDwisthefrequencyerror.Multiplyingwww.hackshp.cns(t)bythelocalcarrieryieldss()t=st()cos()wt+Dwt1c=Amt()cos()wwtcos()t+DwtcccAc=------mt()[]cos()Dwt+cos()2wt+Dwt2cLow-passfilterings1(t)resultsintheoutputAcs()t=------mt()cos()Dwt22Inwords,theeffectoffrequencyerrorDwinthelocaloscillatoristoproduceanewDSB-SCmodulatedsignalwithaneffectivecarrierfrequencyofDw.ItisonlywhenDw=0thatthecoherentdetectorworksproperly.15 5.36Themixerproducesanoutputsignaloffrequencyequaltothesumorthedifferencebetweenitsinputsignalfrequencies.Therangeofsum-ordifference-frequenciesisfrom100kHz(representingthedifferencebetweeninputfrequencies1MHzand900kHz)to9.9MHz(representingthesumofinputfrequencies9MHzand900kHz).Thefrequencyresolutionis100kHz.5.37ThebasicsimilaritybetweenfullAMandPAMisthatinbothcasestheenvelopeofthemodulatedsignalfaithfullyfollowstheoriginalmessage(modulating)signal.Theydifferfromeachotherinthefollowingrespects:1.InAM,thecarrierisasinusoidalsignal;whereasinPAM,thecarrierisaperiodicsequenceofrectangularpulses.2.ThespectrumofanAMsignalconsistsofacarrierplusanuppersidebandandalowersideband.ThespectrumofaPAMsignalconsistsofacarrierplusanuppersidebandandalowersideband,whichrepeatperiodicallyatarateequaltothesamplingrate.5.38(a)gt()=sinc200()tThissincpulsecorrespondstoabandwidthof100Hz.HencetheNyquistrateis200Hz,andtheNyquistintervalis(1/200)seconds.2(b)gt()=sinc()200tThissignalmaybeviewedastheproductofthesincsignalsinc(200t)withitself.Sincemultiplicationinthetimedomaincorrespondstoconvolutioninthefrequencydomain,wefindthatthesignalg(t)hasabandwidthtwicethatofthesincpulsesinc(200t),thatis,200HzandtheNyquistintervalis1/400seconds.2(c)gt()=课后答案网sinc200()t+sinc()200tThebandwidthofg(t)isdeterminedbyitshighestfrequencycomponent.Withsinc(200t)havingabandwidthof100Hzandsincwww.hackshp.cn2(200t)havingabandwidthof200Hz,itfollowsthatthebandwidthofg(t)is200Hz.Correspondingly,theNyquistrateofg(t)is400HzanditsNyquistintervalis1/400seconds.5.39(a)Withasamplingrateof8kHz,thesamplingintervalis1T=-----------------s381´0=125msThereare24voicechannelsand1synchronizingpulse,sothetimeallottedtoeachchannelistT==------5mschannel2516 (b)IfeachvoicesignalissampledattheNyquistrate,thesamplingratewouldbetwicethehighestfrequencycomponent3.4kHz,thatis,6.8kHz.Thesamplingintervalisthen1T=----------------------s36.8´10=147msHence,147T==---------6.68mschannel255.40(a)Thebandwidthrequiredforeachsinglesidebandmodulatedchannelis10kHz.Thetotalbandwidthfor12suchchannelsis12x10=120kHz.(b)TheNyquistrateforeachchannelis2x10=20kHz.For12TDMsignals,thetotaldatarateis12x20=240kHz.Byusingasincpulsewhoseamplitudevariesinaccordancewiththemodulatingsignal,andwithzerocrossingsatmultiplesof(1/240)ms,wewouldneedaminimumbandwidthof120kHz.24005.41(a)TheNyquistratefors1(t)ands2(t)is160Hz.Therefore,------------mustbegreaterthanR2160Hz.Hence,themaximumvalueofRis3.(b)WithR课后答案网=3,wemayusethesignalformatshowninFig.1tomultiplexthesignalss1(t)ands2(t)intoanewsignal,andthenmultiplexs3(t),s4(t)ands5(t)markersforsynchronization:MarkerMarkerwww.hackshp.cn1s3001s2400Figure1.....Times3s4s1s3s4s2s3s4s3s4s3s4s3s4s3s4s3s4s3s4s1s3s4s2(1/7200)s.zerosamples17 Basedonthissignalformat,wemaydevelopthemultiplexingsystemshowninFig.2.2400Hzclock1s1s._..8.2400.2400delaydelayMarkerSamplers1(t)Samplers2(t)Figure2generatorMs5(t)SamplerU.XMMultiplexed1s1ss3(t)SamplerUsignal7200.7200Xdelaydelays4(t)Sampler5.42TheFouriertransformprovidesatoolfordisplayingthespectralcontentofacontinuous-timesignal.Acontinuous-wave(CW)modulatedsignals(t)involvesthemultiplicationofamessagesignalm(t)byasinusoidalcarrierc(t)=Accos(wct)inoneformoranother.TheCWmodulatedsignals(t)maybeviewedasamixturesignalthatinvolves,inoneformoranother,themultiplicationofm(t)byc(t).Multiplicationinthetime-domainistransformedintotheconvolutionoftwospectra,namely,theFouriertransformM(jw)andtheFouriertransformC(jw).TheFouriertransformC(jw)consistsofapairofimpulsesat+wc.Hence,theFouriertransformofthemodulatedsignal,denotedbyS(jw),containsacomponentM(jw-jwc)forpositivefrequencies.Fornegativefrequencies,wehavetheimageofthisspectrumwithrespecttotheverticalaxis.ThepicturesoportrayedteachesusthatthecarrierfrequencywcmustbegreaterthanthehighestfrequencycomponentofthemessagespectrumM(jw).Moreover,giventhatthisconditionissatisfied,Fourieranalysisteachesusthatrecoveryoftheoriginalmessagesignal课后答案网m(t)fromthemodulatedsignals(t)isindeedapracticalreality.Forexample,wemayuseaproductmodulatorconsistingofaproductmodulatorfollowedbyalow-passfilter.Theproductmodulator,suppliedbyalocalcarrieroffrequencywww.hackshp.cnwc,producesareplicaoftheoriginalmessagesignalm(t)plusanewmodulatedsignalwithcarrierfrequency2wc.Bydesigningthelow-passfiltertohaveacutofffrequencyslightlyhigherthanthehighestfrequencyofm(t),recoveryofm(t)exceptforascalingfactorisrealized.Considernextthecaseofpulse-amplitudemodulation(PAM).ThesimplestformofPAMconsistsofmultiplyingthemessagesignalm(t)byaperiodictrainofuniformlyspacedimpulsefunctions,withadjacentimpulsesbeingspacedTssecondsapart.Whatwehavejustdescribedistheinstantaneousformofsampling.Thus,thesamplingprocessrepresentsanotherformofamixturesignal.TheFouriertransformoftheperiodicpulsetrainjustdescribedconsistsofanewuniformlyspacedperiodictrainofimpulsesinthefrequencydomain,witheachpairofimpulsesspacedapartby1/Tshertz.Asalreadymentioned,multiplicationoftwotimefunctionsistransformedintotheconvolutionof18 theirspectrainthefrequencydomain.Accordingly,Fourieranalysisteachesusthattherewillbenospectraloverlapprovidedthatthesamplingfrequency1/Tsisnotlessthantwicethehighestfrequencycomponentoftheoriginalmessagesignalm(t).Hence,providedthatthisconditionissatisfied,therecoveryofareplicaofm(t)isindeedpossibleatthereceiver.Thisrecoverymay,forexample,taketheformofalow-passinterpolationfilterwithacutofffrequencyjustslightlygreaterthanthehighestfrequencycomponentofm(t).AdvancedProblems5.43Thenonlineardeviceisdefinedby2i()t=av()t+av()t(1)01i2iTheinputvi(t)isgivenbyv()t=Acos()wt+Acos()wt(2)iccmmwhereAccos(wct)isthecarrierwaveandAmcos(wmt)isthemodulatingwave.SubstitutingEq.(2)into(1):i()t=a()Acos()wt+Acos()wt01ccmm2+a()Acos()wt+Acos()wt2ccmm=aAcos()wt+aAcos()wt1cc1mm22+aAcos()wt+2aAAcos()wwtcos()t2cc2cmcm22+aAcos()wt(3)2mmUsingthetrigonometricidentity2课后答案网1cosq=---()cos()2q+12wemayrewriteEq.(3)intheequivalentform(afterarearrangementofterms):i()t=2aA()wtwww.hackshp.cn+aAAcos()wwtcos()t01cc2cmcm12+aAcos()wt+---aAcos()2wt1mm22mm121212+---aAcos()2wt++---aA---aA(4)22cc22c22mWemaynowrecognizethefollowingcomponentsintheoutputi0(t):1.Anamplitudemodulatedcomponent(AM)representedby2aAst()=aAæö1+----------------2mcos()wtcos()wt1cèøamc12.AsetofundesirablecomponentsrepresentedbytheremainingcomponentsofEq.(4).19 (a)Inamplitudemodulation,thecarrierfrequencywcistypicallymuchlargerthanthemodulationfrequencywm.Hence,thefrequencycontextofi0(t)maybedepictedasfollows(showingonlythepositivefrequencycomponentssoastosimplifythepresentation):wc0wm2wmwc-wmwcwc+wm2wc(b)Fromthisfigure,weseethatthedesiredAMcomponentoccupiesafrequencybandextendingfromthelowerside-frequencywc-wmtotheuppersidefrequencywc+wm.Toextractthiscomponent,weneedtopasstheoutputofthenonlineardevice,i0(t),throughaband-passfilter.Thefrequencyspecificationsofthisfilterareasfollows:•Passbandofawidthslightlylargerthan2wm,centeredonthecarrierfrequencywc.•Lowerstopband,lyingbelowthelowerside-frequencywc-wmandtherebysuppressingthedccomponentaswellasthefrequencycomponentswmand2wm.•Upperstopband,lyingabovetheupperside-frequencywc+wmandtherebysuppressingthefrequencycomponent2wc.5.44(a)Forthespecialcaseofaninfiniteunipolarsequence(binarysequenceconsistingofasquarewaveofequalmark-to-spaceratio),wehave课后答案网1222mt()=---+---cos()wt+------cos()3wt+------cos()5wt+¼2p03p05p0¥12www.hackshp.cn1=---+---å------------cos()()2k+1w0t2p2k+1k=0wherew0=2pf0=2p(1/2T0)=p/T0.(SeethesolutiontoProblem5.19).Theresulting00Ksignalisgivenby(assumingasinusoidalcarrierofunitamplitudeandfrequencywc)st()=mt()cos()wtc¥121=--2-cos()wct+--p-å2------------k+1cos()()2k+1w0tcos()wct(1)k=020 m(t)1.0......-5T0-3T0-T00T03T05T0Usingtheformula1cos()Acos()B=---[]cos()AB++cos()AB–2wemayrewriteEq.(1)intheequivalentform¥111st()=--2-cos()wct+--p-å2------------k+1[]cos()()()2k+1w0+wct+cos()()2k+1w0-wct(2)k=0Thespectrumofs(t)definedinEq.(2)isdepictedinFig.1(forpositivefrequencies):Figure1(b)FortheBPSKsignal,thebinarysequenceisrepresentedinitspolarnonreturn-to-zero课后答案网sequencewiththefollowingwaveform:www.hackshp.cnm(t)+1-3T0-2T0-T0T02T03T04T0t0-1Correspondingly,m(t)hastheFourierseriesrepresentation:21 ¥41mt()=---å---------------cos()()2k+1w0t(3)p2k+1k=1wherew0=p/T0.ThebinarysequenceofEq.(3)differsfromthatofEq.(1)intworespects:•Ithasnodccomponent•Itisscaledbyafactorof2.TheresultingBPSKsignalisdefinedby(assumingacarrierofunitamplitudeandfrequencywc)st()=mt()cos()wtc¥41=---å---------------cos()()2k+1w0tcos()wctp2k+1k=1¥21=--p-å---------------2k+1[]cos()()()2k+1w0+wct+cos()()()2k+1w0-wct(4)k=1Thespectrumofs(t)definedinEq.(4)isdepictedinFig.2:Figure2课后答案网(c)TheenergyoftheunipolarsignaldefinedinEq.(2)isgivenbywww.hackshp.cn¥1T22---òst()td=åSk()(5)T0k=-¥Theleft-handsideofEq.(5)yields1T21–T0¤22–T0¤22–T02E==---Tòst()td---------2Tòst()dtsò()tdts+ò()ttd00–T0–T0–T0¤21–T02=---------ò()1td2T0–T0¤21=---(6)2Theright-handsideofEq.(5)yields,inlightofEq.(2),theenergy22 22æö1()2p¤æö11E=---+-----------------1+++---------¼èø22èø925()212æö11=---+-----1+++---------¼4p2èø925212p=---+-----×-----4p28111==---+------442whichcheckstheresultofEq.(2).Forthepolarnonreturn-to-zerosignalofEq.(3),theenergyis1T¤22E=---òst()tdT–2T¤1T0¤22T0¤22T02=---------2Tòst()dts+ò()tdts+ò()ttd0–T0–2T0¤–2T0¤1T0¤22T0¤22T02=---------2Tò()–1dt+ò()1dt+ò()–1td0–T0–2T0¤–2T0¤1T0T0==---------------++T------12T2020ApplicationofParseval’stheoremyields()4p¤221212E=-----------------()12++++æö1---æö---æö---¼2èø3èø5èø7课后答案网()28æö111=------1++++---------------¼2èø92549pwww.hackshp.cn28p==-----×-----128p(d)Toraisetheenergyoftheunipolarsignaltoequalthatofthepolarnonreturn-to-zerosignal,theamplitudeoftheformersignalhastobeincreasedbythescalefactor2.(e)ExaminingtheOOKsignalofEq.(2),weseethatitcontainsacarriercomponentinadditiontosidefrequencies,hencethesimilaritytoanAMsignal.Ontheotherhand,examiningEq.(4)fortheBPSKsignal,weseethatitlacksacarriercomponent,hencethesimilaritytoaDSB-SCsignal.23 5.45Themultiplexedsignalisst()=Am()tcos()wt+Am()tsin()wtc1cc2cwherem1(t)andm2(t)aretheincomingmessagesignals.TakingtheFouriertransformofs(t):AcSj()w=------[]M()jw–jw+M()jw+jw21c1cAc+------[]M()jw–jw+M()jw+jw2j2c2cFTFTwherem1()t«M1()jwandm2()t«M2()jw.WithH(jw)denotingthetransferfunctionofthechannel,theFouriertransformofthereceivedsignalisRj()w=Hj()wSj()wAc=------Hj()w[]M()jw–jw+M()jw+jw21c1cAc+------Hj()w[]M()jw–jw+M()jw+jw22c2cFTwherert()«Rj()w.Torecoverm1(t),wemultiplythereceivedsignalr(t)bycos(wct)andthenpasstheresultingoutputthroughalow-passfilterwithcutofffrequencyequaltothemessagebandwidth.Theresultofthisprocessingisasignalwiththespectrum1S()jw=---()Rj()w–jw+Rj()w+jw12ccAc=------Hj()w–jwc[M1()jw–j2wc+M1()jw411+---M()jw–j2w–---M()jw]课后答案网j2cj2Ac+------Hj()w–jwc[M1()jw+M1()jw+j2wc41www.hackshp.cn1+---M()jw–---M()jw+j2w](1)j2j2cForarealchannel,theconditionHj()w+jw=H*()jw–jwccisequivalenttoHj()w+jw=Hj()w–jwccThisequivalencefollowsfromthefactthatforachannelwithreal-valueimpulseresponseh(t),wehaveHj()–w=H*()jw.Hence,substitutingthisconditionintoEq.(1):AcS()jw=------Hj()w–jwM()jw12c1Ac+------Hj()w–jwc[M1()jw–j2wc424 1++M()jw+j2w---M()jw–j2w1cj2c1–---M()jw+j2wj2cPassingthesignaldefinedbyS1(jw)throughalow-passfilterofcutofffrequencyequaltothemessagebandwidth,wm,wegetanoutputwhoseFouriertransformisequaltoAc------Hj()w–jwM()jwfor–w££ww2c1mmRecognizingtheband-passnatureofH(jw),weimmediatelyseethattheoutputsoobtainedisindeedareplicaoftheoriginalmessagesignalm1(t).Similarly,torecoverm2(t),wemultiplyr(t)bysin(wct)andthenpasstheresultingsignalthroughalow-passfilter.Inthiscase,wegetanoutputwithaspectrumequaltoAc------Hj()w–jwM()jw2c25.46Theblockdiagramofthescramblerisasfollows:m(t)Productv1(t)High-passv2(t)Productv3(t)Low-passs(t)modulatorfiltermodulatorfiltercos(wct)cos(wct+wbt)Figure1(a)Thefirstproductmodulatoroutputisv()t=mt()cos()wt1cWhenv1(t)isprocessedbythehigh-passfilter,wegetv2(t).Thesecondproductmodulatoroutputis课后答案网v()t=v()tcos()wt+wt32cbThemagnitudespectraofwww.hackshp.cnm(t),v1(t),v2(t),v3(t)ands(t)areillustratedinFig.2.Fromthisfigure,weobservethatv1(t)isaDSB-SCmodulatedsignal,v2(t)isaSSB-modulatedsignal,v3(t)isaDSB-SCmodulatedsignalwhoselowestfrequencyiszero,ands(t)isalow-passsignalwhosespectrumisuniquelydefinedbym(t).|M(jw)|w-wb-wa0wawb25 |V1(jw)|wc-wbwc+wbw-wcwcFigure2wc-wawc+wa|V2(jw)|w-wcwc-wc-wb-wc-wawc+wawc+wb|V3(jw)|w-wb+wa0wb-wa2wc+2wb|S(jw)|课后答案网w-wb+wa0wb-waTofindanexpressionforthescrambledvoicesignalwww.hackshp.cns(t),weneedtoinvoketheHilberttransform.Specifically,theHilberttransformofm(t)isdefinedby1¥m()tmˆ()t=---ò------------dtp–¥t–tThemˆ()tmaybeviewedastheconvolutionofm(t)with1/(pt).TheFouriertransformof1/(pt)isequalto-jsgn(w),whereì1forw>0ïsgn()w=í0forw=0ïî–for1w<0Hence,Mˆ()jw=–jsgn()wMj()w26 Usingthisrelation,itisastraightforwardmattertoshowthatthespectrumofs(t)definedpreviouslyisthesameasthatofthefollowingexpression:11st()=---mt()cos()wt+---mˆ()tsin()wt4b4b(b)Withs(t)astheinputtothescrambler,theoutputofthefirstproductmodulatorisv()t=st()cos()wt1c11=---mt()cos()wwtcos()t+---mˆ()tsin()wwtcos()t4bc4bc1=---mt()[]cos()wt+wt+cos()wt–wt8cbcb1+---mˆ()t[]sin()wt+wt–sin()wt–wt8cbcbThehigh-passfilteroutputistherefore11v()t=---mt()cos()()w+wt+---mˆ()tsin()()w+wt28cb8cbCorrespondingly,theoutputofthesecondproductmodulatorisv()t=v()tcos()()w+wt32cb121=---mt()cos()()w+wt+---mˆ()tsin()()w+wtcos()()w+wt8cb8cbcb1=------mt()1611+------mt()cos()2()()w+wt+------mˆ()tsin()()w+wt16cb16cbThescrambleroutputistherefore1v()t=------mt()0课后答案网16whichisascaledversionoftheoriginalmessagesignal.www.hackshp.cn27 5.47Consideramessagesignalm(t)whosefrequencycontentliesinsidethebandwl££wwm.Assumingthatthecarrierfrequencywc>wm,wemaydescribethespectrumoftheSSBmodulatedsignals(t)(withonlyitsuppersidebandretained)asfollows:|M(jw)|w-wm-wl0wlwmw-wc-wm-wc0wcwc+wmFigure1-wc-wlwc+wlTodemodulates(t),weapplyittothecoherentdetectorofFig.P5.47.Letv(t)denotetheoutputoftheproductmodulatorinthisfigure,andletv0(t)denotetheoutputofthelow-passfilter.Wemaythendescribethecorrespondingspectraofv(t)andv0(t)ingraphicalformsasshowninFig.2.|S(jw)|w课后答案网-wc-wm-wc0wcwc+wm-wc-wlwc+wlwww.hackshp.cn|V(jw)|wFigure2-2w0c-wm-2wc-wm-wlwlwm2wc2wc+wl-2wc-wl2wc+wmFromthisfigureweseethattheproductmodulatoroutputv(t)consistsoftwocomponents:•Ascaledversionoftheoriginalmessagesignalm(t)•AnewSSBmodulatedsignalwithcarrierfrequency2wc28 Thelattercomponentissuppressedbythelow-passfilter,leavingtheoriginalmessagesignalasthedetectoroutputasshownbythespectrumofFig.3.|V0(jw)|wFigure30-wm-wlwlwm5.48Themultiplexedsignalis4st()=å[]cos()wat+ak-1+cos()wbt+bk-1mk()t(1)k=1wherea0=b0=0.Thecorrespondingoutputoftheproductmodulatorinthecoherentdetectorofthereceiverisv()t=st()[]cos()wt+a+cos()wt+b(2)iai-1bi-1wherei=1,2,3,4.ThereusingEq.(1)in(2):4vi()t=åmk()t[]cos()wat+ak-1+cos()wbt+bk-1k=1´[]cos()wt+a+cos()wt+bai-1ai-1Expandingterms:4vi()t=åmk()t[cos()wat+ak-1cos()wat+ai-1]k=1+cos()wt+acos()wt+bak-1bi-1+cos()wt+bcos()wt+a课后答案网bk-1ai-1[+cos()wt+bcos()wt+b]bk-1bi-141=--2-åmk()twww.hackshp.cn[]cos()ak-1–ai-1+cos()bk-1–bi-1k=1+cos()2wt++aa+cos()2wt++bbak-1i-1bk-1i-1+cos()()w+wt++ababk-1i-1+cos()()w–wt+a–babk-1i-1+cos()()w+wt++ababi-1k-1+cos()()w–wt+a–b]abi-1k-1Thelow-passfilterinthecoherentdetectorremovesthesixhigh-frequencycomponentsofvi(t),leavingtheoutput29 41v¢i()t=--2-åmk()t[]cos()ak-1–ai-1+cos()bk-1–bi-1k=1Therequirementonakandbkisthereforeì2,ik=cos()ak-1–ai-1+cos()bk-1–bi-1=íî0,ik¹where(i,k)=1,2,3,4.5.49ConsideranincomingAMsignalofbandwidth10kHzandcarrierfrequencywcthatliesintherange(0.535-1.605)MHz.Itisrequiredtofrequencytranslatethismodulatedsignaltoafixedbandcenteredat0.455MHz.Theproblemistodeterminetherangeoftuningthatmustbeprovidedinthelocaloscillator.Letwlocaldenotethelocaloscillatorfrequency,whichisrequiredtosatisfythecondition6w–2w=p´10´0.455rad/sclocalorf–f=0.455MHzclocalwherebothfcandflocalareexpressedinMHz.Thatis,f=f–0.455localcWhenfc=0.535MHz,flocal=0.535-0.455=0.08MHz.Whenfc=1.605MHz,flocal=1.605-0.455=1.15MHz.Thustherequiredrangeoftuningofthelocaloscillatoris0.08-1.15MHz,independentoftheAMsignal’sbandwidth.课后答案网5.50Consideraperiodictrainofrectangularpulses,eachofdurationT0.Assumingthatapulseofthetrainiscenteredontheorigin,wemayexpanditasaFourierseries:www.hackshp.cn¥j2pnfstct()=åfssinc()nfsT0en=-¥wherefsisthepulserepetitionfrequencyandtheamplitudeofeachrectangularpulseis1/T0(i.e.,eachpulsehasunitarea).TheassumptionthatfsT0>>1meansthattheimpulsefunctionsinthespectrumoftheperiodicpulsetrainc(t)arewellseparatedfromeachother.Multiplyingamessagesignalg(t)byc(t)yieldsthePAMsignalst()=ct()gt()30 ¥j2pnfst=åfssinc()nfsT0gt()en=-¥HencetheFouriertransformofs(t)isgivenby¥Sj()w=åfssinc()nfsT0Gj()w–jnwsn=-¥wherews=2pfs.Thusthespectrumofanaturallysampledsignalconsistsoffrequency-shiftedreplicasofthemessagespectrumG(jw),withthenthreplicabeingscaledinamplitudebythefactorfssinc(nfsT0),whichdecreaseswithincreasingn.5.51(a)Thespectrumofthecarriercn[]=cos()Wn,n=±12,±,¼cisdefinedbyjW1Ce()=---[]d()jW–jWn+d()jW+jWn,n=±12,±,¼2ccwhereW>WW,withdenotingthehighestfrequencycomponentofm[n]insidecmmtherange0££Wp.Thespectrumofthemodulatedsignalsn[]=cn[]mn[],jWjWjWnamely,S(e),isobtainedbyconvolvingCe()withM()e.(i)n=1:|S(ejW)|...课后答案网...Wwww.hackshp.cn-p+Wc0Wcp+Wc-p+Wc-Wm-p+Wc+WmWc-WmWc+Wm(ii)n=-1:|S(ejW)|......W-p-Wc-Wc0p-Wc-p-Wc-Wm-p-Wc+Wm-Wc-Wm-Wc+Wmp-Wc-Wmp-Wc+Wm31 (iii)n=2|S(ejW)|2Wc-p-Wm2Wc-p+Wm......W02Wc-p2Wc2Wc+p2Wc-Wm2Wc+Wm2Wc+p-Wm2Wc+p+Wm(iv)n=-2|S(ejW)|-p-2Wc+Wm-2Wc+Wm......W-p-2Wc-2Wc0p-2Wc-p-2Wc-Wm-2Wc-Wmp-2Wc-Wmp-2Wc+Wmandsoonforn=+3,...Thetransmitterincludesaband-passdigitalfilterdesignedtopassthespectralbandscenteredon+Wcandrejectallotherbands.