信号与系统分析2022春季作业-参考答案：第十一次作业

§00 作业题目

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  作业要求链接： 信号与系统2022春季作业-第十一次作业 : https://zhuoqing.blog.csdn.net/article/details/124730642

§11 参考答案

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1.1 z变换性质

1.1.1 求序列的z变换

◎ 求解：

  （1） $$x_1\left[n\right]={\left(-3\right)}^n\cdot n\cdot u\left[n\right]$$ Z { ( − 3 ) n ⋅ u [ n ] } = z z + 3 Z\left\{ {\left( { - 3} \right)^n \cdot u\left[ n \right]} \right\} = {z \over {z + 3}} Z{ (−3)n⋅u[n]}=z+3z​ Z { ( − 3 ) n ⋅ n ⋅ u [ n ] } = − z d d z X ( z ) = − z d d z z z + 3 = − 3 z ( z + 3 ) 2 Z\left\{ {\left( { - 3} \right)^n \cdot n \cdot u\left[ n \right]} \right\} = - z{d \over {dz}}X\left( z \right) = - z{d \over {dz}}{z \over {z + 3}} = { { - 3z} \over {\left( {z + 3} \right)^2 }} Z{ (−3)n⋅n⋅u[n]}=−zdzd​X(z)=−zdzd​z+3z​=(z+3)2−3z​

  （2）$$x_2\left[n\right]=\left(n-4\right)\cdot u\left[n\right]$$ Z { n ⋅ u [ n ] } = − z d d z z z − 1 = z ( z − 1 ) 2 Z\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]}=−zdzd​z−1z​=(z−1)2z​ Z { ( n − 4 ) ⋅ u [ n ] } = z ( z − 1 ) 2 − 4 z z − 1 = − 4 z 2 + 5 z ( z − 1 ) 2 Z\left\{ {\left( {n - 4} \right) \cdot u\left[ n \right]} \right\} = {z \over {\left( {z - 1} \right)^2 }} - { {4z} \over {z - 1}} = { { - 4z^2 + 5z} \over {\left( {z - 1} \right)^2 }} Z{ (n−4)⋅u[n]}=(z−1)2z​−z−14z​=(z−1)2−4z2+5z​

  （3）$$x_3\left[n\right]=n\cdot a^{n-2}\cdot u\left[n\right]$$ Z { a n ⋅ u [ n ] } = z z − a Z\left\{ {a^n \cdot u\left[ n \right]} \right\} = {z \over {z - a}} Z{ an⋅u[n]}=z−az​ Z { n ⋅ a n ⋅ u [ n ] } = − z d d z z z − a = a z ( z − a ) 2 Z\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]}=−zdzd​z−az​=(z−a)2az​$$Z\left\{n\cdot a^{n-2}\cdot u\left[n\right]\right\}=\frac{z}{a{\left(z-a\right)}^2}$$

  （4）$$x_4\left[n\right]=2^n\cdot \left(\sum _{k=0}^{+\infty }{\left(-1\right)}^k\cdot u\left[n-k\right]\right)$$ Z { ( − 1 ) n ⋅ u [ n ] } = z z + 1 Z\left\{ {\left( { - 1} \right)^n \cdot u\left[ n \right]} \right\} = {z \over {z + 1}} Z{ (−1)n⋅u[n]}=z+1z​ Z { u [ n ] } = z z − 1 Z\left\{ {u\left[ n \right]} \right\} = {z \over {z - 1}} Z{ u[n]}=z−1z​$$Z\left\{{\left(-1\right)}^k\cdot u\left[n\right]\ast u\left[n\right]\right\}=\frac{z}{z+1}\cdot \frac{z}{z-1}=\frac{z^2}{z^2-1}$$$$Z\left\{2^n\cdot \left[{\left(-1\right)}^{k}u\left[n\right]\ast u\left[n\right]\right]\right\}=\frac{{\left(\frac{z}{2}\right)}^2}{{\left(\frac{z}{2}\right)}^2-1}=\frac{z^2}{z^2-4}$$

  （5）$$x_5\left[n\right]=\frac{a^n}{n+2}\cdot u\left[n+1\right]$$$$Z\left\{\frac{1}{n+1}\cdot u\left[n\right]\right\}=z\cdot \ln \frac{z}{z-1}$$$$Z\left\{\frac{1}{n+2}\cdot u\left[n+1\right]\right\}=z^2\ln \frac{z}{z-1}$$$$Z\left\{\frac{a^n}{n+2}\cdot u\left[n+1\right]\right\}={\left(\frac{z}{a}\right)}^2\ln \frac{\frac{z}{a}}{\frac{z}{a}-1}=\frac{z^2}{a^2}\ln \frac{z}{z-a}$$

