《信号处理导论》 (Sophocles J.Orfanidis 著) 课后习题答案 清华大学出版社 89页

'欢迎光临阳光大学生网,提供最全面的大学生课后习题答案和复习试题免费下载，http://www.sundxs.com/阳光大学生网我们希望呵护您的眼睛，关注您的成长，给您一片绿色的环境，欢迎加入我们，一起分享大学里的学习和生活感悟，免费提供:大学生课后答案，大学考试题及答案，大学生励志书籍。 第一章1.1f=6)1f=8Hzf=fmod(f)=−2sastheobserverperceivethewrongsenseofrotation)2f=12Hzf=fmod(f)=±6sastheobserverlosethesenseofrotation)3f=16Hzf=fmod(f)=6sastheobserverperceivethecorrectsenseofrotation)4f=24Hzf=fmod(f)=6sastheobserverperceivethecorrectsenseofrotation课后答案网www.khdaw.com 1.2f=1Hz,f=4Hz,f=6Hz123)1f=5HzNyquistinterval[:−]5.2,5.2sf=fmod(f)=1mod()5=;11a1sf=fmod(f)=4mod()5=−;12a2sf=fmod(f)=6mod()5=;13a3sx(t)=10sin(2πft)+10sin(2πft)+5sin(2πft)=5sin(2πt)a1a2a3ax(nT)=5sin(2πnT)=5sin(2πn/f)=5sin(2πn)5/asx(nT)=10sin(2πn)5/+10sin(8πn)5/+5sin(12πn)5/=5sin(2πn)5/∴x(nT)=x(nT)a课后答案网www.khdaw.com )2f=10HzNyquistinterval[:−]5,5sf=fmod(f)=1mod(10)=;11a1sf=fmod(f)=4mod(10)=;42a2sf=fmod(f)=6mod(10)=−;43a3sx(t)=10sin(2πft)+10sin(2πft)+5sin(2πft)a1a2a3a=10sin(2πt)+5sin(8πt)x(nT)=10sin(2π/10)+5sin(8π/10)ax(nT)=10sin(2πn/10)+10sin(8πn/10)+5sin(12πn/10)=10sin(2π/10)+5sin(8π/10)∴x(nT)=x(nT)a课后答案网www.khdaw.com 1.3x(t)=2cos(πt)−cos(5πt)f=05kHz,f=5.2kHz12f=3kHzNyquistinterval[:−5.2k5.2,k]s)1f=fmod(f)=5.0mod()3=;5.01a1sf=fmod(f)=5.2mod()3=−;5.02a2sx(t)=2cos(2πft)−cos(2πft)=cos(πt)a1a2a)2x(t)=2cos[2π(f+mf)t]−cos[2π(f+mf)t]mn1s2s课后答案网www.khdaw.com 1.9a)f=5kHz,f=10kHz,f=30kHz,f=45kHz1234f=40kHzNyquistinterval[:−20k,20k]sf=fmod(f)=5k;f=fmod(f)=10k;1a1s2a2sf=fmod(f)=−10k;f=fmod(f)=5k;3a3s4a4sy(t)=2sin(10πt)ab)theoutputoftheprefilter:y(t)=sin(10πt)+sin(20πt)y(t)=y(t)a课后答案网www.khdaw.com