(b)FollowingthecoherentdetectordescribedinSection5.5fordemodulationofaDSB-SCmodulatedsignalofthecontinuous-timevariety,wemaypostulatetheschemeofFig.1fordemodulationofthediscrete-timeDSB-SCmodulatedsignal:课后答案网s[n]Low-passm[n]digitalwww.hackshp.cnfilterFigure1cos(Wcn)Hereitisassumedthat(1)thelocaloscillatorsupplyingcos(Wn)issynchronouswiththecarriergeneratorusedinthetransmitter,and(2)thelow-passdigitalfilterisdesignedwithacutofffrequencyslightlygreaterthanWm.32 ComputerExperiments%Problem5.52clearfm=1e3;fc=2e4;mu=0.75;t=1e-6:1e-6:0.1;y=(1+mu*cos(2*pi*fm*t)).*cos(2*pi*fc*t);figure(1)plot(t,y);xlim([00.002])xlabel(’Time(s)’)ylabel(’Amplitude’)title(’AMWave:fc=2e4Hz,fm=1e3Hz,u=0.75’)L=length(y);Y=fft(y,L);Y=fftshift(Y);Ys=Y.*conj(Y);f=1e6/L*(-(L/2):L/2-1);figure(2)plot(f,Ys)xlim([-0.5e50.5e5])xlabel(’Frequency(Hz)’)ylabel(’Power’)title(’SpectrumAMWave:fc=2e4Hz,fm=1e3Hz,u=0.75’)课后答案网AMWave:fc=2e4Hz,fm=1e3Hz,m=0.7521.5www.hackshp.cn10.50Amplitude−0.5−1−1.5Figure1:ModulatedSignal−200.20.40.60.811.21.41.61.82Time(s)−3x1033 x109SpectrumAMWave:fc=2e4Hz,fm=1e3Hz,m=0.752.521.5Power10.50Figure2:Spectrumof−5−4−3−2−1012345Frequency(Hz)4modulatedsignalx10%Problem5.53fo=1;t=0:0.0001:5;m=sawtooth(2*pi*fo*t,0.5);figure(1)plot(t,m)xlabel(’Time(s)’)ylabel(’Amplitude’)title(’TriangularWave1Hz’)课后答案网mu=0.8;y=(1+mu*m).*cos(2*pi*25*t);figure(2);www.hackshp.cnplot(t,y);xlim([02])xlabel(’Time(s)’)ylabel(’Amplitude’)title(’ModulatedWave’)figure(3)L=length(y);Y=fft(y,L);Y=fftshift(Y);Ys=abs(Y);f=1e4/L*(-(L/2):L/2-1);plot(f,Ys)34 xlim([-5050])xlabel(’Frequency(Hz)’)ylabel(’Magnitude’)title(’SpectrumAMWave:fc=25Hz,fm=1Hz,u=0.8’)TriangularWave1Hz10.80.60.40.20Amplitude−0.2−0.4−0.6−0.8−100.511.522.533.544.55(a)Figure1:WaveformofTime(s)modulatingsignalModulatedWave21.510.50课后答案网Amplitude−0.5−1www.hackshp.cn−1.5(b)Figure2:Modulatedsignal−200.20.40.60.811.21.41.61.82Time(s)35 x104SpectrumAMWave:fc=25Hz,fm=1Hz,m=0.82.521.5Magnitude10.5Figure30−50−40−30−20−1001020304050Frequency(Hz)5.54x104SpectrumAMWave:fc=1Hz,fm=1Hz,m=0.832.5Carrier2MessageOverlap1.5Magnitude10.5课后答案网Figure10−20−15−10−505101520Frequency(Hz)x104SpectrumAMWave:fc=2Hz,fm=1Hz,www.hackshp.cnm=0.832.5Carrier21.5Magnitude1Message0.50Figure2−20−15−10−505101520Frequency(Hz)36 %Problem5.55fo=1;t=0:0.0001:5;m=sawtooth(2*pi*fo*t,0.5);figure(1)plot(t,m)xlabel(’Time(s)’)ylabel(’Amplitude’)title(’TriangularWave1Hz’)mu=0.8;y=m.*cos(2*pi*25*t);figure(2);plot(t,y);xlim([02])xlabel(’Time(s)’)ylabel(’Amplitude’)title(’DSSCModulatedWave’)figure(3)L=length(y);Y=fft(y,L);Y=fftshift(Y);Ys=abs(Y);f=1e4/L*(-(L/2):L/2-1);plot(f,Ys)xlim([-5050])xlabel(’Frequency(Hz)’)ylabel(’Magnitude’)课后答案网title(’SpectrumofDSSCAMWave:fc=25Hz,fm=1Hz,u=0.8’)www.hackshp.cn37 DSSCModulatedWave10.80.60.40.20Amplitude−0.2−0.4−0.6−0.8(a)Figure1:Modulatedsignal−100.20.40.60.811.21.41.61.82Time(s)SpectrumofDSSCAMWave:fc=25Hz,fm=1Hz,m=0.8120001000080006000Magnitude40002000课后答案网0(b)Figure2:Spectrumof−50−40−30−20−1001020304050www.hackshp.cnFrequency(Hz)modulatedsignal38 %Problem5.56clear;clc;wc=0.5*pi;res=0.001;Fs=1;sam=floor(1/(Fs*res));t=0:res:10-res;m=sin(wc*t);f=1/res*([0:1/(length(t)-1):1]-0.5);T=0.5;%Pickpulsedurationvalue(0.05,0.1,0.2,0.3,0.4,0.5)durInSam=floor(T/res);%numberofsamplesinpulsesm=zeros(size(t));%Reservevectorforoutputwaveformfori=0:floor(length(t)/sam)-1;%loopovernumberofsamplessm((i*sam)+1:(i*sam)+durInSam)=m(i*sam+1);endy=fftshift(abs(fft(sm)));figure(1)subplot(2,1,1)plot(t,sm);xlabel(’Time(s)’)ylabel(’Amplitude’)title(’PAMWave:T_0=0.5s’)subplot(2,1,2)plot(f,y)xlabel(’Frequency(Hz)’)课后答案网ylabel(’Amplitude’)xlim([-1010])www.hackshp.cn39 PAMWave:T=0.05s010.50Amplitude−0.5−1012345678910Time(s)200150100Amplitude50Figure1:T0=0.05s0−50−40−30−20−1001020304050Frequency(Hz)PAMWave:T=0.1s010.50Amplitude−0.5−1012345678910Time(s)350300250200课后答案网150Amplitude100500Figure2:T=0.1s−25−20−15www.hackshp.cn−10−505101520250Frequency(Hz)40 PAMWave:T=0.2s010.50Amplitude−0.5−1012345678910Time(s)700600500400300Amplitude200100Figure3:T=0.2s00−15−10−5051015Frequency(Hz)PAMWave:T=0.3s010.50Amplitude−0.5−1012345678910Time(s)1000800600Amplitude400课后答案网2000−15−10−5051015Figure4:T0=0.3swww.hackshp.cnFrequency(Hz)41 PAMWave:T=0.4s010.50Amplitude−0.5−1012345678910Time(s)140012001000800600Amplitude4002000−10−8−6−4−20246810Figure5:T0=0.4sFrequency(Hz)PAMWave:T=0.5s010.50Amplitude−0.5−1012345678910Time(s)200015001000Amplitude课后答案网5000Figure6:T=0.5s−10−8−6−4−202468100www.hackshp.cnFrequency(Hz)42 %Problem5.57clear;clc;res=1e-6;t=0:res:0.01-res;modf=1000;m=sin(modf*2*pi*t);%SamplespersamplingperiodfreqT=1e4;periodT=1/freqT;pulseDur=1e-5;samplesPerPulse=floor(pulseDur/res);samplesPerT=floor(periodT/res);%Numberofcompletesamplingcycleswecangetionthesignalmnum=floor(length(m)/samplesPerT);r=[];fori=1:num,r=[rones(1,samplesPerPulse)zeros(1,samplesPerT-samplesPerPulse)];endfigure(1)subplot(2,1,1)y=r.*m;plot(t,y)title(’NaturallySampledWaveform’)xlabel(’Time(s)’)ylabel(’Amplitude’)课后答案网subplot(2,1,2)plot(t,y)www.hackshp.cntitle(’NaturallySampledWaveform’)xlabel(’Time(s)’)ylabel(’Amplitude’)set(gca,’xlim’,[00.001])figure(2)Fy=fftshift(abs(fft(y)));f=1/res*([0:1/(length(t)-1):1]-0.5);subplot(2,1,1)plot(f,Fy)title(’SpectrumoftheModulatedWaveform’)xlabel(’Frequency(Hz)’)ylabel(’Magnitude’)43 subplot(2,1,2)plot(f,Fy)title(’SpectrumoftheModulatedWaveform’)xlabel(’Frequency(Hz)’)ylabel(’Magnitude’)set(gca,’xlim’,[-1e51e5])NaturallySampledWaveform10.50Amplitude−0.5−100.0010.0020.0030.0040.0050.0060.0070.0080.0090.01Time(s)NaturallySampledWaveform10.50Amplitude−0.5−1(a)Figure100.10.20.30.40.50.60.70.80.91Time(s)−3x10SpectrumoftheModulatedWaveform500400300200Magnitude课后答案网1000−5−4−3−2−1012345Frequency(Hz)5www.hackshp.cnx10SpectrumoftheModulatedWaveform500400300200Magnitude1000−1−0.8−0.6−0.4−0.200.20.40.60.81(b)Figure2Frequency(Hz)5x1044 SolutionstoAdditionalProblems6.26.Asignalx(t)hasLaplacetransformX(s)asgivenbelow.Plotthepolesandzerosinthes-planeanddeterminetheFouriertransformofx(t)withoutinvertingX(s).2(a)X(s)=s+1s2+5s+6(s+j)(s−j)X(s)=(s+3)(s+2)zerosat:±jpolesat:−3,−2X(jω)=X(s)|s=jω−ω2+1=−ω2+5jω+6Pole−ZeroMap10.80.60.40.20ImagAxis−0.2课后答案网−0.4−0.6www.hackshp.cn−0.8−1−3.5−3−2.5−2−1.5−1−0.50RealAxisFigureP6.26.(a)Pole-ZeroPlotofX(s)2(b)X(s)=s−1s2+s+1(s+1)(s−1)X(s)=(s+0.5−j3)(s+0.5+j3)44zerosat:±11 √−1±j3polesat:2X(jω)=X(s)|s=jω−ω2−1=−ω2+jω+1Pole−ZeroMap10.80.60.40.20ImagAxis−0.2−0.4−0.6−0.8−1−1−0.8−0.6−0.4−0.200.20.40.60.81课后答案网RealAxisFigureP6.26.(b)Pole-ZeroPlotofX(s)(c)X(s)=1+2s−4s−2www.hackshp.cn3(s−10)X(s)=3(s−4)(s−2)10zeroat:3polesat:4,2X(jω)=X(s)|s=jω12=+jω−4jω−22 Pole−ZeroMap10.80.60.40.20ImagAxis−0.2−0.4−0.6−0.8−100.511.522.533.54RealAxisFigureP6.26.(b)Pole-ZeroPlotofX(s)6.27.DeterminethebilateralLaplacetransformandROCforthefollowingsignals:(a)x(t)=e−tu(t+2)∞X(s)=x(t)e−stdt−∞∞课后答案网=e−tu(t+2)e−stdt−∞∞=e−t(1+s)dt−2www.hackshp.cne2(1+s)=1+sROC:Re(s)>-1(b)x(t)=u(−t+3)3X(s)=e−stdt−∞−e−3s=sROC:Re(s)<03 (c)x(t)=δ(t+1)∞X(s)=δ(t+1)e−stdt−∞=esROC:alls(d)x(t)=sin(t)u(t)∞1X(s)=ejt−e−jte−stdt02j∞∞1t(j−s)1−t(j+s)=edt−edt02j02j1−11=−2jj−sj+s1=(1+s2)ROC:Re(s)>06.28.DeterminetheunilateralLaplacetransformofthefollowingsignalsusingthedefiningequation:(a)x(t)=u(t−2)∞X(s)=x(t)e−stdt0−∞=u(t−2)e−stdt0−课后答案网∞=e−stdt2e−2swww.hackshp.cn=s(b)x(t)=u(t+2)∞X(s)=u(t+2)e−stdt0−∞=e−stdt0−1=s(c)x(t)=e−2tu(t+1)4 ∞X(s)=e−2tu(t+1)e−stdt0−∞=e−t(s+2)dt0−1=s+2(d)x(t)=e2tu(−t+2)∞X(s)=e2tu(−t+2)e−stdt0−2=et(2−s)dt0−e2(2−s)−1=2−s(e)x(t)=sin(ωot)∞1X(s)=ejωot−e−jωote−stdt0−2j∞∞1t(jωo−s)−t(jωo+s)=edt−edt2j0−0−1−11=−2jjωo−sjωo+sωo=s2+ω2课后答案网o(f)x(t)=u(t)−u(t−2)www.hackshp.cn2X(s)=e−stdt0−1−e−2s=ssin(πt),03(d)x(t)=cos(3t)u(−t)∗e−tu(t)La(t)∗b(t)←−−−→A(s)B(s)s1X(s)=−s2+9s+1ROC:-11(f)x(t)=etdtde−2tu(−t)www.hackshp.cnL−1a(t)=e−2tu(−t)←−−−→A(s)=s+2dLb(t)=a(t)←−−−→B(s)=sA(s)dttL1−sx(t)=eb(t)←−−−→X(s)=B(s−1)=s+1ROC:Re(s)<-16.42.UsethetablesoftransformsandpropertiestodeterminethetimesignalsthatcorrespondtothefollowingbilateralLaplacetransforms:19 (a)X(s)=e5s1withROCRe(s)<−2s+2(left-sided)1LA(s)=←−−−→a(t)=−e−2tu(−t)s+2LX(s)=e5sA(s)←−−−→x(t)=a(t+5)=−e−2(t+5)u(−(t+5))2(b)X(s)=d1withROCRe(s)>3ds2s−3(right-sided)1LA(s)=←−−−→a(t)=e3tu(t)s−3d2LX(s)=A(s)←−−−→x(t)=t2e3tu(t)ds2−s−2s(c)X(s)=s1−e−ewithROCRe(s)<0s2s2s(left-sided)dx(t)=(−tu(−t)+tu(−t−1)+u(−t−2))dtx(t)=−u(−t)+u(−t−1)−δ(t+2)−3s(d)X(s)=s−2dewithROCRe(s)>0dss课后答案网(right-sided)1LA(s)=←−−−→a(t)=u(t)sLB(s)=e−www.hackshp.cn3sA(s)←−−−→b(t)=a(t−3)=u(t−3)dLC(s)=B(s)←−−−→c(t)=−tb(t)=−tu(t−3)dst1LD(s)=←−−−→d(t)=c(τ)dτs−∞tL1←−−−→d(t)=−τdτ=−(t2−9)32t1LX(s)=D(s)←−−−→x(t)=d(τ)dτs−∞tL1←−−−→x(t)=−(τ2−9)dτ23L19←−−−→x(t)=−(t3−27)+(t−3)u(t−3)6220 6.43.UsethemethodofpartialfractionstodeterminethetimesignalscorrespondingtothefollowingbilateralLaplacetransforms:(a)X(s)=−s−4s2+3s+2−32X(s)=+s+1s+2(i)withROCRe(s)<−2(left-sided)x(t)=3e−t−2e−2tu(−t)(ii)withROCRe(s)>−1(right-sided)x(t)=−3e−t+2e−2tu(t)(iii)withROC−2−1(right-sided)x(t)=2e−2t+2e−tcos(2t)−e−tsin(2t)u(t)21 (iii)withROC−2−1(right-sided)x(t)=5e−t−te−tu(t)2(d)X(s)=2s+2s−2s2−111X(s)=2++s+1s−1(i)withROCRe(课后答案网s)<−1(left-sided)www.hackshp.cnx(t)=2δ(t)−e−t+etu(t)(ii)withROCRe(s)>1(right-sided)x(t)=2δ(t)+e−t+etu(t)(iii)withROC−1-1,thesystemisbothstableandcausal.(ii)AllzerosareintheLHP,soastableandcausalinversesystemexists.课后答案网2(b)H(s)=s+2s−3(s+3)(s2+2s+5)www.hackshp.cns−1H(s)=s2+2s+5zeroat:1polesat:−1±2j(i)AllpolesareintheLHP,andwithROC:Re(s)>-1,thesystemisbothstableandcausal.(ii)NotallzerosareintheLHP,sonostableandcausalinversesystemexists.2(c)H(s)=s−3s+2(s+2)(s2−2s+8)s2−3s+2H(s)=s3+4s+1629 zerosat:1,2√polesat:−2,1±j7(i)NotallpolesareintheLHP,sothesystemisnotstableandcausal.(ii)NozerosareintheLHP,sonostableandcausalinversesystemexists.2(d)H(s)=s+2s(s2+3s−2)(s2+s+2)s2+2sH(s)=s4+4s3+3s2+4s−4zerosat:0,−2√−3±177polesat:,−0.5±j24(i)NotallpolesareintheLHP,sothesystemisnotstableandcausal.(ii)Thereisazeroats=0,sonostableandcausalinversesystemexists.6.51.Therelationshipbetweentheinputx(t)andoutputy(t)ofasystemisdescribedbythedifferentialequationd2dd2dy(t)+y(t)+5y(t)=x(t)−2x(t)+x(t)dt2dtdt2dt(a)Doesthissystemhaveastableandcausalinverse?why?Y(s)(s2+s+5)=X(s)(s2−2s+1)(s−1)2H(s)=课后答案网s2+s+5SinceH(s)haszerosintheRHP,thissystemdoesnothaveacausalandstableinverse.www.hackshp.cn(b)Findadifferentialequationdescriptionfortheinversesystem.inv1H(s)=H(s)s2+s+5=s2−2s+1Inversesystem:d2dd2dy(t)−2y(t)+y(t)=x(t)+x(t)+5x(t)dt2dtdt2dt6.52.Astable,causalsystemhasarationaltransferfunctionH(s).Thesystemsatisfiesthefollowingconditions:(i)Theimpulseresponseh(t)isrealvalued;(ii)H(s)hasexactlytwozeros,oneofwhich2isats=1+j;(iii)Thesignaldh(t)+3dh(t)+2h(t)containsanimpulseanddoubletofunknowndt2dt30 strengthsandaunitamplitudestep.FindH(s).d2dh(t)+3h(t)+2h(t)=bδ(1)(t)+aδ(t)+u(t)dt2dt1H(s)s2+3s+2=bs+a+sbs+a+1H(s)=ss2+3s+2bs2+as+1=s(s+2)(s+1)Onezerois1+jandh(t)isrealvalued,whichimplieszerosoccurinconjugatepairs,sotheotherzerois1−j.1s2−s+1H(s)=2s(s+2)(s+1)6.53.Sketchthemagnituderesponseforthesystemsdescribedbythefollowingtransferfunctionsusingtherelationshipbetweenthepoleandzerolocationsandthejωaxisinthes-plane.(a)H(s)=ss2+2s+101|ω|课后答案网|H(jω)|=|j(ω−10)+1||j(ω+10)+1|2(b)H(s)=s+16www.hackshp.cns+1|j(ω+4)||j(ω−4)||H(jω)|=|jω−1|(c)H(s)=s−1s+1|jω−1||H(jω)|=|jω+1|31 Magnituderesponsefor6.53(a)0.80.60.40.20−15−10−5051015Magnituderesponsefor6.53(b)20151050−15−10−5051015Magnituderesponsefor6.53(c)21.510.50−15−10−5051015wFigureP6.53.Magnituderesponseforthesystems6.54.Sketchthephaseresponseforthesystemsdescribedbythefollowingtransferfunctionsusingtherelationshipbetweenthepoleandzerolocationsandthejωaxisinthes-plane.(a)H(s)=s−1s+2ωH(jω)=π−arctan(ω)−arctan()课后答案网2(b)H(s)=s+1s+2www.hackshp.cnωH(jω)=arctan(ω)−arctan()2(c)H(s)=1s2+2s+171H(s)=(s+1)2+421H(jω)=(jω+1+j4)(jω+1−j4)H(jω)=−[arctan(ω+4)+arctan(ω−4)]32 (d)H(s)=s2H(jω)=−ω2H(jω)=π6.54(a)6.54(b)70.460.2540Phase3Phase2−0.210−0.4−20−1001020−20−1001020ww6.54(c)6.54(d)44.5423.50PhasePhase3−22.5−42−20−10课后答案网01020−20−1001020wwFigureP6.54.PhasePlotofwww.hackshp.cnH(s)6.55.SketchtheBodediagramsforthesystemsdescribedbythefollowingtransferfunctions.(a)H(s)=50(s+1)(s+10)20(s+1)(b)H(s)=s2(s+10)(c)H(s)=5(s+1)3(d)H(s)=s+2s2+s+100(e)H(s)=s+2s2+10s+10033 BodeDiagramGm=Inf,Pm=78.63deg(at4.4561rad/sec)200−20Magnitude(dB)−40−600−45−90Phase(deg)−135−180−101210101010Frequency(rad/sec)FigureP6.55.(a)BodePlotofH(s)BodeDiagramGm=Inf,Pm=52.947deg(at2.1553rad/sec)500−50Magnitude(dB)课后答案网−100www.hackshp.cn−120−150Phase(deg)−180−101210101010Frequency(rad/sec)FigureP6.55.(b)BodePlotofH(s)34 BodeDiagramGm=4.0836dB(at1.7322rad/sec),Pm=17.37deg(at1.387rad/sec)200−20Magnitude(dB)−40−600−90Phase(deg)−180−270−101101010Frequency(rad/sec)FigureP6.55.(c)BodePlotofH(s)BodeDiagramGm=Inf,Pm=157.6deg(at10.099rad/sec)100−10−20Magnitude(dB)课后答案网−30−40www.hackshp.cn900Phase(deg)−90−180−101210101010Frequency(rad/sec)FigureP6.55.(d)BodePlotofH(s)35 BodeDiagramGm=Inf,Pm=Inf−15−20−25−30Magnitude(dB)−35−40450Phase(deg)−45−90−101210101010Frequency(rad/sec)FigureP6.55.(e)BodePlotofH(s)6.56.Theoutputofamultipathsystemy(t)maybeexpressedintermsoftheinputx(t)asy(t)=x(t)+ax(t−Tdiff)whereaandTdiffrespectivelyrepresenttherelativestrengthandtimedelayofthesecondpath.(a)Findthetransferfunctionofthemultipathsystem.课后答案网Y(s)=X(s)+ae−sTdiffX(s)www.hackshp.cnH(s)=1+ae−sTdiff(b)Expressthetransferfunctionoftheinversesystemasaninfinitesumusingtheformulaforsummingageometricseries.−111H(s)==H(s)1+ae−sTdiff∞−sTdiffn=−aen=0(c)Determinetheimpulseresponseoftheinversesystem.Whatconditionmustbesatisfiedfortheinversesystemtobebothstableandcausal?36 ∞invnh(t)=(−a)δ(t−nTdiff)n=0|a|<1forthesystemtobebothstableandcausal.(d)Findastableinversesystemassumingtheconditiondeterminedinpart(c)isviolated.Thefollowingsystemisstable,butnotcausal.inv1H(s)=1+ae−sTdiffUselongdivisioninthepositivepowersofesTdiff,i.e.,divide1byae−sTdiff+1.Thisyields:∞ninv1nsTdiffH(s)=−−ean=1∞ninv1h(t)=−−δ(t+nTdiff)an=16.57.InSection2.12wederivedblock-diagramdescriptionsforsystemsdescribedbylinearconstant-coefficientdifferentialequationsbyrewritingthedifferentialequationasanintegralequation.Considerthesecond-ordersystemwiththeintegralequationdescriptiony(t)=−ay(1)(t)−ay(2)(t)+bx(t)+bx(1)(t)+bx(2)(t)10210Recallthatv(n)(t)isthen-foldintegralofv(t)withrespecttotime.UsetheintegrationpropertytotaketheLaplacetransformoftheintegralequationandderivethedirectformIandIIblockdiagramsforthetransferfunctionofthissystem.课后答案网Y(s)Y(s)X(s)X(s)Y(s)=−a1s−a0s2+b2X(s)+b1s+b0s2www.hackshp.cnx(t)y(t)b1−a1b0−a037 FigureP6.57.DirectFormIBythepropertiesoflinearsystems,H1(s)canbeinterchangedwithH2(s),whichleadstodirectformIIaftercombiningtheintegrators.X(s)Y(s)H(s)H(s)12X(s)Y(s)H(s)H(s)21FigureP6.57.PropertiesofLinearSystemsx(t)y(t)−ab11−a0b0FigureP6.57.DirectFormII课后答案网SolutionstoAdvancedProblemswww.hackshp.cn6.58.Provetheinitialvaluetheorembyassumingx(t)=0fort<0andtakingtheLaplacetransformoftheTaylorseriesexpansionofx(t)aboutt=0+.Assumethatthereexistsfunctionf(t)suchthatx(t)=f(t)u(t).TheTaylorseriesexpansionoff(t)is∞f(n)(a)f(t)=(t−a)nn!n=0f(0+)f(0+)=f(0+)+(t−0+)+(t−0+)2+...1!2!Assumingtheexpansionoftisarounda=0+.TheLaplacetransformforx(t)isthus:f(0+)f(0+)x(t)=f(0+)u(t)+(t−0+)u(t)+(t−0+)2u(t)+...1!2!38 f(0+)f(0+)2f(0+)X(s)=+−+...ss2s3f(0+)2f(0+)sX(s)=f(0+)+−+...ss2Thus:limsX(s)=f(0+)s→∞6.59.Thesystemwithimpulseresponseh(t)iscausalandstableandhasarationaltransferfunction.Identifytheconditionsonthetransferfunctionsothatthesystemwithimpulseresponseg(t)isstableandcausal,where(a)g(t)=dh(t)dtdLg(t)=h(t)←−−−→G(s)=sH(s)−h(0−)=sH(s)dtAllpolesofH(s)areinthelefthalfplane,sonoconditionsareneeded.t(b)g(t)=h(τ)dτ−∞t0−L1H(s)H(s)g(t)=h(τ)dτ←−−−→G(s)=h(τ)dτ+=−∞s−∞ssH(s)musthaveatleastonezeroats=0forthetransferfunctiontobestable.6.60.Usethecontinuous-timerepresentationxδ(t)forthediscrete-timesignalx[n]introducedinSec-tion4.4todeterminetheLaplacetransformsofthefollowingdiscrete-timesignals.课后答案网∞L∞x(t)=x[n]δ(t−nT)←−−−→X(s)=x[n]e−snTsδsδn=−∞n=−∞www.hackshp.cn1,−2≤n≤2(a)x[n]=0,otherwiseX(s)=e2sTs+esTs+1+e−sTs+e−2sTsδ(b)x[n]=(1/2)nu[n]∞X(s)=x[n]e−snTsδn=−∞∞n1−snTs=e2n=039 ∞n1−sTs=e2n=01=1−1e−sTs2(c)x[n]=e−2tu(t)t=nT∞X(s)=e−2nTse−snTsδn=0∞n=e−Ts(2+s)n=01=1−e−Ts(2+s)6.61.Theautocorrelationfunctionforasignalx(t)isdefinedas∞r(t)=x(τ)x(t+τ)dτ−∞(a)Writer(t)=x(t)∗h(t).Expressh(t)intermsofx(t).Thesystemwithimpulseresponseh(t)iscalledamatchedfilterforx(t).let:h(t)=x(−t)∞h(t)∗x(t)=h(τ)x(t−τ)dτ−∞∞=x(−τ)x(t−τ)dτ−∞letγ=−τ课后答案网−∞=−x(γ)x(t+γ)dγ∞∞=x(γ)x(t+γ)dγwww.hackshp.cn−∞(b)Usetheresultfrompart(a)tofindtheLaplacetransformofr(t).Lx(−t)←−−−→X(−s)Lr(t)=x(t)∗x(−t)←−−−→R(s)=X(s)X(−s)(c)Ifx(t)isrealandX(s)hastwopoles,oneofwhichislocatedats=σp+jωp,determinethelocationofallthepolesofR(s).Sincex(t)isreal,thenthepolesofX(s)areconjugatesymmetric,thuss=σp±jωp.ThereforethepolesofX(−s)ares=−σp±jωp,whichimpliesthatthepolesofR(s)areats=σp±jωp,−σp±jωp.40 6.62.SupposeasystemhasMpolesatdk=αk+jβkandMzerosatck=−αk+jβk.Thatis,thepoleandzerolocationsaresymmetricaboutthejω-axis.(a)Showthatthemagnituderesponseofanysystemthatsatisfiesthisconditionisunity.Suchasystemistermedanall-passsystemsinceitpassesallfrequencieswithunitgain.Mk=1(s−ck)H(s)=Mk=1(s−dk)Mk=1|jω−αk−jβk||H(jω)|=Mk=1|jω+αk−jβk|Mk=1|j(ω−βk)−αk||H(jω)|=Mk=1|j(ω−βk)+αk|M(ω−β)2−α2|H(jω)|=k=1kkM(ω−β)2+α2k=1kkH(s)=1(b)Evaluatethephaseresponseofasinglerealpole-zeropair,thatis,sketchthephaseresponseofs−αs+αwhereα>0.课后答案网www.hackshp.cns−αH(s)=s+αjω−αH(jω)=jω+αωωH(jω)=π−arctan−arctanααωH(jω)=π−2arctanαForα=1,thenH(jα)=π,andH(−jα)=3π.2241 PlotofÐH(jw)76543210−10−8−6−4−20246810wFigureP6.62.PhasePlotofH(jω)6.63.Considerthenonminimumphasesystemdescribedbythetransferfunction(s+2)(s−1)H(s)=(s+4)(s+3)(s+5)(a)Doesthissystemhaveastableandcausalinversesystem?ThezerosofH(s):s=−2,1.Sinceoneofthezerosisintherighthalfplane,theinversesystemcannotbestableandcausal.课后答案网(b)ExpressH(s)astheproductofaminimumphasesystem,Hmin(s),andanall-passsystem,Hap(s)containingasinglepoleandzero.