• 另外一种求解方法：

$$Z\left\{\frac{a^n}{n+2}\cdot u\left[n\right]\right\}=Z\left\{\frac{a^n}{n+2}\cdot u\left[n\right]+a^{-1}\delta \left[-1\right]\right\}=\frac{-z^2}{a^2}\left[\ln \left(\frac{z-a}{z}\right)+\frac{a}{z}\right]-\frac{z}{a}$$$$=\frac{-z^2}{a^2}\ln \left(\frac{z-a}{z}\right)=\frac{z^2}{a^2}\ln \left(\frac{z-a}{z}\right)$$

  （6） $$x_6\left[n\right]={\left(\frac{1}{3}\right)}^n.\cos \frac{n\pi }{2}\cdot u\left[n\right]$$ Z { cos ⁡ n ω } = z ( z − cos ⁡ ω ) z 2 − 2 z cos ⁡ ω + 1 Z\left\{ {\cos n\omega } \right\} = { {z\left( {z - \cos \omega } \right)} \over {z^2 - 2z\cos \omega + 1}} Z{ cosnω}=z2−2zcosω+1z(z−cosω)​$$Z\left\{\cos \left(\frac{n\pi }{2}\right)\right\}=\frac{z^2}{z^2+1}$$$$Z\left\{{\left(\frac{1}{3}\right)}^n\cos \left(\frac{n\omega }{2}\right)\cdot u\left[n\right]\right\}=\frac{{\left(3z\right)}^2}{{\left(3z\right)}^2+1}=\frac{9z^2}{9z^2+1}$$

1.1.2 初值和终值

◎ 求解：

  （1）解答：
$$X\left(z\right)=\frac{1+z^{-1}+z^{-2}}{\left(1-z^{-1}\right)\cdot \left(1-2z^{-1}\right)}$$

$$x\left[0\right]=X\left(\infty \right)=1$$
  由于存在极点位于2，处于单位圆外，所以$x\left[\infty \right]$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：
$$X\left(z\right)=\frac{1}{\left(1-0.5z^{-1}\right)\left(1+0.5z^{-1}\right)}$$

$$x\left[0\right]=X\left(\infty \right)=1$$$$x\left[\infty \right]=\underset{z\to 1}{\lim }\frac{z-1}{\left(1-0.5z^{-1}\right)\left(1+0.5z^{-1}\right)}=0$$

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

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

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

$$x\left[\infty \right]=\underset{z\to 1}{\lim }\frac{z^{-1}\left(z-1\right)}{1-1.5z^{-1}+0.5z^{-2}}=\underset{z\to 1}{\lim }\frac{z}{z-0.5}=2$$

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

（4）解答：
$$X\left(z\right)=\frac{z^4}{\left(z-1\right)\left(z-0.5\right)\left(z-0.2\right)}$$

$$X\left(z\right)=\frac{z^4}{z^3-1.7z^2+0.8z-0.1}=z+\frac{1.7z^3-0.8z^2+0.1z}{z^3}$$

$$x\left[0\right]=\underset{z\to \infty }{\lim }\frac{1.7z^3-0.8z^2+0.1z}{z^3}=1.7$$

$$x\left[\infty \right]=\underset{z\to 1}{\lim }\frac{z^4\left(z-1\right)}{\left(z-1\right)\left(z-0.5\right)\left(z-0.2\right)}=\frac{1}{0.5\ast 0.8}=2.5$$

>>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

1.1.3 求序列卷积

◎ 求解：

  （1） 根据 z 变换的卷积定理 $y\left[n\right]=x\left[n\right]\ast h\left[n\right]$ ，那么 $y\left[n\right]$ 的 z 变换为： $$Y\left(z\right)=H\left(z\right)\cdot X\left(z\right)=\frac{1}{z-a}\cdot \frac{z}{z-1}=\frac{\frac{1}{a-1}z}{z-a}\cdot \frac{\frac{1}{1-a}z}{z-1}$$ 所以 $$y\left[n\right]=\frac{1}{1-a}u\left[n\right]-\frac{1}{1-a}{a}^n\cdot u\left[n\right]$$

