信号系统 (朱联祥 著) 邮电大学 课后答案 第二章信号与系统系统解答-- 26页

'课后答案网：www.hackshp.cn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课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn1、(1)y+5y+6y=0y(0)=-1,y(0)=1′′′′--2解：特征方程λ+5λ+=60−2t−3t特征根：λ=−2,λ=−∴3.yt()=Ce+Ce12h12⎧y(0)=C+C=1h−12代入初始状态有：⎨解之：C=2,C=−112y′(0)=−2C−3C=−1⎩h−12−2t−3t∴yt()=2e−eh(2)y′′+y′=0y′(0)=0,(0)y=2−−2解：λ+=10λ=±j12，∴yt()=CcostC+sint代入初始状态得：C=2,C=0h1212∴yt()=2costt≥0h2、(1)y(t)+3y(t)+2y(t)=(t),′′′fy′(0)1,(0)=y=0,()ft=ε()t−−对微分方程两端关于t从到作积分有00−+0+0+0+0+∫0ytdt′′()+3∫0ytdt′()+2∫0ytdt()=∫0ε()tdt−−−−y′(0)−y′(0)=,0y(0)−y(0)=0+−+−得y′(0)=y′(0)1,(0)=y=y(0)=0+−+−(2)y+6y+8y=′′′f′y′(0)1,(0)=y=0,f=ε()t−−0+0+0+0+∴∫0ydt′′+6∫0ydt′+8∫0ydt=∫0δ()tdt−−−−得：y′(0)−y′0(=1，)y(0−)y0(=0)+−+−⎧⎪y′(0+)=+1y′0−(=2)∴⎨⎪⎩y(0+=)y课后答案网0−(=0)3）y′′+4y′+3y=f′+f，y′(0=)1,y0(=0,)f=εt()−−上式可写为y′′+4y′+3y=δ()t+εt()t=0时微分方程左端只有www.hackshp.cny′′含冲激，其余均为有限值，故有00000+++++∫0ydt′′+4∫0ydt′+3∫0ydt=∫0δ()tdt+∫0εtdt()−−−−−得y′(0)−y′0(=1,)y0−y(0)=0()+−+−⎧⎪y′(0+)=+1y′0−(=2)∴⎨⎪⎩y(0+=)y0−(=0)−2t4）y′′+4y′+5y=fy′′,(0)=2,y0(=1,)ft=e()εt()−−−2tft′()=δt()−2eεt()若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn−2t原方程可写为y′′+4y′+5y=δ()t−2eεt()0+0+0+0+0+−2t∫0ydt′′+4∫0ydt′+5∫0ydt=∫0δ()tdt−2∫0eε()tdt−−−−−∴y′(0+−)y′0−=(1,0y)+−y0−(=0)()⎧⎪y′(0+=)y′0−(=3)⎨⎪⎩y(0+=)y0−(=1)3.1()y′′+4y′+3y=fy,′(0−=)y0−(=1,ft)=εt()()2解：①求yx()tyx′′+4yx′+3yx=0λ+4λ+=30λ1=−1,λ2=−3−t−3tyt()=ce+ce12⎧y0(_)=C1+C2=1x⎨"y0(_)=−C1−3C2=1⎩x解之:C1=2C2=−1−t−3t∴y(t)=2e−et≥0x②求y(t)−t−3tfyf(t)=课后答案网Cf1e+Cf2e+yp(t)1设yf(t)=P0带如原微分方程有3P=1即P0=3−t−3t1故：y(t)=Ce+Ce+)ff1f2www.hackshp.cn3对原微分方程两端从0−到0+关于t积分有0_0_0_0_"""ydt+4ydt+3ydt=ε(t)dt∫0+f∫0+f∫0+f0∫+"""⎧y(0+)−y(0−)=0⎧y(0+)=0⎪ff⎪f⎨⎨"⎪⎩yf(0+)−yf(0−)=0⎪⎩yf(0+)=0"⎧y(0+)=−C−3C=0ff1f2⎪有：⎨1"⎪yf(0+)=Cf1+Cf2+=0⎩311解之：C=−C=−f1f226若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn1−t1−3t1∴y(t)=(−e+e+)ε(t)f263③求全响应yt)(。