(SeeProblem6.62forthedefinitionofanall-passsystem.)www.hackshp.cn(s+2)(s−1)Hmin(s)=(s+4)(s+3)(s+5)s−1Hap(s)=s+1H(s)=Hmin(s)Hap(s)(c)LetHinv(s)betheinversesystemforH(s).FindHinv(s).Canitbebothstableandcausal?minminmininv(s+4)(s+3)(s+5)Hmin(s)=(s+2)(s−1)ThepolesofHinv(s)are:s=−1,−2.Allareinthelefthalfplane,sohinv(t)canbebothcausalandminminstable.42 (d)SketchthemagnituderesponseandphaseresponseofthesystemH(s)Hinv(s).mininvs−1H(s)Hmin(s)=s+1=Hap(s)jω−1Hap(jω)=jω+1|Hap(jω)|=1Hap(jω)=π−arctan(ω)−arctan(ω)=π−2arctan(ω)MagnitudeandphaseplotofH(w)ap21.51Magnitude0.50−10−8−6−4−202468107654Phase321课后答案网0−10−8−6−4−20246810wFigureP6.63.MagnitudeandPhasePlotofwww.hackshp.cnHap(jω)(e)Generalizeyourresultsfromparts(b)and(c)toanarbitrarynonminimumphasesystemH(s)anddeterminethemagnituderesponseofthesystemH(s)Hinv(s).minGeneralizationforH(s)=H(s)(s−c)whereH(s)istheminimumphasepart,assumeRe(c)>0H(s)=H(s)(s+c)mins−cHap(s)=s+cinv1Hmin(s)=H(s)(s+c)H(s)Hinv(s)=H(s)minapjω−cHap(jω)=jω+c|Hap(jω)|=143 ωHap(jω)=π−arctanc6.64.AnN-thorderlowpassButterworthfilterhassquaredmagnituderesponse21|H(jω)|=1+(jω/jωc)2NTheButterworthfilterissaidtobemaximallyflatbecausethefirst2Nderivativesof|H(jω)|2arezeroatω=0.Thecutofffrequency,definedasthevalueforwhich|H(jω)|2=1/2,isω=ω.Assumingthecimpulseresponseisreal,thentheconjugatesymmetrypropertyoftheFouriertransformmaybeusedtowrite|H(jω)|2=H(jω)H∗(jω)=H(jω)H(−jω).NotingthatH(s)|=H(jω),weconcludethats=jωtheLaplacetransformoftheButterworthfilterischaracterizedbytheequation1H(s)H(−s)=1+(s/jωc)2N(a)FindthepolesandzerosofH(s)H(−s)andsketchtheminthes-plane.Therootsofthedenominatorpolynomialarelocatedatthefollowingpointsinthes-plane:1s=jωc(−1)2N(2k+N−1)jπ=ωce2Nfork=0,1,...,2N−1AssumingN=3andωc=1:Pole−ZeroMap10.80.60.40.2课后答案网0ImagAxis−0.2www.hackshp.cn−0.4−0.6−0.8−1−1.5−1−0.500.511.5RealAxisFigureP6.64.(a)Pole-ZeroPlotofH(s)H(−s)(b)ChoosethepolesandzerosofH(s)sothattheimpulseresponseisbothstableandcausal.NotethatifspisapoleorzeroofH(s),then−spisapoleorzeroofH(−s).Fromthegraphinpart(a),pickingthepolesinthelefthalfplanemakestheimpulseresponsestable44 andcausal.Thisimpliesselectingthepolesfork=0,1,...,N−1.(c)NotethatH(s)H(−s)|=1.FindH(s)forN=1andN=2.s=0ForN=1,thepolesareat:s0=ωcs1=−ωcwhichimplies:1H(s)=s+ωcForN=2,thepolesareat:jπ1s0=ωce4jπ3s1=ωce4jπ5s2=ωce4jπ7s3=ωce4whichimplies:1H(s)=jπ3jπ5(s−ωce4)(s−ωce4)(d)Findthethird-orderdifferentialequationthatdescribesaButterworthfilterwithcutofffrequencyωc=1.ForN=3,the2N=6polesofH(s)H(−s)arelocatedonacircleofunitradiuswithangularspacingof60degrees.HencethelefthalfplanepolesofH(s)are:√13s1=−+j22s2=−1课后答案网√13s3=−−jwww.hackshp.cn22Thetransferfunctionistherefore:1H(s)=√√(s+1)(s+1+j3)(s−1−j3)22221=s3+2s2+2s+1Whichimpliesd3d2dy(t)+2y(t)+2y(t)+y(t)=x(t)dt3dt2dt45 6.65.Itisoftenconvenienttochangethecutofffrequencyofafilter,orchangealowpassfiltertoahighpassfilter.Considerasystemdescribedbythetransferfunction1H(s)=(s+1)(s2+s+1)(a)Findthepolesandzerosandsketchthemagnituderesponseofthissystem.Determinewhetherthissystemislowpassorhighpass,andfindthecutofffrequency(thevalueofωforwhich|H(jω)|=1/(2).√−1±3polesats:=−1,2Thissystemislowpasswithcutofffrequencyatωc=1.(a)MagnituderesponseofH(jw)0−10−20−30)|(dB)w|H(j−40−50−60课后答案网−70−10−8−6−4−20246810www.hackshp.cnwrads/secFigureP6.65.(a)MagnituderesponseofH(s),20log10|H(jω)|.(b)Performthetransformationofvariablesinwhichsisreplacedbys/10inH(s).Repeatpart(a)forthetransformedsystem.1H(s)=ss2s(+1)(++1)101010√polesats:=−10,−5±5Thissystemislowpasswithcutofffrequencyatωc=10.46 (b)MagnituderesponseofH(jw)0−5−10−15)|(dB)−20w|H(j−25−30−35−40−40−30−20−10010203040wrads/secFigureP6.65.(b)MagnituderesponseofH(s),20log10|H(jω)|.(c)Performthetransformationofvariablesinwhichsisreplacedby1/sinH(s).Repeatpart(a)forthetransformedsystem.课后答案网www.hackshp.cn1H(s)=(1+1)(1+1+1)ss2s√−1±3polesats:=−1,2Threezerosats:=0Thissystemisahighpassfilterwithcutofffrequecyωc=1.47 (c)MagnituderesponseofH(jw)0−10−20−30−40)|(dB)−50w|H(j−60−70−80−90−100−3−2−10123wrads/secFigureP6.65.(c)Magnituderesponseof课后答案网H(s),20log10|H(jω)|.(d)FindthetransformationthatconvertsH(s)toahighpasssystemwithcutofffrequencyω=100.Replacesby100:www.hackshp.cns1H(s)=(100+1)(100+100+1)ss2s48 (d)MagnituderesponseofH(jw)0−10−20−30−40)|(dB)−50w|H(j−60−70−80−90−100−200−150−100−50050100150200wrads/secFigureP6.65.(d)MagnituderesponseofH(s),20log10|H(jω)|.SolutionstoComputerExperiments6.66.UsetheMATLABcommandrootstodeterminethepolesandzerosofthefollowingsystems:2(a)H(s)=s+2s3+2s2−s+13(b)H(s)=s+1s4+2s课后答案网2+12(c)H(s)=4s+8s+102s3+8s2+18s+20P6.66:www.hackshp.cn=======Part(a):———-ans=0+1.4142i0-1.4142ians=-2.54680.2734+0.5638i0.2734-0.5638iPart(b):49 ———-ans=-1.00000.5000+0.8660i0.5000-0.8660ians=0.0000+1.0000i0.0000-1.0000i-0.0000+1.0000i-0.0000-1.0000iPart(c):———-ans=-1.0000+1.2247i-1.0000-1.2247ians=-1.0000+2.0000i-1.0000-2.0000i-2.0000课后答案网www.hackshp.cn6.67.UsetheMATLABcommandpzmaptoplotthepolesandzerosforthefollowingsystems:3(a)H(s)=s+1s4+2s2+1121(b)A=,b=,c=01,D=[0]1−6250 P6.67(a)1.5double1pole0.50ImagAxis−0.5double−1pole−1.5−1−0.500.5RealAxisFigureP6.67.(a)Pole-ZeroPlotofH(s)P6.67(b)10.80.60.40.2课后答案网0ImagAxis−0.2www.hackshp.cn−0.4−0.6−0.8−1−7−6−5−4−3−2−1012RealAxisFigureP6.67.(b)Pole-ZeroPlot6.68.UsetheMATLABcommandfreqrespevaluateandplotthemagnitudeandphaseresponsesfor51 Examples6.23and6.24.P6.68:Ex6.23543|H(w)|21−10−8−6−4−20246810w:rad/sP6.68:Ex6.243210arg(H(w)):rad−1−2−10−8−6−4−20246810w:rad/sFigureP6.68.MagnitudeandphaseplotforEx6.23&6.24课后答案网www.hackshp.cn6.69.UsetheMATLABcommandfreqrespevaluateandplotthemagnitudeandphaseresponsesforProblem6.53.52 P6.69(a)0.50.40.3|H(w)|0.20.10−50−40−30−20−1001020304050w:rad/s1.510.50<(H(w)):rad−0.5−1−1.5−50−40−30−20−1001020304050w:rad/sFigureP6.69.MagnitudeandphaseplotforProb6.53(a)P6.69(b)504030|H(w)|2010课后答案网0−50−40−30−20−1001020304050www.hackshp.cnw:rad/s1.510.50<(H(w)):rad−0.5−1−1.5−50−40−30−20−1001020304050w:rad/sFigureP6.69.MagnitudeandphaseplotforProb6.53(b)53 P6.69(c)1.51|H(w)|0.5−50−40−30−20−1001020304050w:rad/s3210<(H(w)):rad−1−2−50−40−30−20−1001020304050w:rad/sFigureP6.69.MagnitudeandphaseplotforProb6.53(c)6.70.Useyourknowledgeoftheeffectofpolesandzerosonthemagnituderesponsetodesignsystemshavingthespecifiedmagnituderesponse.Placepolesandzerosinthes-plane,andevaluatethecorre-spondingmagnituderesponseusingtheMATLABcommandfreqresp.Repeatthisprocessuntilyoufindpoleandzerolocationsthatsatisfythespecifications.课后答案网(a)Designahigh-passfilterwithtwopolesandtwozerosthatsatisfies|H(j0)|=0,0.8≤|H(jω)|≤1.2for|ω|>100π,andhasrealvaluedcoefficients.Twoconjugatepolesareneededaroundthetransition,whichimpliesonepossiblesolutionis:www.hackshp.cns2H(s)=(s+25+j10π)(s+25−j10π)(b)Designalow-passfilterwithrealvaluedcoefficientsthatsatisfies0.8≤|H(jω)|≤1.2for|ω|<π,and|H(jω)|<0.1for|ω|>10π.Onepossiblesolutionis:(s−j50)(s+j50)H(s)=(s+2+jπ)(s+2−jπ)54 P6.70(a):HPF10.80.6|H(w)|0.40.20050100150200250300350400450500w:rad/sP6.70(b):LPF0.80.6|H(w)|0.40.2005101520253035404550w:rad/sFigureP6.70.Magnituderesponse6.71.UsetheMATLABcommandbodetofindthebodediagramsforthesystemsinProblem6.55.BodeDiagramGm=Inf,Pm=78.63deg(at4.4561rad/sec)200课后答案网−20Magnitude(dB)−40www.hackshp.cn−600−45−90Phase(deg)−135−180−101210101010Frequency(rad/sec)55 FigureP6.71.(a)BodePlotofH(s)BodeDiagramGm=Inf,Pm=52.947deg(at2.1553rad/sec)500−50Magnitude(dB)−100−120−150Phase(deg)−180−101210101010Frequency(rad/sec)FigureP6.71.(b)BodePlotofH(s)BodeDiagramGm=4.0836dB(at1.7322rad/sec),Pm=17.37deg(at1.387rad/sec)200−20课后答案网Magnitude(dB)−40−60www.hackshp.cn0−90Phase(deg)−180−270−101101010Frequency(rad/sec)FigureP6.71.(c)BodePlotofH(s)56 BodeDiagramGm=Inf,Pm=157.6deg(at10.099rad/sec)100−10−20Magnitude(dB)−30−40900Phase(deg)−90−180−101210101010Frequency(rad/sec)FigureP6.71.(d)BodePlotofH(s)BodeDiagramGm=Inf,Pm=Inf−15−20−25−30Magnitude(dB)−35课后答案网−4045www.hackshp.cn0Phase(deg)−45−90−101210101010Frequency(rad/sec)FigureP6.71.(e)BodePlotofH(s)6.72.UsetheMATLABcommandsstofindstate-variabledescriptionsforthesystemsinProblem6.48.57 P6.72:=======Part(a):==========a=x1x2x1-3-0x210b=u1x11x20c=x1x2y101d=u1y10Continuous-timemodel.Part(b):==========课后答案网a=x1x2x12-2www.hackshp.cnx240b=u1x12x20c=x1x2y130d=58 u1y10Continuous-timemodel.Part(c):==========a=x1x2x3x1-5-0.875-0.09375x2800x3040b=u1x10.5x20x30c=x1x2x3y100.5-0.25d=u1y10Continuous-timemodel.课后答案网6.73.UsetheMATLABcommandtftofindtransferfunctiondescriptionsforthesystemsinProb-lem6.49.P6.73:www.hackshp.cn=======Part(a):==========Transferfunction:s+3s2+3s+2Part(b):==========Transferfunction:2s−1s2+5s−859 课后答案网www.hackshp.cn60 SolutionstoAdditionalProblems7.17.Determinethez-transformandROCforthefollowingtimesignals:SketchtheROC,poles,andzerosinthez-plane.(a)x[n]=δ[n−k],k>0∞X(z)=x[n]z−nn=−∞=z−k,z=0ImkmultipleReFigureP7.17.(a)ROC(b)x[n]=δ[n+k],k>0课后答案网X(z)=zk,allzImwww.hackshp.cnkmultipleReFigureP7.17.(b)ROC1 (c)x[n]=u[n]∞X(z)=z−nn=01=,|z|>11−z−1Im1ReFigureP7.17.(c)ROC1n课后答案网(d)x[n]=(u[n]−u[n−5])4www.hackshp.cn4n1−1X(z)=z4n=051−1z−1=41−1z−14515z−4=,allzz4(z−1)44polesatz=0,1poleatz=01jk2π5zerosatz=e5k=0,1,2,3,44Notezerofork=0cancelspoleatz=142 Im0.25ReFigureP7.17.(d)ROCn(e)x[n]=1u[−n]40n1−1X(z)=z4n=−∞∞n=(4z)n=011=,|z|<1−4z4Im课后答案网0.25Rewww.hackshp.cnFigureP7.17.(e)ROC(f)x[n]=3nu[−n−1]−1−1nX(z)=3zn=−∞∞n1=z3n=13 1z=31−1z3−1=,|z|<31−3z−1Poleatz=3Zeroatz=0Im3ReFigureP7.17.(f)ROC2|n|(g)x[n]=3−1n∞n3−12−1X(z)=z+z23n=−∞n=0−11=+1−3z−11−2z−123−5z23=6,<|z|<(z−3)(z−2)32课后答案网23Imwww.hackshp.cn2332ReFigureP7.17.(g)ROC1n1n(h)x[n]=u[n]+u[−n−1]244 ∞n−1n1−11−1X(z)=z+z24n=0n=−∞1111=+,|z|>and|z|<1−1z−11−2z−12423Noregionofconvergenceexists.7.18.Giventhefollowingz-transforms,determinewhethertheDTFTofthecorrespondingtimesignalsexistswithoutdeterminingthetimesignal,andidentifytheDTFTinthosecaseswhereitexists:(a)X(z)=5,|z|>11+1z−133ROCincludes|z|=1,DTFTexists.jΩ5X(e)=1+1e−jΩ3(b)X(z)=5,|z|<11+1z−133ROCdoesnotinclude,|z|=1,DTFTdoesnotexist.−1(c)X(z)=z,|z|<1(1−1z−1)(1+3z−1)22ROCdoesnotinclude,|z|=1,DTFTdoesnotexist.−1(d)X(z)=z,1<|z|<3(1−1z−1)(1+3z−1)22ROCincludes|z|=1,DTFTexists.e−jΩ课后答案网X(ejΩ)=(1−1e−jΩ)(1+3e−jΩ)www.hackshp.cn27.19.ThepoleandzerolocationsofX(z)aredepictedinthez-planeonthefollowingfigures.Ineachcase,identifyallvalidROCsforX(z)andspecifythecharacteristicsofthetimesignalcorrespondingtoeachROC.(a)Fig.P7.19(a)Cz(z−3)X(z)=2(z+3)(z−1)43Thereare4possibleROCs(1)|z|>34x[n]isright-sided.5 (2)1<|z|<334x[n]istwo-sided.(3)|z|<13x[n]isleft-sided.(b)Fig.P7.19(b)C(z4−1)X(z)=√π√πj−jz(z−2e4)(z−2e4)Thereare2possibleROCs√(1)|z|>2x[n]isright-sided.√(2)|z|<2x[n]istwo-sided.(c)Fig.P7.19(c)129X(z)=(z−)(z+1)(z+)C,|z|<∞216x[n]isstableandleft-sided.7.20.Usethetablesofz-transformsandz-transformpropertiesgiveninAppendixEtodeterminethez-transformsofthefollowingsignals:课后答案网n(a)x[n]=1u[n]∗2nu[−n−1]2www.hackshp.cnn1z11a[n]=u[n]←−−−→A(z)=,|z|>21−1z−122z1b[n]=2nu[−n−1]←−−−→B(z)=,|z|<21−2z−1zx[n]=a[n]∗b[n]←−−−→X(z)=A(z)B(z)111X(z)=,<|z|<21−1z−11−2z−1221n1n(b)x[n]=nu[n]∗u[n−2]24n1z11a[n]=u[n]←−−−→A(z)=,|z|>21−1z−1226 n1z11b[n]=u[n]←−−−→B(z)=,|z|>41−1z−144zz−2c[n]=b[n−2]←−−−→C(z)=1−1z−14zdx[n]=n[a[n]∗b[n]]←−−−→X(z)=−zA(z)B(z)dz2z−2−3z−31X(z)=4,|z|>1−3z−1+1z−2)2248(c)x[n]=u[−n]1X(z)=1−11−z1=,|z|<11−z(d)x[n]=nsin(πn)u[−n]2πx[n]=−nsin(−n)u[−n]2dz−1X(z)=−zdz1+z−21z=z−z−2z−1(−2z−3)=−z−1+z−2(1+z−2)21z=zz2z3=+1+z2(1+z2)2(e)x[n]=3n−2u[课后答案网n]∗cos(πn+π/3)u[n]6www.hackshp.cn11za[n]=3nu[n]←−−−→A(z)=991−3z−3ππππb[n]=cos(n+)u[n]=cos(n)cos(π3)−sin(n)sin(π3)u[n]6366π−1ππ−1πzcos()(1+zcos())sin()(zsin())b[n]←−−−→B(z)=36−361−2z−1cos(π)+z−21−2z−1cos(π)+z−2661cos(π)(1+z−1cos(π))−sin(π)sin(π)z−1X(z)=A(z)B(z)=936361−3z−31−2z−1cos(π)+z−26z27.21.Giventhez-transformpairx[n]←−−−→zwithROC|z|<4,usethez-transformpropertiesz2−16todeterminethez-transformofthefollowingsignals:(a)y[n]=x[n−2]7 z1y[n]=x[n−2]←−−−→Y(z)=z−2X(z)=z2−16(b)y[n]=(1/2)nx[n]1zz2y[n]=()nx[n]←−−−→Y(z)=X(2z)=2z2−4(c)y[n]=x[−n]∗x[n]z1z2y[n]=x[−n]∗x[n]←−−−→Y(z)=X()X(z)=z257z2−16z4−16(d)y[n]=nx[n]zd32z2y[n]=nx[n]←−−−→Y(z)=−zX(z)=dz(z2−16)2(e)y[n]=x[n+1]+x[n−1]zz3+zy[n]=x[n+1]+x[n−1]←−−−→Y(z)=(z1+z−1)X(z)=z2−16(f)y[n]=x[n]∗x课后答案网[n−3]zzy[n]=x[n]∗x[n−3]←−−−→Y(z)=X(z)z−3X(z)=(z2−16)2www.hackshp.cnz7.22.Giventhez-transformpairn23nu[n]←−−−→X(z),usethez-transformpropertiestodeterminethetime-domainsignalscorrespondingtothefollowingztransforms:(a)Y(z)=X(2z)z11Y(z)=X(2z)←−−−→y[n]=()nx[n]=()nn23nu[n]22(b)Y(z)=X(z−1)1zY(z)=X()←−−−→y[n]=x[−n]=n23−nu[−n]z8 (c)Y(z)=dX(z)dzddzY(z)=X(z)=−z−1−zX(z)←−−−→y[n]=−(n−1)x[n−1]=−(n−1)33n−1u[n−1]dzdz2−2(d)Y(z)=z−zX(z)2z2−z−2z1Y(z)=X(z)←−−−→y[n]=(x[n+2]−x[n−2])221y[n]=(n+2)23n+2u[n+2]−(n−2)23n−2u[n−2]2(e)Y(z)=[X(z)]2zY(z)=X(z)X(z)←−−−→y[n]=x[n]∗x[n]ny[n]=u[n]k23k(n−k)23n−kk=0n=3nu[n]k2n2−2nk3+k4k=07.23.Provethefollowingz-transformproperties:(a)Timereversalz1x[n]←−−−→X()zy[n]=x[−n]∞课后答案网Y(z)=x[−n]z−nn=−∞letl=−n∞1−l=x[l]()www.hackshp.cnzl=−∞1=X()z(b)Timeshiftzx[n−n]←−−−→z−noX(z)oy[n]=x[n−no]∞Y(z)=x[n−n]z−non=−∞letl=n−no9 ∞=x[l]z−(l+no)l=−∞∞=x[l]z−lz−nol=−∞=z−noX(z)(c)Multiplicationbyexponentialsequencezzαnx[n]←−−−→X()αy[n]=αnx[n]∞Y(z)=αnx[n]z−nn=−∞∞z−n=x[n]()αn=−∞z=X()α(d)ConvolutionLetc[n]=x[n]∗y[n]zx[n]∗y[n]←−−−→X(z)Y(z)∞C(z)=(x[n]∗y[n])z−nn=−∞∞∞=x[p]y[n−p]z−nn=−∞p=−∞课后答案网∞∞=x[p]y[n−p]z−(n−p)z−pp=−∞n=−∞www.hackshp.cnY(z)∞=x[p]z−pY(z)p=−∞X(z)=X(z)Y(z)(e)Differentiationinthez-domain.zdnx[n]←−−−→−zX(z)dz∞X(z)=x[n]z−nn=−∞10 Differentiatewithrespecttozandmultiplyby−z.∞dz−zX(z)←−−−→nx[n]z−ndzn=−∞Thereforezdnx[n]←−−−→−zX(z)dz7.24.Usethemethodofpartialfractionstoobtainthetime-domainsignalscorrespondingtothefol-lowingz-transforms:1+7z−1(a)X(z)=6,|z|>1(1−1z−1)(1+1z−1)223x[n]isright-sidedABX(z)=+1−1z−11+1z−1231=A+B711=A−B6322−1X(z)=+1−1z−11+1z−1231n1nx[n]=2()−(−)u[n]231+7z−1(b)X(z)=6,|z|<1(1−1z−1)(1+1z−1)323sameas(a),butx[n]isleft-sided1n1nx[n]=−2()+(−)u[−n−1]课后答案网231+7z−1(c)X(z)=6,1<|z|<1(1−1z−1)(1+1z−1)3223www.hackshp.cnsameas(a),butx[n]istwo-sided1n1nx[n]=−2()u[−n−1]−(−)u[n]232(d)X(z)=z−3z,1<|z|<2z2+3z−122x[n]istwo-sided1−3z−1X(z)=1+3z−1−z−22AB=+1−1z−11+2z−1211 1=A+B1−3=2A−B2−12X(z)=+1+1z−11−2z−121nnx[n]=−(−)u[n]−2(2)u[−n−1]23z2−1z(e)X(z)=4,|z|>4z2−16x[n]isright-sidedABX(z)=+1+4z−11−4z−13=A+B1−=−4A+4B44947X(z)=32+321+4z−11−4z−149n47nx[n]=(−4)+4u[n]3232z3+z2+3z+1(f)X(z)=22,|z|<1z3+3z2+1z222x[n]isleft-sided−12−1X(z)=z++1+z−11+1z−121nn课后答案网x[n]=δ[n−1]+(−)−2(−1)u[−n−1]2432(g)X(z)=2z−2z−2z,|z|>1z2−1www.hackshp.cnx[n]isright-sided1−12X(z)=2++z1+z−11−z−1x[n]=2δ[n+2]+[(−1)n−1]u[n+2]7.25.Determinethetime-domainsignalscorrespondingtothefollowingz-transforms:(a)X(z)=1+2z−6+4z−8,|z|>012 x[n]=δ[n]+2δ[n−6]+4δ[n−8]101−k(b)X(z)=z,|z|>0k=5k101x[n]=δ[n−k]kk=5(c)X(z)=(1+z−1)3,|z|>0X(z)=(δ[n]+δ[n−1])∗(δ[n]+δ[n−1])∗(δ[n]+δ[n−1])x[n]=δ[n]+3δ[n−1]+3δ[n−2]+δ[n−3](d)X(z)=z6+z2+3+2z−3+z−4,|z|>0x[n]=δ[n+6]+δ[n+2]+3δ[n]+2δ[n−3]+δ[n−4]7.26.Usethefollowingcluestodeterminethesignalx[n]andrationalz-transformX(z).(a)X(z)haspolesatz=1/2andz=−1,x[1]=1,x[−1]=1,andtheROCincludesthepointz=3/4.SincetheROCincludesthepointz=3/4,theROCis1<|z|<1.2ABX(z)=+1−1z−11+z−12课后答案网n1nx[n]=Au[n]−B(−1)u[−n−1]21www.hackshp.cnx[1]=1=A2A=2x[−1]=1=−1B(−1)B=1n1nx[n]=2u[n]−(−1)u[−n−1]2(b)x[n]isright-sided,X(z)hasasinglepole,andx[0]=2,x[2]=1/2.x[n]=c(p)nu[n]wherecandpareunknownconstants.0x[0]=2=c(p)13 c=212x[2]==2(p)21p=2n1x[n]=2u[n]2(c)x[n]istwo-sided,X(z)hasonepoleatz=1/4,x[−1]=1,x[−3]=1/4,andX(1)=11/3.ABX(z)=+1−1z−11−cz−14n1nx[n]=Au[n]−B(c)u[−n−1]4x[−1]=1=−Bc−11−3x[−3]==−Bc4c=2B=−211A−2X(1)==+31−11−245A=4n51nx[n]=u[n]+2(2)u[−n−1]447.27.Determinetheimpulseresponsecorrespondingtothefollowingtransferfunctionsif(i)thesystemisstable,or(ii)thesystemiscausal:2−3z−1(a)H(z)=2(1−2z−1)(1+1z−1)2课后答案网11H(z)=+1−2z−11+1z−1www.hackshp.cn2(i)h[n]isstable,ROC1<|z|<2,ROCincludes|z|=1.2n1nh[n]=−(2)u[−n−1]+(−)u[n]2(ii)h[n]iscausal,ROC|z|>2n1nh[n]=2+(−)u[n]214 2(b)H(z)=5zz2−z−632H(z)=+1−3z−11+2z−1(i)h[n]isstable,ROC|z|<2,ROCincludes|z|=1.h[n]=[−(3)n−(−2)n]u[−n−1](ii)h[n]iscausal,ROC|z|>3h[n]=[3n+(−2)n]u[n](c)H(z)=4zz2−1z+14164z−1H(z)=(1−1z−1)24(i)h[n]isstable,ROC|z|>1,ROCincludes|z|=1.41n课后答案网h[n]=16n()u[n]4(ii)h[n]iscausal,ROC|z|>1www.hackshp.cn41nh[n]=16n()u[n]47.28.Useapowerseriesexpansiontodeterminethetime-domainsignalcorrespondingtothefollowingz-transforms:(a)X(z)=1,|z|>11−1z−244∞1−2kX(z)=(z)4k=015 ∞1kx[n]=()δ[n−2k]4k=0n12nevenandn≥0=40nodd(b)X(z)=1,|z|<11−1z−244∞X(z)=−4z2(2z)2kk=0∞=−22(k+1)z2(k+1)k=0∞x[n]=−22(k+1)δ[n+2(k+1)]k=0(c)X(z)=cos(z−3),|z|>0Note:∞(−1)kcos(α)=(α)2k(2k)!k=0∞(−1)kX(z)=(z−3)2k(2k)!k=0∞(−1)k=z−6k(2k)!k=0课后答案网∞(−1)kx[n]=δ[n−6k](2k)!www.hackshp.cnk=0(d)X(z)=ln(1+z−1),|z|>0Note:∞(−1)k−1ln(1+α)=(α)kkk=0∞(−1)k−1X(z)=(z−1)kkk=1∞(−1)k−1x[n]=δ[n−k]kk=116 7.29.Acausalsystemhasinputx[n]andoutputy[n].Usethetransferfunctiontodeterminetheimpulseresponseofthissystem.(a)x[n]=δ[n]+1δ[n−1]−1δ[n−2],y[n]=δ[n]−3δ[n−1]4841−11−1X(z)=1+z−z483−1Y(z)=1−z4Y(z)H(z)=X(z)−25=3+31−1z−11+1z−14211n1nh[n]=5(−)−2()u[n]324nn1n(b)x[n]=(−3)u[n],y[n]=4(2)u[n]−u[n]21X(z)=1+3z−13Y(z)=(1−2z−1)(1−1z−1)2Y(z)H(z)=X(z)10−7=+1−2z−11−1z−12n1nh[n]=10(2)−7()u[n]2n7.30.Asystemhasimpulseresponse课后答案网h[n]=1u[n].Determinetheinputtothesystemiftheoutput2isgivenbywww.hackshp.cn1H(z)=1−1z−12(a)y[n]=2δ[n−4]Y(z)=2z−4Y(z)X(z)=H(z)=2z−4−z−5x[n]=2δ[n−4]−δ[n−5]17 n(b)y[n]=1u[n]+2−1u[n]33212Y(z)=3+31−z−11+1z−12Y(z)X(z)=H(z)141=−+6+321−z−11+1z−121141nx[n]=−δ[n]+u[n]+(−)u[n]26327.31.Determine(i)transferfunctionand(ii)impulseresponserepresentationsforthesystemsde-scribedbythefollowingdifferenceequations:(a)y[n]−1y[n−1]=2x[n−1]21−1−1Y(z)1−z=2zX(z)2Y(z)H(z)=X(z)2z−1=1−1z−121n−1h[n]=2()u[n−1]2(b)y[n]=x[n]−x[n−2]+x[n−4]−x[n−6]Y(z)=1−z−2+z−4−z−6X(z)Y(z)课后答案网H(z)=X(z)=1−z−2+z−4−z−6www.hackshp.cnh[n]=δ[n]−δ[n−2]+δ[n−4]−δ[n−6](c)y[n]−4y[n−1]−16y[n−2]=2x[n]+x[n−1]5254−116−2−1Y(z)1−z−z=(2+z)X(z)525Y(z)H(z)=X(z)213z−1=+51−4z−1(1−4z−1)2554n134nh[n]=2()+n()u[n]54518 7.32.Determine(i)transferfunctionand(ii)difference-equationrepresentationsforthesystemswiththefollowingimpulseresponses:n(a)h[n]=31u[n−1]4n−111h[n]=3u[n−1]443z−1H(z)=41−1z−14Y(z)=X(z)Takingtheinversez-transformyields:13y[n]−y[n−1]=x[n−1]441n1n−2(b)h[n]=u[n]+u[n−1]32nn−111h[n]=u[n]+2u[n−1]321+3z−1−2z−2H(z)=231−5z−1+1z−266Y(z)=X(z)Takingtheinverse课后答案网z-transformyields:5132y[n]−y[n−1]+y[n−2]=x[n]+x[n−1]−x[n−2]www.hackshp.cn66232n1nππ(c)h[n]=2u[n−1]+[cos(n)−2sin(n)]u[n]34664z−11−1z−1cos(π)−1z−1sin(π)H(z)=3+46261−2z−11−z−11cos(π)+1z−232616√√1+5−13z−1+1−13z−2+1z−3=128√64√121−2+13z−1+1+13z−2−1z−33416624Takingthez-transformyields:√√y[n]−2+13y[n−1]+1+13y[n−2]−1y[n−3]34√166√24=x[n]+5−13x[n−1]+1−13x[n−2]+1x[n−3]128641219 (d)h[n]=δ[n]−δ[n−5]H(z)=1−z−5Takingthez-transformyields:y[n]=x[n]−x[n−5]7.33.