信号与系统第十一次作业参考答案

※ 第一题

利用三种逆变方法求下列X(z)X\left( z \right)X(z)的逆变换x[n]x\left[ n \right]x[n]。
X(z)=10z(z−1)(z−2),(∣z∣>2)X\left( z \right) = {{10z} \over {\left( {z - 1} \right)\left( {z - 2} \right)}},\,\,\,\,\left( {\left| z \right| > 2} \right)X(z)=(z−1)(z−2)10z,(∣z∣>2)

■ 求解：

方法1：围线积分方法：
因为∣z∣>2\left| z \right| > 2∣z∣>2，所以信号为右边序列。
x[n]=12πj∮CX(z)⋅zn−1dz=12πj∮C10zn(z−1)(z−2)dzx\left[ n \right] = {1 \over {2\pi j}}\oint\limits_C {X\left( z \right) \cdot z^{n - 1} dz = {1 \over {2\pi j}}\oint\limits_C {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}dz} }x[n]=2πj1C∮X(z)⋅zn−1dz=2πj1C∮(z−1)(z−2)10zndz

x[n]=∑mRes[10zn(z−1)(z−2)]z=zmx\left[ n \right] = \sum\limits_m^{} {{\mathop{\rm Re}\nolimits} s\left[ {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]} _{z = z_m }x[n]=m∑Res[(z−1)(z−2)10zn]z=zm

Res[10zn(z−1)(z−2)]z=1=−10{\mathop{\rm Re}\nolimits} s\left[ {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]_{z = 1} = - 10Res[(z−1)(z−2)10zn]z=1=−10Res[10zn(z−1)(z−2)]z=2=10⋅2n{\mathop{\rm Re}\nolimits} s\left[ {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]_{z = 2} = 10 \cdot 2^nRes[(z−1)(z−2)10zn]z=2=10⋅2n

x[n]=[−10+10⋅2n]⋅u[n]x\left[ n \right] = \left[ { - 10 + 10 \cdot 2^n } \right] \cdot u\left[ n \right]x[n]=[−10+10⋅2n]⋅u[n]

方法2：长除法：

x[n]=10z−1+30z−2+70z−3+⋯x\left[ n \right] = 10z^{ - 1} + 30z^{ - 2} + 70z^{ - 3} + \cdotsx[n]=10z−1+30z−2+70z−3+⋯=10(2n−1)⋅u[n]= 10\left( {2^n - 1} \right) \cdot u\left[ n \right]=10(2n−1)⋅u[n]

>> deconv([10,0,0,0,0,0,0,0,0],[1,-3,2])'
ans=10 30 70 150 310 630 1270

方法3：部分因式分解法：
X(z)z=10(z−1)(z−2)=−10z−1+10z−2{{X\left( z \right)} \over z} = {{10} \over {\left( {z - 1} \right)\left( {z - 2} \right)}} = {{ - 10} \over {z - 1}} + {{10} \over {z - 2}}zX(z)=(z−1)(z−2)10=z−1−10+z−210X(z)=−10zz−1+10zz−2X\left( z \right) = {{ - 10z} \over {z - 1}} + {{10z} \over {z - 2}}X(z)=z−1−10z+z−210z

x[n]=−10⋅u[n]+10⋅2nu[n]x\left[ n \right] = - 10 \cdot u\left[ n \right] + 10 \cdot 2^n u\left[ n \right]x[n]=−10⋅u[n]+10⋅2nu[n]

使用MATLAB对应的变换命令：

>>iztrans(10*z/(z-1)/(z-2))
ans=10*2^n-10

※ 第二题

求下列X(z)X\left( z \right)X(z)的逆变换x[n]x\left[ n \right]x[n]：

（1）X(z)=10(1−0.5z−1)(1−0.25z−1)(∣z∣>0.5)X\left( z \right) = {{10} \over {\left( {1 - 0.5z^{ - 1} } \right)\left( {1 - 0.25z^{ - 1} } \right)}}\,\,\,\,\,\,\,\,\left( {\left| z \right| > 0.5} \right)\;\;\;\;\;X(z)=(1−0.5z−1)(1−0.25z−1)10(∣z∣>0.5)

（2）X(z)=10z2(z−1)(z+1)(∣z∣>1)X\left( z \right) = {{10z^2 } \over {\left( {z - 1} \right)\left( {z + 1} \right)}}\,\,\,\,\,\,\,\,\,\,\left( {\left| z \right| > 1} \right)\;\;\;\;\;X(z)=(z−1)(z+1)10z2(∣z∣>1)
（3）X(z)=1+z−11−2z−1cos⁡ω+z−2(∣z∣>1)X\left( z \right) = {{1 + z^{ - 1} } \over {1 - 2z^{ - 1} \cos \omega + z^{ - 2} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left| z \right| > 1} \right)\;\;\;\;\;X(z)=1−2z−1cosω+z−21+z−1(∣z∣>1)

■ 求解：

（1）求解：
X(z)z=10z(z−0.5)(z−0.25)=20z−0.5+−10z−0.25{{X\left( z \right)} \over z} = {{10z} \over {\left( {z - 0.5} \right)\left( {z - 0.25} \right)}} = {{20} \over {z - 0.5}} + {{ - 10} \over {z - 0.25}}zX(z)=(z−0.5)(z−0.25)10z=z−0.520+z−0.25−10X(z)=20zz−0.5+−10zz−0.25X\left( z \right) = {{20z} \over {z - 0.5}} + {{ - 10z} \over {z - 0.25}}X(z)=z−0.520z+z−0.25−10z∣z∣>0.5\left| z \right| > 0.5∣z∣>0.5x[n]=[20⋅(0.5)n−10⋅(0.25)n]⋅u[n]x\left[ n \right] = \left[ {20 \cdot \left( {0.5} \right)^n - 10 \cdot \left( {0.25} \right)^n } \right] \cdot u\left[ n \right]x[n]=[20⋅(0.5)n−10⋅(0.25)n]⋅u[n]

>>iztrans(10/(1-0.5/z)/(1-0.25/z))'
ans=20*(1/2)^n-10*(1/4)^n

（2）求解：
X(z)z=10z(z−1)(z+1)=5z−1+5z+1{{X\left( z \right)} \over z} = {{10z} \over {\left( {z - 1} \right)\left( {z + 1} \right)}} = {5 \over {z - 1}} + {5 \over {z + 1}}zX(z)=(z−1)(z+1)10z=z−15+z+15X(z)=5zz−1+5zz+1X\left( z \right) = {{5z} \over {z - 1}} + {{5z} \over {z + 1}}X(z)=z−15z+z+15z∣z∣>1\left| z \right| > 1∣z∣>1x[n]=5⋅[1+(−1)n]⋅u[n]x\left[ n \right] = 5 \cdot \left[ {1 + \left( { - 1} \right)^n } \right] \cdot u\left[ n \right]x[n]=5⋅[1+(−1)n]⋅u[n]

>>iztrans(10*z*z/(z-1)/(z+1))'
ans=5*(-1)^n+5

（3）求解： 根据正弦、余弦单边序列的z变换：
Z[cos⁡(ω0n)u[n]]=z(z−cos⁡ω0)z2−2zcos⁡ω0+1Z\left[ {\cos \left( {\omega _0 n} \right)u\left[ n \right]} \right] = {{z\left( {z - \cos \omega _0 } \right)} \over {z^2 - 2z\cos \omega _0 + 1}}Z[cos(ω0n)u[n]]=z2−2zcosω0+1z(z−cosω0)Z[sin⁡(ω0n)u[n]]=zsin⁡ω0z2−2zcos⁡ω0+1Z\left[ {\sin \left( {\omega _0 n} \right)u\left[ n \right]} \right] = {{z\sin \omega _0 } \over {z^2 - 2z\cos \omega _0 + 1}}Z[sin(ω0n)u[n]]=z2−2zcosω0+1zsinω0

X(z)=z2+zz2−2zcos⁡ω+1X\left( z \right) = {{z^2 + z} \over {z^2 - 2z\cos \omega + 1}}X(z)=z2−2zcosω+1z2+z=z(z−cos⁡ω)z2−2zcos⁡ω+1+1+cos⁡ωsin⁡ωzsin⁡ωz2−2zcos⁡ω+1= {{z\left( {z - \cos \omega } \right)} \over {z^2 - 2z\cos \omega + 1}} + {{1 + \cos \omega } \over {\sin \omega }}{{z\sin \omega } \over {z^2 - 2z\cos \omega + 1}}=z2−2zcosω+1z(z−cosω)+sinω1+cosωz2−2zcosω+1zsinω
x[n]=(cos⁡ωn+1+cos⁡ωsin⁡ωsin⁡ωn)⋅u[n]x\left[ n \right] = \left( {\cos \omega n + {{1 + \cos \omega } \over {\sin \omega }}\sin \omega n} \right) \cdot u\left[ n \right]x[n]=(cosωn+sinω1+cosωsinωn)⋅u[n]=sin⁡(nω)+sin⁡(n+1)ωsin⁡ω⋅u[n]= {{\sin \left( {n\omega } \right) + \sin \left( {n + 1} \right)\omega } \over {\sin \omega }} \cdot u\left[ n \right]=sinωsin(nω)+sin(n+1)ω⋅u[n]

>>iztrans((1+1/z)/(1-2*z^-1*cos(w)+z^-2))'
ans=(sin(n*w)*(cos(w)+1))/sin(w)-(cos(n*w)*(cos(w)+1))/cos(w)+(cos(n*w)*(2*cos(w)+1))/cos(w)

>>iztrans((z*z+z)/(z*z-2*z*cos(w)+1))'
ans=(sin(n*w)*(cos(w)+1))/sin(w)-(cos(n*w)*(cos(w)+1))/cos(w)+(cos(n*w)*(2*cos(w)+1))/cos(w)

sin⁡nω(cos⁡ω+1)sin⁡ω−cos⁡nω(cos⁡ω+1)cos⁡ω+cos⁡nω(2cos⁡+1)cos⁡ω{{\sin n\omega \left( {\cos \omega + 1} \right)} \over {\sin \omega }} - {{\cos n\omega \left( {\cos \omega + 1} \right)} \over {\cos \omega }} + {{\cos n\omega \left( {2\cos + 1} \right)} \over {\cos \omega }}sinωsinnω(cosω+1)−cosωcosnω(cosω+1)+cosωcosnω(2cos+1)
(sin⁡nω⋅cos⁡ω−sin⁡ω⋅cos⁡nω)⋅(cos⁡ω+1)+sin⁡ω⋅cos⁡nω⋅(2cos⁡ω+1)sin⁡ω⋅cos⁡ω{{\left( {\sin n\omega \cdot \cos \omega - \sin \omega \cdot \cos n\omega } \right) \cdot \left( {\cos \omega + 1} \right) + \sin \omega \cdot \cos n\omega \cdot (2\cos \omega + 1)} \over {\sin \omega \cdot \cos \omega }}sinω⋅cosω(sinnω⋅cosω−sinω⋅cosnω)⋅(cosω+1)+sinω⋅cosnω⋅(2cosω+1)

※ 第三题

求下面两个函数的拉普拉斯变换：

■ 求解：

（a）求解：
f1(t)=(1−t)[u(t)−u(t−1)]f_1 \left( t \right) = \left( {1 - t} \right)\left[ {u\left( t \right) - u\left( {t - 1} \right)} \right]f1(t)=(1−t)[u(t)−u(t−1)]F1(s)=∫01(1−t)⋅e−stdt=e−s+s−1s2F_1 \left( s \right) = \int_0^1 {\left( {1 - t} \right) \cdot e^{ - st} dt} = {{e^{ - s} + s - 1} \over {s^2 }}F1(s)=∫01(1−t)⋅e−stdt=s2e−s+s−1F(s)=F1(s)1−e−s=e−s+s−1(1−e−s)⋅s2=−1s2+1s(1−e−s)F\left( s \right) = {{F_1 \left( s \right)} \over {1 - e^{ - s} }} = {{e^{ - s} + s - 1} \over {\left( {1 - e^{ - s} } \right) \cdot s^2 }} = - {1 \over {s^2 }} + {1 \over {s\left( {1 - e^{ - s} } \right)}}F(s)=1−e−sF1(s)=(1−e−s)⋅s2e−s+s−1=−s21+s(1−e−s)1

（b）求解：
f1(t)=δ(t)−δ(t−12)f_1 \left( t \right) = \delta \left( t \right) - \delta \left( {t - {1 \over 2}} \right)f1(t)=δ(t)−δ(t−21)F1(s)=∫01f1(t)e−stdt=1−e−12sF_1 \left( s \right) = \int_0^1 {f_1 \left( t \right)e^{ - st} dt} = 1 - e^{ - {1 \over 2}s}F1(s)=∫01f1(t)e−stdt=1−e−21sF(s)=F1(s)1−e−s=1−e−12s1−e−s=11+e−s2F\left( s \right) = {{F_1 \left( s \right)} \over {1 - e^{ - s} }} = {{1 - e^{ - {1 \over 2}s} } \over {1 - e^{ - s} }} = {1 \over {1 + e^{ - {s \over 2}} }}F(s)=1−e−sF1(s)=1−e−s1−e−21s=1+e−2s1

>>laplace((1-t)*(heaviside(t)-heaviside(t-1)))'
ans=(exp(-s)*(s*exp(s)-exp(s)+1))/s^2

※ 第四题

已知下列X(s)X\left( s \right)X(s)，求各自的拉普拉斯反变换的初值和终值：
（1）s+3(s+4)(s+5){{s + 3} \over {\left( {s + 4} \right)\left( {s + 5} \right)}}(s+4)(s+5)s+3
（2）2s2+2s+3(s+1)(s2+4){{2s^2 + 2s + 3} \over {\left( {s + 1} \right)\left( {s^2 + 4} \right)}}(s+1)(s2+4)2s2+2s+3
（3）s+4(s+1)2(s+2){{s + 4} \over {\left( {s + 1} \right)^2 \left( {s + 2} \right)}}(s+1)2(s+2)s+4
（4）e−ss2(s−2)4{{e^{ - s} } \over {s^2 \left( {s - 2} \right)^4 }}s2(s−2)4e−s

■ 求解：

（1）求解：x(0+)=1,x(∞)=0x\left( {0_ + } \right) = 1,\,\,\,x\left( \infty \right) = 0x(0+)=1,x(∞)=0

>>ilaplace((s+3)/(s+4)/(s+5))'
ans=2*exp(-5*t)-exp(-4*t)

（2）求解： x(0+)=0,x(∞)=0x\left( {0_ + } \right) = 0,\,\,\,x\left( \infty \right) = 0x(0+)=0,x(∞)=0

>>ilaplace((s+4)/((s+1)^2*(s+2)))'
ans= 2*exp(-2*t)-2*exp(-t)+3*t*exp(-t)

（3）求解： x(0+)=2,x(∞)=x\left( {0_ + } \right) = 2,\,\,\,\,x\left( \infty \right) =x(0+)=2,x(∞)=不存在

>>ilaplace((2*s*s+2*s+3)/((s+1)*(s*s+4)))'
ans=(7*cos(2*t))/5 +(3*exp(-t))/5 +(3*sin(2*t))/10

（4）求解： x(0+)=0,x(∞)=x\left( {0_ + } \right) = 0,\,\,\,\,x\left( \infty \right) =x(0+)=0,x(∞)=不存在

>>ilaplace(exp(-s)/((s*s*(s-2).^4)))'
ans=heaviside(t-1)*(t/16 -exp(2*t-2)/8 -(exp(2*t-2)*(t-1)^2)/8 +(exp(2*t-2)*(t-1)^3)/24 +(3*exp(2*t-2)*(t-1))/16 +1/16)

※ 第五题

求下列函数的拉普拉斯变换：
（1） (1−e−at)⋅u(t)\left( {1 - e^{ - at} } \right) \cdot u\left( t \right)(1−e−at)⋅u(t)
（2） (sin⁡t+2cos⁡t)⋅u(t)\left( {\sin t + 2\cos t} \right) \cdot u\left( t \right)(sint+2cost)⋅u(t)
（3） t⋅e−2t⋅u(t)t \cdot e^{ - 2t} \cdot u\left( t \right)t⋅e−2t⋅u(t)
（4） e−tu(t−2)e^{ - t} u\left( {t - 2} \right)e−tu(t−2)

■ 求解：

这里给出了完整版本的答案：

（1）解答： L[1−e−at]=L[1]−L[e−at]=1s−1s+aL\left[ {1 - e^{ - at} } \right] = L\left[ 1 \right] - L\left[ {e^{ - at} } \right] = {1 \over s} - {1 \over {s + a}}L[1−e−at]=L[1]−L[e−at]=s1−s+a1

>>laplace(1-exp(-a*t))'
ans=1/s-1/(a+s)

（2） 解答：L[sin⁡t+2cos⁡t]=2ss2+1+1s2+1=2s+1s2+1L\left[ {\sin t + 2\cos t} \right] = {{2s} \over {s^2 + 1}} + {1 \over {s^2 + 1}} = {{2s + 1} \over {s^2 + 1}}L[sint+2cost]=s2+12s+s2+11=s2+12s+1

>>laplace(sin(t)+2*cos(t))'
ans=(2*s)/(s^2 +1)+1/(s^2 +1)

（3）解答： L[t⋅e−2t]=−ddsL[e−2t]=−dds(1s+2)=1(s+2)2L\left[ {t \cdot e^{ - 2t} } \right] = - {d \over {ds}}L\left[ {e^{ - 2t} } \right] = - {d \over {ds}}\left( {{1 \over {s + 2}}} \right) = {1 \over {\left( {s + 2} \right)^2 }}L[t⋅e−2t]=−dsdL[e−2t]=−dsd(s+21)=(s+2)21

>>laplace(t*exp(-2*t))'
ans=1/(s+2)^2

（4）解答：
L[e−t⋅u(t−2)]=e−2L[e−(t−2)⋅u(t−2)]=e−2⋅e−2s⋅1s+1L\left[ {e^{ - t} \cdot u\left( {t - 2} \right)} \right] = e^{ - 2} L\left[ {e^{ - \left( {t - 2} \right)} \cdot u\left( {t - 2} \right)} \right] = e^{ - 2} \cdot e^{ - 2s} \cdot {1 \over {s + 1}}L[e−t⋅u(t−2)]=e−2L[e−(t−2)⋅u(t−2)]=e−2⋅e−2s⋅s+11

>>laplace(exp(-t)*heaviside(t-2))'
ans=(exp(-2*s)*exp(-2))/(s+1)

（5）解答： L[e−t⋅sin⁡(2t)]=L[sin⁡(2t)]∣s→s+1=2(s2+4)∣s→s+1=2(s+1)2+4L\left[ {e^{ - t} \cdot \sin \left( {2t} \right)} \right] = \left. {L\left[ {\sin \left( {2t} \right)} \right]} \right|_{s \to s + 1} = \left. {{2 \over {\left( {s^2 + 4} \right)}}} \right|_{s \to s + 1} = {2 \over {\left( {s + 1} \right)^2 + 4}}L[e−t⋅sin(2t)]=L[sin(2t)]∣s→s+1=(s2+4)2∣∣∣∣s→s+1=(s+1)2+42

>>laplace(exp(-t)*sin(2*t))'
ans=2/((s+1)^2 +4)

（6）解答： L[(1+2t)⋅e−t]=L[1+2t]∣s→s+1=(1s+2s2)∣s→s+1=1s+1+2(s+1)2L\left[ {\left( {1 + 2t} \right) \cdot e^{ - t} } \right] = \left. {L\left[ {1 + 2t} \right]} \right|_{s \to s + 1} = \left. {\left( {{1 \over s} + {2 \over {s^2 }}} \right)} \right|_{s \to s + 1} = {1 \over {s + 1}} + {2 \over {\left( {s + 1} \right)^2 }}L[(1+2t)⋅e−t]=L[1+2t]∣s→s+1=(s1+s22)∣∣∣∣s→s+1=s+11+(s+1)22

>>laplace((1+2*t)*exp(-t))'
ans=(2*(s/2 +3/2))/(s+1)^2

（7）解答： L[cos⁡(at)]=ss2+a2L\left[ {\cos \left( {at} \right)} \right] = {s \over {s^2 + a^2 }}L[cos(at)]=s2+a2s
L{[1−cos⁡(at)]⋅e−bt}=L[1−cos⁡(at)]∣s→s+b=1s+b−s+b(s+b)2+a2L\left\{ {\left[ {1 - \cos \left( {at} \right)} \right] \cdot e^{ - bt} } \right\} = \left. {L\left[ {1 - \cos \left( {at} \right)} \right]} \right|_{s \to s + b} = {1 \over {s + b}} - {{s + b} \over {\left( {s + b} \right)^2 + a^2 }}L{[1−cos(at)]⋅e−bt}=L[1−cos(at)]∣s→s+b=s+b1−(s+b)2+a2s+b

>>laplace(cos(a*t))'
ans=s/(a^2 +s^2)

>>laplace((1-cos(a*t))*exp(-b*t))'
ans=1/(b+s)-(b+s)/((b+s)^2 +a^2)

（8）解答： L[t2+2t]=2s2+2s3L\left[ {t^2 + 2t} \right] = {2 \over {s^2 }} + {2 \over {s^3 }}L[t2+2t]=s22+s32

>>laplace(t*t+2*t)'
ans=2/s^2 +2/s^3

（9）解答： L[2δ(t)−3e−7t]=2−3s+7L\left[ {2\delta \left( t \right) - 3e^{ - 7t} } \right] = 2 - {3 \over {s + 7}}L[2δ(t)−3e−7t]=2−s+73

>>laplace(2*dirac(t)-3*exp(-7*t))'
ans=2 -3/(s+7)

（10）解答： L[e−at⋅sin⁡(bt)]=b(s+a)2+b2L\left[ {e^{ - at} \cdot \sin \left( {bt} \right)} \right] = {b \over {\left( {s + a} \right)^2 + b^2 }}L[e−at⋅sin(bt)]=(s+a)2+b2b

>>laplace(exp(-a*t)*sin(b*t))'
ans=b/((a+s)^2 +b^2)

（11）解答：cos⁡2(Ωt)=12[1+cos⁡(2Ωt)]\cos ^2 \left( {\Omega t} \right) = {1 \over 2}\left[ {1 + \cos \left( {2\Omega t} \right)} \right]cos2(Ωt)=21[1+cos(2Ωt)]

L[cos⁡2(Ωt)]=L[12(1+cos⁡(2Ωt))]=s2+2Ω2s(s2+4Ω2)L\left[ {\cos ^2 \left( {\Omega t} \right)} \right] = L\left[ {{1 \over 2}\left( {1 + \cos \left( {2\Omega t} \right)} \right)} \right] = {{s^2 + 2\Omega ^2 } \over {s\left( {s^2 + 4\Omega ^2 } \right)}}L[cos2(Ωt)]=L[21(1+cos(2Ωt))]=s(s2+4Ω2)s2+2Ω2

>>laplace(cos(w*t).^2)'
ans=(s^2 +2*w^2)/(s*(s^2 +4*w^2))

（12）解答： L[e−(t+a)⋅cos⁡(ωt)]=e−a{L[e−t⋅cos⁡(ωt)]}=e−a⋅L[cos⁡(ωt)]∣s→s+1=e−a⋅(s+1)(s+1)2+ω2L\left[ {e^{ - \left( {t + a} \right)} \cdot \cos \left( {\omega t} \right)} \right] = e^{ - a} \left\{ {L\left[ {e^{ - t} \cdot \cos \left( {\omega t} \right)} \right]} \right\} = e^{ - a} \cdot \left. {L\left[ {\cos \left( {\omega t} \right)} \right]} \right|_{s \to s + 1} = {{e^{ - a} \cdot \left( {s + 1} \right)} \over {\left( {s + 1} \right)^2 + \omega ^2 }}L[e−(t+a)⋅cos(ωt)]=e−a{L[e−t⋅cos(ωt)]}=e−a⋅L[cos(ωt)]∣s→s+1=(s+1)2+ω2e−a⋅(s+1)

>>laplace(exp(-(t+a))*cos(w*t))'
ans=(s+1)/(exp(a)*s^2 +2*exp(a)*s+exp(a)*w^2 +exp(a))

（13）解答： L[1b−a(e−at−ebt)]=1s+a−1s−bb−a=a+ba−bs2+(a−b)s−abL\left[ {{1 \over {b - a}}\left( {e^{ - at} - e^{bt} } \right)} \right] = {{{1 \over {s + a}} - {1 \over {s - b}}} \over {b - a}} = {{{{a + b} \over {a - b}}} \over {s^2 + \left( {a - b} \right)s - ab}}L[b−a1(e−at−ebt)]=b−as+a1−s−b1=s2+(a−b)s−aba−ba+b

>>laplace((exp(-a*t)-exp(b*t))/(b-a))'
ans=-(1/(a+s)+1/(b-s))/(a-b)

（14）解答： L[1b−a(e−at−e−bt)]=1s+a−1s+bb−a=1(s+a)(s+b)L\left[ {{1 \over {b - a}}\left( {e^{ - at} - e^{ - bt} } \right)} \right] = {{{1 \over {s + a}} - {1 \over {s + b}}} \over {b - a}} = {1 \over {\left( {s + a} \right)\left( {s + b} \right)}}L[b−a1(e−at−e−bt)]=b−as+a1−s+b1=(s+a)(s+b)1

>>laplace((exp(-a*t)-exp(-b*t))/(b-a))'
ans=-(1/(a+s)-1/(b+s))/(a-b)

※ 第六题

已知因果序列x[n]x\left[ n \right]x[n]的z变换X(z)X\left( z \right)X(z)，求序列的初值x[n]x\left[ n \right]x[n]与终值x[∞]x\left[ \infty \right]x[∞]。

（1）X(z)=1+z−1+z−2(1−z−1)⋅(1−2z−1)X\left( z \right) = {{1 + z^{ - 1} + z^{ - 2} } \over {\left( {1 - z^{ - 1} } \right) \cdot \left( {1 - 2z^{ - 1} } \right)}}X(z)=(1−z−1)⋅(1−2z−1)1+z−1+z−2
（2）X(z)=1(1−0.5z−1)(1+0.5z−1)X\left( z \right) = {1 \over {\left( {1 - 0.5z^{ - 1} } \right)\left( {1 + 0.5z^{ - 1} } \right)}}X(z)=(1−0.5z−1)(1+0.5z−1)1
（3）X(z)=z−11−1.5z−1+0.5z−2X\left( z \right) = {{z^{ - 1} } \over {1 - 1.5z^{ - 1} + 0.5z^{ - 2} }}X(z)=1−1.5z−1+0.5z−2z−1
（4）X(z)=z4(z−1)(z−0.5)(z−0.2)X\left( z \right) = {{z^4 } \over {\left( {z - 1} \right)\left( {z - 0.5} \right)\left( {z - 0.2} \right)}}X(z)=(z−1)(z−0.5)(z−0.2)z4

■ 求解：

（1）解答：
X(z)=1+z−1+z−2(1−z−1)⋅(1−2z−1)X\left( z \right) = {{1 + z^{ - 1} + z^{ - 2} } \over {\left( {1 - z^{ - 1} } \right) \cdot \left( {1 - 2z^{ - 1} } \right)}}X(z)=(1−z−1)⋅(1−2z−1)1+z−1+z−2

x[0]=X(∞)=1x\left[ 0 \right] = X\left( \infty \right) = 1x[0]=X(∞)=1
由于存在极点位于2，处于单位圆外，所以x[∞]x\left[ \infty \right]x[∞]不存在。

>>iztrans((1+1/z+1/z/z)/((1-1/z)*(1-2/z)))'
ans=(7*2^n)/2 +kroneckerDelta(n,0)/2 -3

（2）解答：
X(z)=1(1−0.5z−1)(1+0.5z−1)X\left( z \right) = {1 \over {\left( {1 - 0.5z^{ - 1} } \right)\left( {1 + 0.5z^{ - 1} } \right)}}X(z)=(1−0.5z−1)(1+0.5z−1)1

x[0]=X(∞)=1x\left[ 0 \right] = X\left( \infty \right) = 1x[0]=X(∞)=1x[∞]=lim⁡z→1z−1(1−0.5z−1)(1+0.5z−1)=0x\left[ \infty \right] = \mathop {\lim }\limits_{z \to 1} {{z - 1} \over {\left( {1 - 0.5z^{ - 1} } \right)\left( {1 + 0.5z^{ - 1} } \right)}} = 0x[∞]=z→1lim(1−0.5z−1)(1+0.5z−1)z−1=0

>>iztrans(1/((1-0.5/z)*(1+0.5/z)))'
ans=(-1/2)^n/2 +(1/2)^n/2

（3）解答：
X(z)=z−11−1.5z−1+0.5z−2X\left( z \right) = {{z^{ - 1} } \over {1 - 1.5z^{ - 1} + 0.5z^{ - 2} }}X(z)=1−1.5z−1+0.5z−2z−1

x[0]=X(∞)=0x\left[ 0 \right] = X\left( \infty \right) = 0x[0]=X(∞)=0

x[∞]=lim⁡z→1z−1(z−1)1−1.5z−1+0.5z−2=lim⁡z→1zz−0.5=2x\left[ \infty \right] = \mathop {\lim }\limits_{z \to 1} {{z^{ - 1} \left( {z - 1} \right)} \over {1 - 1.5z^{ - 1} + 0.5z^{ - 2} }} = \mathop {\lim }\limits_{z \to 1} {z \over {z - 0.5}} = 2x[∞]=z→1lim1−1.5z−1+0.5z−2z−1(z−1)=z→1limz−0.5z=2

>>iztrans(1/z/(1-1.5/z+0.5/z/z))'
ans=2 -2*(1/2)^n

（4）解答：
X(z)=z4(z−1)(z−0.5)(z−0.2)X\left( z \right) = {{z^4 } \over {\left( {z - 1} \right)\left( {z - 0.5} \right)\left( {z - 0.2} \right)}}X(z)=(z−1)(z−0.5)(z−0.2)z4

X(z)=z4z3−1.7z2+0.8z−0.1=z+1.7z3−0.8z2+0.1zz3X\left( z \right) = {{z^4 } \over {z^3 - 1.7z^2 + 0.8z - 0.1}} = z + {{1.7z^3 - 0.8z^2 + 0.1z} \over {z^3 }}X(z)=z3−1.7z2+0.8z−0.1z4=z+z31.7z3−0.8z2+0.1z

x[0]=lim⁡z→∞1.7z3−0.8z2+0.1zz3=1.7x\left[ 0 \right] = \mathop {\lim }\limits_{z \to \infty } {{1.7z^3 - 0.8z^2 + 0.1z} \over {z^3 }} = 1.7x[0]=z→∞limz31.7z3−0.8z2+0.1z=1.7

x[∞]=lim⁡z→1z4(z−1)(z−1)(z−0.5)(z−0.2)=10.5∗0.8=2.5x\left[ \infty \right] = \mathop {\lim }\limits_{z \to 1} {{z^4 \left( {z - 1} \right)} \over {\left( {z - 1} \right)\left( {z - 0.5} \right)\left( {z - 0.2} \right)}} = {1 \over {0.5*0.8}} = 2.5x[∞]=z→1lim(z−1)(z−0.5)(z−0.2)z4(z−1)=0.5∗0.81=2.5

>>iztrans(z^4/((z-1)*(z-0.5)*(z-0.2)))'
ans=(1/5)^n/30 -(5*(1/2)^n)/6 +iztrans(z,z,n)+5/2

※ 第七题

利用z变换的性质求以下序列的z变换，表明收敛域：

（1） x1[n]=(−2)nn⋅u[n]x_1 \left[ n \right] = \left( { - 2} \right)^n n \cdot u\left[ n \right]x1[n]=(−2)nn⋅u[n]

（2） x2[n]=(n−1)⋅u[n]x_2 \left[ n \right] = \left( {n - 1} \right)^{} \cdot u\left[ n \right]x2[n]=(n−1)⋅u[n]

（3） x3[n]=nan−1⋅u[n]x_3 \left[ n \right] = na^{n - 1} \cdot u\left[ n \right]x3[n]=nan−1⋅u[n]

（4） x4[n]=2n⋅∑k=0∞(−2)k⋅u[n−k]x_4 \left[ n \right] = 2^n \cdot \sum\limits_{k = 0}^\infty {\left( { - 2} \right)^k \cdot u\left[ {n - k} \right]}x4[n]=2n⋅k=0∑∞(−2)k⋅u[n−k]

（5） x5[n]=ann+2⋅u[n+1]x_5 \left[ n \right] = {{a^n } \over {n + 2}} \cdot u\left[ {n + 1} \right]x5[n]=n+2an⋅u[n+1]

（6） x6[n]=(13)n.cos⁡(nπ2)⋅u[n]x_6 \left[ n \right] = \left( {{1 \over 3}} \right)^n .\cos \left( {{{n\pi } \over 2}} \right) \cdot u\left[ n \right]x6[n]=(31)n.cos(2nπ)⋅u[n]

■ 求解：

（1）求解：
x1[n]=(−2)nn⋅u[n]x_1 \left[ n \right] = \left( { - 2} \right)^n n \cdot u\left[ n \right]x1[n]=(−2)nn⋅u[n]

Z{(−2)n}=zz+2Z\left\{ {\left( { - 2} \right)^n } \right\} = {z \over {z + 2}}Z{(−2)n}=z+2z

Z{x[n]⋅n}=−zddzX(z)Z\left\{ {x\left[ n \right] \cdot n} \right\} = - z{d \over {dz}}X\left( z \right)Z{x[n]⋅n}=−zdzdX(z)

Z{(−2)n⋅n}=−zddzzz+2=−2z(z+2)2Z\left\{ {\left( { - 2} \right)^n \cdot n} \right\} = - z{d \over {dz}}{z \over {z + 2}} = {{ - 2z} \over {\left( {z + 2} \right)^2 }}Z{(−2)n⋅n}=−zdzdz+2z=(z+2)2−2z

>> ztrans((-2)^n*n)'
ans = -(2*z)/(z + 2)^2

（2）求解：
x2[n]=(n−1)⋅u[n]x_2 \left[ n \right] = \left( {n - 1} \right)^{} \cdot u\left[ n \right]x2[n]=(n−1)⋅u[n]

Z{n⋅u[n]}=−zddzzz−1=z(z−1)2Z\left\{ {n \cdot u\left[ n \right]} \right\} = - z{d \over {dz}}{z \over {z - 1}} = {z \over {\left( {z - 1} \right)^2 }}Z{n⋅u[n]}=−zdzdz−1z=(z−1)2z

Z{(n−1)⋅u[n]}=z(z−1)2−zz−1=−z2+2z(z−1)2Z\left\{ {\left( {n - 1} \right) \cdot u\left[ n \right]} \right\} = {z \over {\left( {z - 1} \right)^2 }} - {z \over {z - 1}} = {{ - z^2 + 2z} \over {\left( {z - 1} \right)^2 }}Z{(n−1)⋅u[n]}=(z−1)2z−z−1z=(z−1)2−z2+2z

>> ztrans((n-1))'
ans = z/(z - 1)^2 - z/(z - 1)

（3）求解：
x3[n]=nan−1⋅u[n]x_3 \left[ n \right] = na^{n - 1} \cdot u\left[ n \right]x3[n]=nan−1⋅u[n]

Z{an⋅u[n]}=zz−aZ\left\{ {a^n \cdot u\left[ n \right]} \right\} = {z \over {z - a}}Z{an⋅u[n]}=z−az

Z{n⋅an⋅u[n]}=−zddzzz−a=az(z−a)2Z\left\{ {n \cdot a^n \cdot u\left[ n \right]} \right\} = - z{d \over {dz}}{z \over {z - a}} = {{az} \over {\left( {z - a} \right)^2 }}Z{n⋅an⋅u[n]}=−zdzdz−az=(z−a)2az

Z{n⋅an−1⋅u[n]}=ddzzz−a=z(z−a)2Z\left\{ {n \cdot a^{n - 1} \cdot u\left[ n \right]} \right\} = {d \over {dz}}{z \over {z - a}} = {z \over {\left( {z - a} \right)^2 }}Z{n⋅an−1⋅u[n]}=dzdz−az=(z−a)2z

>> ztrans(n*a^(n-1))'
ans = z/(a - z)^2

（4）求解：
x4[n]=2n⋅∑k=0∞(−2)k⋅u[n−k]x_4 \left[ n \right] = 2^n \cdot \sum\limits_{k = 0}^\infty {\left( { - 2} \right)^k \cdot u\left[ {n - k} \right]}x4[n]=2n⋅k=0∑∞(−2)k⋅u[n−k]

Z{(−2)n⋅u[n]}=zz+2Z\left\{ {\left( { - 2} \right)^n \cdot u\left[ n \right]} \right\} = {z \over {z + 2}}Z{(−2)n⋅u[n]}=z+2z

Z{u[n]}=zz−1Z\left\{ {u\left[ n \right]} \right\} = {z \over {z - 1}}Z{u[n]}=z−1z

Z{(−2)ku[n]∗u[n]}=zz+2⋅zz−1=z2(z+2)(z−1)Z\left\{ {\left( { - 2} \right)^k u\left[ n \right] * u\left[ n \right]} \right\} = {z \over {z + 2}} \cdot {z \over {z - 1}} = {{z^2 } \over {\left( {z + 2} \right)\left( {z - 1} \right)}}Z{(−2)ku[n]∗u[n]}=z+2z⋅z−1z=(z+2)(z−1)z2

Z{2n⋅(−2)ku[n]∗u[n]}=(z2)2(z2+2)(z2−1)=z2(z+4)(z−2)Z\left\{ {2^n \cdot \left( { - 2} \right)^k u\left[ n \right] * u\left[ n \right]} \right\} = {{\left( {{z \over 2}} \right)^2 } \over {\left( {{z \over 2} + 2} \right)\left( {{z \over 2} - 1} \right)}} = {{z^2 } \over {\left( {z + 4} \right)\left( {z - 2} \right)}}Z{2n⋅(−2)ku[n]∗u[n]}=(2z+2)(2z−1)(2z)2=(z+4)(z−2)z2

（5）求解：
x5[n]=ann+2⋅u[n+1]x_5 \left[ n \right] = {{a^n } \over {n + 2}} \cdot u\left[ {n + 1} \right]x5[n]=n+2an⋅u[n+1]

Z{1n+1⋅u[n]}=z⋅ln⁡zz−1Z\left\{ {{1 \over {n + 1}} \cdot u\left[ n \right]} \right\} = z \cdot \ln {z \over {z - 1}}Z{n+11⋅u[n]}=z⋅lnz−1z

Z{1n+2⋅u[n+1]}=z2ln⁡zz−1Z\left\{ {{1 \over {n + 2}} \cdot u\left[ {n + 1} \right]} \right\} = z^2 \ln {z \over {z - 1}}Z{n+21⋅u[n+1]}=z2lnz−1z

Z{ann+2⋅u[n+1]}=(za)2ln⁡zaza−1=z2a2ln⁡zz−aZ\left\{ {{{a^n } \over {n + 2}} \cdot u\left[ {n + 1} \right]} \right\} = \left( {{z \over a}} \right)^2 \ln {{{z \over a}} \over {{z \over a} - 1}} = {{z^2 } \over {a^2 }}\ln {z \over {z - a}}Z{n+2an⋅u[n+1]}=(az)2lnaz−1az=a2z2lnz−az

Z{ann+2⋅u[n+1]}=Z{ann+2⋅u[n]+a−1δ[−1]}=−z2a2[ln⁡(z−az)+az]−za=−z2a2ln⁡(z−az)=z2a2ln⁡(zz−a)Z\left\{ {{{a^n } \over {n + 2}} \cdot u\left[ {n + 1} \right]} \right\} = Z\left\{ {{{a^n } \over {n + 2}} \cdot u\left[ n \right] + a^{ - 1} \delta \left[ { - 1} \right]} \right\} = {{ - z^2 } \over {a^2 }}\left[ {\ln \left( {{{z - a} \over z}} \right) + {a \over z}} \right] - {z \over a}\,\, = {{ - z^2 } \over {a^2 }}\ln \left( {{{z - a} \over z}} \right) = {{z^2 } \over {a^2 }}\ln \left( {{z \over {z - a}}} \right)Z{n+2an⋅u[n+1]}=Z{n+2an⋅u[n]+a−1δ[−1]}=a2−z2[ln(zz−a)+za]−az=a2−z2ln(zz−a)=a2z2ln(z−az)

（6）求解：
x6[n]=(13)n.cos⁡(nπ2)⋅u[n]x_6 \left[ n \right] = \left( {{1 \over 3}} \right)^n .\cos \left( {{{n\pi } \over 2}} \right) \cdot u\left[ n \right]x6[n]=(31)n.cos(2nπ)⋅u[n]

Z{cos⁡(nω)}=z(z−cos⁡ω)z2−2cos⁡ω⋅z+1Z\left\{ {\cos \left( {n\omega } \right)} \right\} = {{z\left( {z - \cos \omega } \right)} \over {z^2 - 2\cos \omega \cdot z + 1}}Z{cos(nω)}=z2−2cosω⋅z+1z(z−cosω)

Z{cos⁡(nπ2)}=z2z2+1Z\left\{ {\cos \left( {{{n\pi } \over 2}} \right)} \right\} = {{z^2 } \over {z^2 + 1}}Z{cos(2nπ)}=z2+1z2

Z{(13)ncos⁡(nπ2)}=(3z)2(3z)2+1=9z29z2+1Z\left\{ {\left( {{1 \over 3}} \right)^n \cos \left( {{{n\pi } \over 2}} \right)} \right\} = {{\left( {3z} \right)^2 } \over {\left( {3z} \right)^2 + 1}} = {{9z^2 } \over {9z^2 + 1}}Z{(31)ncos(2nπ)}=(3z)2+1(3z)2=9z2+19z2

>>ztrans((1/3).^n*cos(n*pi/2))'
ans=(9*z^2)/(9*z^2 +1)

※ 第八题

利用z变换的性质求以下序列的卷积，已知：

（1） x[n]=an−1⋅u[n−1],h[n]=u[n]x\left[ n \right] = a^{n - 1} \cdot u\left[ {n - 1} \right],\,\,h\left[ n \right] = u\left[ n \right]x[n]=an−1⋅u[n−1],h[n]=u[n]

（2） x[n]=2u[n−1],h[n]=∑k=0∞(−1)kδ[n−k]x\left[ n \right] = 2u\left[ {n - 1} \right],\,\,\,h\left[ n \right] = \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \delta \left[ {n - k} \right]}x[n]=2u[n−1],h[n]=k=0∑∞(−1)kδ[n−k]

■ 求解：

（1）求解：
序列x[n],h[n]x\left[ n \right],h\left[ n \right]x[n],h[n]的z变换分别是：X(z)=zz−a⋅z−1=1z−a,H(z)=zz−1X\left( z \right) = {z \over {z - a}} \cdot z^{ - 1} = {1 \over {z - a}},\,\,\,\,H\left( z \right) = {z \over {z - 1}}X(z)=z−az⋅z−1=z−a1,H(z)=z−1z

根据z变换的卷积定理，y[n]=x[n]∗h[z]y\left[ n \right] = x\left[ n \right] * h\left[ z \right]y[n]=x[n]∗h[z]，那么序列y[n]y\left[ n \right]y[n]的z变换为：Y(z)=H(z)⋅X(z)=1z−a⋅zz−1=1a−1zz−a+11−azz−1Y\left( z \right) = H\left( z \right) \cdot X\left( z \right)\, = {1 \over {z - a}} \cdot {z \over {z - 1}} = {{{1 \over {a - 1}}z} \over {z - a}} + {{{1 \over {1 - a}}z} \over {z - 1}}Y(z)=H(z)⋅X(z)=z−a1⋅z−1z=z−aa−11z+z−11−a1z
则：y[n]=11−au[n]−11−aan⋅u[n]y\left[ n \right] = {1 \over {1 - a}}u\left[ n \right] - {1 \over {1 - a}}a^n \cdot u\left[ n \right]y[n]=1−a1u[n]−1−a1an⋅u[n]

（2）求解：
序列x[n],h[n]x\left[ n \right],h\left[ n \right]x[n],h[n]的z变换分别是：X(z)=2z−1X\left( z \right) = {2 \over {z - 1}}X(z)=z−12

h[n]=∑k=0∞(−1)kδ[n−k]=(−1)n⋅u[n]h\left[ n \right] = \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \delta \left[ {n - k} \right]} = \left( { - 1} \right)^n \cdot u\left[ n \right]h[n]=k=0∑∞(−1)kδ[n−k]=(−1)n⋅u[n]

H(z)=zz+1H\left( z \right) = {z \over {z + 1}}H(z)=z+1z

那么，根据z变换的卷积定理，y[n]=x[n]∗h[n]y\left[ n \right] = x\left[ n \right] * h\left[ n \right]y[n]=x[n]∗h[n]对应的z变换为：
Y(z)=X(z)⋅H(z)=2z−1⋅zz+1=zz−1+−zz+1Y\left( z \right) = X\left( z \right) \cdot H\left( z \right) = {2 \over {z - 1}} \cdot {z \over {z + 1}} = {z \over {z - 1}} + {{ - z} \over {z + 1}}Y(z)=X(z)⋅H(z)=z−12⋅z+1z=z−1z+z+1−z

则：y[n]=u[n]−(−1)n⋅u[n]y\left[ n \right] = u\left[ n \right] - \left( { - 1} \right)^n \cdot u\left[ n \right]y[n]=u[n]−(−1)n⋅u[n]

※ 第九题

已知X(z)X\left( z \right)X(z)和H(z)H\left( z \right)H(z)如下面所示，利用z域的卷积定理求Z{x[n]⋅h[n]}Z\left\{ {x\left[ n \right] \cdot h\left[ n \right]} \right\}Z{x[n]⋅h[n]}。

（1） X(z)=11−12z−1,∣z∣>0.5;H(z)=11−2z,∣z∣<0.5X\left( z \right) = {1 \over {1 - {1 \over 2}z^{ - 1} }},\,\,\left| z \right| > 0.5;\,\,\,\,H\left( z \right) = {1 \over {1 - 2z}},\,\,\left| z \right| < 0.5X(z)=1−21z−11,∣z∣>0.5;H(z)=1−2z1,∣z∣<0.5
（2）X(z)=zz−e−b,∣z∣>e−b;H(z)=z⋅sin⁡ω0z2−2zcos⁡ω0+1,∣z∣>1X\left( z \right) = {z \over {z - e^{ - b} }},\,\,\left| z \right| > e^{ - b} ;\;\;\;H\left( z \right) = {{z \cdot \sin \omega _0 } \over {z^2 - 2z\cos \omega _0 + 1}},\,\,\left| z \right| > 1X(z)=z−e−bz,∣z∣>e−b;H(z)=z2−2zcosω0+1z⋅sinω0,∣z∣>1

■ 求解：

（1）解答：
Z{x[n]⋅h[n]}=12πj⋅∮CX(v)⋅H(zv)⋅v−1dvZ\left\{ {x\left[ n \right] \cdot h\left[ n \right]} \right\} = {1 \over {2\pi j}} \cdot \oint_C {X\left( v \right) \cdot H\left( {{z \over v}} \right) \cdot v^{ - 1} dv}Z{x[n]⋅h[n]}=2πj1⋅∮CX(v)⋅H(vz)⋅v−1dv=12πj⋅∮C11−12v−1⋅11−2zvv−1dv= {1 \over {2\pi j}} \cdot \oint_C {{1 \over {1 - {1 \over 2}v^{ - 1} }} \cdot {1 \over {1 - 2{z \over v}}}v^{ - 1} dv}=2πj1⋅∮C1−21v−11⋅1−2vz1v−1dv=12πj⋅∮Cv(v−12)(v−2z)dv= {1 \over {2\pi j}} \cdot \oint_C {{v \over {\left( {v - {1 \over 2}} \right)\left( {v - 2z} \right)}}dv}=2πj1⋅∮C(v−21)(v−2z)vdv

根据X(z),H(z)X\left( z \right),H\left( z \right)X(z),H(z)的收敛域，可知：∣v∣>0.5,∣zv∣<0.5\left| v \right| > 0.5,\,\,\,\left| {{z \over v}} \right| < 0.5∣v∣>0.5,∣∣vz∣∣<0.5。
所以上述积分函数包含的极点为：12,2z{1 \over 2},\,\,2z21,2z。

Z{x[n]⋅h[n]}=Res[v(v−12)(v−2z)]v=12+Res[v(v−12)(v−2z)]v=2z=1Z\left\{ {x\left[ n \right] \cdot h\left[ n \right]} \right\} = {\mathop{\rm Re}\nolimits} s\left[ {{v \over {\left( {v - {1 \over 2}} \right)\left( {v - 2z} \right)}}} \right]_{v = {1 \over 2}} + \,{\mathop{\rm Re}\nolimits} s\left[ {{v \over {\left( {v - {1 \over 2}} \right)\left( {v - 2z} \right)}}} \right]_{v = 2z} = 1Z{x[n]⋅h[n]}=Res[(v−21)(v−2z)v]v=21+Res[(v−21)(v−2z)v]v=2z=1

所以x[n]⋅h[n]=δ[n]x\left[ n \right] \cdot h\left[ n \right] = \delta \left[ n \right]x[n]⋅h[n]=δ[n]

（2）解答：
Z{x[n]⋅h[n]}=12πj⋅∮CX(v)⋅H(zv)⋅v−1dvZ\left\{ {x\left[ n \right] \cdot h\left[ n \right]} \right\} = {1 \over {2\pi j}} \cdot \oint_C {X\left( v \right) \cdot H\left( {{z \over v}} \right) \cdot v^{ - 1} dv}Z{x[n]⋅h[n]}=2πj1⋅∮CX(v)⋅H(vz)⋅v−1dv=12πj⋅∮Cvv−e−b⋅zv⋅sin⁡ω0(zv)2−2(zv)cos⁡ω0+1⋅v−1dv= {1 \over {2\pi j}} \cdot \oint_C {{v \over {v - e^{ - b} }} \cdot {{{z \over v} \cdot \sin \omega _0 } \over {\left( {{z \over v}} \right)^2 - 2\left( {{z \over v}} \right)\cos \omega _0 + 1}} \cdot v^{ - 1} dv}=2πj1⋅∮Cv−e−bv⋅(vz)2−2(vz)cosω0+1vz⋅sinω0⋅v−1dv=12πj⋅∮C2sin⁡ω0v(v−e−b)(v2−2zcos⁡0v+z2)dv= {1 \over {2\pi j}} \cdot \oint_C {{{2\sin \omega _0 v} \over {\left( {v - e^{ - b} } \right)\left( {v^2 - 2z\cos _0 v + z^2 } \right)}}dv}=2πj1⋅∮C(v−e−b)(v2−2zcos0v+z2)2sinω0vdv

根据X(z),H(z)X\left( z \right),H\left( z \right)X(z),H(z)的收敛域，可得：∣v∣>e−b,∣v∣<∣z∣\left| v \right| > e^{ - b} ,\,\,\left| v \right| < \left| z \right|∣v∣>e−b,∣v∣<∣z∣

所以上述围线积分包含的极点为：e−be^{ - b}e−b。
Z{x[n]⋅h[n]}=Res[2sin⁡ω0v(v−e−b)(v2−2zcos⁡ω0v+z2)]v=e−b=z⋅sin⁡ω0⋅e−be−2b−2e−bcos⁡ω0z+z2Z\left\{ {x\left[ n \right] \cdot h\left[ n \right]} \right\} = {\mathop{\rm Re}\nolimits} s\left[ {{{2\sin \omega _0 v} \over {\left( {v - e^{ - b} } \right)\left( {v^2 - 2z\cos \omega _0 v + z^2 } \right)}}} \right]_{v = e^{ - b} } = {{z \cdot \sin \omega _0 \cdot e^{ - b} } \over {e^{ - 2b} - 2e^{ - b} \cos \omega _0 z + z^2 }}Z{x[n]⋅h[n]}=Res[(v−e−b)(v2−2zcosω0v+z2)2sinω0v]v=e−b=e−2b−2e−bcosω0z+z2z⋅sinω0⋅e−b
因此：
x[n]⋅h[n]=e−nbcos⁡ω0n⋅u[n]x\left[ n \right] \cdot h\left[ n \right] = e^{ - nb} \cos \omega _0 n \cdot u\left[ n \right]x[n]⋅h[n]=e−nbcosω0n⋅u[n]

※ 第十题

已知Z{x[n]}=X(z)Z\left\{ {x\left[ n \right]} \right\} = X\left( z \right)Z{x[n]}=X(z)，试证明：Z[∑k=−∞nx(k)]=zz−1X(z)Z\left[ {\sum\limits_{k = - \infty }^n {x\left( k \right)} } \right] = {z \over {z - 1}}X\left( z \right)\;\;\;\;\;Z[k=−∞∑nx(k)]=z−1zX(z)

■ 求解：

证明方法1：利用卷积定理证明：
由于对序列的累加和，可以看成序列与u[n]u\left[ n \right]u[n]的卷积：∑k=−∞nx[k]=x[n]∗u[n]\sum\limits_{k = - \infty }^n {x\left[ k \right]} = x\left[ n \right] * u\left[ n \right]k=−∞∑nx[k]=x[n]∗u[n]。所以在根据z变换的卷积定理可知，序列的累加和的z变换等于序列的z变换乘以u[n]u\left[ n \right]u[n]的z变换。而：Z{u[n]}=zz−1Z\left\{ {u\left[ n \right]} \right\} = {z \over {z - 1}}Z{u[n]}=z−1z，所以：Z[∑k=−∞nx(k)]=zz−1X(z)Z\left[ {\sum\limits_{k = - \infty }^n {x\left( k \right)} } \right] = {z \over {z - 1}}X\left( z \right)\;\;\;\;\;Z[k=−∞∑nx(k)]=z−1zX(z)

证明方法2：

Z[∑k=−∞nx[k]]=∑n=−∞∞(∑k=−∞nx[k])⋅z−nZ\left[ {\sum\limits_{k = - \infty }^n {x\left[ k \right]} } \right] = \sum\limits_{n = - \infty }^\infty {\left( {\sum\limits_{k = - \infty }^n {x\left[ k \right]} } \right) \cdot z^{ - n} }Z[k=−∞∑nx[k]]=n=−∞∑∞(k=−∞∑nx[k])⋅z−n=∑n=−∞∞(∑k=−∞∞x[k]⋅u[n−k])⋅z−n= \sum\limits_{n = - \infty }^\infty {\left( {\sum\limits_{k = - \infty }^\infty {x\left[ k \right] \cdot u\left[ {n - k} \right]} } \right) \cdot z^{ - n} }=n=−∞∑∞(k=−∞∑∞x[k]⋅u[n−k])⋅z−n=∑k=−∞n∑n=−∞∞x[k]⋅u[n−k]⋅z−n= \sum\limits_{k = - \infty }^n {\sum\limits_{n = - \infty }^\infty {x\left[ k \right] \cdot u\left[ {n - k} \right] \cdot z^{ - n} } }=k=−∞∑nn=−∞∑∞x[k]⋅u[n−k]⋅z−n=∑k=−∞∞x[k]∑n=−∞∞u[n−k]⋅z−(n−k)⋅z−k= \sum\limits_{k = - \infty }^\infty {x\left[ k \right]\sum\limits_{n = - \infty }^\infty {u\left[ {n - k} \right] \cdot z^{ - \left( {n - k} \right)} } \cdot z^{ - k} }=k=−∞∑∞x[k]n=−∞∑∞u[n−k]⋅z−(n−k)⋅z−k=∑k=−∞∞x[k]z−k∑n=−∞∞u[n−k]⋅z−(n−k)=zz−1X(z)= \sum\limits_{k = - \infty }^\infty {x\left[ k \right]z^{ - k} \sum\limits_{n = - \infty }^\infty {u\left[ {n - k} \right] \cdot z^{ - \left( {n - k} \right)} } } = {z \over {z - 1}}X\left( z \right)=k=−∞∑∞x[k]z−kn=−∞∑∞u[n−k]⋅z−(n−k)=z−1zX(z)

※ 第十一题

已知Z{x[n]}=X(z)Z\left\{ {x\left[ n \right]} \right\} = X\left( z \right)Z{x[n]}=X(z)，并且：x1[n]=x[n3],n=3k;x1[n]=0,n=3k+1,3k+2x_1 \left[ n \right] = x\left[ {{n \over 3}} \right],\,\,\,\,n = 3k;\,\,\,\,x_1 \left[ n \right] = 0,\,\,n = 3k + 1,3k + 2x1[n]=x[3n],n=3k;x1[n]=0,n=3k+1,3k+2
x2[n]=x[3n]x_2 \left[ n \right] = x\left[ {3n} \right]x2[n]=x[3n]
求：x1[n],x2[n]x_1 \left[ n \right],\,\,x_2 \left[ n \right]x1[n],x2[n]的z变换。

■ 求解：

（1）解答：

X1(z)=∑n=−∞∞x1[n]z−n=∑k=−∞∞x[k]z−3kX_1 \left( z \right) = \sum\limits_{n = - \infty }^\infty {x_1 \left[ n \right]z^{ - n} } = \sum\limits_{k = - \infty }^\infty {x\left[ k \right]z^{ - 3k} }X1(z)=n=−∞∑∞x1[n]z−n=k=−∞∑∞x[k]z−3k=∑k=−∞∞x[k](z3)−k=X(z3)= \sum\limits_{k = - \infty }^\infty {x\left[ k \right]\left( {z^3 } \right)^{ - k} } = X\left( {z^3 } \right)=k=−∞∑∞x[k](z3)−k=X(z3)

（2）解答：

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2020年春季学习信号与系统课程作业参考答案-第十一次作业

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