  （2） 序列 $x\left[n\right],h\left[n\right]$ 的 z 变换分别为：$$X\left(z\right)=\frac{2}{z-1}$$ H ( z ) = Z { ∑ k = 0 ∞ ( − 1 ) k δ [ n − k ] } = Z { ( − 1 ) n u [ n ] } = z z + 1 H\left( z \right) = Z\left\{ {\sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \delta \left[ {n - k} \right]} } \right\} = Z\left\{ {\left( { - 1} \right)^n u\left[ n \right]} \right\} = {z \over {z + 1}} H(z)=Z{ k=0∑∞​(−1)kδ[n−k]}=Z{ (−1)nu[n]}=z+1z​ 那么，根据 z 变换的卷积定理可得： $$Y\left(z\right)=X\left(z\right)\cdot H\left(z\right)=\frac{2}{z-1}\cdot \frac{z}{z+1}=\frac{z}{z-1}+\frac{-z}{z+1}$$ 所以 $$y\left[n\right]=u\left[n\right]-{\left(-1\right)}^n\cdot u\left[n\right]$$

1.1.4 求序列的乘积z变换

◎ 求解：

  （1）解答：
Z { x [ n ] ⋅ h [ n ] } = 1 2 π j ⋅ ∮ C X ( v ) ⋅ H ( z v ) ⋅ v − 1 d v Z\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​⋅∮C​X(v)⋅H(vz​)⋅v−1dv$$=\frac{1}{2\pi j}\cdot {\oint }_C\frac{1}{1-\frac{1}{2}{v}^{-1}}\cdot \frac{1}{1-2\frac{z}{v}}{v}^{-1}dv$$$$=\frac{1}{2\pi j}\cdot {\oint }_C\frac{v}{\left(v-\frac{1}{2}\right)\left(v-2z\right)}dv$$

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

Z { x [ n ] ⋅ h [ n ] } = R e s [ v ( v − 1 2 ) ( v − 2 z ) ] v = 1 2 +   R e s [ v ( v − 1 2 ) ( v − 2 z ) ] v = 2 z = 1 Z\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} = 1 Z{ x[n]⋅h[n]}=Res[(v−21​)(v−2z)v​]v=21​​+Res[(v−21​)(v−2z)v​]v=2z​=1

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

  （2）解答：
Z { x [ n ] ⋅ h [ n ] } = 1 2 π j ⋅ ∮ C X ( v ) ⋅ H ( z v ) ⋅ v − 1 d v Z\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​⋅∮C​X(v)⋅H(vz​)⋅v−1dv$$=\frac{1}{2\pi j}\cdot {\oint }_C\frac{v}{v-e^{-b}}\cdot \frac{\frac{z}{v}\cdot \sin {\omega }_0}{{\left(\frac{z}{v}\right)}^2-2\left(\frac{z}{v}\right)\cos {\omega }_0+1}\cdot v^{-1}dv$$$$=\frac{1}{2\pi j}\cdot {\oint }_C\frac{2\sin {\omega }_{0}v}{\left(v-e^{-b}\right)\left(v^2-2z{\cos }_{0}v+z^2\right)}dv$$

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

  所以上述围线积分包含的极点为：$e^{-b}$。
Z { x [ n ] ⋅ h [ n ] } = R e s [ 2 sin ⁡ ω 0 v ( v − e − b ) ( v 2 − 2 z cos ⁡ ω 0 v + z 2 ) ] v = e − b = z ⋅ sin ⁡ ω 0 ⋅ e − b e − 2 b − 2 e − b cos ⁡ ω 0 z + z 2 Z\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)(v2−2zcosω0​v+z2)2sinω0​v​]v=e−b​=e−2b−2e−bcosω0​z+z2z⋅sinω0​⋅e−b​
因此：
$$x\left[n\right]\cdot h\left[n\right]=e^{-nb}\cos {\omega }_{0}n\cdot u\left[n\right]$$

1.1.5 证明z变换累加性质

◎ 证明：

  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$u\left[n\right]$的卷积：$\sum _{k=-\infty }^{n}x\left[k\right]=x\left[n\right]\ast u\left[n\right]$。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$u\left[n\right]$的z变换。而： Z { u [ n ] } = z z − 1 Z\left\{ {u\left[ n \right]} \right\} = {z \over {z - 1}} Z{ u[n]}=z−1z​，所以：$$Z\left[\sum _{k=-\infty }^{n}x\left(k\right)\right]=\frac{z}{z-1}X\left(z\right)$$