(a)Takethez-transformofthestate-updateequationEq.(2.62)usingthetime-shiftpropertyEq.(7.13)toobtainq˜(z)=(zI−A)−1bX(z)whereQ1(z)Q2(z)q˜(z)=...QN(z)isthez-transformofq[n].UsethisresulttoshowthatthetransferfunctionofaLTIsystemisexpressedintermsofthestate-variabledescriptionasH(z)=c(zI−A)−1b+Dq[n+1]=Aq[n]+bx[n]q˜(z)=Aq˜(z)+bX(z)q˜(z)(zI−A)=bX(z)q˜(z)=(zI−A)−1bX(z)课后答案网y[n]=cq[n]+Dx[n]www.hackshp.cnY(z)=cq˜(z)+DX(z)Y(z)=c(zI−A)−1bX(z)+DX(z)H(z)=c(zI−A)−1b+D(b)Determinetransferfunctionanddifference-equationrepresentationsforthesystemsdescribedbythefollowingstate-variabledescriptions.Plotthepoleandzerolocationsinthez-plane.1−00(i)A=2,b=,c=1−1,D=[1]0122H(z)=c(zI−A)−1b+Dz2−2z−5=4z2−1420 1.510.50ImaginaryPart−0.5−1−1.5−1−0.500.511.522.5RealPartFigureP7.33.(b)-(i)Pole-ZeroPlot11−1(ii)A=22,b=,c=21,D=[0]−1−10242zH(z)=z2−1z−34810.8课后答案网0.60.4www.hackshp.cn0.20ImaginaryPart−0.2−0.4−0.6−0.8−1−1−0.500.51RealPartFigureP7.33.(b)-(ii)Pole-ZeroPlot21 11−2(iii)A=48,b=,c=01,D=[0]−732242z−13H(z)=2z2−1z+1241.510.50ImaginaryPart−0.5−1−1.5−1−0.500.511.522.53RealPartFigureP7.33.(b)-(iii)Pole-ZeroPlot7.34.Determinewhethereachofthesystemsdescribedbeloware(i)causalandstableand(ii)mini-mumphase.(a)H(z)=2z+3z2+z−5课后答案网163zeroat:z=−2www.hackshp.cn51polesat:z=−,44(i)Notallpolesareinside|z|=1,thesystemisnotcausalandstable.(ii)Notallpolesandzerosareinside|z|=1,thesystemisnotminimumphase.(b)y[n]−y[n−1]−1y[n−2]=3x[n]−2x[n−1]42zerosat:z=0,3√1±2polesat:z=222 (i)Notallpolesareinside|z|=1,thesystemisnotcausalandstable.(ii)Notallpolesandzerosareinside|z|=1,thesystemisnotminimumphase.(c)y[n]−2y[n−2]=x[n]−1x[n−1]2z(z−1)H(z)=2z2−21zerosat:z=0,2√polesat:z=±2(i)Notallpolesareinside|z|=1,thesystemisnotcausalandstable.(ii)Notallpolesandzerosareinside|z|=1,thesystemisnotminimumphase.7.35.Foreachsystemdescribedbelow,identifythetransferfunctionoftheinversesystem,anddeter-minewhetheritcanbebothcausalandstable.−1−2(a)H(z)=1−8z+16z1−1z−1+1z−224(z−4)2H(z)=(z−1)22(z−1)2Hinv(z)=2(z−4)2polesat:z=4(double)Fortheinversesystem,notallpolesareinside|z|=1,sothesystemisnotcausalandstable.z2−81课后答案网(b)H(z)=100z2−1z2−1www.hackshp.cnHinv(z)=z2−811009polesat:z=(double)10Fortheinversesystem,allpolesareinside|z|=1,sothesystemcanbecausalandstable.−1n−1n(c)h[n]=10u[n]−9u[n]24z(z−2)H(z)=(z+1)(z+1)24(z+1)(z+1)Hinv(z)=24z(z−2)23 polesat:z=0,2Fortheinversesystem,notallpolesareinside|z|=1,sothesystemcannotbebothcausalandstable.1n1n(d)h[n]=24u[n−1]−30u[n−1]232(z+1)H(z)=2(z−1)(z−1)23(z−1)(z−1)Hinv(z)=232(z+1)21poleat:z=−2Fortheinversesystem,allpolesareinside|z|=1,sothesystemcanbebothcausalandstable.(e)y[n]−1y[n−2]=6x[n]−7x[n−1]+3x[n−2]4invX(z)H(z)=Y(z)(z−1)(z+1)=226z2+7z+3√7±j23polesat:z=12Fortheinversesystem,allpolesareinside|z|=1,sothesystemcanbebothcausalandstable.(f)y[n]−1y[n−1]=x[n]2课后答案网z−1Hinv(z)=2zwww.hackshp.cnpoleat:z=0Fortheinversesystem,allpolesareinside|z|=1,sothesystemcanbebothcausalandstable.7.36.AsystemdescribedbyarationaltransferfunctionH(z)hasthefollowingproperties:1)thesystemiscausal;2)h[n]isreal;3)H(z)hasapoleatz=j/2andexactlyonezero;4)theinversesystem∞−nhastwozeros;5)h[n]2=0;6)h[0]=1.n=0By2)and3)24 jpolesatz=±2By4)inv1H=H1±jπHhastwopoles.z=e22A(1−Cz−1)H(z)=1−z−1cos(π)+1z−224A(1−Cz−1)=1+1z−24By5)∞∞−n−nh[n]2=h[n]z=H(z)n=0n=0z=2A(1−Cz−1)SinceH(z)=,1+1z−24H(z)=0impliesC=2nn1π1πh[n]=Acos(n)−2Asin(n)u[n]2222h[0]=1=A1−2z−1H(z)=1+1z−24nn1π1πh[n]=cos(n)−2sin(n)u[n]2222(a)Isthissystemstable?Thepolesareinside课后答案网|z|=1,sothesystemisstable.(b)Istheinversesystembothstableandcausal?No,theinversesystemhasapoleatwww.hackshp.cnz=2,whichisnotinside|z|=1.(c)Findh[n].nn1π1πh[n]=cos(n)−2sin(n)u[n]2222(d)Findthetransferfunctionoftheinversesystem.inv1H(z)=H(z)1+1z−2=41−2z−125 7.37.Usethegraphicalmethodtosketchthemagnituderesponseofthesystemshavingthefollowingtransferfunctions:−2(a)H(z)=z1+49z−2641H(z)=(z+j7)(z−j7)88jΩ1H(e)=(ejΩ+j7)(ejΩ−j7)88Im781ReFigureP7.37.(a)Graphicalmethod.P7.37(a)MagnitudeResponse4.543.5课后答案网3)|jW2.5www.hackshp.cn|H(e21.510.5−4−3−2−101234WFigureP7.37.(a)MagnitudeResponse−1−2(b)H(z)=1+z+z326 z2+z1+1H(z)=3z2ej2Ω+ejΩ+1H(ejΩ)=3ej2Ωpolesat:z=0,(double)±j2πzerosat:z=e3Im31ReFigureP7.37.(b)Graphicalmethod.P7.37(b)MagnitudeResponse10.90.80.7课后答案网0.6)|jW0.5|H(ewww.hackshp.cn0.40.30.20.10−4−3−2−101234WFigureP7.37.(b)MagnitudeResponse−1(c)H(z)=1+z1+(18/10)cos(π)z−1+(81/100)z−2427 1+z−1H(z)=9j3π−19−j3π−1(1−e4z)(1−e4z)1010ej2Ω+ejΩH(ejΩ)=ejΩ+(18/10)cos(π)ejΩ+(81/100)4zeros:z=−19±jπjπpoles:z=e4e10Im341Re910FigureP7.37.(c)Graphicalmethod.P7.37(c)MagnitudeResponse654课后答案网)|jW3|H(ewww.hackshp.cn210−4−3−2−101234WFigureP7.37.(c)MagnitudeResponse7.38.Drawblock-diagramimplementationsofthefollowingsystemsasacascadeofsecond-ordersec-28 tionswithreal-valuedcoefficients:1jπ−11−jπ−11jπ−11−jπ−1(1−e4z)(1−e4z)(1+e8z)(1+e8z)(a)H(z)=44441jπ1−jπ3j7π3−j7π(1−e3z−1)(1−e3z−1)(1−e8z−1)(1−e8z−1)2244H(z)=H1(z)H2(z)1−1cos(π)z−1+1z−2H(z)=241611−cos(π)z−1+1z−2341+1cos(π)z−1+1z−2H(z)=281621−3cos(7π)z−1+9z−22816X(z)Y(z)−1−1zz371cos()−1cos()cos()cos()3242828−1−1zz−11−914161616FigureP7.38.(a)Blockdiagram.−121jπ−11−jπ−1(1+2z)(1−e2z)(1−e2z)(b)H(z)=223−13jπ3jπ3−1(1−z)(1−e3z−1)(1−e3z−1)(1+z)8884H(z)=H1(z)H2(z)1+4z−1+4z−2H1(z)=1+3z−1−9z−28321+1z−2课后答案网H(z)=421−3cos(π)z−1+9z−24364X(z)www.hackshp.cnY(z)−1−1zz−33cos()8443−1−1zz9−91324644FigureP7.38.(b)Blockdiagram.29 7.39.Drawblockdiagramimplementationsofthefollowingsystemsasaparallelcombinationofsecond-ordersectionswithreal-valuedcoefficients:1njn−jn−1n(a)h[n]=2u[n]+u[n]+u[n]+u[n]22222111H(z)=+++1−1z−11−jz−11+jz−11+1z−122223+1z−12=2+1−1z−21+1z−244H(z)=H1(z)+H2(z)X(z)3−1z0.5−1Y(z)z0.252−1z课后答案网−1zwww.hackshp.cn−0.25FigureP7.39.(a)Blockdiagram.1jπn1jπn1−jπn1−jπn(b)h[n]=2e4u[n]+e3u[n]+e3u[n]+2e4u[n]24422112H(z)=1−1ejπ4z−1+1−1ejπ3z−1+1−1e−jπ3z−1+1−1e−jπ4z−12442√4−2z−12−1z−1=+41−√1z−1+1z−21−1z−1+1z−224416H(z)=H1(z)+H2(z)30 X(z)41−1z22−1Y(z)z−0.252−1z0.25−0.25−1z−1/16FigureP7.39.(b)Blockdiagram.7.40.DeterminethetransferfunctionofthesystemdepictedinFig.P7.40.X(z)Y(z)H(z)H(z)12课后答案网H(z)www.hackshp.cn3FigureP7.40.Systemdiagram1−2z−1H1(z)=1−1z−24z−2H2(z)=1+3z−1−1z−2481H3(z)=1+2z−13H(z)=H1(z)H2(z)+H1(z)H3(z)31 7.41.Letx[n]=u[n+4].(a)Determinetheunilateralz-transformofx[n].∞zux[n]=u[n+4]←−−−→X(z)=x[n]z−nn=0∞X(z)=z−nn=01=1−z−1(b)Usetheunilateralz-transformtime-shiftpropertyandtheresultof(a)todeterminetheunilateralz-transformofw[n]=x[n−2].zuw[n]=x[n−2]←−−−→W(z)=x[−2]+x[−1]z−1+z−2X(z)z−2W(z)=1+z−1+1−z−17.42.Usetheunilateralz-transformtodeterminetheforcedresponse,thenaturalresponse,andthecompleteresponseofthesystemsdescribedbythefollowingdifferenceequationswiththegiveninputsandinitialconditions.(a)y[n]−1y[n−1]=2x[n],y[−1]=1,x[n]=(−1)nu[n]321X(z)=1+1z−121课后答案网Y(z)−z−1Y(z)+1=2X(z)31−11Y(z)1−z=+2X(z)33www.hackshp.cn111Y(z)=+2X(z)31−1z−11−1z−133Y(n)(z)Y(f)(z)NaturalResponse(n)11Y(z)=31−1z−13n(n)11y[n]=u[n]33ForcedResponse64Y(f)(z)=5+51+1z−11−1z−123nn(f)6141y[n]=−+u[n]525332 CompleteResponsenn61171y[n]=−+u[n]52153(b)y[n]−1y[n−2]=x[n−1],y[−1]=1,y[−2]=0,x[n]=2u[n]92X(z)=1−z−11Y(z)−z−2Y(z)+z−1=z−1X(z)91−21−1−1Y(z)1−z=z+zX(z)991z−1z−1X(z)Y(z)=+91−1z−21−1z−299Y(n)(z)Y(f)(z)NaturalResponse1z−1Y(n)(z)=91−1z−291111=−61−1z−161+1z−133nn(n)111y[n]=−−u[n]633ForcedResponse933Y(f)(z)=4−4−21−z−11+1z−11−1z−133nn(f)93131y[n]=−−−u[n]44323CompleteResponsennnn93131111课后答案网y[n]=−−−u[n]+−−u[n]44323633(c)y[n]−1y[n−1]−1y[n−2]=x[n]+x[n−1],y[−1]=1,y[−2]=−1,x[n]=3nu[n]48www.hackshp.cn1X(z)=1−3z−111Y(z)−z−1Y(z)+1−z−2Y(z)+z−1−1=X(z)+z−1X(z)481−11−211−1−1Y(z)1−z−z=+z+(1+z)X(z)488811+z−1(1+z−1)X(z)Y(z)=+8(1−1z−1)(1+1z−1)(1−1z−1)(1+1z−1)2424Y(n)(z)Y(f)(z)NaturalResponse(n)1111Y(z)=−41−1z−181+1z−12433 nn(n)111y[n]=2−−u[n]824ForcedResponse96−2−1Y(f)(z)=65+5+131−3z−11−1z−11+1z−124nn(f)96n2111y[n]=(3)−−−u[n]6552134CompleteResponsenn96n31211y[n]=(3)−−−u[n]652021044SolutionstoAdvancedProblems7.43.Usethez-transformofu[n]andthedifferentiationinthez-domainpropertytoderivetheformulaforevaluatingthesum∞n2ann=0assuming|a|<1.∞X(z)=x[n]z−nn=0letx[n]=anu[n]∞d−(n−1)X(z)=−nx[n]zdz课后答案网n=−∞d2∞X(z)=n(n−1)x[n]z−(n−2)dz2n=−∞∞∞www.hackshp.cn=n2x[n]z−(n−2)−nx[n]z−(n−2)n=−∞n=−∞2∞∞d12n−(n−2)n−(n−2)=naz−nazdz21−az−1n=0n=0Evaluateatz=12∞∞d1=n2an−nandz21−az−1z=1n=0n=0dX(z)|dzz=1∞d21d1n2an=−dz21−az−1dz1−az−1n=0z=1z=12aa=+(1−a)3(1−a)234 3a−a2=(1−a)37.44.Acontinuous-timesignaly(t)satisfiesthefirst-orderdifferentialequationdy(t)+2y(t)=x(t)dtUsetheapproximationdy(t)≈[y(nT)−y((n−1)T)]/Ttoshowthatthesampledsignaly[n]=dtsssy(nTs)satisfiesthefirst-orderdifferenceequationy[n]+αy[n−1]=v[n]Expressαandv[n]intermsofTsandx[n]=x(nTs).y[n]−y[n−1]+2y[n]=x[n]Ts11y[n]+2−y[n−1]=x[n]TsTs1Tsy[n]−y[n−1]=x[n]1+2Ts1+2Ts1α=−1+2TsTsv[n]=x[n]1+2Ts7.45.Theautocorrelationsignalforareal-valuedcausalsignalx[n]isdefinedas∞rx[n]=x[l]x[n+l]课后答案网l=0Assumethez-transformofrx[n]convergesforsomevaluesofz.Findx[n]ifwww.hackshp.cn1Rx(z)=(1−αz−1)(1−αz)where|α|<1.rx[n]=x[n]∗x[−n]Lety[n]=x[−n]∞rx[n]=y[k]x[n−k]k=−∞∞=x[−k]x[n−k]k=−∞letp=−k35 ∞rx[n]=x[p]x[n+p]p=−∞∞=x[l]x[n+l]l=−∞∞=x[l]x[n+l]l=0sincex[l]=0,forl<01Rx(z)=X()X(z)z11=1−αz1−αz−1Implies1X(z)=1−αz−1x[n]=αnu[n]7.46.Thecross-correlationoftworeal-valuedsignalsx[n]andy[n]isexpressedas∞rxy(n)=x[l]y[n+l]l=−∞(a)Expressrxy[n]asaconvolutionoftwosequences.rxy[n]=x[n]∗y[−n],seepreviousproblem(b)Findthez-transformof课后答案网rxy[n]asafunctionofthez-transformsofx[n]andy[n].1Rxy(z)=X(z)Y()www.hackshp.cnz7.47.Asignalwithrationalz-transformhasevensymmetry,thatis,x[n]=x[−n].(a)Whatconstraintsmustthepolesofsuchasignalsatisfy?zx[n]←−−−→X(z)z1x[−n]←−−−→X()zImplies1X(z)=X()z36 Thismeansifthereisapoleatz,theremustalsobeapoleat1.Hencepolesoccurinreciprocalpairs.ozo(b)Showthatthez-transformcorrespondstoastablesystemifandonlyif∞x[n]<∞.n=−∞Thereciprocalpolesarez,1assumetheytakethefollowingform:ozoz=rejθooo11−jθo=ezoroIfzisinside|z|=1,itsz-transformisright-sidedandstable.Forthepoleat1,itscorrespondingozoz-transformiseitherright-sidedunstable,orleft-sidedstable.Forconvergence,theROCmustincludetheunitcircle,|z|=1,whichmeansthez-transformsareexponentiallydecayingastheyapproach∞,−∞respectively.(c)Suppose2−(17/4)z−1X(z)=(1−(1/4)z−1)(1−4z−1)DeterminetheROCandfindx[n].Forthesystemtobestable,thepoleatz=1mustberight-sided,andthepoleatz=4mustbe4leftsidedsotheirz-transformsareexponentiallydecayingastheyapproach∞,−∞respectively.ThisimpliestheROCis1<|z|<447.48.ConsideraLTIsystemwithtransferfunction1−a∗zH(z)=,|a|<1课后答案网z−aHerethepoleandzeroareaconjugatereciprocalpair.(a)Sketchapole-zeroplotforthissysteminthez-plane.Imjwww.hackshp.cnLeta=|a|e11jthen=ea*|a|a11ReaFigureP7.48.(a)Pole-Zeroplot.37 (b)Usethegraphicalmethodtoshowthatthemagnituderesponseofthissystemisunityforallfre-quency.Asystemwiththischaracteristicistermedanall-passsystem.1−a∗ejΩH(ejΩ)=ejΩ−a1=1−a∗ejΩ|ejΩ−a|Asshownbelow,H(ejΩ)=1forallΩ.ImjH(e)11+a1−aa11Re+ajH(e)2课后答案网Im11−awww.hackshp.cn11+aa11Re+aFigureP7.48.(b)MagnitudeResponse.38 jjH(e)H(e)121+FigureP7.48.(b)MagnitudeResponse.(c)Usethegraphicalmethodtosketchthephaseresponseofthissystemfora=1.21−1ejΩH(ejΩ)=2ejΩ−12"#"#!11argH(ejΩ)=arg1−ejΩ−argejΩ−22"#!1=π+argejΩ−2−argejΩ−2课后答案网ImImwww.hackshp.cnpolezero0.52ReReFigureP7.48.(c)Pole/Zerographicalmethod.39 zero22FigureP7.48.(c)Zerophaseresponse.poleFigureP7.48.(c)Polephaseresponse.jH(e)课后答案网www.hackshp.cnFigureP7.48.(c)Phaseresponse.(d)Usetheresultfrom(b)toprovethatanysystemwithatransferfunctionoftheform$P1−a∗zH(z)=k|a|<1kz−akk=140 correspondstoastableandcausalall-passsystem.Since|ak|<1,ifthesystemiscausal,thenthesystemisstable.$P1−a∗ejΩH(ejΩ)=kejΩ−akk=11−a∗ejΩ1−a∗ejΩ1−a∗pejΩH(ejΩ)=12···ejΩ−a1ejΩ−a2ejΩ−ap$P1−a∗ejΩH(ejΩ)=kejΩ−akk=11−a∗ejΩ1−a∗ejΩ1−a∗ejΩH(ejΩ)=12···pejΩ−a1ejΩ−a2ejΩ−ap=1Thesystemisall-pass.(e)Canastableandcausalall-passsystemalsobeminimumphase?Explain.Forastableandcausalall-passsystem,|ak|<1forallk.UsingP=1:1−a∗z1H(z)=z−a11Thezeroiszz=∗a1Whichimplies1|zz|=∗>1课后答案网|a1|Thissystemcannotalsobeminimumphase.www.hackshp.cn7.49.LetH(z)=F(z)(z−a)andG(z)=F(z)(1−az)where01.(a)ShowthatH(z)canbefactoredintheformH(z)=Hmin(z)Hap(z)whereHmin(z)isminimumphaseandHap(z)isall-pass(seeProblemP7.48).%M−1−1−1b0k=1(1−ckz)H(z)=(1−cMz)%N(1−dkz−1)k=1%M−1−11−1−1b0k=1(1−ckz)(1−cM∗z)1−cMz=%N(1−dz−1)1−1z−1k=1kcM∗Hmin(z)Hap(z)=Hmin(z)Hap(z)(b)FindaminimumphaseequalizerwithtransferfunctionH(z)chosensothat|H(ejΩ)H(ejΩ)|=1eqeqanddeterminethetransferfunctionofthecascadeH(z)Heq(z).1Heq(z)=Hmin(z)soH(z)Heq(z)=Hap(z)|H(z)Heq(z)|=|Hap(z)|=1课后答案网1−cz−1MHap(z)=1−1z−1cM∗7.51.Averyusefulstructureforimplementingnonrecursivesystemsistheso-calledlatticestucture.www.hackshp.cnThelatticeisconstructedasacascadeoftwoinput,twooutputsectionsoftheformdepictedinFig.P7.51(a).AnMthorderlatticestructureisdepictedinFig.P7.51(b).(a)Findthetransferfunctionofasecond-order(M=2)latticehavingc=1andc=−1.1224A(z)−1−1zzX(z)0.5−0.250.5−0.25Y(z)B(z)FigureP7.51.(a)LatticeDiagram.43 1−1Y(z)=−A(z)z+B(z)4−11A(z)=X(z)(z+)21−1B(z)=X(z)(z+1)21−23−1Y(z)=−z+z+1X(z)48Y(z)H(z)=X(z)3−11−2=1+z−z84(b)Wemaydeterminetherelationshipbetweenthetransferfunctionandlatticestructurebyexaminingtheeffectofaddingasectiononthetransferfunction,asdepictedinFig.P7.51(c).HerewehavedefinedHi(z)asthetransferfunctionbetweentheinputandtheoutputofthelowerbranchintheithsectionandH˜(z)asthetransferfunctionbetweentheinputandtheoutputoftheiupperbranchintheithsection.Writetherelationshipbetweenthetransferfunctionstothe(i−1)standithstagesasH˜i(z)H˜i−1(z)=T(z)Hi(z)Hi−1(z)whereT(z)isatwo-by-twomatrix.ExpressT(z)intermsofcandz−1.iH˜(z)=H˜(z)z−1+cH(z)ii−1ii−1H(z)=H˜(z)z−1c+H(z)ii−1ii−1H˜(z)z−1cH˜(z)iii−1=课后答案网H(z)z−1c1H(z)iii−1z−1ciT(z)=z−1c1www.hackshp.cni(c)UseinductiontoprovethatH˜(z)=z−iH(z−1).iii=1H˜i−1(z)=Hi−1(z)=1H˜(z)=z−1+c11H(z)=z−1c+111H(z−1)=1+zc11H˜(z)=z−1H(z−1)11i=kassumeH˜(z)=z−kH(z−1)kki=k+1H˜(z)=z−1H˜(z)+cH(z)k+1kk+1k44 H˜(z)=z−1cH˜(z)+H(z)k+1k+1kksubstituteH˜(z)=z−kH(z−1)kkH˜(z)=z−(k+1)H(z−1)+cH(z)k+1kk+1kH(z)=z−(k+1)cH(z−1)+H(z)k+1k+1kkH(z−1)=z(k+1)cH(z)+H(z−1)k+1k+1kkH˜(z)=z−(k+1)z(k+1)cH(z)+H(z−1)k+1k+1kkH˜(z)=z−(k+1)H(z−1)k+1k+1ThereforeH˜(z)=z−iH(z−1)ii(d)Showthatthecoefficientofz−iinH(z)isgivenbyc.iiH˜z−1cH˜iii−1=Hz−1c1Hiii−1H=z−1cH˜+Hiii−1i−1H˜(z)=z−(i−1)+...+ci−1i−1Thehighestorderof(z−1)inH(z)is(i),andthecoefficientsofz−iis(c),sinceH(z)doesnotiii−1contributetoz−i,thereforethecoefficientofz−iisH(z)isgivenby(c).ii(e)Bycombiningtheresultsof(b)-(d)wemayderiveanalgorithmforfindingthecirequiredbythelatticestructuretoimplementanarbitraryorderMnonrecursivetransferfunctionH(z).Startwithi=MsothatH(z)=H(z).Theresultof(d)impliesthatcisthecoefficientofz−MinH(z).MMBydecreasingi,continuethisalgorithmtofindtheremainingci.Hint:Usetheresultof(b)tofindatwo-by-twomatrix课后答案网A(z)suchthatH˜i−1(z)H˜i(z)=A(z)www.hackshp.cnHi−1(z)Hi(z)H˜i(z)H˜i−1(z)=THi(z)Hi−1(z)H˜i−1(z)H˜i−1(z)=A(1)Hi−1(z)Hi−1(z)whereA=T−111−ciA=z−1(1−c2)−z−1ciiletMH(z)=bz−kkk=045 =HM(z)WhereH(z)isgiven.FromthiswehavecM=bM.Thefollowingisanalgorithmtoobtainallci’s.(1)cM=bM(2)fori=Mto2,descendingcomputeH˜i−1(z)andHi−1(z)from(1)getci−1fromHi−1(z)endThuswewillhavec1,c2,...cM.7.52.Causalfiltersalwayshaveanonzerophaseresponse.Onetechniqueforattainingzerophasere-sponsefromacausalfilterinvolvesfilteringthesignaltwice,onceintheforwarddirectionandthesecondtimeinthereversedirection.Wemaydescribethisoperationintermsoftheinputx[n]andfilterimpulseresponseh[n]asfollows.Lety1[n]=x[n]∗h[n]representfilteringthesignalintheforwarddirection.Nowfiltery1[n]backwardstoobtainy2[n]=y1[−n]∗h[n].Theoutputisthengivenbyreversingy2[n]toobtainy[n]=y2[−n].(a)Showthatthissetofoperationsisequivalentlyrepresentedbyafilterwithimpulseresponseho[n]asy[n]=x[n]∗ho[n]andexpressho[n]intermsofh[n].y[n]=y1[n]∗h[n]=(x[n]∗h[n])∗h[−n]=x[n]∗(h[n]∗h[−n])=x[n]∗ho[n]ho[n]=h[n]∗h[−n](b)Showthatho课后答案网[n]isanevensignalandthatthephaseresponseofanysystemwithanevenimpulseresponseiszero.www.hackshp.cnho[−n]=h[−n]∗h[n]=h[n]∗h[−n]ho[−n]=ho[n]Whichshowsthatho[n]isanevensignal.Sinceho[n]iseven,thephaseresponsecanbefoundfromthefollowing:∞H(ejΩ)=h[n]e−jΩnoon=−∞∞H∗(ejΩ)=h[n]ejΩnoon=−∞46 ∞=h[−n]e−jΩnon=−∞∞=h[n]e−jΩnon=−∞H∗(ejΩ)=H(ejΩ)oo!argH(ejΩ)=0o(c)Foreverypoleorzeroatz=βinh[n],showthath[n]hasapairofpolesorzerosatz=βandz=1.oβho[n]=h[n]∗h[−n]soH(z)=H(z)H(z−1)oletz−cH(z)=z−βz−cz−1−cHo(z)=z−βz−1−β(z−c)(1−cz)=(z−β)(1−z)ββTherefore,H(z)hasapairofpolesatz=β,1oβletz−βH(z)=z−p课后答案网z−βz−1−βHo(z)=z−pz−1−p(z−β)(z−1)ββ=www.hackshp.cnp(z−p)(z−1)βpTherefore,H(z)hasapairofzerosatz=β,1oβ7.53.Thepresentvalueofaloanwithinterestcompoundedmonthlymaybedescribedintermsofthefirst-orderdifferenceequationy[n]=ρy[n−1]−x[n]r/12whereρ=1+,ristheannualinterestrateexpressedasapercent,x[n]isthepaymentcredited100attheendofthenthmonth,andy[n]istheloanbalanceatthebeginningofthen+1stmonth.Thebeginningloanbalanceistheinitialconditiony[−1].Ifuniformpaymentsof$caremadeforLconsecutivemonths,thenx[n]=c{u[n]−u[n−L]}.47 (a)Usetheunilateralz-transformtoshowthatL−1−ny[−1]ρ−czY(z)=n=01−ρz−1Hint:Uselongdivisiontoshowthat1−z−LL−1=z−n1−z−1n=0y[−1]ρ−X(z)Y(z)=1−ρz−11−z−LX(z)=c1−z−1=c1+z−1+z−2+...+z−L+1L−1=cz−nn=0whichimpliesL−1−ny[−1]ρ−czY(z)=n=01−ρz−1(b)Showthatz=ρmustbeazeroofY(z)iftheloanistohavezerobalanceafterLpayments.L−1−ny[−1]p+czY(z)=n=01−ρz−1Thepoleatz=presultsinaninfinitelengthy[n]ingeneral.IftheloanreacheszeroaftertheLthpayment,wehave:课后答案网y[n]=0,n≥L−1L−2So,Y(z)=y[n]z−nwww.hackshp.cnn=0Thuswemusthave:L−1L−2y[−1]ρ−cz−n=(1−ρz−1)y[n]z−nn=0n=0tocancelthepoleatz=pFrompolynomialtheory,thefirsttermiszeroiff(z=ρ)=0,orρ=1+risazeroofY(z)12(c)Findthemonthlypayment$casafunctionoftheintitialloanvaluey[−1]andtheinterestraterassumingtheloanhaszerobalanceafterLpayments.L−1y[−1]ρ−cρ−n=0n=048 1−ρ−Lc=y[−1]ρ1−ρ−1ρ−1c=y[−1]1−ρ−LSolutionstoComputerExperiments7.54.UsetheMATLABcommandzplanetoobtainapole-zeroplotforthefollowingsystems:−2(a)H(z)=1+z2+z−1−1z−2+1z−324P7.54(a)10.80.60.40.2课后答案网0ImaginaryPart−0.2www.hackshp.cn−0.4−0.6−0.8−1−1−0.500.51RealPartFigureP7.54.