−t−3t1−t1−3t1(y)t=y(t)+y(t)=2e−e−e+e+xf263t≥03−t5−3t1=e−e+263""""−t（2）y+4y+4y=f+3f，y"(0_)=,2y(0_)=,1f=eε(t)2解：①λ+4λ+4=0λ=−2。1,2−2ty()t=(Cx0+Cx2)ex⎧⎪y"(0)=−2C+C=2x−x0x1⎨得C=1,C=4x0x1⎪y(0)=C=1⎩x−x0−2t∴y()t=(14)+tet≥0x(2)求yt()f−2tyt()=(C+Cte)+y()tffffp01−2t设y()t=pe并代入原微分方程，有fp1−2t−2t−2t−t−t(pe)""4(+pe)"4(+pe)=(e)"3+e111得p−4p+4p=−+13即p=21111−2t−t故yt()=(C+Cte)+2eff0f1课后答案网0+0+0+0+−t0+−t由y""dt+4y"dt+4ydt=[()δt−eε()]tdt+3eε()tdt∫0f∫0f∫0f∫0∫0−−−−−⎧⎪y"(0)y"(0)1−=⎧⎪y"(0)1y"(0)1=+=f+f−f+f−有⎨⎨⎪y(0)y(0)−=0⎪y(0)=0⎩f+f−www.hackshp.cn⎩f+⎧⎪y"(0)=−2C+C−=21f+f0f1∴⎨解之：C=−2,C=−1ff1y(0)=C+=200⎪⎩f+f0−t−2t∴yt()[2=e−(2+te)]()εtf(3)求yt()−t−2tyt()=yt()+yt()=2e+(3t−1)et≥0fx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cny""2"2+y+y=f","(0)1,(0)y=y=0,f=ε()t−−解:1.求y()tx2λ+2λ+=20,λ=−±1j1,2−t∴yt()=e(CcostC+sin)txx1x2−t−ty"()t=e(CcostC−sin)t−e(CcostC+sin)txx2x1x1x2代入初始状态：y(0)=C=0,"(0)y=C=1x−x1x−x2−t∴yt()=esintt≥0x2.求yt()首先确定y"(0)与y(0)ff+f+0+0+0+0+y""dt+2y"dt+2ydt=δ()tdt∫0f∫0f∫0f∫0−−−−可得y"(0)−y"(0)1,=y(0)−y(0)=0;f+f−f+f+⎧y"(0)1=f+则⎨y""2+y"2+y=δ()tfff⎩y(0)=0f+当t≥1时，y""2+y"2+y=0fff−t∴yt()=e(cosAt+sin)tf代入初始条件：y"(0)=B=1,y(0)=A=0f+f+−t∴yt()=esinttε()f3.求全响应yt()−tyt()=y+y=2esintt≥0xf2.4（1）y(k+2)+3y(k+1)+2y(k)=0,y(0)=,2y(1)=1xx2解：特征方程r+3r+2=课后答案网0(r+1)(r+2)=0特征根：r=−,1r=−212kkwww.hackshp.cnkky(k)=Cr+Cr=C(−1)+C(−2)x11x22x1x2⎧Cx+Cx=212代入初始条件⎨−C−2C=1解得Cx1=,5Cx2=−3⎩x1x2kk∴y(k)=5(−1)−3(−2)k≥0x(2)y(k+2)+2y(k+1)+2y(k)=0.y(0)=,0y(1)=.1xx解：若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn2r+2r+2=0→r=−1±j1,2kky(k)=C(−1+j)+C(−1−j)xxx12⎧yx(0)=Cx+Cx=012⎨y(1)=(−1+j)C+(−1−j)C=1⎩xx1x2jjC=−,C=xx1222jkjk∴y(k)=(−)(−1+j)+(−1−j)x223πkjkk3π=(2)e2=(2)sink4k≥0(3)y(k+2)+2y(k+1)+y(k)=0y(0)=y(1)=1xx解：22r+2r+1=0(r+1)=0∴r=r=−112ky(k)=(C+CK)(−1)xxx12⎧yx(0)=Cx=11⎨y()1=(C+C)(−1)=1⎩xx1x2⎧Cx=1∴1⎨C=−2⎩x2kyx(k)=(1−2k)(−1)课后答案网k≥0(4)y(k)+2y(k−1)=0y(0)=2x解：γ−2=0γ=2ywww.hackshp.cn(k)=C(2)kxxky(0)=C=2故y(k)=2(2)k>=0xxx(5)y(k)+2y(k−)1+4y(k−)2=0y(0)=,0y(1)=2xx解：22γ+2γ+4=0即(γ+1)+3=0特征根γ=−1±3j1,2kky(k)=C(−1+3j)+C(−1−3j)xx1x2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn1*1C=C=C=−x1x2x1j3j3kk1(−1+j3)(−1−j3)故y(k)=[−]x3jj2π2πjk−jk2k+1e3−e322πk==2sinkk>=032j33(6)y(k)−7y(k−1)+16y(k−2)−12y(k−3)=0y(0)=,0y(1)=−1,y(2)=−3xxx322解：γ−7γ+16γ−12=0即(γ−3)(γ−2)=0γ1=3γ2,3=2kky(k)=C(3)+(C+Ck)(2)xx0x1x2带入初始条件有⎧y(0)=c+c=0c=−c⎪xx0x1x0x1⎪⎨yx(1)=3cx0+2cx1+2cx2=−1−cx1+2cx2=−1⎪⎪⎩yx(2)=9cx0+4cx1+8cx2=−3−5cx1+8cx2=−3解之得：cx0=,1c课后答案网x1=−1，cx2=−1kk故：y(k)=3+(1+k)(2)k>=0x2.5(1)y(k)+3y(k−1)+2y(k−2)=f(k),y(−1)=,0y(−2)=12www.hackshp.cn解：γ+3γ+2=0γ=−,1γ=−212kky(k)=1(−)1+2(−)2xcxcx−1−1⎧⎪y(−1)=c(−)1+c(−)2=0⎧4+=4xx1x2即：cx1cx2⎨−2−2⎨⎪yx(−2)=cx1(−)1+cx2(−)2=1⎩−2cx1−cx2=0⎩⎧c=2kk解之得：x1故：()⎡24⎤0⎨yk=⎢⎣(−)1−(−)2⎥⎦k>=⎩cx2=−4（2）y(k)+2y(k−)1+y(k−)2=f(k)−f(k−)1y(−1)=1,y(−2)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn解：y(k)+2y(k−1)+y(k−2)=0xxx22γ+2γ+1=0(γ+)1=0γ=−11,2ky(k)=(c+cx2k)(−)1xx1⎧⎪yx(−1)=−cx1+cx2=1⎧c=1x1⎨⎨⎪⎩yx(−2)=−cx1−2cx2=3⎩cx2=2k故：y(k)=(1+2k)(−)1k>=0x（3）y(k)+y(k−)2=f(k−2),y(−)1=−2,y(−)2=−12解：γ+1=0;γ=±j1,2ππy(k)=Acosk+Bsinkx22ππy(k)=(−cosk+2sink)y(−1)=−B=−2x22y(−2)=A=−1π�=5cos(k−634.)k≥022.6(1)y(k)−2y(k−1)=f(k),y(−1)=−,1f(k)=2ε(k)kk解：γ−2=,0γ=2y(k)=C(2)+y(k)=C2−2课后答案网py(k)=p,p−2p=,2p=−2p0000令k=,0y(0)−2y(−1)=,2y(0)=0y)0(=C−2,C=2www.hackshp.cnk所以y(k)=2(2)=,2k≥0k其中yx(k)=Cx(2)Cx=−,1C=−2kx=−2(2),k≥02ky(k)=C(2)+y(k)fffpkk=y(k)−y=2(2)−2−[−2(2)]xk=[4(2)−2]ε(k)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn(2)yk()3(+yk−1)2(+yk−2)=fk()y(1)1,(2)−=y−=0,()fk=ε()k2解：γ+3γ+=⇒20γ=−1,γ=−212kkyk()=C(1)−+C(2)−xxx12⎧1y(1)−=y(1)−=−C−C⎪⎪xx1x2⎪⎧Cx=121⎨⇒⎨1⎪C=−4⎪y(2)−=y(2)−=C+C⎩x2xxx⎪⎩142kk∴yk()=−(1)−4(2)−≥k0x1令y=P.