c)通过prefilter后信号在频率f处的幅度值：330−28.078/2048log()=28.078dB⇒H(f)=10=.003952320通过prefilter后信号在频率f处的幅度值：445−56.156/2048log()=56.156dB⇒H(f)=10=.000162420∴theoutputoftheprefilteris:y(t)=sin(10πt)+sin(20πt)+.00395sin(60πt)+.00016sin(90πt)f=fmod(f)=5k;f=fmod(f)=10k;1a1s2a2sf=fmod(f)=−10k;f=fmod(f)=5k;3a3s4a4s∴theoutputoftheanalogreconstructoris:y(t)=sin(10πt)+sin(20πt)+.00395(−20πt)+.00016sin(10πt)a=.10016sin(10πt)+.09605sin(20πt)课后答案网www.khdaw.com 第二章2.1R16Q===2B3221)x=3.7;truncation;two’scomplementbipolarADCtestb1b2b3mxQ=mQc=u(x−xQ)240b2010b001121312001x=2Q课后答案网www.khdaw.com 2)x=－3.7;truncation;two’scomplementbipolarADCtestb1b2b3mxQ=mQc=u(x−xQ)-2-41b1102-1-20b1113-2-4110x=−4Q课后答案网www.khdaw.com 3)x=3.7;rounding;offsetbinarybipolarADCQy=x+2testb1b2b3mxQ=mQc=u(y−xQ)001b1001b1102412b111360311024x=4Q课后答案网www.khdaw.com 4)x=－3.7;rounding;offsetbinarybipolarADCQy=x+2testb1b2b3mxQ=mQc=u(y−xQ)000b1001b010－2－412b011－1－203010－2－4x=−4Q课后答案网www.khdaw.com 2.3)1Q−3−3e=≤1×10⇒Q≤12×10rms12R−3⇒=Q≤12×10⇒B≥115.⇒B=12B2)2QR10e====7.0millivoltrmsB1212212212)3R20log()=6B=72dB10Q课后答案网www.khdaw.com 2.43444×10×60×16×2=8448×10bit1Megabytes=1024×1024×8=8388608bit48448×10∴=10.07Megabytes8388608课后答案网www.khdaw.com 第三章3.1a.nonlineartime-invariantb.nonlineartime-variantc.nonlineartime-invariantd.lineartime-variante.nonlineartime-variant3.23.3.22h(n)=3δ(n)−2δ(n−)1+4δ(n−)3h(n)=3[−20]4课后答案网www.khdaw.com 3.3nh(n)=(−)9.0u(n)3.4y(n)=9.0y(n−)1+x(n)课后答案网www.khdaw.com 3.16+∞y(n）=∑h(m)x(n−m)m=−D+∞+∞yD(n）=∑hD(m)x(n−m)=∑h(m−D)x(n−m)m=0m=0令m−D=l+∞=∑h(l)x(n−l−D)=y(n−D)l=−D∴y(n）=y(n−D)D课后答案网www.khdaw.com 第四章4.1h=1[121],x=1[211211]1(a)theconvolutiontabley=1[357767643]1(b)theLTIformofconvolutionarrangedinatableform1×11×12×11×111×21×22×21×221×11×12×11×111×11×12×11×111×21×22×21×221×11×12×11×111×11×12×11×111×11×12×11×11课后答案网www.khdaw.com (c)length-3n012345678910y1356411y1356412y312331y13577676431课后答案网www.khdaw.com length-5n012345678910y113577552y2124431y13577676431课后答案网www.khdaw.com 4.3a.Theoverallindexrangenfortheoutputy(n)⎧3≤m≤6⎧10≤m≤20⎨或⎨⇒13≤n≤26⎩10≤n−m≤20⎩3≤n−m≤6directform:y(n)=∑h(m)x(n-m)⎧3≤m≤6⎧3≤m≤6⎨⇒⎨⇒max(,3n−20)≤m≤min(,6n−10)⎩10≤n−m≤20⎩n−20≤m≤n−10n−10⎧⎪13≤n≤16y(n)=∑h(m)x(n-m)⎪m=3⎪6⎨17≤n≤23y(n)=∑h(m)x(n-m)⎪m=3⎪6⎪23≤n≤26y(n)=∑h(m)x(n-m)⎩m=n−20课后答案网www.khdaw.com LTIform:y(n)=∑x(m)h(n-m)⎧10≤m≤20⎧10≤m≤20⎨⇒⎨⇒max(10,n−)6≤m≤min(20,n−)3⎩3≤n−m≤6⎩n−6≤m≤n−3n−3⎧⎪13≤n≤16y(n)=∑x(m)h(n-m)⎪m=10⎪n−3⎨17≤n≤23y(n)=∑x(m)h(n-m)⎪m=n−6⎪20⎪23≤n≤26y(n)=∑x(m)h(n-m)⎩m=n−6b.y(n)=[12344444444321]13≤n≤26课后答案网www.khdaw.com n＝13~15:input-ontransient;1313~15:input-ontransient;~15:input-ontransient;n=16~23steadystate;n=n=16~23steadystate;16~23steadystate;n=24~26input-offtransientn=n=24~26input-offtransient24~26input-offtransient课后答案网www.khdaw.com 4.4a.aa..my(n)=∑au(m)u(n−m)m⎧m≥0⎨⇒n≥m≥0,0≤m≤n⎩n−m≥0nn+1m1−a∴y(n)=∑a=⋅u(n)m=01−a1thesteady−stateresponse:;1−an+11−athetransientresponse:⋅u(n)课后答案网1www.khdaw.com−a b.mn−my(n)=∑au(m)(−)1u(n−m)m⎧m≥0⎨⇒n≥m≥0,0≤m≤n⎩n−m≥0nn+1mn−mn1−(−a)∴y(n)=∑a(−)1=(−)1⋅u(n)m=01+an(−)1thesteady−stateresponse:;1−an+1n1−(−a)thetransientresponse(:−)1⋅u(n)1+a课后答案网www.khdaw.com 4.5n−my(n)=∑[u(m)−u(m−L)]au(n−m)m⎧0≤m≤L−1⎨⇒n≥m≥;00≤m≤min(n,L−)1⎩n−m≥0nn+1n−m1−a0≤n≤L−1y(n)=∑a=m=01−aL−1n+1−Ln+1n−ma−an≥Ly(n)=∑a=m=01−a课后答案网www.khdaw.com 4.15a.aa..(hn)=δ(n)−δ(n−)3h=1[00−]1b.x(n)y(n)⎧w0=x+⎪⎨y=w0−w3−1⎪z⎩delay(w)3,w1−1⎧y=x−w3z-1⎪w2⎪w3=w2−1或⎨zw=w⎪21w3⎪w=x⎩1课后答案网www.khdaw.com c.h=1[00−]1LTIformtable11×100−1×111×100−1×121×20021×2041×4y11213也可以用其它方法求出结果课后答案网www.khdaw.com d.nx(n)w(n)w(n)w(n)w(n)y=w−w01230301100011111001222110232221114442213课后答案网www.khdaw.com 