(a)Pole-ZeroplotofH(z)1+z−1+3z−2+1z−3(b)H(z)=221+3z−1+1z−22249 P7.54(b)10.80.60.40.20ImaginaryPart−0.2−0.4−0.6−0.8−1−1.5−1−0.500.511.5RealPartFigureP7.54.(b)Pole-ZeroplotofH(z)7.55.UsetheMATLABcommandresidueztoobtainthepartialfractionexpansionsrequiredtosolveProblem7.24(d)-(g).P7.55:=======Part(d):==========课后答案网r=2www.hackshp.cn-1p=-2.00000.5000k=0Part(e):50 ==========r=1.46881.5312p=4-4k=0Part(f):==========r=2-1p=-1.0000-0.5000课后答案网k=01www.hackshp.cnPart(g):==========r=1-1p=-151 1k=2007.56.UsetheMATLABcommandtf2sstofindstate-variabledescriptionsforthesystemsinProblem7.27.P7.56:=======Part(a):==========A=1.50001.00001.00000B=10C=1.50002.0000D=课后答案网2Part(b):www.hackshp.cn==========A=1610B=1052 C=530D=5Part(c):==========A=0.2500-0.06251.00000B=10C=40D=0课后答案网7.57.UsetheMATLABcommandss2tftofindthetransferfunctionsinProblem7.33.P7.57:www.hackshp.cn=======Part(b)-(i):==========Num=1.0000-2.0000-1.2500Den=1.00000-0.250053 Part(b)-(ii):==========Num=020Den=1.0000-0.2500-0.3750Part(b)-(iii):==========Num=02.0000-6.5000Den=1.0000-0.50000.2500课后答案网www.hackshp.cn7.58.UsetheMATLABcommandzplanetosolveProblem7.35(a)and(b).54 P7.58(a)Poles−Zerosoftheinversesystem21.510.520ImaginaryPart−0.5−1−1.5−2−1−0.500.511.522.533.54RealPartFigureP7.58.(a)Pole-Zeroplotsoftheinversesystem.P7.58(b)Poles−Zerosoftheinversesystem10.80.60.4课后答案网0.20www.hackshp.cnImaginaryPart−0.2−0.4−0.6−0.8−1−1−0.500.51RealPartFigureP7.58.(b)Pole-Zeroplotsoftheinversesystem.55 7.59.UsetheMATLABcommandfreqztoevaluateandplotthemagnitudeandphaseresponseofthesystemgiveninExample7.21.P7.59121086Magnitude42−3−2−10123Omega210Phase(rad)−1−2−3−2−10123OmegaFigureP7.59.MagnitudeResponseforExample7.21课后答案网www.hackshp.cn7.60.UsetheMATLABcommandfreqztoevaluateandplotthemagnitudeandphaseresponseofthesystemsgiveninProblem7.37.56 P7.60(a)43.532.5Magnitude21.51−3−2−10123Omega3210Phase(rad)−1−2−3−3−2−10123OmegaFigureP7.60.(a)MagnitudeandphaseresponseP7.60(b)10.80.60.4Magnitude0.2课后答案网−3−2−10123www.hackshp.cnOmega210Phase(rad)−1−2−3−2−10123OmegaFigureP7.60.(b)Magnitudeandphaseresponse57 P7.60(c)543Magnitude21−3−2−10123Omega1.510.50Phase(rad)−0.5−1−1.5−3−2−10123OmegaFigureP7.60.(c)Magnitudeandphaseresponse7.61.UsetheMATLABcommandsfilterandfiltictoplottheloanbalanceatthestartofeachmonthn=0,1,...L+1forProblem7.53.Assumethat课后答案网y[−1]=$10,000,L=60,r=0.1andthemonthlypaymentischosentobringtheloanbalancetozeroafter60payments.FromProblem7.53:www.hackshp.cnρ−1c=y[−1]1−ρ−Ly[−1]=10,000L=60rρ=1+=1.0083312c=0.02125b=[y[−1]ρ−c,−c(ones(1,59))]a=[1,ρ]58 P7.611000090008000700060005000Balance400030002000100000102030405060MonthFigureP7.61.Monthlyloanbalance.7.62.UsetheMATLABcommandzp2sostodetermineacascadeconnectionofsecond-ordersectionsforimplementingthesystemsinProblem7.38.=======Part(a):==========课后答案网sos=1.0000-0.35360.0625www.hackshp.cn1.0000-0.50000.25001.00000.46190.06251.00001.38580.5625Part(b):==========sos=1.00004.00004.00001.00000.3750-0.28131.0000-0.00000.25001.0000-0.37500.140659 7.63.Acausaldiscrete-timeLTIsystemhasthetransferfunction0.0976(z−1)2(z+1)2H(z)=(z−0.3575−j0.5889)(z−0.3575+j0.5889)(z−0.7686−j0.3338)(z−0.7686+j0.3338)(a)Usethepoleandzerolocationstosketchthemagnituderesponse.Im−10.3338a=tan10.7686−10.5889a=tan20.3575adouble1adouble2ReFigureP7.63.(a)PoleZeroplot课后答案网symmetricwww.hackshp.cnaa122FigureP7.63.(a)SketchoftheMagnitudeResponse.(b)UsetheMATLABcommandszp2tfandfreqztoevaluateandplotthemagnitudeandphaseresponse.60 P7.63(b)0.80.60.4|H(Omega)|0.20−3−2−10123Omega3210−10.Wemaythenwrite1Hs()=------------------------------()sa+D¢()s2 whereD¢()sistheremainderofthedenominatorpolynomial.ForaButterworthlow-passfilterofcutofffrequencywc,allthepolesofH(s)lieonacircleofradiuswcintheleft-halfplane.Hence,wemusthavea=-wc.(b)ForaButterworthlow-passfilterofevenorderN,allthepolesofthetransferfunctionH(s)arecomplex.Theyalllieonacircleofradiuswcintheleft-halfplane.Lets=-a-jb,witha>0andb>0,denoteacomplexpoleofH(s).AllthecoefficientsofH(s)arereal.Thisconditioncanonlybesatisfiedifwehaveacomplexconjugatepoleats=-a+jb.Wemaythenexpressthecontributionofthispairofpolesas11---------------------------------------------------------=------------------------------()saj++b()saj+–b()sa+2+b2whosecoefficientsareallreal.WethereforeconcludethatforevenfilterorderN,allthepolesofH(s)occurincomplex-conjugatepairs.8.20ThetransferfunctionofaButterworthlow-passfilteroforder5is1Hs()=----------------------------------------------------------------------------------------------------(1)22()s+1()s++0.618s1()s++1.618s1Thelow-passtohigh-passtransformationisdefinedby1s®---swhereitisassumedthatthecutofffrequencyofthehigh-passfilterisunity.Hencereplacingswith1/sinEq.(1),wefindthatthetransferfunctionofaButterworthhigh-passfilteroforder5is1Hs()=--------------------------------------------------------------------------------------------------æö1---1+æö----1++0.618-------------1æö----1++0.618-------------1èø课后答案网sèøs2sèøs2s5s=----------------------------------------------------------------------------------------------------22()s+1()swww.hackshp.cn++0.618s1()s++1.618s1Themagnituderesponseofthishigh-passfilterisplottedinFig.1.1.41.210.80.6MagnitudeResponse0.4Figure10.2000.511.522.533.544.55NormalizedFrequencyw/wc3 8.21Wearegiventhetransferfunction1Hs()=--------------------------------------------------------------------------------------------------(1)2()s+1()s++0.618s1()s++1.618s1Tomodifythislow-passfiltersoastoassumeacutofffrequencywc,weusethetransformationss®------wcHence,replacingswiths/wcinEq.(1),weobtain:1Hs()=-----------------------------------------------------------------------------------------------------------------------sæö2sæös2sæö------1+ç÷------s+0.618------1+ç÷------+1.618------1+èøwèø2wèø2wcwcwccc5wc=----------------------------------------------------------------------------------------------------------------------------2222()s+w()s++0.618wsw()s+1.618ws+wccccc8.22Wearegiventhetransferfunction1Hs()=--------------------------------------------(1)2()s+1()s++s1Totransformthislow-passfilterintoaband-passfilterwithbandwidthBcenteredonw0,weusethefollowingtransformation:22s+w0s®-----------------课后答案网BsWithw0=1andBwww.hackshp.cn=0.1,thetransformationtakesthevalue2s+1s®--------------(2)0.1sSubstitutingEq.(2)into(1):1Hs()=---------------------------------------------------------------------------------------------2222æös-------------+1-1+æöæös-------------+1-++æö-------------s+1-1èø0.1sèøèø0.1sèø0.1s30.001s=-------------------------------------------------------------------------------------------------------------------------------24232()s++0.1s1(s+++++2s10.1s0.1s0.01s)4 30.001s=----------------------------------------------------------------------------------------------------------------(3)2432()s++0.1s1()s++++0.1s2.01s0.1s1Themagnituderesponseofthisband-passfilteroftheButterworthtype,obtainedbyputtings=jwinEq.3,isplottedinFig.1.BandpassResponse1.41.210.80.6SquaredMagnitudeResponse0.4Figure10.2000.20.40.60.811.21.41.61.82NormalizedFrequencyw/wc8.23Forthecounterparttothelow-passfilteroforderoneinFig.8.14(a),wehave1Wooo课后答案网+v+IFv2(t)1(t)--www.hackshp.cnoooInputOutputFilter5 Forthecounterparttothelow-passfilteroforderthreeinFig.8.14(b),wehave1W1Hoooo+v+IFv2(t)1(t)-IF-ooooInputOutputFilterNotethatinbothofthesefigures,the1Wresistancemaybeusedtoaccountforthesourceresistance.Notealsothattheloadresistanceisinfinitelylarge.8.24(a)ForFig.8.14(a),thecapacitorhasthevalue1C=---------------------------------------F45()10()2p´10310=--------pF2p=159pF(b)ForFig.8.14(a),thetwocapacitorsare159PFeach.Theinductorhasthevalue4()10L=--------------------------H5()2p´10100=---------mH课后答案网2p=15.9mHwww.hackshp.cn8.25LetthetransferfunctionoftheFIRfilterbedefinedby–1Hz()=()1–zAz()(1)whereA(z)isanarbitrarypolynomialinz-1.TheH(z)ofEq.(1)hasazeroatz=1asprescribed.Letthesequencea[n]denotetheinversez-transformofA(z).ExpandingEq.(1):–1Hz()=Az()–zAz()(2)6 whichmayberepresentedbytheblockdiagram:-ºA(z).z-1SH(z)+FromEq.(2)wereadilyfindthattheimpulseresponseofthefiltermustsatisfytheconditionhn[]=an[]–an[]–1wherea[n]istheinversez-transformofanarbitrarypolynomialA(z).8.26LetthetransferfunctionofthefilterbedefinedbyM¤2jW-jnWH¢()e=åhd[]nen=-M¤2Replacingnwithn-M/2soastomakethefiltercausal,wemaythuswriteM-jnæö–----M-WjWMèø2H¢()e=åhdn–-----e2n=0MjMW¤2M-jnW=eåhdn–-----e(1)2n=0WearegiventhatMMhn[]=h[]nfor–-----££n-----课后答案网d22ThisconditionisequivalenttoMhn–-----=hn[]for0££nMd2www.hackshp.cnHence,wemayrewriteEq.(1)intheformNjWjMW¤2-jnWH¢()e=eåhn[]en=0jMW¤2jW=eHe()EquivalentlywehavejW–jMW¤2jWHe()=eH¢()ewhichisthedesiredresult.7 8.27AccordingtoEqs.(8.64)and(8.65),themagnituder=|z|andphaseq=arg{z}aredefinedby2212¤æö()1+s+wr=ç÷---------------------------------(1)22èø()1–s+w–1w–1q=tanæö-------------–tanæö------------w(2)èø1+sèø1–sThesetworelationsarebasedonthetransformationjq1+szr==e-----------,s=s+jw1–sForthemoregeneralcaseofasamplingrate1/Tsforwhichwehave1z–1s=----------------Tz+1sorTs1+-----s2z=------------------Ts1–-----s2TTsswemayrewriteEqs.(1)and(2)byreplacingwwith-----wandswith-----s,obtaining2212¤æöæö222ç÷èø-----+s+wTç÷------------------------------------sr=ç÷æö222ç÷-----–s+wèøèøT课后答案网sæöæö–1ç÷w–1ç÷wq=tanç÷----------------www.hackshp.cn–tanç÷---------------ç÷-----2+sç÷-----2–sèøTèøTss8.28(a)FromSection1.10,werecalltheinput-outputrelationyn[]=xn[]r+yn[]–1Takingz-transforms:–1Yz()=Xz()r+zYz()Thetransferfunctionofthefilteristherefore8 Yz()1Hz()==------------------------------(1)Xz()–11–rzForr=1,wehave1Hz()=----------------(2)–11–zForz=jW,thefrequencyresponseofthefilterisdefinedbyjW1He()=-------------------–jW1–ewithjW1He()=----------------------–jW1–e1=---------------------------------------------------------------2212¤[]()1–cosW+sinW1=--------------------------------------12¤()22–cosWwhichisplottedinFig.1for0££Wp.FromthisfigureweseethatthefilterdefinedinEq.(2)doesnotdeviatefromtheidealintegratorbymorethan1%for0££W0.49.r=14课后答案网3.53www.hackshp.cn2.51Percenterroroccursatw=0.492Magnitude1.51Filtered0.5Figure1Ideal000.511.522.53NormalizedFrequency9 (b)Forr=0.99,theuseofEq.(1)yields1Hz()=---------------------------(3)–11–0.99zforwhichthefrequencyresponseisdefinedbyjW1He()=------------------------------–jW1–0.99eThatis,jW1He()=---------------------------------–jW1–0.99e1=------------------------------------------------------------------------------------------2212¤[]()1–0.99cosW+()0.99sinW1»-----------------------------------------------------12¤()1.98–1.98cosWwhichisplottedinFig.2.FromthissecondfigureweseethattheusablerangeofthefilterofEq.(3)asanintegratorisreducedto0.35.r=0.9943.531Percenterroroccursatw=0.362.5课后答案网2Magnitude1.5www.hackshp.cn1Figure2Filtered0.5Ideal000.511.522.53NormalizedFrequency8.29ThetransferfunctionofthedigitalIIRfilteris30.0181()z+1Hz()=----------------------------------------------------------------------------------------------2()z–0.50953()z–1.2505z+0.39812ExpandingthenumeratoranddenominatorpolynomialsofH(z)inascendingpowersof10 z-1,wemaywrite–1–2–30.01811()+++3z3zzHz()=---------------------------------------------------------------------------------------------–1–2–31–1.7564z+1.0308z–0.2014zHence,thefiltermaybeimplementedindirectformIIusingthefollowingconfiguration:0.0205.+InputSSOutput+x[n]-+y[n]z-11.7564.3SSz-1-1.0308.3SSz-10.2014.8.30(a)Thereceivedsignal,ignoringchannelnoise,isgivenbyyt()=xt()++0.1xt()–100.2xt()–15wherex(t)isthetransmittedsignal,andtimetismeasuredinmicroseconds.Supposey(t)issampleduniformlywithasamplingperiodof5ms,yieldingyn[]=xn[]++0.1xn[]–20.2xn[]–3(1)(b)Takingthez-transformsofEq.(1):–2–3Yz()=Xz()++0.1zXz()0.2zXz()which,inturn,yieldsthetransferfunctionforthechannel:课后答案网Yz()–2–3Hz()==-----------1++0.1z0.2zXz()Thecorrespondingequalizerisdefinedbythetransferfunctionwww.hackshp.cn11H()z==----------------------------------------------------------eqHz()–2–31++0.1z0.2zwhichisrealizedbytheIIRfilter:y[n]+S.Equalizeroutput-z.-2S0.1z-10.211 ForH()ztobestable,allofitsthreepolesor,equivalently,allthreerootsoftheeqcubicequation–2–31++0.1z0.2z=0,orequivalently,3z++00.1z0.2=mustlieinsidetheunitcircleinthez-plane.UsingMATLAB,wefindtherootsarez=0.2640+j0.5560z=0.2640-j0.5560z=-0.5280whichconfirmstabilityoftheIIRequalizer.[Note:ThestabilityofH()zmayalsobeexploreduysinganindirectapproach,eqnamely,theRouth-Hurwitzcriterionwhichavoidshavingtocomputetheroots.First,weusethebilineartransformation1+sz=-----------1–sandthenconstructtheRoutharrayasdescribedinSection9.12.ThestabilityofH()zisconfirmedbyexaminingthecoefficientsofthefirstcolumnoftheRoutheqarray.ThefactthatallthesecoefficientsarefoundtobepositiveassuresthestabilityofH()z.]eqRealizingtheequalizerbymeansofanFIRstructure,wehave–2–3–1H()z=()1++0.1z0.2zeq–2–3–2–32–2–33=1–()0.1z+0.2z++()0.1z+0.2z–()0.1z+0.2z¼–2–3–4–5–6=1–()0.1z+0.2z+()0.01z++0.04z0.04z课后答案网–6–7–8–9–()0.001z+++0.006z0.012z0.008z+¼Ignoringcoefficientssmallerthan1%asspecified,wehavetheapproximateresult:www.hackshp.cn–2–3–4–5–6–8H()z»1–0.1z–0.2z+++0.01z0.04z0.04z–0.012zeqwhichisrealizedusingthefollowingFIRstructure:y[n].z-2.z-1.z-1.z-1.z-1.z-2-0.1-0.20.010.040.04-0.012SSSSSSEqualizeroutput12 AdvancedProblems8.31Theintegratoroutputistyt()=òx()ttd(1)t-T0FTLetxt()«Xj()w.Wemaythereforereformulatetheexpressionfory(t)asæötç÷æö1¥jwtyt()=òç÷èø------2pòXj()wedwdtt-T–¥0ç÷èøìïïïïíïïïïîx()tInterchangingtheorderofintegration:¥1æötjwtyt()=ò------2pXj()wèøòedtdw–¥t-T0æöT01¥TwTjwèøt–-----2-=------Xj()w×w------0sincæö----------0ed(2)2pò–¥2pèø2p(a)InvokingtheformulafortheinverseFouriertransform,weimmediatelydeducefromEq.(2)thattheFouriertransformoftheintegratoroutputy(t)isgivenbyT0æöwT0–jwT0¤2Yj()w=------sinc----------e(3)2pèø2pExaminingthisformula,wealsoreadilyseethaty(t)canbeequivalentlyobtainedbypassingtheinputsignal课后答案网x(t)throughafilterwhosefrequencyresponseisdefinedbyT0æöwT0–jwT0¤2Hj()w=------sinc----------e2pèø2pThemagnituderesponseofthefilterisdepictedinFig.1:www.hackshp.cn0.20.18T/2*pi0.160.14IdealLowPassFilter0.120.10.080.06Figure10.040.020−6*pi/T−4*pi/T−2*pi/T02*pi/T4*pi/T6*pi/T13 (b)Figure1alsoincludesthemagnituderesponseofan“approximating”ideallow-passfilter.Thislatterfilterhasacutofffrequencyw0=2p/T0andpassbandgainofT0/2p.Moreover,thefilterhasaconstantdelayofT0/2.Theresponseofthisidealfiltertoastepfunctionappliedattimet=0isgivenby2pæöT0T------èøt–------0T02sinly¢()t=------ò-----------dlp–¥lAtt=T0,wethereforehaveT0psinly¢()T=-----------------dl0pòl–¥T0æö0sinlpsinl=-----p-èøò----------l-dl+ò----------l-dl–¥0T0=------()S()¥+S()ppii=1.09T0FromEq.(1)wefindthattheidealintegratoroutputattimet=T0inresponsetothestepfunctionx(t)=u(t)isgivenbyT0yT()=u()ttd0ò0Itfollowsthereforethattheoutputofthe“approximating”ideallow-passfilterexceedstheoutputoftheidealintegratorby9%.ItisnoteworthythatthisovershootisindeedamanifestationoftheGibb’sphenomenon.8.32Totransformaprototypelow-passfilterintoabandstopfilterofmidbandrejectionfrequency课后答案网w0andbandwidthB,wemayusethetransformationBss®-----------------(1)22s+w0www.hackshp.cnAnasillustrativeexample,considerthelow-passfilter:1Hs()=-----------(2)s+1UsingthetransformationofEq.(1)inEq.(2),weobtainabandstopfilterdefinedby1Hs()=---------------------------Bs-----------------+122s+w014 22s+w0=------------------------------22s++Bsw0whichischaracterizedasfollows:Hs()==Hs()1s=0s=¥Hs()=0sj=±w08.33AnFIRfilteroftype1hasanevenlengthMandissymmetricaboutM/2inthatitscoefficientssatisfytheconditionhn[]=hMn[]–forn=01,,,¼MThefrequencyresponseofthefilterisMjW–jnWHe()=åhn[]en=0whichmaybereformulatedasfollows:M------12MjW–jnWM–jMW¤2–jnWHe()=åhn[]e++h-----eåhn[]e2n=0Mn=-----+12MM------1------122–jnWM–jMW¤2–jW()M-n=åhn[]e++h-----eåhMn[]–e2n=0n=0MM------1------1课后答案网22–jnWM–jMW¤2–jW()M-n=åhn[]e++h-----eåhn[]e(1)2n=0www.hackshp.cnn=0DefineMMak[]==2h-----–k,k12,,,¼-----22Ma[]0=h-----2andletMn=-----–k2WemaythenrewriteEq.(1)intheequivalentform:15 Mìü-----2jW–jMW¤2ïïMjkW–jkWMHe()=eíýåh-----–k()e+e+h-----22ïïk=1îþMìü-----2–jMW¤2ïïjkW–jkW=eíýå2ak[]()e+e+a()0ïïk=1îþMìü-----2–jMW¤2ïï=eíýåak[]cos()kW(2)ïïk=0îþFromEq.(2)wemaymakethefollowingobservationsforanFIRfilteroftypeI:jW1.ThefrequencyresponseHe()hasalinearphasecomponentexemplifiedbythe–jMW¤2exponentiale.2.AtW=0,M¤2j0He()=åak[]k=0AtW=p,M¤2----M-+kjp2He()课后答案网=åak[]()–1k=0jWTheimplicationsofthesetworesultsarethattherearenorestrictionsonwww.hackshp.cnHe()atW=0andW=p.8.34ForanFIRfilteroftypeII,thefilterlengthMisevenanditisantisymmetricinthatitscoefficientssatisfytheconditionMhn[]=0–hMn[]–,££n-----1–2ThefrequencyresponseofthefilterisMjW–jnWHe()=åhn[]en=0whichmaybereformulatedasfollows:16 M------12MjW–jnWM–jMW¤2–jnWHe()=åhn[]e++h-----eåhn[]e2n=0Mn=-----+12MM------1------122–jnWM–jMW¤2–j()Mn–W=åhn[]e++h-----eåhMn[]–e2n=0n=0MM------1------122–jnWM–jMW¤2–j()Mn–W=åhn[]e+h-----e–åhn[]e(1)2n=0n=0DefineMMak[]=2h-----–k,k=12,,,¼-----22Ma[]0=h-----,2andletMk=-----–n2WemaythenrewriteEq.(1)intheequivalentformM¤2M¤2jW–jMW¤2MjkWM–jMW¤2–jMW¤2M–jkWHe()=eåh-----–ke+h-----e–eåh-----–ke222n=0k=1M¤2–jMW¤2ìüjkW–jkW=课后答案网eíý2åak[]()e–e+a[]0îþk=1M¤2–jMWwww.hackshp.cn¤2ìü=eíýjaå[]ksin()kW+a[]0îþk=1M¤2–jMW¤2=eåak[]sin()kW(2)k=0FromEq.(2)wemaymakethefollowingobservationsonthefrequencyresponseofanFIRfilterofTypeII1.Thephaseresponseincludesalinearcomponentexemplifiedbytheexponential–jMW¤2e.2.AtW=0,17 j0He()=0AtW=p,sin(kp)=0forintegerkandtherefore,jpHe()=08.35AnFIRfilteroftypeIIIischaracterizedasfollows:•ThefilterlengthMisanoddinteger.•Thefilterissymmetricaboutthenonintegermidpointn=M/2inthatitscoefficientssatisfytheconditionhn[]=0hMn[]–for££nMThefrequencyresponseofthefilterisMjW–jnWHe()=åhn[]en=0whichmaybereformulatedasfollows:M-1----------2MjW–jnW–jnWHe()=åhn[]e+å[]nen=0M+1n=------------2M-1M-1----------n=----------22–jnW–j()Mn–W课后答案网=åhn[]e+åhMn[]–en=0n=0M-1-----------2www.hackshp.cn–jnW–j()Mn–W=åhn[]()e+en=0M-1----------2-æöMæöMj-----–nW–j-----–nW–jMW¤2æöèø2èø2=eåhn[]ç÷e+e(1)èøn=0DefineM+1M+1bk[]=2h------------–kfork=12,,,¼------------22andlet18 M+1n=------------–k2WemaythenrewriteEq.(1)intheequivalentformM+1-----------2-æö1æö1jWk–---–jWk–---jW–jMW¤21æöèø2èø2He()=eå---bk()ç÷e+e2èøk=1M+1------------2=e–jMW¤2bk()cosæöWæök–1---(2)åèøèø2k=1FromEq.(2)wemaymakethefollowingobservationsonthefrequencyresponseofanFIRfilteroftypeIII:1.Thephaseresponseofthefilterislinearasexemplifiedbytheexponentialfactor–jMW¤2e.2.AtW=0,M+1------------2j0He()=åbk()k=1j0whichshowsthatthereisnorestrictiononHe().AtW=p,M+1------------课后答案网2He()jp=e–jMp¤2bk()cosæöpæök–1---åèøèø2k=1www.hackshp.cn-----------M+1-2–jMp¤2=eåbk()sin()pkk=1whichiszerosincesin(pk)=0forallintegervaluesofk.8.36AnFIRfilteroftypeIVischaracterizedasfollows:•ThefilterlengthMisanoddinteger.•Thefilterisantisymmetricaboutthenonintegermidpointn=M/2inthatitscoefficientssatisfytheconditionhn[]=0–hMn[]–for££nMThefrequencyresponseofthefilteris19 MjW-jnWHe()=åhn[]en=0whichmaybereformulatedasfollows:M-1-----------2MjW-jnW-jnWHe()=åhn[]e+åhn[]en=0M+1n=------------2M-1M-1----------------------22-jnW-jMn()–W=åhn[]e+åhMn[]–en=0n=0M-1M-1----------------------22-jnW-jMn()–W=åhn[]e–åhn[]en=0n=0M-1----------2-æöMæöM–jn–-----Wjn–-----W–jMW¤2æöèø2èø2=eåhn[]ç÷e–e(1)èøn=0DefineM+1M+1bk[]=2h------------–kfork=12,,,¼------------22andletM+1k=------------–n2WemaythenrewriteEq.(1)intheequivalentform课后答案网M+1-----------2-æö1æö1jk–---W-jk–---WjW–jMWwww.hackshp.cn¤2M+1æöèø2èø2He()=eåh------------–kç÷e–e2èøk=1M+1------------2=je–jMW¤22h-----------M+1-–ksinæöæök–1---Wå2èøèø2k=1M+1------------2=je–jMW¤2bk[]sinæöæök–1---W(2)åèøèø2k=1FromEq.(2)wemaymakethefollowingobservationsontheFIRfilteroftypeIV:20 1.Thephaseresponseofthefilterincludesalinearcomponentexemplifiedbythe–jMW¤2exponentiale.2.AtW=0,j0He()=0AtW=p,M+1------------2He()jp=je–jMW¤2bk()sinæöæök–1---påèøèø2k=1M+1------------2–jMW¤2k+1=jeå()–1bk[]k=1jpwhichshowsthatHe()canassumeanarbitraryvalue.8.37TheFIRdigitalfilterusedasadiscrete-timedifferentiatorinExample8.6exhibitsthefollowingproperties:•ThefilterlengthMisanoddinteger.•Thefrequencyresponseofthefiltersatisfiestheconditions:1.AtW=0,j0He()=02.AtW=p,jWHe()=0ThesepropertiesarebasicpropertiesofanFIRfilteroftypeIIIdiscussedinProblem8.34.课后答案网WethereforeimmediatelydeducethattheFIRfilterofExample8.6isantisymmetricaboutthenonintegerpointwww.hackshp.cnn=M/2.8.38ForadigitalIIRfilter,thetransferfunctionH(z)maybeexpressedasNz()Hz()=------------Dz()whereN(z)andD(z)arepolynomialsinz-1.ThefilterisunstableifanypoleofH(z)or,equivalently,anyzeroofthedenominatorpolynomialD(z)liesoutsidetheunitcircleinthez-plane.Accordingtothebilineartransform,Hz()=H()saz-1s=---------z+121 whereHa(s)isthetransferfunctionofananalogfilterusedasthebasisfordesigningthedigitalIIRfilter.ThepolesofH(z)outsidetheunitcircleinthez-planecorrespondtocertainpolesofH(s)intherighthalfofthes-plane.Conversely,thepolesofH(s)intherighthalfofthes-planearemappedontotheoutsideoftheunitcircleinthez-plane.NowifanypoleofHa(s)liesintheright-halfplane,theanalogfilterisunstable.Henceifanysuchfilterisusedinthebilineartransform,theresultingdigitalfilterislikewiseunstable.8.39WearegivenananalogfilterwhosetransferfunctionisdefinedbyNAkHa()s=å-------------sd–kk=1RecalltheLaplacetransformpairdktL1e«-------------sd–kItfollowsthereforethattheimpulseresponseoftheanalogfilterisNdktha()t=åAke(1)k=1Accordingtothemethodofimpulseinvariance,theimpulseresponseofadigitalfilterderivedfromtheanalogfilterofEq.(1)isdefinedbyhn[]=Th()nTsaswhereTsisthesamplingperiod.Hence,fromEq.(1)wefindthatNndkTshn[]=å课后答案网TsAke(2)k=1Nowrecallthez-transformpair:www.hackshp.cnndkTsz1e«---------------------------dkTs–11–ezHence,thetransferfunctionofthedigitalfilterisdeducedfromEq.(2)tobeNTAskHz()=å---------------------------dkTs–1k=11–ez8.40Consideradiscrete-timesystemwhosetransferfunctionisdenotedbyH(z).Bydefinition,22 Yz()Hz()=-----------Xz()whereY(z)andX(z)arerespectivelythez-transformsoftheoutputsequencey[n]andinputsequencex[n].LetHeq(z)denotethez-transformoftheequalizerconnectedincascadewithH(z).Letx¢[]ndenotetheequalizeroutputinresponsetoy[n]astheinput.Ideally,x¢[]n=xnn[]–0wheren0isanintegerdelay.Hence,–n0–n0X¢()zzXz()zH()z==---------------------------------=------------eqYz()Yz()Hz()jWPuttingze=,wemaythuswrite–jn0WjWeH()e=------------------eqjWHe()8.41ThephasedelayofanFIRfilterofevenlengthMandantisymmetricimpulseresponseislinearwithfrequencyWasshownbyqW()=–MWHence,suchafilterusedasanequalizerintroducesaconstantdelay¶qW()tW()==–----------------Msamples¶WTheimplicationofthisresultisthataswemakethefilterlengthMlarger,theconstantdelayintroducedbytheequalizeriscorrespondinglyincreased.Fromapractical课后答案网perspective,suchatrendishighlyundesirable.www.hackshp.cn23 ComputerExperiments%Solutionto8.42b=fir1(22,1/3,hamming(23));subplot(2,1,1)plot(b);title(’ImpulseResponse’)ylabel(’Amplitude’)xlabel(’Time(s)’)grid[H,w]=freqz(b,1,512,2*pi);subplot(2,1,2)plot(w,abs(H))title(’MagnitudeResponse’)ylabel(’Magnitude’)xlabel(’Frequency(w)’)gridImpulseResponse0.40.30.2Amplitude0.10−0.10510152025课后答案网Time(s)MagnitudeResponse1.41.210.8www.hackshp.cn0.6Magnitude0.40.2000.511.522.533.5Frequency(w)%Solutiontoproblem8.43clear;M=100;n=0:M;f=(n-M/2);24 %Integrationbyparts(seeexample8.6)h=cos(pi*f)./f-sin(pi*f)./(pi*f.^2);k=isnan(h);h(k)=0;h_rect=h;h_hamm=h.*hamming(length(h))’;[H,w]=freqz(h_rect,1,512,2*pi);figure(1)subplot(2,1,1)plot(h_rect);title(’RectangularWindowedDifferentiator’)xlabel(’Step’)ylabel(’Amplitude’)gridsubplot(2,1,2)plot(w,abs(H))title(’MagnitudeResponse’)ylabel(’Magnitude’)xlabel(’Frequency(w)’)grid[H,w]=freqz(h_hamm,1,512,2*pi);figure(2);subplot(2,1,1)plot(h_hamm);title(’HammingWindowedDifferentiator’)xlabel(’Step’)课后答案网ylabel(’Amplitude’)gridsubplot(2,1,2)www.hackshp.cnplot(w,abs(H))title(’MagnitudeResponse’)ylabel(’Magnitude’)xlabel(’Frequency(w)’)grid25 RectangularWindowedDifferentiator10.50Amplitude−0.5−1020406080100120StepMagnitudeResponse432Magnitude1000.511.522.533.5Frequency(w)HammingWindowedDifferentiator10.50Amplitude−0.5−1020406080100120StepMagnitudeResponse3.532.521.5Magnitude课后答案网10.5000.511.522.533.5www.hackshp.cnFrequency(w)8.44ForaButterworthlow-passfilteroforderN,thesquaredmagnituderesponseis21Hj()w=---------------------------(1)æöw2N1+------èøwcwherewcisthecutofffrequency.(a)Wearegiventhefollowingspecifications:(i)wc=2px800rad/s(ii)Atw=2px1,500rad/s,wehave26 210log10Hj()w=–15dBor,equivalently21Hj()w=-------------------31.6228SubstitutingthesevaluesinEq.(1):2p´1200,2N31.6228=1+æö---------------------------èø2p´8002N=1+()1.5SolvingforthefilterorderN:1N=---()log30.6228¤log1.52=4.2195SowechooseN=5.%SolutiontoProblem8.44omegaC=0.2;N=5;wc=tan(omegaC/2);coeff=[13.23615.23615.23613.23611];%(seetable8.1)ns=wc^N;ds=coeff.*(wc.^[0:N]);课后答案网[nz,dz]=bilinear(ns,ds,0.5);[H,W]=freqz(nz,dz,512);www.hackshp.cnsubplot(2,1,1)plot(W,abs(H))title(’MagnitudeResponseofIIRlow-passfilter’)xlabel(’rad/s’)ylabel(’Magnitude’)gridphi=180/pi*angle(H);subplot(2,1,2)plot(W,phi)title(’PhaseResponse’)xlabel(’rad/s’)ylabel(’degrees’)grid27 %set(gcf,’name’,[’LowPass:order=’num2str(N)’wc=’num2str(omegaC)])MagnitudeResponseofIIRlow−passfilter1.41.210.80.6Magnitude0.40.2000.511.522.533.5rad/sPhaseResponse2001000degrees−100−20000.511.522.533.5rad/s%SolutiontoProblem8.45omegaC=0.6;N=5;wc=tan(omegaC/2);coeff=[13.23615.23615.23613.23611];%(seetable8.1)ns=[1/wc^Nzeros(1,N)];ds=fliplr(coeff./(wc.^[0:N]));课后答案网[nz,dz]=bilinear(ns,ds,0.5);[H,W]=freqz(nz,dz,512);www.hackshp.cnsubplot(2,1,1)plot(W,abs(H))title(’MagnitudeResponseofIIRhigh-passfilter’)xlabel(’rad/s’)ylabel(’Magnitude’)gridphi=180/pi*angle(H);subplot(2,1,2)plot(W,phi)title(’PhaseResponse’)xlabel(’rad/s’)ylabel(’degrees’)28 gridMagnitudeResponseofIIRhigh−passfilter10.80.60.4Magnitude0.2000.511.522.533.5rad/sPhaseResponse2001000degrees−100−20000.511.522.533.5rad/s%Solutiontoproblem8.46w=0:pi/511:pi;den=sqrt(1+2*((w/pi).^2)+(w/pi).^4);Hchan=1./den;taps=95;M=taps-1;n=0:M;f=n-M/2;课后答案网%Term1hh1=fftshift(ifft(ones(taps,1))).*hamming(taps);www.hackshp.cn%Term2h=cos(pi*f)./f-sin(pi*f)./(pi*f.^2);k=isnan(h);h(k)=0;hh2=2*(hamming(taps)’.*h)/pi;%Term3hh3a=-(1./(pi*f)).*sin(pi*f);hh3b=-(2./(pi*f).^2).*cos(pi*f);hh3c=(2./(pi*f).^3).*sin(pi*f);hh3=hamming(taps).*(hh3a+hh3b+hh3c)’;hh=hh1’+hh2+hh3’;hh(48)=2/3;29 [Heq,w]=freqz(hh,1,512,2*pi);p=0.7501;Hcheq=(p*abs(Heq)).*Hchan’;plot(w,p*abs(Heq),’b--’)holdonplot(w,abs(Hchan),’g-.’)plot(w,abs(Hcheq),’r-’)legend(’Heq’,’Hcan’,’Hcheq’,1)holdoffxlabel(’Frequency(mega)’)ylabel(’MagnitudeResponse’)1.5HeqHcanHcheq1MagnitudeResponse0.5000.511.522.533.5课后答案网Frequency(W)www.hackshp.cn30 CHAPTER9AdditionalProblems9.21(a)Theclosed-loopgainofthefeedbackamplifierisgivenbyAT=----------------(1)1+bAWearegiventhefollowingvaluesfortheforwardamplificationAandfeedbackfactorb:A=2,500b=0.01SubstitutingthesevaluesinEq.(1):25002500T==--------------------------------------------------=92.151+0.01´250026(b)ThesensitivityofthefeedbackamplifiertochangesinAisgivenbyADTT¤11S===-------------------------------------TDAA¤1+bA26With(DA/A)=10%=0.10,wethushaveDTAæöD-------A-------=STTèøA1===------()0.100.00380.38%269.22(a)Theclosed-loopgainofthefeedbacksystemisGGapT=----------------------------1+HGG课后答案网ap(b)Thereturndifferenceofthesystemiswww.hackshp.cnF=1+HGGapHencethesensitivityofTwithrespecttochangesinGpisGpDTT¤S=----------------------TDG¤Gpa1=---F=----------------------------1+HGGapGp(c)Wearegiven:H=1andGp=1.5.HenceforST=1%=0.01,werequireF=1001 ThecorrespondingvalueofGaisthereforeF–1G=------------aHGp99==-------661.59.23Thelocalfeedbackaroundthemotorhastheclosed-loopgainGp/(1+HGp).Theclosed-loopgainofthewholesystemisthereforeKGG¤()1+HGrcppT=------------------------------------------------------------1+KGG¤()1+HGrcppKGGrcp=------------------------------------------------1++HGKGGprcp9.24Theclosed-loopgainoftheoperationalamplifierisV()sZ()s22--------------–=-------------V()sZ()s11WearegivenZ1(s)=R1andZ2(s)=R2.Theclosed-loopgainortransferfunctionoftheoperationalamplifierinFig.P9.24isthereforeV()sR22--------------–=------V()sR119.25(a)FromFig.P9.25,wehave课后答案网1Z()s=R+---------11sC1Z()s=R22www.hackshp.cnThetransferfunctionofthisoperationalamplifieristhereforeV()sZ()s22--------------–=-------------V()sZ()s11R2=–----------------------1R+---------1sC1sCR12=–-------------------------(1)1+sCR11(b)Forpositivevaluesoffrequencywthatsatisfythecondition2 1----------->>RwC11wemayapproximateEq.(1)asV()s2--------------»–sCRV()s121Thatis,theoperationalamplifieractsasadifferentiator.9.26Throughoutthisproblem,H(s)=1,inwhichcasetheopen-looptransferequalsG(s).Inanyevent,theopen-looptransferfunctionofthefeedbackcontrolsystemmaybeexpressedastherationalfunctionP(s)/(spQ(s)),whereneitherthepolynomialP(s)nor1Q1(s)hasazeroats=0.Since1/sisthetransferfunctionofanintegrator,itfollowsthatpisthenumberoffreeintegratorsinthefeedbackloop.Theorderpisreferredtoasthetypeofthefeedbackcontrolsystem.(a)Fortheproblemathand,wearegiven15Gs()=---------------------------------()s+1()s+3Thecontrolsystemisthereforetype0.Thesteady-stateerrorforunitstepis1Î=----------------ss1+KpwhereK=limGs()Hs()ps®0Thatis,15K=lim---------------------------------p()s+1()s+3课后答案网s®015==------53Hence,www.hackshp.cn11Î=------------=---ss15+6Forbothrampandparabolicinputs,thesteady-stateerrorisinfinitelylarge.(b)For5Gs()=------------------------------------ss()+1()s+4thecontrolsystemistype1.Thesteady-stateerrorforastepinputiszero.Forarampofunitslope,thesteady-stateerroris1Î=------ssKvwhere3 5K=lim---------------------------------v()s+1()s+4s®05=---4Thesteady-stateerroristherefore4Î=---ss5Foraparabolicinputthesteady-stateerrorisinfinitelylarge.(c)For5()s+1Gs()=---------------------2s()s+3thecontrolsystemistype2.Hence,thesteady-stateerroriszeroforbothstepandrampinputs.Foraunitparabolicinputthesteady-stateerroris1Î=------ssKawhere5()s+1K=lim-------------------as+3s®05=---3Thesteady-stateerroristherefore3/5.Foraunitparabolicinputthesteady-stateerrorisinfinitelylarge.(d)For5()s+1()s+2Gs()=------------------------------------2s()s+3thecontrolsystemistype2.Thesteady-stateerroristhereforezeroforbothstepand课后答案网rampinputs.Foraunitparabolicinputthesteady-stateerroris1Î=------ssKwww.hackshp.cnnwhere5()s+1()s+2K=lim------------------------------------ns+3s®010=------3Thesteady-stateerroristherefore3/10.9.27Fortheresultsonsteady-stateerrorscalculatedinProblem9.26tohold,allthefeedbackcontrolsystemsinparts(a)to(d)oftheproblemhavetobestable.4 15(a)Gs()Hs()=---------------------------------()s+1()s+3ThecharacteristicequationisAs()=()s+1()s+3+152=s++4s18bothrootsofwhichareintheleft-halfplane.Thesystemisthereforestable.5(b)Gs()Hs()=------------------------------------ss()+1()s+45=-------------------------------32s++5s4s32As()=s+++5s4s5ApplyingtheRouth-Hurwitzcriterion:s314s255s115/50s050Therearenosignchangesinthefirstcolumncoefficientsofthearray;thesystemisthereforestable.5()s+1(c)Gs()Hs课后答案网()=---------------------2s()s+33www.hackshp.cn2As()=s+++3s5s5TheRouth-Hurwitzarrayis:s315s235s110/30s050Hereagaintherearenosignchangesinthefirstcolumnofcoefficients,andthecontrolsystemisthereforestable.5 5()s+1()s+2(d)Gs()Hs()=------------------------------------2s()s+3322As()=s++++3s5s15s1032=s+++8s15s10TheRouth-Hurwitzarrayiss3115s2810s1110/80s0100Hereagaintherearenosignchangesinthefirstcolumnofarraycoefficients,andthecontrolsystemisthereforestable.429.28(a)s++02s1=Equivalently,wemaywrite22()s+1=0whichhasdoublerootsats=+j.Thesystemisthereforeonthevergeofinstability.43(b)s+++ss0.5=0Byinspection,wecansaythatthisfeedbackcontrolsystemisunstablebecausetheterms2ismissingfromthecharacteristicequation.WecanverifythisobservationbyconstructingtheRoutharray:课后答案网s4100.5s3110www.hackshp.cns2-10.50s1+1.50s09.50Therearetwosignchangesinthefirstcolumnofarraycoefficients,indicatingthatthecharacteristicequationhasapairofcomplex-conjugaterootsintheright-halfofthes-plane.Thesystemisthereforeunstableaspreviouslyobserved.432(c)s++++02s2s2s4=6 TheRoutharrayiss4134s3220s2240s1-20s0-40Therearetwosignchangesinthefirstcolumnofarraycoefficients,indicatingthepresenceoftworootsofthecharacteristicequationintheright-halfofthes-plane.Thecontrolsystemisthereforeunstable.9.29Thecharacteristicequationofthecontrolsystemis32s+++ssK=0ApplyingtheRouth-Hurwitzcriterion:s311s21Ks11-K0s0K0ThecontrolsystemisthereforestableprovidedthattheparameterKsatisfiesthecondition:0<0,a>0,a>0and320aa–aa2130---------------------------->0(1)a2(b)Forthecharacteristicequation32s+++ssK=0wehavea3=1,a2=1,anda1=1anda0=K.Hence,applyingthecondition(1):111´–´K------------------------------->01orK<1Also,werequirethatK>0sincewemusthavea0>0.Hence,Kmustsatisfythecondition0>K<1forstability.9.31(a)WearegiventhelooptransferfunctionKLs()=------------------------------,K>02ss()++s217L(s)hasthreepoles,oneats=0andtheothertwoats=–---±j-------.22Hence,therootlocusofL(s)has3branches.Itstartsatthese3polelocationsandterminateatthezerosofL(s)ats=¥.Thestraight-lineasymptoticsoftherootlocusaredefinedbytheangles(seeEq.(9.69))()2k+1pq==------------------------,k012,,k3orp/3,课后答案网p,and5p/3.Theirlocationpointontherealaxisofthes-planeisdefinedby(seeEq.(9.70))æö1+------7-æö1–------7-0+–---j–---jèø2www.hackshp.cn2èø22s=----------------------------------------------------------------------031=–---3SinceL(s)hasapairofcomplex-conjugatepoles,weneed(inadditiontotherulesdescribedinthetext)aruleconcerningtheangleatwhichtherootlocusleavesacomplexpole(i.e.,angleofdeparture).Specifically,wewishtodeterminetheangleqdepindicatedinFig.1fortheproblemathand:8 jws-planexqdepq1s0q2xFigure1Theangleq1andq2aredefinedby–1æö72¤–1q===tan--------------tan()2.6458111.7°1èø–12¤q=90°2Applyingtheanglecriterion(Eq.(9..........)):q+90°+111.7°=180depHence,q=–21.70°depWemaythussketchtherootlocusofthegivensystemasinFig.2:课后答案网jwxwww.hackshp.cns-planexs0xFigure2(b)Thecharacteristicequationofthesystemis32s+++s2sK=09 ApplyingtheRouth-Hurwitzcriterion:s312s21Ks12-K0s0K0ThesystemisthereforeonthevergeofinstabilitywhenK=2.ForthisvalueofKtherootlocusintersectsthejw-axisats2+K=0orsj==±K±j2,asindicatedinFig.2.9.32Theclosed-looptransferfunctionofacontrolsystemwithunityfeedbackisLs()Ts()=--------------------1+Ls()WearegivenKKLs()==---------------------------------ss()+1s2+3HenceKTs()=------------------------2s++3K(a)ForK=0.1thesystemisoverdamped0.1Ts()=---------------------------2课后答案网s++s0.1wherew==0.1andz0.5¤0.1.n(b)ForK=0.25thesystemiscriticallydamped:www.hackshp.cn0.25Ts()=------------------------------2s++s0.25wherew==0.5andz1.n(c)ForK=2.5thesystemisunderdamped:0.25Ts()=---------------------------2s++s2.5wherew==2.5andz0.5¤2.5.n10 Figure1plotstherespectivestepresponsesofthesethreespecialcasesofthefeedbacksystem.%Solutiontoproblem9.32t=0:0.1:40;clc;%UnderdampedSystemK=0.1;w=sqrt(K);z=0.5/w;fac=w*sqrt(1-z^2)*t+atan(sqrt(1-z^2)/z);y1=1-1/sqrt(1-z^2).*exp(-z*w*t).*sin(fac);figure(1);clf;plot(t,y1,’-’)%ylim([01.5])xlabel(’Time(s)’)ylabel(’Response’)%title(’UnderdampedK=0.1’)holdon%CriticallyDampedSystemK=0.25;w=sqrt(K);z=0.5/wtau=1/wy2=1-exp(-t/tau)-t.*exp(-t/tau);课后答案网plot(t,y2,’-.’)%OverDampedSystemwww.hackshp.cnK=2.50000;w=sqrt(K);z=0.5/w;t1=1/(z*w-w*sqrt(z^2-1));t2=1/(z*w+w*sqrt(z^2-1));k1=0.5*(1+z/sqrt(z^2+1));k2=0.5*(1-z/sqrt(z^2+1));y3=1-k1*exp(-t/t1)-k2*exp(-t/t2);plot(t,y3,’--’)holdofflegend(’K=0.1’,’K=0.25’,’K=2.5’)11 1.4K=0.1K=0.25K=2.51.210.80.6Response0.40.20Figure1−0.2−0.40510152025303540Time(s)课后答案网www.hackshp.cn12 9.33WearegiventhelooptransferfunctionKs()+0.5Ls()=-------------------------4()s+1Figure1showstherootlocusdiagramoftheclosed-loopfeedbacksystemforvaryingpositivevaluesofK.%Solutiontoproblem9.33num=[10.5];den=[14641];rlocus(num,den)axis([-1.5.1-2.22.2])RootLocus21.510.50ImagAxis−0.5−1课后答案网−1.5−2−1.5www.hackshp.cn−1−0.50RealAxis13 9.34WearegiventhelooptransferfunctionKLs()=-----------------------------------2()s+1()s+5(a)Figure1showstherootlocusdiagramoftheclosed-loopfeedbacksystemforvaryingK.Next,puttings=jw,KLj()w=---------------------------------------------2()jw+1()jw+5Parts(a),(b)and(c)ofFig.2showplotsoftheNyquistlocusofthesystemforK=50,72,and100.Onthebasisofthesefigureswecanmakethefollowingstatements:1.ForK=50thelocusdoesnotencirclethecriticalpoint(-1,0)andthesystemisstable.2.ForK=100thelocusenclosesthecriticalpoint(-1,9)andthesystemisunstable.3.ForK=72theNyquistlocuspassesthroughthecriticalpoint(-1,0)andthesystemisonthevergeofinstability.(b)ThecriticalvalueK=72isdeterminedinaccordancewiththeconditionKcriticalLj()w=-----------------------------------------------------p2212¤()w+1()w+25ppwherewpisthephasecrossoverfrequencydefinedby–1–1w180°=2tan()w+tanæö------p课后答案网pèø5Foragainmarginof5dB:20logK=20logK-510www.hackshp.cn10criticalHence7272K==-----------------------------------------------=40.49antilog(0.25)1.7783(c)ForK=40.49thegain-crossoverfrequencyis2222()40.49=()1+w()25+wgg2Letw=x,sorewritethisequationintheformofacubicequationinx:g32x++27x51x–1614=0Theonlypositiverootofxis6.2429.Hencew==6.24292.4986g14 ThephaseofL(jw)atw=2.4986is–1–1æö2.49862tan()2.4986+tan----------------=2´68.2°+26.5°èø2=162.9°Thephasemarginisthereforej==180°–162.9°17.1°m%Solutiontoproblem9.34clear;clc;num=[1];den=[17115];figure(1);clf;rlocus(num,den)figure(2);clf;nyquist(num,den)xlim([-1.10.22])RootLocus105K=73.560ImagAxis课后答案网−5www.hackshp.cn−10Figure1−15−10−505RealAxis15 NyquistDiagram0.10.050ImaginaryAxis−0.05−0.1−1−0.8−0.6−0.4−0.200.2Figure2RealAxis课后答案网www.hackshp.cn16 9.35WearegivenKLs()=-------------------ss()+1Fors=jw,KLj()w=----------------------------jw()jw+1fromwhichwededuce:KLj()w=---------------------------------(1)212¤ww()+1–1arg{}Lj()w=–90°–tan()w(2)FromEq.(2)weobservethatthephaseresponsearg{L(jw)}isconfinedtotherange[-90o,-180o].Thevalueof-180oisattainedonlyatw=¥.AtthisfrequencyweseefromEq.(1)thatthemagnituderesponse|L(jw)|iszero.Hence,theNyquistlocuswillneverencirclethecriticalpoint(-1,0)forallpositivevaluesofK.ThefeedbacksystemisthereforestableforallK>0.9.36WearegivenKLs()=---------------------2s()s+1Fors=jw,KLj()w=------------------------------2课后答案网-w()jw+1fromwhichwededuce:KLj()w=-----------------------------------2www.hackshp.cn212¤w()w+1–1arg{}Lj()w=180°w–tan()HencetheNyquistlocuswillencirclethecriticalpoint(-1,0)forallK>0;thatis,thesystemisunstableforallK>0.WemayalsoverifythisresultbyapplyingtheRouth-Hurwitzcriterion:s310s21Ks1-K017 s0K0Thereare2signchangesinthefirstcolumnofcoefficientsinthearray,assumingK>0.HencethesystemisunstableforallK>0.9.37WearegiventhelooptransferfunctionKLs()=------------------------------------ss()+1()s+2(a)Fors=jw,KLj()w=-----------------------------------------------------jw()jw+1()jw+2¶HenceKLj()w=-------------------------------------------------------------(1)212¤212¤ww()+1()w+4–1–1æöwarg{}Lj()w=–90°–tan()w–tan----(2)èø2SettingK=6inEq.(1)underthecondition|L(jwp)|=1:22236=w()w+1()w+4ppp2wherewpisthephase-crossoverfrequency.Puttingwp=xandrearrangingterms:32x++5课后答案网x4x–36=0Solvingthiscubicequationforxwefindthatithasonlyonepositiverootatx=2.Hence,www.hackshp.cnw=2pSubstitutingthisvalueinEq.(2):–1–1æö2arg{}Lj()w=–290°–tan()–tan-------pèø2==–90°–54.8°–35.2°–180°Thisresultconfirmswpasthephase-crossoverfrequency,andK=6asthecriticalvalueofthescalingfactorforwhichtheNyquistlocusofL(jw)passesthroughthecriticalpoint(-1,0).18 WemayalsoverifythisresultbyapplyingtheRouth-Hurwitzcriteriontothecharacteristicequation:32s+++03s2sK=Specifically,wewrites312s23Ks16-K0----------3s0KThesystemisthereforeonthevergeofinstabilityforK=6.(b)ForK=2thegainmarginis20log6–20log2==20log39.542dB101010TocalculatethephasemarginforK=2weneedtoknowthegain-crossoverfrequencywg.Atw=wg,|L(jw)|=1.Hencefortheproblemathand21=----------------------------------------------------------------212¤212¤w()w+1()w+4ggg2Letw=x,forwhichwemaythenrewritethisequationasg32x++5x4x–4=0Theonlypositiverootofthisequationisx=0.5616.Thatis,w==0.56160.7494g课后答案网Thephasemarginistherefore–1–1180°–()90°+www.hackshp.cntan()0.7494+tan()0.3749=90°–36.9°–20.5°=32.6°(c)Letwgdenotethegain-crossoverfrequencyfortherequiredphasemarginjm=20°.WemaythenwriteLj()w=1g–1w–1æö------garg{}Lj()w=–90°–tan()w–tanggèø2Witharg{}Lj()w=180°–20°gwethushave19 –1w–1æö------g70°w=tan()+tangèø2Throughaprocessoftrialanderror,weobtainthesolutionw=0.972gForLj()w=1wethusrequireg212¤212¤K=w()w+1()w+4ggg12¤12¤=0.9721.9448()()4.9448=3.0143Thegainmarginisthereforeæö620log----------------»20log2=6dB10èø3.0143109.38Forthepurposeofillustration,supposetheloopfrequencyresponseL(jw)hasafinitemagnitudeatw=0.SupposealsothefeedbacksystemrepresentedbyL(jw)isstable.WemaythensketchtheNyquistlocusofthesystemfor0>w,wemayapproximateEq.(1)asVs()sK0Hs()()1¤Kv--------------»----------------------------------------F()sKHs()10s=------KvCorrespondingly,wemaywrite1dvt()»-----------f()tKdt1v9.47Theerrorsignale(t)isdefinedasthedifferencebetweentheactuatingortargetsignalyd(t)andthecontrolledsignal(i.e.,actualresponse)y(t).ExpressingthesesignalsintermsoftheirrespectiveLaplacetransforms,wemaywriteEs()=Y()s–Ys()课后答案网d=[]1–Ts()Y()s(1)dGs()Hs()=1–---------------------------------www.hackshp.cnYd()s1+Gs()Hs()1=---------------------------------Y()s(2)1+Gs()Hs()dThesteady-stateerrorofafeedbackcontrolsystemisdefinedasthevalueoftheerrorsignale(t)whentimetapproachesinfinity.DenotingthisquantitybyÎ,wemaythussswriteÎ=limet()(3)sss®0UsingthefinalvaluetheoremofLaplacetransformtheorydescribedinChapter6,wemayredefinethesteady-stateerrorÎintheequivalentformss45 Î=limsEs()(4)sss®0Hence,substitutingEq.(2)into(4),wegetsY()sdÎ=lim---------------------------------(5)sss®01+Gs()Hs()Equation(5)showsthatthestead-stateerrorÎofafeedbackcontrolsystemdependsonsstwoquantities:•Theopen-looptransferfunctionG(s)H(s)ofthesystem•TheLaplacetransformYd(s)ofthetargetsignalyd(t)However,forEq.(5)tobevalid,theclosed-loopcontrolsystemofFig.9.14mustbestable.Ingeneral,G(s)H(s)maybewrittenintheformofarationalfunctionasfollows:Ps()Gs()Hs()=-------------------(6)psQ()s1whereneitherthepolynomialP(s)norQ1(s)hasazeroats=0.Since1/sisthetransferfunctionofanintegrator,itfollowsthatpisthenumberoffreeintegratorsinthelooptransferfunctionG(s)H(s).Theorderpisreferredtoasthetypeofthefeedbackcontrolsystem.Wethusspeakofafeedbackcontrolsystembeingoftype0,type1,type2,andsoonforp=0,1,2,...,respectively.Inlightofthisclassification,wenextconsiderthesteady-stateerrorforthreedifferentinputfunctions.StepInputForthestepinputyd(t)=u(t),wehaveYd(s)=1/s.Hence,Eq.(5)yieldsthesteady-stateerror课后答案网1Î=lim---------------------------------sss®01+Gs()Hs()1=----------------www.hackshp.cn(7)1+KpwhereKpiscalledthepositionerrorconstant,definedbyK=limGs()Hs()ps®0Ps()=lim-------------------(8)ps®0sQ()s1Forp>1,KpisunboundedandthereforeÎss=0.Forp=0,Kpisfiniteandthereforeι0.Accordingly,wemaystatethatthesteady-stateerrorforastepinputiszeroforssafeedbackcontrolsystemoftype1orhigher.Ontheotherhand,forasystemoftype0,thesteady-stateerrorisnotzero,anditsvalueisgivenbyEq.(7).46 RampInputFortherampinputy(t)-tu(t),wehaveY=1/s2.Inthiscase.Eq.(5)yieldsdd1Î=lim-----------------------------------sss®0ss+Gs()Hs()1=------(9)KvwhereKvisthevelocityerrorconstant,definedbyK=limsGs()Hs()vs®0Ps()=lim-----------------------(10)p-1s®0sQ()s1Forp>2,KvisunboundedandthereforeÎss=0.Forp=1,KvisfiniteandÎss¹0.Forp=0,KviszeroandÎ¥ss=.Accordingly,wemaystatethatthesteady-stateerrorforarampinputiszeroforafeedbackcontrolsystemoftype2orhigher.Forasystemoftype1,thesteady-stateerrorisnotzero,anditsvalueisgivenbyEq.(9).Forasystemoftype0,thesteady-stateerrorisunbounded.ParabolicInputFortheparabolicinputy(t)=(t2/2)u(t),wehaveY=1/s3.TheuseofEq.(5)yieldsdd1Î=lim----------------------------------------ss22s®0s+sGs()Hs()1=------(11)KawhereKa课后答案网iscalledtheaccelerationerrorconstant,definedbyPs()K=lim-----------------------(12)ap-2s®0sQ()swww.hackshp.cn1Forp>3,KaisunboundedandthereforeÎss=0.Forp=2,KaisfiniteandthereforeÎss=0.Forp=0,1,KaiszeroandthereforeÎ¥ss=.Accordingly,wemaystatethatforaparabolicinput,thesteady-stateerroriszeroforatype3systemorhigher.Forasystemoftype2,thesteady-stateerrorisnotzero,anditsvalueisgivenbyEq.(11).Forasystemoftype0ortype1,thesteady-stateerrorisunbounded.InTable1,wepresentasummaryofthesteady-stateerrorsaccordingtosystemtypeasdeterminedabove.47 AdditionalNotesAtype0systemisreferredtoasaregulator.Theobjectofsuchasystemistomaintainaphysicalvariableofinterestatsomeprescribedconstantvaluedespitethepresenceofexternaldisturbances.Examplesofregulatorsystemsinclude:•Controlofthemoisturecontentofpaper,aproblemthatarisesinthepaper-makingprocess•Controlofthechemicalcompositionofthematerialoutletproducedbyareactor•Biologicalcontrolsystem,whichmaintainsthetemperatureofthehumanbodyatapproximately37oCdespitevariationsinthetemperatureofthesurroundingenvironmentTable1:Steady-StateErrorsAccordingtoSystemTypeStepRampParabolicType01/(1+Kp)¥¥Type101/Kv¥Type2001/KaType1andhighersystemsarereferredtoasservomechanisms.Theobjectivehereistomakeaphysicalvariablefollow,ortrack,somedesiredtime-varyingfunction.Examplesofservomechanismsinclude:•Controlofarobot,inwhichtherobotmanipulatorismadetofollowapresettrajectoryinspace•Controlofamissile,guidingittofollowaspecifiedflightpath•Trackingofamaneuveringtarget(e.g.,enemyaircraft)byaradarsystemExample:FigureP9.47showstheblockdiagramofafeedbackcontrolsysteminvolvingadcmotor.Thedcmotorisanelectromechanicaldevicethatconvertsdcelectricalenergy课后答案网intorotationalmechanicalenergy.ItstransferfunctionisapproximatelydefinedbyKGs()=-------------------------s()ts+1Lwww.hackshp.cnwhereKisthegainandtListheloadtimeconstant.Hereitisassumedthatthefieldtimeconstantofthedcmotorissmallcomparedtotheloadtimeconstant.Thetransferfunctionofthecontrollerisats+1Hs()=------------------ts+1Therequirementistofindthesteady-stateerrorsofthiscontrolsystem.ThelooptransferfunctionofthecontrolsysteminFig.P9.47isK()ats+1Gs()Hs()=--------------------------------------------(13)s()ts+1()ts+1Lwhichrepresentsatype1system.ComparingEq.(13)withEq.(6),wehavefortheexampleathand48 aKæö1Ps()=--------s+------tèøatLæö1æö1Q()s=s+-----s+---1èøtèøtLp=1HencetheuseofthesevaluesinEqs.(8),(10),and(12)yieldsthefollowingsteady-stateerrorsforthecontrolsystemofFig.P9.47:(a)Stepinput:Î=0.ss(b)Rampinput:Î=1/K=constant.ss(c)Parabolicinput:Î¥=.ssTheeffectofmakingthegainKlargeistoreducethesteady-stateerrorofthesystemduetoarampinput.This,inturn,improvesthebehaviorofthesystemasavelocitycontrolsystem.ControllerDCmotor+ats+1K.InputSOutputts+1s(tLs+1)-FigureP9.479.48FromFig.P9.20theclosed-looptransferfunctionofthefeedbacksystemis2Ks¤()+2sTs()=---------------------------------------21+Ks¤()+2sK=---------------------------(1)2s课后答案网++2sKIngeneral,wemayexpressT(s)intheform2wT()0nwww.hackshp.cnTs()=----------------------------------------(2)22s++2zwswnnHence,comparingEqs.(1)and(2):T()0=1(3)w=K(4)nz==1¤w1¤K(5)nWearegivenK=20,forwhichtheuseofEqs.(4)and(5)yieldsw==204.472rad/secondnz=0.22449 Thetimeconstantofthesystemis1t=---------=1secondzwn9.49ThelooptransferfunctionofthefeedbackcontrolsystemisLs()=Gs()Hs()0.2KP=---------------------------------()s+1()s+3Theclosed-looptransferfunctionofthesystemisLs()Ts()=--------------------1+Ls()0.2KP=----------------------------------------------------(1)2s++4s()3+0.2KPForasecond-ordersystemdefinedby2T()w0nTs()=----------------------------------------(2)22s++2zwswnnthedampingfactoriszandthenaturalfrequencyiswn.ComparingEqs.(1)and(2):2zw=4n2w=3+0.2KnP2T()w0=0.2KnPHence,课后答案网w=3+0.2K(3)nPz=2¤3+0.2K(4)www.hackshp.cnPT()0=0.2K¤()3+0.2K(5)pPWearerequiredtohavew=2rad/snHencetheuseofEq.(3)yields3+20.2K=PThatis,43–K==------------5p0.250 Next,theuseofEq.(4)yields2z=------------------------------3+0.2´52==-------14whichmeansthatthesystemiscriticallydamped.Thetimeconstantofthesystemisdefinedby1t=---------zwn1==------------0.5seconds12´(Note:TheuseofEq.(5)yields0.2´51T()0==-----------------------------3+0.2´549.50ThelooptransferfunctionofthesystemisKLs()=æöK+------Iæö-----------0.25èøPsèøs+1K¤K=0.25Kæö1+-----------------IPæö-----------1Pèøsèøs+1WithKI/KP=0.1wehave0.25K()s+0.1Ls()=---------------------------------------课后答案网P(1)ss()+1TherootlocusofthesystemisthereforeasshowninFig.1:www.hackshp.cnjws-planex0xs-1-0.10Figure151 Theclosed-looptransferfunctionofthesystemisLs()Ts()=--------------------(2)1+Ls()SubstitutingEq.(1)into(2):0.25K()5+0.1PTs()=-----------------------------------------------------------------(3)ss()+1+0.25K()s+0.1PForaclosed-loopats=-5werequirethatthedenominatorofT(s)satisfythefollowingequation()s+5()sa+=0(4)wheres=-aistheotherclosed-looppoleofthesystem.ComparingthedenominatorofEq.(3)withEq.(4):5+a=1+0.25KP5a=0.1´0.25KPSolvingthispairofequationsforKPanda,weget20K==-------------16.33P1.225a==0.005´K0.08165P9.51ThelooptransferfunctionofthePDcontrollersystemisæö1Ls()=()K+Ks-------------------PDèøss()+1K=K课后答案网æös+-------Pæö------------------1-DèøKèøss()+1DWearegivenKP/Kwww.hackshp.cnD=4.Hence,K()s+4DLs()=------------------------ss()+1Wemaythereforesketchtherootlocusofthesystemasfollows:52 %Solutiontoproblem9.51clear;clc;figure(1)K=1;num=K*[14];den=[1(1+K)K*4];rlocus(num,den)figure(2)K=3;num=K*[14];den=[1(1+K)K*4];rlocus(num,den)RootLocus32+j2sqrt(2)210ImagAxis−1−22−j2sqrt(2)−3课后答案网−14−12−10−8−6−4−20www.hackshp.cnRealAxis53 Theclosed-looptransferfunctionofthesystemisLs()Ts()=--------------------1+Ls()K()s+4D=--------------------------------------------(1)2s++sK()s+4DWearerequiredtochooseKDsoastolocatetheclosed-looppolesats=–2±j22.Thatis,thecharacteristicequationofthesystemistobe()s++2j22()s+2–j22=0or2s++4s12=0(2)ComparingthedenominatorofEq.(1)withEq.(2):1+4K=D4K=12DBothoftheseconditionsaresatisfiedbychoosingK=3D9.52(a)ForafeedbacksystemusingPIcontroller,thelooptransferfunctionisKLs()=æöK+------IL¢()sèøPswhereL¢()sistheuncompensatedlooptransferfunction.Fors=jwwemaywrite课后答案网KI1K+------=------()K+jKwPjwjwIPThecontributionofthePIcontrollertotheloopphaseresponseofthefeedbacksystemiswww.hackshp.cn–1Kwf=–90°+tanæö-----------PèøKIForallpositivevaluesofK/K,theangletan-1(Kw/K)islimitedtotherange[0,90o].PIPIItfollowsthereforethattheuseofaPIcontrollerintroducesaphaselagintotheloopphaseresponseofthefeedbacksystem,asshowninFig.1.(b)ForafeedbacksystemusingPDcontroller,thelooptransferfunctionofthesystemmaybeexpressedasLs()=()K+KsL¢()sPDwhereL¢()sistheuncompensatedlooptransferfunction.Fors=jw,thecontributionofthePDcontrollertotheloopphaseresponseofthesystemistan-1(Kw/K).ForallDP54 positivevaluesofK/K,thiscontributionislimitedtotherange[0,90o].WethereforeDPconcludethattheuseofPDcontrollerhastheeffectofintroducingaphaseleadintotheloopphaseresponseofthesystem,asillustratedinFig.2.%Solutiontoproblem9.52clear;clc;w=0:0.01:5*pi;Kp=1.2;Ki=1;y=-pi/2+atan(w*Kp/Ki);plot(w,y)xlabel(’mega’)ylabel(’Phasehi’)title(’-i/2+tan^{-1}(megaKp/Ki)’)figure(2)y=-pi/2+atan(w*Kp/Ki);plot(w,y)xlabel(’mega’)ylabel(’Phasehi’)title(’tan^{-1}(megaKp/Ki)’)−1−p/2+tan(wKp/Ki)0−0.2课后答案网−0.4www.hackshp.cnKp/Ki=1.2−0.6f−0.8Phase−1−1.2−1.4Figure1−1.60246810121416w55 −1tan(wKp/Ki)0−0.2−0.4Kp/Ki=1.2−0.6f−0.8Phase−1−1.2−1.4Figure2−1.60246810121416wKI9.53ThetransferfunctionofthePIDcontrollerisdefinedbyK++------Ks.TherequirementPsDistousethiscontrollertointroducezerosats=-1+j2intothelooptransferfunctionofthefeedbacksystem.Hence,222s++()K¤KsK()¤K=()s+1+()2PDID2=s++2s5Comparingterms:课后答案网KP-------=2www.hackshp.cnKDKI------5=DThelooptransferfunctionofthecompensatedfeedbacksystemisthereforeKIK++------KsPsDLs()=-------------------------------------()s+1()s+22K()s++2s5D=---------------------------------------ss()+1()s+256 Figure1displaystherootlocusofL(s)forvaryingKD.Fromthisfigureweseethattherootlocusisconfinedtothelefthalfofthes-planeforallpositivevaluesofKD.ThefeedbacksystemisthereforestableforKD>0.%Solutiontoproblem9.53clear;clc;num=[125];den=[1320];rlocus(num,den)RootLocus21.510.50ImagAxis−0.5−1−1.5−2−3.5−3−2.5−2−1.5−1−0.50Figure1RealAxis9.54LetKPdenotetheactionoftheproportionalcontroller.Wemaythenexpresstheloop课后答案网transferfunctionofthecontrolledinvertedpendulumasLs()=-------------------------------------------------KP()s+3.1www.hackshp.cn()s–3.12s()s+4.4()s–4.4whichhaszerosats=+3.1andpolesats=0(order2)ands=+4.4.TheresultingrootlocusissketchedinFig.1.ThisdiagramshowsthatforallpositivevaluesofKP,the57 closed-looptransferfunctionofthesystemwillhaveapoleintheright-halfplaneandapairofpolesonthejw-axis.s-planedoublepolex0x0xs-4.4-3.13.14.40Figure1Accordingly,theinvertedpendulumcannotbestabilizedusingaproportionalcontrollerforKP>0.ThereaderisinvitedtoreachasimilarconclusionforKP<0.ThestabilizationofaninvertedpendulumismadedifficultbythepresenceoftwofactorsinthelooptransferfunctionL(s):1.Adoublepoleats=0.2.Azeroats=3.1intheright-halfplane.Wemaycompensatefor(1)butnothingcanbedoneabout(2)ifthecompensatorisitselftobestable.Moreover,wehavetomakesurethatthetransferfunctionofthecompensatorisproperforittoberealizable.Wemaythusproposetheuseofacompensator(controller)whosetransferfunctionis2Ks()s–4.4Cs()=------------------------------------------------------课后答案网()s+1()s+2()s+3.1Thecompensatedlooptransferfunctionistherefore(afterperformingpole-zerowww.hackshp.cncancellation)L¢()s=Cs()Ls()Ks()–3.1=------------------------------------------------------()s+1()s+2()s+4.4Figure2showsasketchoftherootlocusofL¢()s.FromthisfigureweseethatthecompensatedsystemisstableprovidedthatthegainfactorKsatisfiesthecondition42´0<1(i.e.,thecriticallydampedoroverdampedcase).Weareinterestedintheunderdampedcase.Thefirstmaximumofy(t)occursforn=1;hencethepeaktimeispT=-------------------------p2w1–znThefirstmaximumofy(t)alsodefinesthepercentageovershoot.Thus,puttingt=TpinEq.(...)andsimplifyingtheresult,weget课后答案网2–pz¤()1–zy=1+emaxThepercentageovershootisthereforewww.hackshp.cn2–pz¤()1–zP.O.=100eFinally,todeterminethesettlingtimeTsweseekthetimetforwhichthestepresponsey(t)definedinEq.