则有P+3P+2P=1,P=fp00006由yk()=fk()3(−yk−1)2(−yk−2)得：⎧y(0)=f(0)−y(1)2−−y(2)1−=⎧y(0)1=ffff⎨⇒⎨y(1)=f(1)3−y(0)2−y(1)−=−2y(1)=−2⎩fff⎩f⎧1⎧1y(0)=C+C+=1C=−⎪⎪ff1f26⎪⎪f12⎨解之得：⎨14⎪y(1)=−C−2C+=−2⎪C=ff1f2f2⎪⎩6⎪⎩31k4k1∴y()[k=−(1)−+(2)−+]()εkf236kk1k4k1yk()=yk()+yk()[(1)=−−4(2)−−(1)−+(2)−+]xf2361k8k1=[(1)−−(2)−+]≥k0236课后答案网1−stg(t)=(1−e)ε(t)i9dgi(t)5−st2.7(a)解：h(t)==eε(t)idtwww.hackshp.cn910−sth(t)=Rh(t)=eε(t)ui3(b)解:由图知i+i+i=icrls2dudiuLdiclll其中：i=c=Lci==c2RdtdtRRdtL12故有：LCi′′+i′+i=i即：i′′+i′+i=iLLLSLLLsR5555故i′′+i2′+5i=5iH(p)==LLLS22(p+2p+5)(p+1)+4若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5−th(t)=esin2tε(t)iL2dhiL1d5−th=L=×[esin2tε(t)]uLdt5dt21−t−t=[−esin2tε(t)+2ecos2tε(t)]2−t1−t=[ecos2t−esin2t]ε(t)2diL∵u=LLdtdhiL∴h=LuLdty′+2y=f(′t)−f(t)p−1p+2−33H(p)===1−p+2p+2p+22．8（1）−2th(t)=δ(t)−3eε(t)ttt3−2τ−2t(g)t∫=0−(hτ)dτ=∫o−δ(τ)dτ−3∫0−edτ=ε(t)+(1−e)ε(t)2（2）yt"()2()+yt=ft""()22pp+2p−2p−+444Hp()===−+p2p+2p+2p+2−2tht()=δ"()2()4t−δt+eε()tgt()=∫0th()ττd=∫0tδττ"()课后答案网d−2∫0tδττ()d+4∫0te−2τdτεi()t−−−−−2t=δ()2()2()2t−εt+εt−eε()t−2t=δ()2t−eε()twww.hackshp.cn2.9，求ht()111(1)2""8y+y=fHp()==222p+82p+41ht()=sin2()ttε4(2)y""+y"+y=f"+f1113p++p+p+122212Hp()===+p2+p+11231233123(p+)+(p+)+(p+)+242424若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cntt−31−3ht()=e2costtε()+e2sinttε()232(3)y""2"+y+y=f"2+fp+2p+211Hp()===+222p+2p+1(p+1)(p+1)(p+1)−t−tht()=(e+te)()εt(4)y"""6""11"6+y+y+y=f"2+fp+2p+21H(p)===32p+6p+11p+6(p+1)(p+2)(p+3)(p+1)(p+2)ptptee1−t1−3th(t)=+=(e−e)ε(t)p+3p+1p=−322p=−12.10求h(k)(1)y(k)+