4.18a.aa..h=1[0−1]2b.bb..11×10−1×12×121×10−1×12×111×20−1×22×211×20−1×22×221×20−1×22×211×20−1×22×211×10−1×1011×101×11×1y11132433312课后答案网www.khdaw.com c.n012345678910y11113024y222−13312y11132433312课后答案网www.khdaw.com d.x(n)y(n)+z−1⎧y=x−w2+2w3-1⎪w1⎪w3=w2−1⎨z-12w=w⎪21w2⎪w=x−1⎩1zw3课后答案网www.khdaw.com 第五章5.25X(z)=zROC:z≠∞5.3n10n−10x(n)=(−)5.0u(n)−(−)5.0(−)5.0u(n−10)−10−110110z1−5.0(z)X(z)=−(−)5.0=−1−1−11+5.0z1+5.0z1+5.0zROC:z≠0课后答案网www.khdaw.com 5.4nπnπj−jx(n)=)9.0(n[e2+e2]u(n)=[(9.0j)n+(−9.0j)n]u(n)11X(z)=+−1−11−9.0jz1+9.0jzROC:z>9.0课后答案网www.khdaw.com 5.511a)X(z)=+ROC:z>4−1−11−.025z1−4z11b)X(z)=+ROC.0:254两者ROC的交集为空集，所以不存在z变换课后答案网www.khdaw.com 5.7(a)−1−2−3X(z)=1+3z−4z−12zx(n)=δ(n)+3δ(n−)1−4δ(n−)2−12δ(n−)3因果，稳定(b)x(n)=5δ(n)+3δ(n+)3+2δ(n−)2−12δ(n−)3非因果，稳定课后答案网www.khdaw.com 5.842(b)X(z)=8−+−1−11−5.0z1+5.0znnz>5.0x(n）=8δ(n)−)5.0(4u(n)+(2−)5.0u(n)因果稳定nnz<5.0x(n）=8δ(n)+)5.0(4u(−n−)1−(2−)5.0u(−n−)1非因果非稳定411(e)X(z)=++−1−1−11−z1+5.0z1−5.0znnROC:z>1x(n）=4u(n)+(−)5.0u(n)+)5.0(u(n)因果临界稳定课后答案网www.khdaw.com (c)<法一>：15.05.0设W(z)==+−1−1−11−.064z1−8.0z1+8.0znnz>8.0w(n）=5.0⋅)8.0(u(n)+5.0⋅(−)8.0u(n)nnz<8.0w(n）=−5.0⋅)8.0(u(−n−)1−5.0⋅(−)8.0u(−n−)1z>8.0时nnn−5x(n)=3⋅)8.0(u(n)+3⋅(−)8.0u(n)+5.0⋅)8.0(u(n−)5n−5+5.0⋅(−)8.0u(n−)5z<8.0时nnx(n)=−3⋅)8.0(u(−n−)1−3⋅(−)8.0u(−n−)1n−5n−5−5.0⋅)8.0(u(−n+)4−5.0⋅(−)8.0u(−n+)4课后答案网www.khdaw.com <法二>−1z−3−1+6zz.0642X(z)=−++2−1.064.0641−.064z−3−1zzAA12=−+++2−1−1.064.0641+8.0z1−8.0zA≈.453;A≈.14712这种方法计算太麻烦，不可取。课后答案网www.khdaw.com 第六章6.1−1−21+z−6z)1(H(z)=−1−21−9.0z+2.0z−105136)2(H(z)=−30++−1−11−5.0z1−4.0znnh(n)=−30δ(n)−105⋅)5.0(u(n)+136⋅)4.0(u(n)−jω−2jω1+e−6e)3(H(ω)=−jω−2jω1−9.0e+2.0e)4(零点：z=−,3z=2极点：z=,5.0z=4.0课后答案网www.khdaw.com (5)见黑板课后答案网www.khdaw.com (6)directformx(n)y(n)+−1z−1z0.9wv11−1-0.2z−−11zz-6wv22⎧y=9.0v1−2.0v2+x+w1−6w2⎪⎨v2=v1;v1=y⎪w=w;w=x⎩211课后答案网www.khdaw.com (7)canonicalformwx(n)+0+y(n)0.9−1zw1-0.2z−1-6w2⎧w0=x+9.0w1−2.0w2⎪⎪y=w0+w1−6w2⎨w=w⎪21⎪w=w⎩10课后答案网www.khdaw.com 6.5(e)−18−jω81−8.0(z)1−8.0(e)H(z)=H(ω)=−1−jω1+8.0z1+8.0e2nπj极点：z=−,8.0z=0零点：z=8.0e8n=0L7iH(ω)的图课后答案网www.khdaw.com 6.12111X(z)=;Y(z)=+1−11−1−11−7.0z1−7.0z1−5.0z−1Y(z)1−7.0z1∴H(z)==1+−1X(z)1−5.0z11又有Y(z)=2−11−5.0zY2(z)111n∴X(z)==⋅;x(n)=)6.0(u(n)2−12H(z)21−6.0z2nn−1h(n)=δ(n)+)5.0(u(n)−)5.0(7.0u(n−)1课后答案网www.khdaw.com 6.132πfπ2π∆fπ0)1(ω==;∆ω==0f10f125ss∆ω⎧a1=−2Rcosω0=.18782∆ω=1(2−R)⇒R=1−≈.09874⇒⎨22a=R=.0975⎩2neff∆ωnlnε)2(R=ε⇒1(−)eff=ε⇒n=eff2∆ωln(1−)2∆ω∆ω又Q当∆ω很小时ln(1−)=−[ln(1+x)=x]222lnε∴n=−eff∆ω课后答案网www.khdaw.com −2)3(ε=10−2−2ln10approximaten:n==366.4678effeffπ125−2ln10exactn:n==364.0545effeffln(.09874)课后答案网www.khdaw.com 6.21Nyquistinterval,0[:240)进入到Nyquistinterval的频率成分为：f=ifi=3,2,1,0i1−41−z∴H(z)=ρ小于近似等于14−41−ρz课后答案网www.khdaw.com 6.22(a)pole/zeropattern(a(a)pole/zeropattern)pole/zeropattern(b)magnituderesponse(b(b)magnituderesponse)magnituderesponse课后答案网www.khdaw.com (c)−16−16H(z)−azH(z)=1−z⇒h(n)=ah(n−16)+δ(n)−δ(n−16)h)0(=1h)1(=0h)2(=0Mh(16)=ah)0(+δ(16)−δ)0(=a−1h(17)=ah)1(=0h(18)=ah)2(=0Mh(32)=ah(16)=a(a−)1h(33)=0M课后答案网www.khdaw.com ⎧1n=0⎪k−1∴h(n)=⎨a(a−)1n=16k(k=1L)⎪⎩0n为其它值课后答案网www.khdaw.com 第七章7.1a)h(n)−h(n−)5=δ(n−)1+2δ(n−)2+3δ(n−)3+4δ(n−)4⎧h)0(=0⎪h)1(=1⎪⎪⎪h)2(=2∴⎨h)3(=3⎪⎪h)4(=4⎪⎪⎩h(n)=h(n−)5n≥5 b)directform:⎧v0=x⎪⎪w0=w5+v1+2v2+3v3+4v4⎨⎪delay(w)5,⎪⎩delay(v)4,canonicalform:⎧w0=w5+x⎪⎨y=w1+2w2+3w3+4w4⎪⎩delay(w)4, c)−1−1−2−3z1(+2z+3z+4z)H(z)=−1−1−2−3−41(−z)(1+z+z+z+z)−1−1−2−3z1+2z+3z+4z=⋅−1−1−2−3−41−z1+z+z+z+z−1zH(z)=1−11−z−1−2−31+2z+3z+4zH(z)=2−1−2−3−41+z+z+z+z ⎧w00=x+w01⎪y=w⎪001⎪⎪w01=w00⎨w=y−w−w−w−w⎪10011121314⎪y=w+2w+3w+4w10111213⎪⎪w=w；w=w；w=w；w=w⎩1413131212111110 7.9−31+z1)H(z)=1−41−.064z−1−1−22)1+z1−z+zH(z)=⋅−2−21−8.0z1+8.0z3)cascaderealization 4)sample-by-samplealgorithm⎧w00=x+8.0w02⎪y=w+w⎪00001⎪⎪w02=w01；w01=w00⎨w=y−8.0w⎪10012⎪y=w−w+w101112⎪⎪w=w；w=w⎩12111110 第八章−1−28.1z+2zH(z)=−31−z8.2h(n)=,3,2,1{−,4−,3,2,1,5−,4−3,2,1,5L}y(n)=y(n−)5+x(n)+2x(n−)1+3x(n−)2−4x(n−)3−5x(n−)4directformrealization⎧v0=x⎪⎪w0=v0+2v1+3v2−4v3−5v4+w5⎨⎪delay(w)5,⎪⎩delay(v)4,课后答案网www.khdaw.com canonicalformrealization⎧w0=w5+x⎪⎨y=w0+2w1+3w2−4w3−5w4⎪⎩delay(w)5,课后答案网www.khdaw.com 8.31b1)令H(z)=1⇒=1⇒b=1+az=−11+a22)NRR=∑hn⎫⎪11+a2n⎬⇒NRR=1(+a)⋅=0(1⇒σ>σ2yvvσ1−av∴噪声被放大b)从图解释noiseisamplified从图中可以看出滤波器在ω=π处幅度为1，而在其它频率处幅度大于1。信号频谱集中在ω=π处，所以信号可以无失真地通过，但噪声为白噪声，频谱在ω∈,0[π]恒值分布，那么位于ω=π处以外的白噪声就被放大了。课后答案网www.khdaw.com 8.32N−1−n设HighpassFIRfilter的传递函数为H(z)=∑hnzn=0N−1N−2nN−1n令H(－)1=∑(−)1hn=1⇒hN−1=(−)11[−∑(−)1hn]n=0n=0N−1N−2N−2N−22222n2∴NRR=∑hn=∑hn+hN−1=∑hn+1[−∑(−)1hn]n=0n=0n=0n=0要NRR最小，则应有N−2∂NRRnii+N−1=2hi−1[2−∑(−)1hn]⋅(−)1=2hi−(2−)1hN−1=0∂hin=0对上式分以下几种情况讨论：课后答案网www.khdaw.com i为偶数：h=hiN−1)1N为奇数i为奇数：h=−hiN−1⎧1h=i为偶数N−1in1⎪⎪N∴∑(−)1hn=1⇒N⋅hN−1=1⇒hN−1=⇒⎨n=0N⎪1h=−i为奇数⎪i⎩Ni为偶数：h=−hiN−1)2N为偶数i为奇数：h=hiN−1⎧1h=i为偶数N−1in1⎪⎪N∴∑(−)1hn=1⇒−N⋅hN−1=1⇒hN−1=−⇒⎨n=0N⎪1h=−i为奇数⎪i⎩Nn1∴h=(−)1n=1,0LN−1nN课后答案网www.khdaw.com 第九章9.2f=10KHz；f=80KHz0sffssk=f⇒k=f=8N−k=64−8=5600NN9.31f=5KHz；f=40KHz；T=T⋅16=⋅16=2.3mes0sL周f0TL1）L==T⋅f=2.3×40＝128LsT采Tf周s或==(8说明一个周期有8个采样)⇒L=8×16＝128Tf采0f02）N=L=128；k=⋅N=16；N−k=128−16=112fs Tf周s3）设L个采样含c个周期的采样⇒L=N=c⋅=c⋅Tf采0fff00sk=⋅N=⋅c⋅=c=整数fffss0∴ThenumberofperiodscontainedintheNsamplesisthesameastheDFTindexatwhichyougetapeak.9.9∆f=1209−941=268min3ff8×10ss∆f≥⇒L≥==29.85∴L=30minL∆f268minffsshammingwindow:∆f≥2⇒L≥2⋅=597.