  证明方法2：

$$Z\left[\sum _{k=-\infty }^{n}x\left[k\right]\right]=\sum _{n=-\infty }^{\infty }\left(\sum _{k=-\infty }^{n}x\left[k\right]\right)\cdot z^{-n}$$$$=\sum _{n=-\infty }^{\infty }\left(\sum _{k=-\infty }^{\infty }x\left[k\right]\cdot u\left[n-k\right]\right)\cdot z^{-n}$$$$=\sum _{k=-\infty }^n\sum _{n=-\infty }^{\infty }x\left[k\right]\cdot u\left[n-k\right]\cdot z^{-n}$$$$=\sum _{k=-\infty }^{\infty }x\left[k\right]\sum _{n=-\infty }^{\infty }u\left[n-k\right]\cdot z^{-\left(n-k\right)}\cdot z^{-k}$$$$=\sum _{k=-\infty }^{\infty }x\left[k\right]z^{-k}\sum _{n=-\infty }^{\infty }u\left[n-k\right]\cdot z^{-\left(n-k\right)}=\frac{z}{z-1}X\left(z\right)$$

1.1.6 序列尺度特性

◎ 求解：

  （1）解答：

$$X_1\left(z\right)=\sum _{n=-\infty }^{+\infty }{x}_1\left[n\right]z^{-n}=\sum _{k=-\infty }^{+\infty }x\left[k\right]z^{-4k}=X\left(z^4\right)$$

  （2）解答：

$$X_2\left(z\right)=\sum _{n=-\infty }^{+\infty }{x}_2\left[n\right]z^{-n}=\sum _{k=4n,n=-\infty }^{+\infty }x\left[k\right]z^{-\frac{k}{4}}$$$$=\sum _{n=-\infty }^{+\infty }\frac{1}{4}\left[1^n+j^n+{\left(-1\right)}^n+{\left(-j\right)}^n\right]x\left[k\right]z^{-\frac{k}{4}}$$$$=\frac{1}{4}\left[X\left(z^{\frac{1}{4}}+{\left(-jz\right)}^{\frac{1}{4}}+{\left(-z\right)}^{\frac{1}{4}}+{\left(jz\right)}^{\frac{1}{4}}\right)\right]$$

1.2 求解微分方程

◎ 求解：

  （1） 对微分方程两边进行拉普拉斯变换，根据拉普拉斯变换的微分定理可以将系统的初始条件代入方程：

$$s^{2}Y\left(s\right)-sy\left(0_{-}\right)-y^{\prime }\left(0_{-}\right)+2\left[sY\left(s\right)-y\left(0_{-}\right)\right]+Y\left(s\right)=1+3s$$$$s^{2}Y\left(s\right)-2s-1+2\left[sY\left(s\right)-2\right]+Y\left(s\right)=1+3s$$$$\left(s^2+2s+1\right)Y\left(s\right)=5s+6$$$$Y\left(s\right)=\frac{5s+6}{s^2+2s+1}$$

  进行拉普拉斯反变换可以得到方程的解：
$$y\left(t\right)=\left(t+5\right)e^{-t},t\ge 0$$

from head import *
from sympy                  import symbols,simplify,expand,print_latex,Heaviside
from sympy                  import *

#------------------------------------------------------------
s = symbols('s')
t = symbols('t', positive=True)
expression = (5*s+6)/(s**2+2*s+1)

result = inverse_laplace_transform(expression, s,t)

#------------------------------------------------------------
print_latex(result)
_=tspexecutepythoncmd("msg2latex")
clipboard.copy(str(result))

  （2） 根据上题相同步骤，可得：$$s^{2}Y\left(s\right)-sy\left(0_{-}\right)-y^{\prime }\left(0_{-}\right)+5\left[sY\left(s\right)-y\left(0_{-}\right)\right]+6Y\left(s\right)=3X\left(s\right)$$$$X\left(s\right)=\frac{1}{s+1}$$$$\left(s^2+5s+6\right)Y\left(s\right)=\frac{1}{s+1}+1$$$$Y\left(s\right)=\frac{s+2}{\left(s+1\right)\left(s^2+5s+6\right)}$$

  进行拉普拉斯反变换，可以得到方程的解
$$y\left(t\right)=\frac{1}{2}{e}^{-t}-\frac{1}{2}{e}^{-3t},t\ge 0$$

s = symbols('s')
t = symbols('t', positive=True)
expression = (s+2)/(s**2+5*s+6)/(s+1)

result = inverse_laplace_transform(expression, s,t)