2y(k-1)=f(k-1)−1E1k−1=←⎯→h(k)=(−2)ε(k−1)解：H(E)=1+2E−1E+2(2)y(k+2)+3y(k+1)+2y(k)=f(k+1)+f(k)E+1E+11H(E)===2E+3E+2课后答案网(E+1)(E+2)E+2k−1←⎯→h(k)=(−2)ε(k−1)1(3)y(k)+y(k-1)+y(k-2)=f(k)4www.hackshp.cn22EE解：H(E)==2112E+E+(E+)42d12k−1dk+1h(k)=(E+)H(E)E=(E)ε(k)11dE2E=dEE=22k1k=(k+1)E1ε(k)=(k+1)(−)ε(k).E=−22(4)y(k)-4y(k-1)+8y(k-2)=f(k)解:h(k)-4y(k-1)+8y(k-2)=δ(k)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnk>0时,有h(k)-4h(k-1)+8h(k-2)=02πγ−4γ+8=0γ12=2+j2=22∠−4h(c)=δ(c)+4δ(k-1)-8h(k-2)h(0)=δ(0)+4h(-1)-8h(-2)=1=p1ph(1)=δ(1)+4h(0)-8h(-1)=4=22Q+2222故：p=1,Q=1.kkπkπH(k)=(22)(sin+sin)ε(k)44kkπkπ=2(22)(sin+sin)ε(k)44(5)y(k+2)+2y(k+1)+2y(k)=f(k+1)+2f(k)解：h(k+2)+2h(k+1)+2h(k)=δ(k)000k3kπ3kπh(k)=()(pcos+Qsin)o2443kπ3kπ1h(2)=2[pcos+Qsin]=-2Q=1所以Q=-o2223kπ3kπ111h(1)=2[pcos+Qsin]=2[p−]o4422211=p-=0p=+22k13kπ13kπ所以h(k)=（课后答案网）[-sin+cos]ε(k-1)o22424k−13kπ3π=2sin(-)ε(k-1)44h(k)=h(k+1)+2h(k)o0www.hackshp.cnk3π3πk−13kπ3π=2sin[(k+1)-]ε(k)+22sin(-)ε(k-1)4444k3π3πk−13kπ3π=2sin[(k+1)-]ε(k)+22sin(-)ε(k-1)4444k3kπ=-2cosε(k-1)42.11(1)yk(+2)+yk(+1)=2(fk+1)+fk()(由图得)移序得：yk(+1)+yk()=2()fk−fk(−1)设hk(+1)+hk()=δ(),k有γ=−100k∴hk()=C(1)−ε(k−1).0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn又∵h(1)1=∴C=−10kk−1∴hk()=−−(1)ε(k−1)=−(1)ε(k−1)0hk()=2()hk−hk(−1)00k−1k−2=2(1)−ε(k−1)(1)+−ε(k−2)(2)由图可得差分方程yk(+2)−yk(+1)0.24()+yk=fk(+2)0.5(−fk+1)22E−0.5EE−0.5EHE()==2E−E+0.24(E−0.4)(E−0.6)HE()1111=×+×E2E−62E−0.41E1E∴HE()=×+×2E−62E−0.41k1k↔hk()[(0.6)=+(0.4)]()εk222.12图示法求解下列卷积t<0ft()=ft()*ft()=0121t1t10≤=)0xkk⎛⎞1IIyf()k=(Cf+Ckf(−1)+)yfpk()令yfp()k=P0⎜⎟带入原差分方程得01⎝⎠2kk−1k−2k⎛⎞1⎛⎞1⎛⎞1⎛⎞1P⎜⎟+2P⎜⎟+P⎜⎟=3⎜⎟000⎝⎠2⎝⎠2⎝⎠2⎝⎠2课后答案网P+4P+4P=30001∴P=03∵yk()=fk()2(−yk−1)www.hackshp.cn−yk(−2)∴y(0)=−32y(1)−−y(2)−=3fff339y(1)=−2y(0)−y(1)−=−=−6fff222⎧1y(0)=C+