∴L=60minL∆fmin 9.4f8s)1频域采样间隔为∆f===5.0binN16±f=∆f⋅m=5.0⋅m=±18⇒m=±36mbink=mmod(16)=4和122）先求进入Nyquist间隔的f=2和6af=∆f⋅k⇒k=4和12abin 9.12)1directDFT⎡1⎤⎢⎥2⎢⎥⎡11111111⎤⎢2⎥⎡12⎤⎢⎥⎢⎥⎢⎥1−j−1j1−j−1j10X=⎢⎥⎢⎥=⎢⎥⎢1−11−11−11−1⎥⎢2⎥⎢0⎥⎢⎥⎢⎥⎢⎥⎣1j−1−j1j−1−j⎦⎢1⎥⎣0⎦⎢1⎥⎢⎥⎢⎣2⎥⎦ )2wrappedDFT⎡1111⎤⎡3⎤⎡12⎤⎢⎥⎢⎥⎢⎥1−j−1j30X=⎢⎥⎢⎥=⎢⎥⎢1−11−1⎥⎢3⎥⎢0⎥⎢⎥⎢⎥⎢⎥⎣1j−1−j⎦⎣3⎦⎣0⎦)3用DFT求IDFT∗∗⎧⎡1111⎤⎡12⎤⎫⎡3⎤⎪⎢⎥⎢⎥⎪⎢⎥~⎪1⎢1−j−1j⎥⎢0⎥⎪⎢3⎥x=⎨⋅⋅⎬=⎪4⎢1−11−1⎥⎢0⎥⎪⎢3⎥⎢⎥⎢⎥⎢⎥⎪⎩⎣1j−1−j⎦⎣0⎦⎪⎭⎣3⎦ 9.21)1L=16,N=16152π−jnkX=∑x(n)e16k=1,0L,15kn=0)2L=16,N=8152π−jnlXˆ=∑x(n)e8l=1,0L7,ln=0⎧Xˆ=X00⎪⎪Xˆ=X12∴⎨⎪M⎪ˆX=X⎩714 9.22x(t)=cos(24πt)+2sin(12πt)cos(8πt)=cos(24πt)+sin(20πt)+sin(4πt)∴f=12k,f=10k,f=2k,f=8k⇒f=4k,f=2k,f=2k123s1a2a3a)1x(t)=cos(8πt)+2sin(4πt)a1j24πt1−j24πt1j20πt1−j20πt1j4πt1−j4πt)2x(t)=e+e+e−e+e−e222j2j2j2jj24πn−j24πnj20πn−j20πn1111x(nT)=x(n)=e8+e8+e8−e8222j2jj4πn−j4πn11+e8−e82j2jjπnj3πnjπnj3πnx(n)=1ejπn+1ejπn+1e2−1e2+1e2−1e2222j2j2j2j jπnj3πnx(n)=ejπn−je2+je2式)1(QL=,8N=872π7π~1jnk1jnk∴x(n)=x(n)=∑X(k)⋅e8=∑X(k)⋅e4式)2(8k=08k=0式)1(和式)2(比较可得：X(k)=,0,0[−8j8,0,8,0,j]0, 9.13)1碟形图见黑板π2）x(n)=4cos(n)+cos(πn)2ππjn−jn11=2e2+2e2+ejπn+e−jπn22ππjn−jn112j2πn2jπnj2πn−jπn=2e+2e⋅e+e+e⋅e22π3πjnjn=2e2+2e2+ejπn8−pointFFT计算的x(n)的频谱为0[01608016]0所以8−pointFFT计算的x(n)的频谱与x(n)的真实频谱相比谱线的位置一样，幅度是8倍的关系 π3）x(n)=4cos(n)+cos(πn)2π3πω=,ω=,ω=πn1n2n3222πω=kk=0L7k8∴在ω，ω，ω处看到x(n)的频谱kk=2kk=6kk=4 9.14∗)1X=0[44j404−4j]4碟形图见黑板x=2[10−1−210−]12）L=,8N=872π7π~1jnk1jnk∴x(n)=x(n)=∑X(k)⋅e8=∑X(k)⋅e48k=08k=0jπnj2πnj3πnj5πnj6πnj7πn1∴x(n)=（4e4−4je4+4e4+4e4+4je4+4e4）8jπnjπnj3πnj3πnjπnjπn1−−−=（4e4−4je2+4e4+4e4+4je2+4e4）123142431231231231238)1()2()3()3()2()1(1πππ=8[cos(πn)+8sin(πn)+8cos(πn)]8422πππ=cos(πn)−sin(πn)+cos(πn)422 9.18)1x=0[2345]1或其它)2x=0[23451]0或x=0[03451]2或其它本题考查两点基本概念：1）一序列先进行mod－Nreduction后再作N－pointDFT与其直接作N－pointDFT结果相等。