1.3 求解差分方程

◎ 求解：

  （1） 第一小题求解：

  方程两边同时进行z变换：
$$Y\left(z\right)+3z^{-1}Y\left(z\right)+3\cdot y\left[-1\right]=X\left(z\right)$$$$\left(1+3z^{-1}\right)Y\left(z\right)=\frac{z}{z-\frac{1}{2}}-3$$$$Y\left(z\right)=\frac{z\left(-2z+\frac{3}{2}\right)}{\left(z-\frac{1}{2}\right)\left(z+3\right)}=\frac{\frac{1}{7}z}{z-\frac{1}{2}}+\frac{-\frac{15}{7}z}{z+3}$$
$$y\left[n\right]=\frac{1}{7}{\left(\frac{1}{2}\right)}^{n}u\left[n\right]-\frac{15}{7}{\left(-3\right)}^{n}u\left[n\right]$$

  零输入响应：
$$Y_{zi}\left(z\right)=\frac{-3}{1+3z^{-1}}=\frac{-3z}{z+3}$$$$y_{zi}\left[n\right]=-3{\left(-3\right)}^n\cdot u\left[n\right]={\left(-3\right)}^{n+1}u\left[n\right]$$

  零状态响应：
$$Y_{zs}\left(z\right)=\frac{z^2}{\left(z-\frac{1}{2}\right)\left(z+3\right)}=\frac{\frac{1}{7}z}{z-\frac{1}{2}}+\frac{\frac{6}{7}z}{z+3}$$$$y_{zs}\left[n\right]=\frac{1}{7}{\left(\frac{1}{2}\right)}^{n}u\left[n\right]+\frac{6}{7}{\left(-3\right)}^{n}u\left[n\right]$$

  （2）第二小题求解：

  方程两边同时进行z变换：
$$Y\left(z\right)-\frac{1}{2}{z}^{-1}Y\left(z\right)-\frac{1}{2}y\left[-1\right]=\left(1-\frac{1}{2}{z}^{-1}\right)\cdot \frac{z}{z-1}$$$$Y\left(z\right)=\frac{z\left(\frac{3}{2}z-1\right)}{\left(z-1\right)\left(z-\frac{1}{2}\right)}=\frac{z}{z-1}+\frac{\frac{1}{2}}{z-\frac{1}{2}}$$$$y\left[n\right]=u\left[n\right]+{\left(\frac{1}{2}\right)}^{n+2}u\left[n\right]$$

  系统的零输入响应为：
$$Y_{zi}\left(z\right)=\frac{\frac{1}{2}z}{z-\frac{1}{2}}$$$$y_{zi}\left[n\right]={\left(\frac{1}{2}\right)}^{n+1}u\left[n\right]$$

  系统的零状态响应为：
$$Y_{zs}\left(z\right)=\frac{z}{z-1}$$$$y_{zs}\left[n\right]=u\left[n\right]$$

1.4 系统的单位冲激响应

◎ 求解：

  分别对输入信号$x\left(t\right)$和系统的零状态响应$y\left(t\right)$进行Laplace变换：
$$X\left(s\right)=LT\left[e^{-t}\cdot u\left(t\right)\right]=\frac{1}{s+1}$$

$$Y\left(s\right)=LT\left[\frac{1}{2}{e}^{-t}-e^{-2t}+2e^{-3t}\right]=\frac{1}{2}\frac{1}{s+1}-\frac{1}{s+2}+\frac{2}{s+3}$$

$$=\frac{\frac{3}{2}{s}^2+\frac{9}{2}s+4}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}$$

  根据线性是不变系统的性质，系统的零状态输出等于系统的输入信号与系统的单位冲击响应信号的卷积：$$y\left(t\right)=x\left(t\right)\ast h\left(t\right)$$

  根据Laplace变换的卷积定理：$$Y\left(s\right)=X\left(s\right)\cdot H\left(s\right)$$

  因此：$$H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}=\frac{\frac{3}{2}{s}^2+\frac{9}{2}s+4}{{\left(s+1\right)}^2\left(s+2\right)\left(s+3\right)}$$

>>ilaplace((1.5*s*s+4.5*s+4)/((s+1)^2*(s+2)*(s+3)))'
ans=exp(-2*t)-exp(-3*t)+(t*exp(-t))/2

则系统的单位冲击响应信号$h\left(t\right)$等于：$$h\left(t\right)=e^{-2t}-e^{-3t}+\frac{1}{2}t\cdot e^{-t},t\ge 0$$

>>ilaplace((1.5*s*s+4.5*s+4)/((s+1)^2*(s+2)*(s+3)))'
ans=exp(-2*t)-exp(-3*t)+(t*exp(-t))/2

则系统的单位冲击响应信号$h\left(t\right)$等于：$$h\left(t\right)=e^{-2t}-e^{-3t}+\frac{1}{2}t\cdot e^{-t},t\ge 0$$

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• 信号与系统2022春季作业-第十一次作业