=3⎧8ff⎪⎪0⎪C=3f0⎨⇒⎨319⎪y(1)=(C+C)(1)−+=−⎪C=2⎪⎩ff0f1⎩f1628k11k故yk()[(=+2)(1)k−+()]()εkf332yk()=yk()+yk()xf若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5k11k=(2k+)(1)−+()≥k03321k2.33yk()=()ε()k2解：hk()=∇gk()=gk()−gk(−1)1k1k−1=()ε()()k−ε(k−1)22112.34()ayk()=fk()−yk(−1)即yk()+yk(−1)=fk()22E1kkHE()=↔hk()=−()ε(),()kfk=(2)ε()k12E+2k1kyk()=fk()*()hk=(2)ε()*(k−)ε()kf2∞n1kn−=∑(2)ε()(n−)ε(k−n)n=−∞2k1knn4k11k=−()∑22i=[(2)+(−)]()εk2n=0552课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn(b)y(k)=f(k)+1.5y(k−)1−0.5y(k−)2y(k)−1.5y(k−)1+0.5y(k−)2=f(k)22EEH(E)==2E−1.5E+0.5(E−1)(E−0.)5k+1kh(k)=αε(k)−α0(.)5ε(k)=αε(k)+0(.)5ε(k)nk−nyf(k)=f(k)*h(k)=∑αε(n)[ε(k−n)−0(.)5ε(k)]k1knknn=(α∑α)ε(k)−()∑αα∗ε(k)n=02n=0k+1k+1⎡1−α1k1−4⎤=⎢α*−()⎥ε(k)⎣1−224−1⎦⎡k⎤8k1⎛1⎞=⎢α+⎜⎟−2⎥ε(k)⎢⎣33⎝2⎠⎥⎦2.35解：设第k个月的本利之和是y()k，则y(k)=f()k+fk(−1+)βfk(−1)y(k)−y(k−1)−0.003yk(−1=)f(k).........1+β=1.003y(k)−1.003y(k−1)=fk().............γ=1.003y(k+)1−1.003y(k)=f(k+1)ky(k)=C1(.003)+y(k)............令y=Ppp0P+1.003P=100010∴P=−0课后答案网0.003k10y(k)=C0(.003)−0.00310y)0(=C−..........www.hackshp.cn....C=33530.003ky(k)=33531(.003)−3333y(12)=142.(7元)2.36解：设k条直线分成y(k)块平面.k=0时，y(0)=1;则由题意得：y(k)-y(k-1)=f(k).其中f(k)=kε(k)则y(k+1)-y(k)=(k+1)ε(k)两边求算子有若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnEEEy(E)-Ey(0)-y(E)=+2(E−1)E−1EEEy(E)=++32(E−1)(E−1)E−1y(k)=0.52k+0.5k+1k≥02.37解：()yk=(h)k[∗(f)k∗h()k−f(k)∗h(k)]312(h)k=h(k)∗[δ(k)*h(k)−δ(k)∗h(k)]:312=(ε)k[∗δ(k)−δ(k−N)]=ε(k)−ε(k−N)2.38解：(h)k=(h)k[∗h()k∗h(k)−h(k)]3121k=()8.0ε(k)∗[ε(k)∗δ(k−3)−ε(k)]kk=8.0ε(k)∗ε(k−3)−8.0ε(k)∗ε(k)∞∞nn=∑8.0ε(n)ε(k−n−3)−∑8.0ε(n)ε(k−n)n=−∞n=−∞k−3knnhk()=(∑(0.8))(εk−3)(−∑(0.8))()εkn=0课后答案网n=0k−2k+1=−5[(0.8)+1](εk−3)5[1(0.8)+−]()εkwww.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn'