作业要求链接： 信号与系统2022春季作业-第十次作业 : https://zhuoqing.blog.csdn.net/article/details/124464837
§10 参考答案
1.1 拉普拉斯反变换
1.1.1 求下列函数的拉普拉斯逆变换
◎ 求解：
（1） L − 1 [ 1 s + 2 ] = e − 2 t , t ≥ 0 L^{ - 1} \left[ {
{1 \over {s + 2}}} \right] = e^{ - 2t} ,\,\,t \ge 0 L−1[s+21]=e−2t,t≥0
（2） L − 1 [ 4 2 s + 5 ] = L − 1 [ 2 s + 5 2 ] = 2 ⋅ e − 5 2 t , t ≥ 0 L^{ - 1} \left[ {
{4 \over {2s + 5}}} \right] = L^{ - 1} \left[ {
{2 \over {s + {5 \over 2}}}} \right] = 2 \cdot e^{ - {5 \over 2}t} ,\,\,t \ge 0 L−1[2s+54]=L−1[s+252]=2⋅e−25t,t≥0
（3） L − 1 [ 2 s ( 2 s + 3 ) ] = L − 1 ( 2 3 ⋅ 1 s − 2 3 ⋅ 1 s + 3 2 ) = 2 3 − 2 3 e − 3 2 t , t ≥ 0 L^{ - 1} \left[ {
{2 \over {s\left( {2s + 3} \right)}}} \right] = L^{ - 1} \left( {
{2 \over 3} \cdot {1 \over s} - {2 \over 3} \cdot {1 \over {s + {3 \over 2}}}} \right) = {2 \over 3} - {2 \over 3}e^{ - {3 \over 2}t} ,\,\,\,t \ge 0 L−1[s(2s+3)2]=L−1(32⋅s1−32⋅s+231)=32−32e−23t,t≥0
（4） L − 1 [ 1 s ( s 2 + 3 ) ] = L − 1 ( 1 3 ⋅ 1 s − 1 3 ⋅ s s 2 + 3 ) = 1 3 − 1 3 cos 3 t , t ≥ 0 L^{ - 1} \left[ {
{1 \over {s\left( {s^2 + 3} \right)}}} \right] = L^{ - 1} \left( {
{1 \over 3} \cdot {1 \over s} - {1 \over 3} \cdot {s \over {s^2 + 3}}} \right) = {1 \over 3} - {1 \over 3}\cos \sqrt 3 t,\,\,\,\,t \ge 0 L−1[s(s2+3)1]=L−1(31⋅s1−31⋅s2+3s)=31−31cos3t,t≥0
（5） L − 1 [ 3 ( s + 4 ) ( s + 3 ) ] = L − 1 ( 3 s + 3 − 4 s + 4 ) = 3 e − 3 t − 3 e − 4 t , t ≥ 0 L^{ - 1} \left[ {
{3 \over {\left( {s + 4} \right)\left( {s + 3} \right)}}} \right] = L^{ - 1} \left( {
{3 \over {s + 3}} - {4 \over {s + 4}}} \right) = 3e^{ - 3t} - 3e^{ - 4t} ,\,\,t \ge 0 L−1[(s+4)(s+3)3]=L−1(s+33−s+44)=3e−3t−3e−4t,t≥0
（6） L − 1 [ 3 s ( s + 4 ) ( s + 3 ) ] = L − 1 ( − 9 s + 3 + 12 s + 4 ) = − 9 e − 3 t + 12 e − 4 t , t ≥ 0 L^{ - 1} \left[ {
{
{3s} \over {\left( {s + 4} \right)\left( {s + 3} \right)}}} \right] = L^{ - 1} \left( {
{
{ - 9} \over {s + 3}} + {
{12} \over {s + 4}}} \right) = - 9e^{ - 3t} + 12e^{ - 4t} ,\,\,t \ge 0 L−1[(s+4)(s+3)3s]=L−1(s+3−9+s+412)=−9e−3t+12e−4t,t≥0
（7） L − 1 ( 1 s 2 + 1 + 1 ) = sin t + δ ( t ) , t ≥ 0 L^{ - 1} \left( {
{1 \over {s^2 + 1}} + 1} \right) = \sin t + \delta \left( t \right),\,\,\,t \ge 0 L−1(s2+11+1)=sint+δ(t),t≥0
（8） L − 1 ( 1 s 2 − 3 s + 2 ) = L − 1 ( − 1 s − 1 + 1 s − 2 ) = − e t + e 2 t , t ≥ 0 L^{ - 1} \left( {
{1 \over {s^2 - 3s + 2}}} \right) = L^{ - 1} \left( {
{
{ - 1} \over {s - 1}} + {1 \over {s - 2}}} \right) = - e^t + e^{2t} ,\,\,t \ge 0 L−1(s2−3s+21)=L−1(s−1−1+s−21)=−et+e2t,t≥0
（9） L − 1 [ 1 s ( R C s + 1 ) ] = L − 1 ( 1 s − 1 s + 1 R C ) = 1 − e − 1 R C t , t ≥ 0 L^{ - 1} \left[ {
{1 \over {s\left( {RCs + 1} \right)}}} \right] = L^{ - 1} \left( {
{1 \over s} - {1 \over {s + {1 \over {RC}}}}} \right) = 1 - e^{ - {1 \over {RC}}t} ,\,\,t \ge 0 L−1[s(RCs+1)1]=L−1(s1−s+RC11)=1−e−RC1t,t≥0
（10） L − 1 [ 1 − R C s s ( 1 + R C s ) ] = L − 1 ( 1 s − 2 s + 1 R C ) = 1 − 2 e − t R C , t ≥ 0 L^{ - 1} \left[ {
{
{1 - RCs} \over {s\left( {1 + RCs} \right)}}} \right] = L^{ - 1} \left( {
{1 \over s} - {2 \over {s + {1 \over {RC}}}}} \right) = 1 - 2e^{ - {t \over {RC}}} ,\,\,t \ge 0 L−1[s(1+RCs)1−RCs]=L−1(s1−s+RC12)=1−2e−RCt,t≥0
（11） 首先对于 ω ( s 2 + ω 2 ) ⋅ 1 ( R C s + 1 ) {\omega \over {\left( {s^2 + \omega ^2 } \right)}} \cdot {1 \over {\left( {RCs + 1} \right)}} (s2+ω2)ω⋅(RCs+1)1 进行因式分解。根据所知道的分母存在的二阶和一阶多项式，可以将因式分解成两个有理分式， ω ( s 2 + ω 2 ) ⋅ 1 ( R C s + 1 ) = m s + n s 2 + ω 2 + l s + 1 R C {\omega \over {\left( {s^2 + \omega ^2 } \right)}} \cdot {1 \over {\left( {RCs + 1} \right)}} = {
{ms + n} \over {s^2 + \omega ^2 }} + {l \over {s + {1 \over {RC}}}} (s2+ω2)ω⋅(RCs+1)1=s2+ω2ms+n+s+RC1l 其中 m , n , l m,n,l m,n,l 是待定系数。 将分解因式合并，使用多项式系数匹配方法，求出 m , n , l m,n,l m,n,l 。 m s 2 + m R C s + n s + n R C + l s 2 + l ω 2 ( s 2 + ω 2 ) ( s + 1 R C ) = ω R C ( s 2 + ω 2 ) ( s + 1 R C ) {
{ms^2 + {m \over {RC}}s + ns + {n \over {RC}} + ls^2 + l\omega ^2 } \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}} = {
{
{\omega \over {RC}}} \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}} (s2+ω2)(s+RC1)ms2+RCms+ns+RCn+ls2+lω2=(s2+ω2)(s+RC1)RCω ( m + l ) s 2 + ( n + m R C ) s + l ω 2 ( s 2 + ω 2 ) ( s + 1 R C ) = ω R C ( s 2 + ω 2 ) ( s + 1 R C ) {
{\left( {m + l} \right)s^2 + \left( {n + {m \over {RC}}} \right)s + l\omega ^2 } \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}} = {
{
{\omega \over {RC}}} \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}} (s2+ω2)(s+RC1)(m+l)s2+(n+RCm)s+lω2=(s2+ω2)(s+RC1)RCω 所以有 { m + l = 0 n + m R C = 0 n R C + l ω 2 = ω R C \left\{ \begin{matrix} {m + l = 0}\\{n + {m \over {RC}} = 0}\\{
{n \over {RC}} + l\omega ^2 = {\omega \over {RC}}}\\\end{matrix} \right. ⎩⎨⎧m+l=0n+RCm=0RCn+lω2=RCω 求解可得 { m = − R C ω 1 + ( R C ω ) 2 n = ω 1 + ( R C ω ) 2 l = R C ω 1 + ( R C ω ) 2 \left\{ \begin{matrix} {m = {
{ - RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}}\\{n = {\omega \over {1 + \left( {RC\omega } \right)^2 }}}\\{l = {
{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}}\\\end{matrix} \right. ⎩⎪⎨⎪⎧m=1+(RCω)2−RCωn=1+(RCω)2ωl=1+(RCω)2RCω
所以 ω s 2 + ω 2 ⋅ 1 R C s + 1 = − R C ω 1 + ( R C ω ) 2 s + ω 1 + ( R C ω ) 2 s 2 + ω 2 + R C ω 1 + ( R C ω ) 2 s + 1 R C {\omega \over {s^2 + \omega ^2 }} \cdot {1 \over {RCs + 1}} = {
{
{
{ - RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}s + {\omega \over {1 + \left( {RC\omega } \right)^2 }}} \over {s^2 + \omega ^2 }} + {
{
{
{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}} \over {s + {1 \over {RC}}}} s2+ω2ω⋅RCs+11=s2+ω21+(RCω)2−RCωs+1+(RCω)2ω+s+RC11+(RCω)2RCω L − 1 ( ω s 2 + ω 2 ⋅ 1 R C s + 1 ) = − R C 1 + ( R C ω ) 2 cos ω t + sin ω t 1 + ( R C ω ) 2 + R C ω 1 + ( R C ω ) 2 e − t R C , t ≥ 0 L^{ - 1} \left( {
{\omega \over {s^2 + \omega ^2 }} \cdot {1 \over {RCs + 1}}} \right) = {
{ - RC} \over {1 + \left( {RC\omega } \right)^2 }}\cos \omega t + {
{\sin \omega t} \over {1 + \left( {RC\omega } \right)^2 }} + {
{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}e^{ - {t \over {RC}}} ,\,\,t \ge 0 L−1(s2+ω2ω⋅RCs+11)=1+(RCω)2−RCcosωt+1+(RCω)2sinωt+1+(RCω)2RCωe−RCt,t≥0
（12） L − 1 ( 4 s + 5 s 2 + 5 s + 6 ) = L − 1 ( − 3 s + 2 + 7 s + 3 ) = − e − 2 t + 7 e − 3 t , t ≥ 0 L^{ - 1} \left( {
{
{4s + 5} \over {s^2 + 5s + 6}}} \right) = L^{ - 1} \left( {
{
{ - 3} \over {s + 2}} + {7 \over {s + 3}}} \right) = - e^{ - 2t} + 7e^{ - 3t} ,\,\,t \ge 0 L−1(s2+5s+64s+5)=L−1(s+2−3+s+37)=−e−2t+7e−3t,t≥0
（13） L − 1 [ 100 ( s + 50 ) s 2 + 201 s + 200 ] = L − 1 [ 4900 199 s + 1 + 15000 199 s + 200 ] = 4900 199 e − t + 15000 199 e − 200 t , t ≥ 0 L^{ - 1} \left[ {
{
{100\left( {s + 50} \right)} \over {s^2 + 201s + 200}}} \right] = L^{ - 1} \left[ {
{
{
{
{4900} \over {199}}} \over {s + 1}} + {
{
{
{15000} \over {199}}} \over {s + 200}}} \right] = {
{4900} \over {199}}e^{ - t} + {
{15000} \over {199}}e^{ - 200t} ,\,\,t \ge 0 L−1[s2+201s+200100(s+50)]=L−1[s+11994900+s+20019915000]=1994900e−t+19915000e−200t,t≥0
（14） 首先对于逆变换表达式进行因式分解： s + 3 ( s + 1 ) 3 ⋅ ( s + 2 ) = A ( s + 1 ) 3 + B ( s + 1 ) 2 + C s + 1 + D s + 2 {
{s + 3} \over {\left( {s + 1} \right)^3 \cdot \left( {s + 2} \right)}} = {A \over {\left( {s + 1} \right)^3 }} + {B \over {\left( {s + 1} \right)^2 }} + {C \over {s + 1}} + {D \over {s + 2}} (s+1)3⋅(s+2)s+3=(s+1)3A+(s+1)2B+s+1C+s+2D D = s + 3 ( s + 1 ) 3 ∣ s = − 2 = − 1 D = \left. {
{
{s + 3} \over {\left( {s + 1} \right)^3 }}} \right|_{s = - 2} = - 1 D=(s+1)3s+3∣∣∣∣∣s=−2=−1 A = s + 3 s + 2 ∣ s = − 1 = 2 A = \left. {
{
{s + 3} \over {s + 2}}} \right|_{s = - 1} = 2 A=s+2s+3∣∣∣∣s=−1=2 B = d d s ( s + 3 s + 2 ) ∣ s = − 1 = 1 s + 2 − s + 3 ( s + 2 ) 2 ∣ s = − 1 = − 1 B = \left. {
{d \over {ds}}\left( {
{
{s + 3} \over {s + 2}}} \right)} \right|_{s = - 1} = {1 \over {s + 2}} - \left. {
{
{s + 3} \over {\left( {s + 2} \right)^2 }}} \right|_{s = - 1} = - 1 B=dsd(s+2s+3)∣∣∣∣s=−1=s+21−(s+2)2s+3∣∣∣∣∣s=−1=−1 C = 1 2 ⋅ d 2 d s 2 ( s + 3 s + 2 ) ∣ s = − 1 = 1 2 2 ( s + 3 ( s + 2 ) 2 − 2 ( s + 2 ) 2 ∣ s = − 1 = 1 C = {1 \over 2} \cdot \left. {
{
{d^2 } \over {ds^2 }}\left( {
{
{s + 3} \over {s + 2}}} \right)} \right|_{s = - 1} = {1 \over 2}{
{2(s + 3} \over {\left( {s + 2} \right)^2 }} - \left. {
{2 \over {\left( {s + 2} \right)^2 }}} \right|_{s = - 1} = 1 C=21⋅ds2d2(s+2s+3)∣∣∣∣s=−1=21(s+2)22(s+3−(s+2)22∣∣∣∣∣s=−1=1 L − 1 [ s + 3 ( s + 1 ) 3 ( s + 2 ) ] = t 2 e − t − t e − t + e − t − e − 2 t , t ≥ 0 L^{ - 1} \left[ {
{
{s + 3} \over {\left( {s + 1} \right)^3 \left( {s + 2} \right)}}} \right] = t^2 e^{ - t} - te^{ - t} + e^{ - t} - e^{ - 2t} ,\,\,t \ge 0 L−1[(s+1)3(s+2)s+3]=t2e−t−te−t+e−t−e−2t,t≥0
◎ 利用MATLAB进行求解
( 1 ) 1 s + 2 \left( 1 \right)\,\,\,{1 \over {s + 2}} (1)s+21
ilaplace(1/(s+2))’
ans=exp(-2*t)
( 2 ) 4 2 s + 5 \left( 2 \right)\,\,\,{4 \over {2s + 5}} (2)2s+54
ilaplace(4/(2s+5))’
ans=2exp(-(5*t)/2)
( 3 ) 2 s ( 2 s + 3 ) \left( 3 \right)\,\,\,{2 \over {s\left( {2s + 3} \right)}} (3)s(2s+3)2
ilaplace(2/(s*(2s+3)))’
ans=2/3-(2exp(-(3*t)/2))/3
( 4 ) 1 s ( s 2 + 3 ) \left( 4 \right)\,\,\,{1 \over {s\left( {s^2 + 3} \right)}} (4)s(s2+3)1
ilaplace(1/(s*(s*s+3)))’
ans=1/3-cos(3^(1/2)*t)/3
( 5 ) 3 ( s + 4 ) ( s + 3 ) \left( 5 \right)\,\,{3 \over {\left( {s + 4} \right)\left( {s + 3} \right)}} (5)(s+4)(s+3)3
ilaplace(3/(s+4)/(s+3))’
ans=3exp(-3t)-3exp(-4t)
( 6 ) 3 s ( s + 4 ) ( s + 3 ) \left( 6 \right)\,\,{ {3s} \over {\left( {s + 4} \right)\left( {s + 3} \right)}} (6)(s+4)(s+3)3s
ilaplace(3s/(s+4)/(s+3))’
ans=12exp(-4t)-9exp(-3*t)
( 7 ) 1 s 2 + 1 + 1 \left( 7 \right)\,\,\,{1 \over {s^2 + 1}} + 1 (7)s2+11+1
ilaplace(1/(s*s+1)+1)’
ans=dirac(t)+sin(t)
( 8 ) 1 s 2 − 3 s + 2 \left( 8 \right)\,\,\,{1 \over {s^2 - 3s + 2}} (8)s2−3s+21
ilaplace(1/(ss-3s+2))’
ans=exp(2*t)-exp(t)
( 9 ) 1 s ( R C s + 1 ) \left( 9 \right)\,\,{1 \over {s\left( {RCs + 1} \right)}} (9)s(RCs+1)1
ilaplace(1/(s*(RCs+1)))’
ans=1-exp(-t/(C*R))
( 10 ) 1 − R C s s ( 1 + R C s ) \left( {10} \right)\,\,\,{ {1 - RCs} \over {s\left( {1 + RCs} \right)}} (10)s(1+RCs)1−RCs
ilaplace(1/(s*(RCs+1)))’
ans=1-exp(-t/(C*R))
( 11 ) ω ( s 2 + ω 2 ) ⋅ 1 ( R C s + 1 ) \left( {11} \right)\,\,{\omega \over {\left( {s^2 + \omega ^2 } \right)}} \cdot {1 \over {\left( {RCs + 1} \right)}} (11)(s2+ω2)ω⋅(RCs+1)1
ilaplace(w/(ss+ww)/(RCs+1))’
ans=(sin(tw)-CRwcos(tw))/(C2*R2w2+1)+(C*R*w*exp(-t/(C*R)))/(C2*R2*w2+1)
( 12 ) 4 s + 5 s 2 + 5 s + 6 \left( {12} \right)\,\,\,\,{ {4s + 5} \over {s^2 + 5s + 6}} (12)s2+5s+64s+5
ilaplace((4s+5)/(ss+5s+6))’
ans=7exp(-3t)-3exp(-2*t)
( 13 ) 100 ( s + 50 ) s 2 + 201 s + 200 \left( {13} \right)\,\,\,{ {100\left( {s + 50} \right)} \over {s^2 + 201s + 200}} (13)s2+201s+200100(s+50)
ilaplace(100*(s+50)/(ss+201s+200))’
ans=(4900exp(-t))/199+(15000exp(-200*t))/199
( 14 ) s + 3 ( s + 1 ) 3 ⋅ ( s + 2 ) \left( {14} \right)\,\,\,{ {s + 3} \over {\left( {s + 1} \right)^3 \cdot \left( {s + 2} \right)}} (14)(s+1)3⋅(s+2)s+3
ilaplace((s+3)/(s+1)^3/(s+2))’
ans=exp(-t)-exp(-2t)-texp(-t)+t^2*exp(-t)
1.1.2 求对数函数拉普拉斯逆变换
◎ 求解： 根据拉普拉冲变换公示 F ( s ) = ∫ 0 − + ∞ f ( t ) e − s t d t F\left( s \right) = \int_{0_ - }^{ + \infty } {f\left( t \right)e^{ - st} dt} F(s)=∫0−+∞f(t)e−stdt 两边同时对 s 求导 d d s F ( s ) = d d s ∫ 0 − + ∞ f ( t ) e − s t d t = ∫ 0 − + ∞ − t ⋅ f ( t ) e − s t d t {d \over {ds}}F\left( s \right) = {d \over {ds}}\int_{0_ - }^{ + \infty } {f\left( t \right)e^{ - st} dt} = \int_{0_ - }^{ + \infty } { - t \cdot f\left( t \right)e^{ - st} dt} dsdF(s)=dsd∫0−+∞f(t)e−stdt=∫0−+∞−t⋅f(t)e−stdt 所以，可以得到 s 域 微分性质 t ⋅ f ( t ) = − L − 1 ( d F ( s ) d s ) t \cdot f\left( t \right) = - L^{ - 1} \left( { { {dF\left( s \right)} \over {ds}}} \right) t⋅f(t)=−L−1(dsdF(s)) 那么 f ( t ) = L − 1 [ ln s s + 9 ] f\left( t \right) = L^{ - 1} \left[ {\ln {s \over {s + 9}}} \right] f(t)=L−1[lns+9s] − t ⋅ f ( t ) = L − 1 [ d d s ln s s + 9 ] = L − 1 ( − 1 s + 9 + 1 s ) = 1 − e − 9 t , t ≥ 0 - t \cdot f\left( t \right) = L^{ - 1} \left[ { {d \over {ds}}\ln {s \over {s + 9}}} \right] = L^{ - 1} \left( { - {1 \over {s + 9}} + {1 \over s}} \right) = 1 - e^{ - 9t} ,\,t \ge 0 −t⋅f(t)=L−1[dsdlns+9s]=L−1(−s+91+s1)=1−e−9t,t≥0 所以 f ( t ) = − 1 t + 1 t ⋅ e − 9 t , t ≥ 0 f\left( t \right) = { { - 1} \over t} + {1 \over t} \cdot e^{ - 9t} ,\,\,t \ge 0 f(t)=t−1+t1⋅e−9t,t≥0
对应MATLAB求解命令：
>> ilaplace(log(s/(s+9))
ans=(exp(-9*t)-1)/t
1.2 Z反变换
1.2.1 求函数对应的序列
（1） x [ n ] = Z − 1 ( 1 ) = δ [ n ] x\left[ n \right] = Z^{ - 1} \left( 1 \right) = \delta \left[ n \right] x[n]=Z−1(1)=δ[n]
（2） x [ n ] = Z − 1 ( z 3 ) = δ [ n + 3 ] x\left[ n \right] = Z^{ - 1} \left( {z^3 } \right) = \delta \left[ {n + 3} \right] x[n]=Z−1(z3)=δ[n+3]
（3） x [ n ] = Z − 1 ( z − 1 ) = δ [ n − 1 ] x\left[ n \right] = Z^{ - 1} \left( {z^{ - 1} } \right) = \delta \left[ {n - 1} \right] x[n]=Z−1(z−1)=δ[n−1]
（4） x [ n ] = Z − 1 ( − 2 z − 2 + 2 z + 1 ) = − 2 δ [ n − 2 ] + δ [ n ] + 2 δ [ n + 1 ] x\left[ n \right] = Z^{ - 1} \left( { - 2z^{ - 2} + 2z + 1} \right) = - 2\delta \left[ {n - 2} \right] + \delta \left[ n \right] + 2\delta \left[ {n + 1} \right] x[n]=Z−1(−2z−2+2z+1)=−2δ[n−2]+δ[n]+2δ[n+1]
（5） 根据收敛域判断序列为右边序列。 x [ n ] = Z − 1 ( 1 1 − a z − 1 ) = a n u [ n ] x\left[ n \right] = Z^{ - 1} \left( {
{1 \over {1 - az^{ - 1} }}} \right) = a^n u\left[ n \right] x[n]=Z−1(1−az−11)=anu[n]
（6） 根据收敛域判断序列为左边序列。 x [ n ] = Z − 1 ( 1 1 − a z − 1 ) = − a n ⋅ u [ − n − 1 ] x\left[ n \right] = Z^{ - 1} \left( {
{1 \over {1 - az^{ - 1} }}} \right) = - a^n \cdot u\left[ { - n - 1} \right] x[n]=Z−1(1−az−11)=−an⋅u[−n−1]
1.2.2 求函数X(z)的逆变换
（1） 根据 z 变换收敛域 ∣ z ∣ > 5 \left| z \right| > 5 ∣z∣>5 判断序列为右边序列，所以 x [ n ] = Z − 1 ( 1 1 + 0.5 z − 1 ) = ( − 0.5 ) n ⋅ u [ n ] x\left[ n \right] = Z^{ - 1} \left( { {1 \over {1 + 0.5z^{ - 1} }}} \right) = \left( { - 0.5} \right)^n \cdot u\left[ n \right] x[n]=Z−1(1+0.5z−11)=(−0.5)n⋅u[n]
iztrans(1/(1+0.5/z))
ans=(-1/2)^n
（2） 根据 z 变换收敛域 ∣ z ∣ > 1 2 \left| z \right| > {1 \over 2} ∣z∣>21 判断序列是右边序列。 1 − 0.5 z − 1 1 + 3 4 z − 1 + 1 8 z − 2 = z ( z − 0.5 ) ( z + 1 4 ) ⋅ ( z + 1 2 ) = − 3 z z + 1 4 + 4 z + 1 2 { {1 - 0.5z^{ - 1} } \over {1 + {3 \over 4}z^{ - 1} + {1 \over 8}z^{ - 2} }} = { {z\left( {z - 0.5} \right)} \over {\left( {z + {1 \over 4}} \right) \cdot \left( {z + {1 \over 2}} \right)}} = { { - 3z} \over {z + {1 \over 4}}} + {4 \over {z + {1 \over 2}}} 1+43z−1+81z−21−0.5z−1=(z+41)⋅(z+21)z(z−0.5)=z+41−3z+z+214 所以 x [ n ] = [ − 3 ( − 1 4 ) n + 4 ( − 1 2 ) n ] ⋅ u [ n ] = ( − 1 4 ) n ( − 3 + 2 n + 2 ) ⋅ u [ n ] x\left[ n \right] = \left[ { - 3\left( { { { - 1} \over 4}} \right)^n + 4\left( { { { - 1} \over 2}} \right)^n } \right] \cdot u\left[ n \right] = \left( { - {1 \over 4}} \right)^n \left( { - 3 + 2^{n + 2} } \right) \cdot u\left[ n \right] x[n]=[−3(4−1)n+4(2−1)n]⋅u[n]=(−41)n(−3+2n+2)⋅u[n]
iztrans((1-0.5/z)/(1+3/4/z+1/8/z/z))
ans=4*(-1/2)n-3*(-1/4)n
（3） X ( z ) = 1 − 1 2 z − 1 1 − 1 4 z − 2 = z z + 1 2 X\left( z \right) = { {1 - {1 \over 2}z^{ - 1} } \over {1 - {1 \over 4}z^{ - 2} }} = {z \over {z + {1 \over 2}}} X(z)=1−41z−21−21z−1=z+21z x [ n ] = ( − 1 2 ) n ⋅ u [ n ] x\left[ n \right] = \left( { - {1 \over 2}} \right)^n \cdot u\left[ n \right] x[n]=(−21)n⋅u[n]
iztrans((1-1/2/z)/(1-1/4/z/z))’
ans=(-1/2)^n
（4） X ( z ) z = z − a z ( 1 − a z ) = − 1 a ( z − a ) z ( z − 1 a ) = − a z + a − 1 a z − 1 a { {X\left( z \right)} \over z} = { {z - a} \over {z\left( {1 - az} \right)}} = { { - {1 \over a}\left( {z - a} \right)} \over {z\left( {z - {1 \over a}} \right)}} = { { - a} \over z} + { {a - {1 \over a}} \over {z - {1 \over a}}} zX(z)=z(1−az)z−a=z(z−a1)−a1(z−a)=z−a+z−a1a−a1 x [ n ] = − a δ [ n ] + ( a − 1 a ) ⋅ ( 1 a ) n ⋅ u [ n ] x\left[ n \right] = - a\delta \left[ n \right] + \left( {a - {1 \over a}} \right) \cdot \left( { {1 \over a}} \right)^n \cdot u\left[ n \right] x[n]=−aδ[n]+(a−a1)⋅(a1)n⋅u[n]
iztrans((1-a/z)/(1/z-a))’
ans=piecewise([a~=0,((a2-1)*((1/a)n-kroneckerDelta(n,0)))/a-kroneckerDelta(n,0)/a])
（5） X ( z ) z = 10 z ( z − 0.5 ) ( z − 0.25 ) = 20 z − 0.5 + − 10 z − 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−0.25−10 X ( z ) = 20 z z − 0.5 + − 10 z z − 0.25 X\left( z \right) = { {20z} \over {z - 0.5}} + { { - 10z} \over {z - 0.25}} X(z)=z−0.520z+z−0.25−10z 根据收敛域判断序列为右边序列。所以 x [ n ] = [ 20 × 0. 5 n − 10 × 0.2 5 n ] ⋅ u [ n ] x\left[ n \right] = \left[ {20 \times 0.5^n - 10 \times 0.25^n } \right] \cdot u\left[ n \right] x[n]=[20×0.5n−10×0.25n]⋅u[n]
iztrans(10/(1-0.5/z)/(1-0.25/z))’
ans=20*(1/2)n-10*(1/4)n
（6） X ( z ) z = 10 z ( z − 1 ) ( z + 1 ) = 5 z − 1 + 5 z + 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+15 X ( z ) = 5 z z − 1 + 5 z z + 1 X\left( z \right) = { {5z} \over {z - 1}} + { {5z} \over {z + 1}} X(z)=z−15z+z+15z 根据序列的收敛域判断序列为右边序列，所以 x [ n ] = 5 × [ 1 + ( − 1 ) n ] ⋅ u [ n ] x\left[ n \right] = 5 \times \left[ {1 + \left( { - 1} \right)^n } \right] \cdot u\left[ n \right] x[n]=5×[1+(−1)n]⋅u[n]
iztrans(10zz/(z-1)/(z+1))’
ans=5*(-1)^n+5
（7） 根据正弦、布线单边序列的 z 变换公示： Z { cos ω 0 n ⋅ u [ n ] } = z ( z − cos ω 0 ) z 2 − 2 z cos ω 0 + 1 Z\left\{ {\cos \omega _0 n \cdot u\left[ n \right]} \right\} = { {z\left( {z - \cos \omega _0 } \right)} \over {z^2 - 2z\cos \omega _0 + 1}} Z{ cosω0n⋅u[n]}=z2−2zcosω0+1z(z−cosω0) Z { sin ω 0 n ⋅ u [ n ] } = z sin ω 0 z 2 − 2 z cos ω 0 + 1 Z\left\{ {\sin \omega _0 n \cdot u\left[ n \right]} \right\} = { {z\sin \omega _0 } \over {z^2 - 2z\cos \omega _0 + 1}} Z{ sinω0n⋅u[n]}=z2−2zcosω0+1zsinω0 将 X(z) 分解成对应的有理分式： X ( z ) = z 2 + z z 2 − 2 z cos ω + 1 = z ( z − cos ω ) z 2 − 2 z cos ω + 1 + 1 + cos ω sin ω ⋅ z sin ω z 2 − 2 z cos ω + 1 X\left( z \right) = { {z^2 + z} \over {z^2 - 2z\cos \omega + 1}} = { {z\left( {z - \cos \omega } \right)} \over {z^2 - 2z\cos \omega + 1}} + { {1 + \cos \omega } \over {\sin \omega }} \cdot { {z\sin \omega } \over {z^2 - 2z\cos \omega + 1}} X(z)=z2−2zcosω+1z2+z=z2−2zcosω+1z(z−cosω)+sinω1+cosω⋅z2−2zcosω+1zsinω 所以 x [ n ] = ( cos ω n + 1 + cos ω sin ω ⋅ sin ω n ) ⋅ u [ n ] = sin n ω + sin ( n + 1 ) ω sin ω u [ n ] x\left[ n \right] = \left( {\cos \omega n + { {1 + \cos \omega } \over {\sin \omega }} \cdot \sin \omega n} \right) \cdot u\left[ n \right] = { {\sin n\omega + \sin \left( {n + 1} \right)\omega } \over {\sin \omega }}u\left[ n \right] x[n]=(cosωn+sinω1+cosω⋅sinωn)⋅u[n]=sinωsinnω+sin(n+1)ωu[n]
iztrans((1+1/z)/(1-2z-1*cos(w)+z-2))’
ans=(sin(nw)(cos(w)+1))/sin(w)-(cos(nw)(cos(w)+1))/cos(w)+(cos(nw)(2cos(w)+1))/cos(w)
iztrans((zz+z)/(zz-2zcos(w)+1))’
ans=(sin(nw)(cos(w)+1))/sin(w)-(cos(nw)(cos(w)+1))/cos(w)+(cos(nw)(2*cos(w)+1))/cos(w)
1.2.3 利用三种方法求解X(z)的逆变换
◎ 求解：
方法1：围线积分方法（留数法）
根据 z 变换的收敛域 ∣ z ∣ > 2 \left| z \right| > 2 ∣z∣>2 可知序列为右边序列。
x [ n ] = 1 2 π j ∮ C X ( z ) ⋅ z n − 1 d z = 1 2 π j ∮ C 10 z n ( z − 1 ) ( z − 2 ) d z x\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 ] = ∑ m R e s [ 10 z n ( z − 1 ) ( z − 2 ) ] z = p m x\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 = p_m } } x[n]=m∑Res[(z−1)(z−2)10zn]z=pm R e s [ 10 z n ( z − 1 ) ( z − 2 ) ] z = 1 = − 10 , R e s [ 10 z n ( z − 1 ) ( z − 2 ) ] z = 2 = 10 ⋅ 2 n {\mathop{\rm Re}\nolimits} s\left[ {
{
{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]_{z = 1} = - 10,\,\,\,{\mathop{\rm Re}\nolimits} s\left[ {
{
{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]_{z = 2} = 10 \cdot 2^n Res[(z−1)(z−2)10zn]z=1=−10,Res[(z−1)(z−2)10zn]z=2=10⋅2n x [ n ] = ( − 10 + 10 × 2 n ) ⋅ u [ n ] x\left[ n \right] = \left( { - 10 + 10 \times 2^n } \right) \cdot u\left[ n \right] x[n]=(−10+10×2n)⋅u[n]
方法2：长除法
由于序列为右边序列，所以按照 z 降幂长除：
▲ 图1.2.1
deconv([10,0,0,0,0,0,0,0,0],[1,-3,2])
asn=10 30 70 150 310 630 1270
方法3：因式分解方法
X ( z ) z = 10 ( z − 1 ) ( x − 2 ) = − 10 z − 1 + 10 z − 2 { {X\left( z \right)} \over z} = { {10} \over {\left( {z - 1} \right)\left( {x - 2} \right)}} = { { - 10} \over {z - 1}} + { {10} \over {z - 2}} zX(z)=(z−1)(x−2)10=z−1−10+z−210 X ( z ) = − 10 z z − 1 + 10 z z − 2 X\left( z \right) = { { - 10z} \over {z - 1}} + { {10z} \over {z - 2}} X(z)=z−1−10z+z−210z x [ n ] = − 10 ⋅ u [ n ] + 10 ⋅ 2 n ⋅ u [ n ] x\left[ n \right] = - 10 \cdot u\left[ n \right] + 10 \cdot 2^n \cdot u\left[ n \right] x[n]=−10⋅u[n]+10⋅2n⋅u[n]
iztrans(10*z/(z-1)/(z-2)
ans=10*2^n-10
1.3 拉普拉斯变换性质
1.3.1 拉普拉斯变换
（1）求函数的拉普拉斯变换
（1） L [ 1 − e − a t ] = L [ 1 ] − L [ e − a t ] = 1 s − 1 s + a L\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） L [ sin t + 2 cos t ] = 2 s s 2 + 1 + 1 s 2 + 1 = 2 s + 1 s 2 + 1 L\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） L [ t ⋅ e − 2 t ] = − d d t L ( e − 2 t ) = − d d s ( 1 s + 2 ) = 1 ( s + 2 ) 2 L\left[ {t \cdot e^{ - 2t} } \right] = - {d \over {dt}}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]=−dtdL(e−2t)=−dsd(s+21)=(s+2)21
>>laplace(t*exp(-2*t))
ans=1/(s+2)^2
（4） L [ e − t ⋅ u ( t − 2 ) ] = e − 2 ⋅ L [ e − ( t − 2 ) ⋅ u ( t − 2 ) ] = e − 2 ⋅ e − 2 s ⋅ 1 s + 1 = e − 2 ( 1 + s ) s + 1 L\left[ {e^{ - t} \cdot u\left( {t - 2} \right)} \right] = e^{ - 2} \cdot L\left[ {e^{ - \left( {t - 2} \right)} \cdot u\left( {t - 2} \right)} \right] = e^{ - 2} \cdot e^{ - 2s} \cdot {1 \over {s + 1}} = { {e^{ - 2\left( {1 + s} \right)} } \over {s + 1}} L[e−t⋅u(t−2)]=e−2⋅L[e−(t−2)⋅u(t−2)]=e−2⋅e−2s⋅s+11=s+1e−2(1+s)
>>laplace(exp(-t)*heaviside(t-2))
ans=(exp(-2*s)*exp(-2))/(s+1)
（2）求周期信号拉普拉斯变换
◎ 求解
（1） 周期三角脉冲信号：
f 1 ( 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)] F 1 ( s ) = ∫ 0 1 ( 1 − t ) ⋅ e − s t d t = e − s + s − 1 s 2 F_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−1
F ( s ) = F 1 ( s ) 1 − e − s = e − s + s − 1 ( 1 − e − s ) ⋅ s 2 = − 1 s 2 + 1 s ( 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
>>laplace((1-t)*(heaviside(t)-heaviside(t-1)))'
ans=(exp(-s)*(s*exp(s)-exp(s)+1))/s^2
（2） 周期脉冲序列信号：
f 1 ( t ) = δ ( t ) − δ ( t − 1 2 ) f_1 \left( t \right) = \delta \left( t \right) - \delta \left( {t - {1 \over 2}} \right) f1(t)=δ(t)−δ(t−21) F 1 ( s ) = ∫ 0 1 f 1 ( t ) e − s t d t = 1 − e − 1 2 s F_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−21s F ( s ) = F 1 ( s ) 1 − e − s = 1 − e − 1 2 s 1 − e − s = 1 1 + e − s 2 F\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
1.3.2 求信号的初值和终值
◎ 求解：
（1） x ( 0 + ) = 1 , x ( ∞ ) = 0 x\left( {0_ + } \right) = 1,\,\,x\left( \infty \right) = 0 x(0+)=1,x(∞)=0
>>ilaplace((s+3)/(s+4)/(s+5))'
ans=2*exp(-5*t)-exp(-4*t)
（2） x ( 0 + ) = 0 , x ( ∞ ) = 0 x\left( {0_ + } \right) = 0,\,\,x\left( \infty \right) = 0 x(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） $x\left( {0_ + } \right) = 2,,,x\left( \infty \right) = $ 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） 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)
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