2）不同序列有可能有相同的mod－Nreduction结果 9.301）y=x∗h=4[47−18711−2421−10−1−]3~y=8[68−2868−]2)2碟形图见黑板 9.411）overlap−savemethodQ要做length−8circularconvolutions∴每段长度取8Qh=1[−1−11],阶数为3∴每段间重叠3，即M=3⎧x1=0[001111]3⎪x=1[133331]1⎪2⎪∴⎨x3=3[111222]2⎪x=2[221111]0⎪4⎪x=1[100000]0⎩5~~y=[−32310−102],y=2[020−20−20],12~~y=1[−2−1210−10],y=2[1−2−1010−1],34~y=1[0−20100]05y=1[0−1020−20−20210−10−1010−10]1y=x∗h=1[0−1020−20−20210−10−1010−10]1 )2overlap−addmethodQ要做length−8circularconvolutions∴每段卷积输出长度L≤8yi即N+M≤(8N为每子段x的长度，M为滤波器阶数）∴取N=5x=1[1113],x=3[3311],x=1[222]2123x=1[111]04~y=y=1[0−102−3−23],11~y=y=3[0−3−20101],22~y=y=1[1−1−10−202],33~y=y=1[0−10−101]044以上y到y重叠相加得14y=1[0−1020−20−20210−10−1010−10]1 第十章10.1+∞−jωkπjωkdω式(10)1.1.D(ω)=∑d(k)e⇔d(k)=∫D(ω)e−π2πk=−∞)1(d(k)isevenandreald(k)为实数+∞k=−l+∞d(k)为偶函数+∞∗jωk−jωl−jωlD(ω)=∑d(k)e=∑d(−l)e=∑d(l)e=D(ω)k=−∞l=−∞l=−∞∴D(ω)isreal+∞k=−l+∞d(k)为偶函数+∞jωk−jωl−jωlD(−ω)=∑d(k)e=∑d(−l)e=∑d(l)e=D(ω)k=−∞l=−∞l=−∞∴D(ω)iseven )2(d(k)isoddandreald(k)为实数+∞k=−l+∞d(k)为奇函数+∞∗jωk−jωl−jωlD(ω)=∑d(k)e=∑d(−l)e=−∑d(l)ek=−∞l=−∞l=−∞=−D(ω)∴D(ω)isimaginary+∞k=−l+∞d(k)为奇函数+∞jωk−jωl−jωlD(−ω)=∑d(k)e=∑d(−l)e=−∑d(l)e=−D(ω)k=−∞l=−∞l=−∞∴D(ω)isodd)3(d(k)isonlyreal−valuebutnotsymmetryd(k)为实数+∞+∞∗jωk−j(−ω)kD(ω)=∑d(k)e=∑d(k)e=D(−ω)k=−∞k=−∞ 10.18截止频率为ω的理想低通滤波器的单位脉冲响应为：csin(ωk)cd(k)=−∞1，N≥NexactΩpassΩstop2NApass/10∴A(Ω)=10lg[1+()⋅(10−1)]stopΩpassΩstop2NexactApass/10>10lg[1+()⋅(10−1)]=AstopΩpass∴A(Ω)>AstopstopΩΩstopstop同理得Ω=，Ω=010exact1Astop/10−2NAstop/102Nexact(10)1(10−)1Ωpass2NexactAstop/1010lg[1+()⋅(10−1)]=ApassΩstop ΩpassΩ()<1，N≥Nexactpass2NAstop/10ΩstopA(Ω)=10lg[1+()⋅(10−1)]⎯⎯⎯⎯⎯⎯⎯→passΩstopΩpass2NexactAstop/10A(Ω)<10lg[1+()⋅(10−1)]=AstoppassΩstop∴A(Ω)