(...)decreasesandstayswithin+d%ofthefinalvaluey(¥)=1.0.For–zwntd=1,thistimeiscloselyapproximatedbythedecayingexponentialfactore,asshownby–zwnTe»0.01or4.6T»---------szwn61 9.56Weoftenfindthatthepolesandzerosoftheclosed-looptransferfunctionT(s)ofafeedbacksystemaregroupedinthecomplexs-planeroughlyinthemannerillustratedinFig.P9.56.Inparticular,dependingonhowclosethepolesandzerosaretothejw-axiswemayidentifytwogroupings:1.Dominantpolesandzeros,whicharethosepolesandzerosofT(s)thatlieclosetothejw-axis.Theyaresaidtobedominantbecausetheyexertaprofoundinfluenceonthefrequencyresponseofthesystem.Anotherwayofviewingthissituationistorecognizethatpolesclosetothejw-axiscorrespondtolargetimeconstantsofthesystem.Thecontributionsmadebythesepolestothetransientresponseofthesystemareslowandthereforedominant.2.Insignificantpolesandzeros,whicharethosepolesandzerosofT(s)thatarefarremovedfromthejw-axis.Theyaresaidtobeinsignificantbecausetheyhaverelativelylittleinfluenceonthefrequencyresponseofthesystem.Intermsoftime-domainbehavior,polesthatarefarawayfromthejw-axiscorrespondtosmalltimeconstants.Thecontributionsofsuchpolestothetransientresponseofthesystemaremuchfasterandthereforeinsignificant.Lets=-aands=-bdefinetheboundariesofthedominantpolesandtheinsignificantpoles,asindicatedinFig.P9.56.Asaruleofthumb,thegroupingofpolesintodominantpolesandinsignificantpolesisjustifiediftheratiob/aisgreaterthan4.Giventhatwehaveahigh-orderfeedbacksystemwhoseclosed-looptransferfunctionfitsthepictureportrayedinFig.P9.56,wemaythenapproximatethesystembyareduced-ordermodelsimplybyretainingthedominantpolesandzerosofT(s).Theuseofareduced-ordermodelinplaceoftheoriginalsystemismotivatedbythefollowingconsiderations:•Low-ordermodelsaresimple;theyarethereforeintuitivelyappealinginsystem课后答案网analysisanddesign.•Low-ordermodelsarelessdemandingincomputationaltermsthanhigh-orderones.www.hackshp.cnAcaseofparticularinterestiswhentheuseofafirst-orderorsecond-ordermodelasthereduced-ordermodelisjustified,forthenwecanexploitthewealthofinformationavailableonsuchlow-ordermodels.Whendiscardingpolesandzerosinthederivationofareduced-ordermodel,itisimportanttorescalethefainofthesystem.Specifically,thereduced-ordermodelandtheoriginalsystemshouldhavethesamegainatzerofrequency.Example:Consideragainthelinearfeedbackamplifier.AssumingthatK=8,approximatethissystemusingasecond-ordermodel.Usethestepresponsetoassessthequalityoftheapproximation.Solution:ForK=8,thecharacteristicequationofthefeedbackamplifierisgivenby62 32s+++6s11s54=0Usingthecomputer,therootsofthisequationarefoundtobes=–5.7259,s=–0.1370±j3.0679ThelocationsofthesethreerootsareplottedinFig.2.Weimmediatelyobservefromthepole-zeromapofFig.2thatthepolesats=-0.1370+j3.0679arethedominantpoles,andthepolesats=-5.7259isaninsignificantpole.Thenumeratoroftheclosed-looptransferfunctionT(s)consistssimplyof6K.Accordingly,theclosed-loopgainofthefeedbackamplifiermaybeapproximatedas6K¢T¢()s»---------------------------------------------------------------------------------------------------------------()s++0.1370j3.0679()s+0.1370–j3.06796K¢=-----------------------------------------------------2s++0.2740s9.4308TomakesurethatT¢()sisscaledproperlyrelativetotheoriginalT(s),thegainK¢ischosenasfollows:9.4308K¢==----------------´81.397254forwhichwethushaveT¢()0=T(0).InlightofEq.(9.41),weset2zw=0.2740n2w=9.4308nfromwhichwereadilyfindthatz=0.0446w=3.0710nThestepresponseofthefeedbackamplifieristhereforeunderdamped.ThetimeconstantoftheexponentiallydampedresponseisfromEq.(9.44)1课后答案网1t===----------------------8.2993szw0.137nThefrequencyoftheexponentiallydampedresponseiswww.hackshp.cn22w1–z=3.07101–()0.0446n=3.0679rad/sFigure3showstwoplots,onedisplayingthestepresponseoftheoriginalthird-orderfeedbackamplifierwithK=8,andtheotherdisplayingthestepresponseofthe63 approximatingsecond-ordermodelwithK¢=1.3972.Thetwoplotsareverysimilar,whichindicatesthatthereduced-ordermodelisadequatefortheexampleathand.ImaginaryDominantpolexs-planeInsignificantpolexRealDominantpolexFigure1:Pole-zeromapforfeedbackamplifierwithgainK=8.RootLocus6420ImagAxis−2−4−6−9课后答案网−8−7−6−5−4−3−2−101RealAxisFigure2:Rootlocusoforiginal3rdordersystemwww.hackshp.cn64 1.83rdOrderReduced2ndOrder1.61.41.21Response0.80.60.40.20051015202530354045TimeFigure3:Stepresponseforthird-ordersystemshownasdashedcurve,andstepresponsefor(reduced)second-ordersystemshownassolidcurve.%Solutiontoproblem9.56%Original3rdordersystemnum=6*8;den=[161154];figure(2)rlocus(num,den)课后答案网figure(3)[order3,T]=step(tf(num,den));www.hackshp.cn%Reduced2ndordersystemnum=6*1.3972;den=[10.27409.4308];[order2,T]=step(tf(num,den));plot(T,order3,’b-’,T,order2,’r--’)legend(’3rdOrder’,’Reduced2ndOrder’)xlabel(’Time’)ylabel(’Response’)65 9.57Considerfirsttheoriginalsystemdescribedbythetransferfunction48Ts()=----------------------------------------------------------------------------------------2()s+5.7259()s++0.2740s9.4308ThecorrespondingBodediagram,obtainedbyputtings=jw,isplottedinFig.1onthenextpage.Considernextthereduced-ordermodeldescribedbythetransferfunction8.3832T¢()s=-----------------------------------------------------2s++0.2740s9.4308whichisobtainedfromT(s)byignoringthedistantpoleats=-5.7259andreadjustingtheconstantgainfactor.Figure2,onthepageafterthenextone,plotstheBodediagramoftheapproximatingsystem.Comparingthesetwofigures,1and2,weseethatthefrequencyresponsesoftheoriginalfeedbacksystemanditsreduced-orderapproximatemodelareclosetoeachotheroverthefrequencyrange08,whichfollowsfromthelastentryofthefirstcolumnofarraycoefficients.ForK=8,thesystemisonthevergeofinstability.Fromthefourthrow,itfollowsthatthisoccursforthelastrow:s0.2.Fromthefourthrowofthearray,italsofollowsthatKmustsatisfythecondition251.2+0.2K–0.2K>0forthesystemtobestable.Thatis,()K–16.508()K+15.508<0fromwhichitfollowsthatforstabilityK<16.508Ifthisconditionissatisfied,then5.6-0.2K>0.WhenK=16.508thesystemisagainonthevergeofinstability.FromthethirdrowoftheRoutharray,thisoccurswhen2()5.6–课后答案网0.2Ks+0()K–8=or8.508sj==±-------------www.hackshp.cn±j1.9232.302Accordingly,wemaystatethatthefeedbacksystemisconditionallystableprovidedthatKsatisfiesthecondition816.5,thecomplexclosed-looppolesofthesystemmovetotheright-halfplane.Henceforstability,wemusthave816.508,namely,K=20.ThelocusofFig.3(d)encirclesthecriticalpoint(-1,0)andthesystemisthereforeunstable.Finally,Fig.3(c)showstheNyquistlocusforK=16.508.Hereagainweseethatthelocuspassesthroughthecriticalpoint(-1,0)exactly,andsothesystemisonthevergeofinstability.CarehastobeexercisedintheinterpretationoftheseNyquistloci,hencethe课后答案网reasonfortheuseofshading.%SolutiontoProblem9.62www.hackshp.cnfigure(1);clf;K=1;num=K*[11];den=[1562-8];rlocus(num,den);Kmin=rlocfind(num,den)Kmax=rlocfind(num,den)figure(2);clf;Kmid=(Kmin+Kmax)/2;num=Kmid*[11];margin(num,den)75 figure(3);clf;nyquist(num,den)RootLocus43210ImagAxis−1−2−3−4Figure1−7−6−5−4−3−2−101RealAxisBodeDiagramGm=2.6725dB(at1.924rad/sec),Pm=15.132deg(at1.5257rad/sec)200−20−40Magnitude(dB)−60−80−100−135−180Phase(deg)−225−270课后答案网−1012Figure210101010Frequency(rad/sec)www.hackshp.cnNyquistDiagram0.80.60.40.20ImaginaryAxis−0.2−0.4−0.6Figure3−0.8−2−1.8−1.6−1.4−1.2−1−0.8−0.6−0.4−0.20RealAxis76 %SolutiontoProblem9.63figure(1);clf;K=1;num1=0.5*K*[12];den1=[116480];subplot(3,1,1)rlocus(num1,den1)%Makefigure9.22K=1;num2=K;den2=[110];subplot(3,1,2)rlocus(num2,den2)%Makefigure9.28K=1;num3=0.5*K*[12];den3=[18-480];subplot(3,1,3)rlocus(num3,den3)figure(2);clf;subplot(2,2,1)nyquist(num1,den1)subplot(2,2,3)nyquist(num2,den2)subplot(2,2,4)nyquist(num3,den3)课后答案网RootLocus10www.hackshp.cn0ImagAxis−10−12−10−8−6−4−20RealAxisRootLocus0.50ImagAxis−0.5−1−0.9−0.8−0.7−0.6−0.5−0.4−0.3−0.2−0.10RealAxisRootLocus20Figure10ImagAxis−20−12−10−8−6−4−2024RealAxis77 NyquistDiagram210ImaginaryAxis−1−2−1−0.8−0.6−0.4−0.20RealAxisNyquistDiagramNyquistDiagram20210100ImaginaryAxisImaginaryAxis−10−1Figure2−2−20−1−0.8−0.6−0.4−0.20−1−0.8−0.6−0.4−0.20RealAxisRealAxis9.64WearegivenaunityfeedbacksystemwithlooptransferfunctionKLs()=-------------------ss()+1K=-------------(1)2s+sTheclosed-looptransferfunctionofthesystemisLs()Ts()=--------------------1+Ls()K=------------------------(2)2s课后答案网++sK(a)WithK=1,T(s)takesthevalue1Ts()=----------------------(3)2www.hackshp.cns++s1Ingeneral,thetransferfunctionofasecond-ordersystemisdescribedas2wnTs()=------------------------------------------(4)22s++2zwswnnnComparingEqs.(3)and(4):w=1nz=0.5nTheclosed-looppolesoftheuncompensatedsystemwithK=1arelocatedat13s=–---±j---.2278 Figure1showstherootlocusoftheuncompensatedsystem.jwclosed-looppolefor.K=1jÖ3/2s-planesxx0-1.-jÖ3/2Figure1(b)Thefeedbacksystemistobecompensatedsoastoproduceadominantpairofclosed-looppoleswithwn=2andzn=0.5.Thatis,thedampingfactorisleftunchangedbutthenaturalfrequencyisdoubled.Thissetofspecificationsisequivalenttopolelocationsats=–1±j3,asindicatedinFig.2,shownbelow.ItisclearfromtherootlocusofFig.1thatthisrequirementcannotbesatisfiedbyachangeinthegainKoftheuncompensatedlooptransferfunctionL(s).Rather,wemayhavetouseaphase-leadcompensatorasindicatedintheproblemstatement.Closed-loopjwpole.jÖ3s-plane120oq2q1xoxxs1-1-10-课后答案网tat.-jÖ3Figure2www.hackshp.cnWemayrestatethedesignrequirement:•Designaphase-leadcompensatedsystemsothattheresultingrootlocushasadominantpairofclosed-looppolesats=–1±j3asindicatedinFig.2.Forthistohappen,theanglecriterionoftherootlocusmustbesatisfiedats=–1±j3.Withtheopen-looppolesoftheuncompensatedsystemats=0ands=-1,wereadilyseefromFig.2thatthesumofcontributionsofthesetwopolestotheanglecriterionis–9120°–0°=–210°Wethereforerequireaphaseadvanceof30otosatisfytheanglecriterion.79 Figure2alsoincludesthepole-zeropatternofthephase-leadcompensator,wheretheanglesq1andq2arerespectivelydefinedbyæö–1ç÷3q=tan----------------(5)1ç÷1èø------–1atæö–1ç÷3q=tan------------(6)2ç÷1èø---1–twheretheparametersaandtpertaintothecompensator.Torealizeaphaseadvanceof30,werequireq–3q=0°12orq=30°q+12Takingthetangentsofbothsidesofthisequation:tan30°q+tan2tanq=----------------------------------------113–tan0°qtan21-------+tanq23=------------------------------(7)11–-------tanq23UsingEqs.(5)and(6)in(7)andrearrangingterms,wefindthatwitha>1thetimeconstanttisconstrainedbytheequation2t–0.95课后答案网t+00.025=Solvingthisequationfort:t=0.923orwww.hackshp.cnt=0.027Thesolutiont=0.923correspondstoacompensatorwhosetransferfunctionhasazerototherightofthedesireddominantpolesandapoleontheirleft.Ontheotherhand,forthesolutiont=0.027boththepoleandthezeroofthecompensatorlietotheleftofthedominantclosed-looppoles,whichconformstothepictureportrayedinFig.2.Sewechooset=0.027,assuggestedintheproblemstatement.(c)Thelooptransferfunctionofthecompensatedfeedbacksystemfora=10isL()s=G()sLs()ccats+1K=-------------------------------------ts+1ss()+180 K()0.27s+1=--------------------------------------------------ss()+1()0.027s+1Figure3(a)showsacompleterootlocusofL(s).AnexpandedversionoftherootlocusaroundtheoriginisshowninFig.3(b).UsingtheRLOCFINDcommandofMATLAB,thepoint–1+j3islocatedontherootlocus.ThevalueofKcorrespondingtothisclosed-looppoleis3.846.%SolutiontoProblem9.64figure(1);clf;K=1;t=0.027;num=K*[10*t1];den=[t1+t10];rlocus(num,den)figure(2);clf;rlocus(num,den)axis([-51-44])K=rlocfind(num,den)RootLocus403020100ImagAxis−10Figure1−20课后答案网−30−40−35−30−25−20−15−10−50RealAxiswww.hackshp.cnRootLocus432K=3.89710ImagAxis−1Figure2−2−3−4−5−4−3−2−101RealAxis81 9.65ThelooptransferfunctionofthecompensatedfeedbacksystemisLs()=Gs()Hs()10K()ats+1=-----------------------------------------------,a<1(1)s()0.2s+1()ts+1Westartthesystemdesignwiththerequirementforasteady-stateerrorof0.1toarampinputofunitslope.Thesystemunderstudyisatype-1system;seeProblem9.47.ThevelocityerrorconstantofsuchasystemisgivenbyK=limsGs()Hs()vs®0Hence,10K()ats+1K=lim-----------------------------------------------v()0.25s+1()ts+1s®0=10KFromthedefinitionofKv:1------=0.1orK=10KvvHence,K=1.Next,weusetheprescribedvalueofpercentageovershoottocalculatetheminimumpermissiblephasemargin.Wedothisintwosteps:1.WeusetherelationbetweenpercentageovershootP.O.anddampingratioz(seeProblem9.55)2–pz¤1–zP.O.课后答案网=100e(2)withP.O.=10%inresponsetoastepinput,theuseofEq.(2)yieldsz1------------------==---log100.73292pe1–zwww.hackshp.cnHence,solvingforz:z=0.5911Wemaythussetz=0.6.2.Weusetherelationbetweenphasemarginjanddampingratioz(seeProblemm9.55)–1æö2zjm=tanç÷-----------------------------------------èø424z+21–zUsingthevaluez=0.6:82 –1æö1.2jm=tanç÷-----------------------------------------------------------èø424´0.6+1–2´0.6–1=tan()1.6767=59.19°Hence,wemaysetj=59°.mThenextstepinthedesignistocalculatetheminimumvalueofthegain-crossoverfrequencyw.Todothis,weusetherequirementthatthe5%settlingtimeofthestepqresponseshouldbelessthan2s.Wemayagainproceedintwostages:1.UsingtheformulaforthesettlingtimeTsettingwithd=5%=0.05(seeProblem9.55)–zwnTse=0.05SolvingthisequationforwnwithTsetting=2secondsandz=0.6:log20ew=----------------n2´0.6=2.5rad/s2.Usingtheformulaforthegain-crossoverfrequency(seeProblem9.58):42w=w4z+21–zgn42=2.54´0.6+1–2´0.6=1.79rad/sSowemayset课后答案网wg=1.8rad/s.Thestageisnowsetforcalculatingtheparametersofthecompensatingnetwork.Figures1(a)and1(b)showtheuncompensatedloopgainresponseandloopphaseresponse,respectively.FromFig.1(a)weseethatatwww.hackshp.cnw=wg=1.8rad/s,theuncompensatedloopgainis14.4dB,whichcorrespondstothenumericalvalue5.23.Hence,1a==----------0.195.23Wealsonotethatthecornerfrequency1/atshouldcoincidewithwg/10,andsowemaysetwg1------=------10atWithwg=1.8rad/sanda=0.19,wethusget83 10t=----------awg10=------------------------0.19´1.8=29secondsThetransferfunctionofthecompensatoristhereforeæöats+1Hs()=K------------------èøts+15.6s+1=-------------------20s+1PerformanceevaluationandFineTuningThecompensatedlooptransferfunctionisLs()=Gs()Hs()105.6()s+1=--------------------------------------------------s()0.2s+1()29s+1Figure2showsthecompensatedloopresponseofthefeedbacksystem.Accordingtothisfigure,thephasemarginjis65.46oandthegain-crossoverfrequencywis1.823rad/s,mgbothofwhicharewithinthedesignspecifications.Figure3showstheresponseoftheclosed-loopsystemtoastepinput.Theovershootislessthantheprescribedvalueof10%.Butthe5%settlingtimeisgreaterthantheprescribedvalueof2seconds.Wethereforeneedtofinetunethecompensatordesign.Weproposetomovethepole-zeropatternofthecompensator’stransferfunction课后答案网H(s)awayfromthejw-axissoastoreducethesettlingtime.Specifically,wemodifyH(s)as5.2s+1Hs()=-------------------20s+1www.hackshp.cnHencethemodifiedlooptransferfunctionis105.2()s+1Ls()=-----------------------------------------(3)s()0.2s()20s+1Figures4and5showthemodifiedloopfrequencyresponseandclosed-loopstepresponseofthefeedbacksystem,respectively.Fromthesefiguresweobservethefollowing:1.Thephasemarginis61.3oandthegain-crossoverfrequencyis2.359,bothofwhicharewithinthedesignspecifications.2.Thepercentageovershootofthestepresponseisjustunder10%andthe5percentsettlingtimeisjustunder2seconds.84 WecanthereforesaythatthemodifiedtransferfunctionofEq.(3)doesindeedmeetalltheprescribeddesignspecifications.DesignoftheCompensatorFigure6showsanoperationalamplifiercircuitforimplementingthephase-leadcompensator,characterizedbythetransferfunction5.2s+1Hs()=-------------------20s+1ThetransferfunctionofthiscircuitisV()sRRCs+1-------------2-=-----2-æö------------------------------------------11V()sRèø()R+RCs+113121C1R1.R2.R3.o-.+oV2(s)Figure6Hence,R2------1=R3RC=5.211()R+RC=2012课后答案网1ChooseC1=10mF.Wemaythensolvethesethreeequationsfortheresistiveelementsofthecircuit:www.hackshp.cnR=0.52MW1R=1.48MW2R=1.48MW3Withthis,thesystemdesigniscompleted.85 %SolutiontoProblem9.65%Uncompensatedfigure(1);clf;num=[10];den=[0.210];margin(num,den)%Compensatedfigure(2);clf;a=5.6;b=0.2;num=[10*a10];den=[29*b(b+29)10];margin(num,den)%FineTunedfigure(4);clf;a=5.2;b=0.2;num=[10*a10];den=[20*b(b+20)10];margin(num,den)BodeDiagramGm=Inf,Pm=38.668deg(at6.2481rad/sec)2010UncompensatedLoopResponse014.4dB−10−20−30Magnitude(dB)−40wg=1.8课后答案网−50−60−90−120www.hackshp.cnPhase(deg)Figure1−150−180012101010Frequency(rad/sec)86 BodeDiagramGm=Inf,Pm=65.46deg(at1.8226rad/sec)100CompensatedLoopResponse500Magnitude(dB)−50−100−90−135Figure2Phase(deg)−180−3−2−1012101010101010Frequency(rad/sec)BodeDiagramGm=Inf,Pm=61.298deg(at2.3587rad/sec)6040200Magnitude(dB)−20−40−60−90−135Phase(deg)wg=2.359−180Figure3−2课后答案网−10121010101010Frequency(rad/sec)www.hackshp.cn9.66TheMATLABcodeisattached.(a)WearegiventhelooptransferfunctionKs()–1Ls()=--------------------------------------------2()s+1()s++s1ThisisthesameastheL(s)consideredinProblem9.61,exceptforthefactthatthistimethegainfactorKisnegative.LetKK=–¢K¢³0WemaythenrewriteL(s)as87 K¢()–1s+Ls()=--------------------------------------------2()s+1()s++s1Nowwecanproceedinexactlythesamewayasbefore,treatingK¢asapositivegainfactor.Figure1showstherootlocusofL(s).Usingthecommandrlocfind,wefindthatthesystemisonthevergeofinstability(i.e.,theclosed-looppolesresideexactlyonthejw-axis)whenK¢=1orK=-1.WemayverifythisspecialvalueofK¢usingtheRouth-Hurwitzcriterion.Thecharacteristicequationofthesystemis32s++2s()2–K¢s+()1+K¢=0Hence,constructingtheRoutharray:3s12–K¢2s21+K¢131()–K¢s-----------------------020s1+0K¢ThethirdelementofthefirstcolumnofarraycoefficientsiszerowhenK¢=1orK=-1.Thisoccurswhen22s+0()1+K¢=or课后答案网sj=±asshownontherootlocusinFig.1.www.hackshp.cn(b)WearegiventhelooptransferfunctionKs()+1Ls()=-------------------------------------------------------432s+++5s6s2s–8whichisthesameastheL(s)consideredinProblem9.62,exceptthistimeKisnegative.SetKK=–¢,whereK¢isnonnegative.HencewemayrewriteL(s)as–K¢()s+1Ls()=-------------------------------------------------------432s+++5s6s2s–8andthusproceedinthesamewayasbefore.88 Figure3showstherootlocusofL(s)forsixdifferentvaluesofK,namely,K=-1,-2,-3andK=-0.1,-0.2,-0.3.WeseethatforallthesevaluesofKorthecorrespondingvaluesofK¢,weseethattherootlocushasapoleintheright-halfplane.Indeed,thefeedbacksystemdescribedherewillalwayshaveaclosed-looppoleintheright-halfplane.Hence,thesystemisunstableforallK<0.%SolutiontoProblem9.66figure(1);clf;num=-1*[1-1];den=[1221];rlocus(num,den)figure(2)K=[-3-2-1-0.4-0.2-0.1];forn=1:length(K),num=K(n)*[11];den=[1562-8];subplot(2,3,n)rlocus(num,den)title([’K=’num2str(K(n))])endRootLocus1K=−10.80.60.40.20课后答案网ImagAxis−0.2−0.4−0.6www.hackshp.cn−0.8−1Figure1−10−8−6−4−20246RealAxis89 RootLocusRootLocusRootLocus444222000ImagAxisImagAxisImagAxis−2−2−2−4−4−4−4−2024−4−2024−4−2024RealAxisRealAxisRealAxisRootLocusRootLocusRootLocus444222000ImagAxisImagAxisImagAxisFigure2−2−2−2−4−4−4−4−2024−4−2024−4−2024RealAxisRealAxisRealAxis课后答案网www.hackshp.cn90'