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

信号与系统第十一次作业参考答案  

※ 第一题

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

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■ 求解：

方法1：围线积分方法：
因为 $\left| z \right| > 2$，所以信号为右边序列。
$x\left[ n \right] = {1 \over {2\pi j}}\oint\limits_C {X\left( z \right) \cdot z^{n - 1} dz = {1 \over {2\pi j}}\oint\limits_C {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}dz} }$

$x\left[ n \right] = \sum\limits_m^{} {{\mathop{\rm Re}\nolimits} s\left[ {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]} _{z = z_m }$

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

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

方法2：长除法：

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

>> 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：
${{X\left( z \right)} \over z} = {{10} \over {\left( {z - 1} \right)\left( {z - 2} \right)}} = {{ - 10} \over {z - 1}} + {{10} \over {z - 2}}$ $X\left( z \right) = {{ - 10z} \over {z - 1}} + {{10z} \over {z - 2}}$

$x\left[ n \right] = - 10 \cdot u\left[ n \right] + 10 \cdot 2^n u\left[ n \right]$

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

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

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

（2） $X\left( z \right) = {{10z^2 } \over {\left( {z - 1} \right)\left( {z + 1} \right)}}\,\,\,\,\,\,\,\,\,\,\left( {\left| z \right| > 1} \right)\;\;\;\;\;$
（3） $X\left( z \right) = {{1 + z^{ - 1} } \over {1 - 2z^{ - 1} \cos \omega + z^{ - 2} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left| z \right| > 1} \right)\;\;\;\;\;$

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■ 求解：

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

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

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

>>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：
$Z\left[ {\cos \left( {\omega _0 n} \right)u\left[ n \right]} \right] = {{z\left( {z - \cos \omega _0 } \right)} \over {z^2 - 2z\cos \omega _0 + 1}}$ $Z\left[ {\sin \left( {\omega _0 n} \right)u\left[ n \right]} \right] = {{z\sin \omega _0 } \over {z^2 - 2z\cos \omega _0 + 1}}$

$X\left( z \right) = {{z^2 + z} \over {z^2 - 2z\cos \omega + 1}}$ $= {{z\left( {z - \cos \omega } \right)} \over {z^2 - 2z\cos \omega + 1}} + {{1 + \cos \omega } \over {\sin \omega }}{{z\sin \omega } \over {z^2 - 2z\cos \omega + 1}}$
$x\left[ n \right] = \left( {\cos \omega n + {{1 + \cos \omega } \over {\sin \omega }}\sin \omega n} \right) \cdot u\left[ n \right]$ $= {{\sin \left( {n\omega } \right) + \sin \left( {n + 1} \right)\omega } \over {\sin \omega }} \cdot u\left[ n \right]$

>>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)

${{\sin n\omega \left( {\cos \omega + 1} \right)} \over {\sin \omega }} - {{\cos n\omega \left( {\cos \omega + 1} \right)} \over {\cos \omega }} + {{\cos n\omega \left( {2\cos + 1} \right)} \over {\cos \omega }}$
${{\left( {\sin n\omega \cdot \cos \omega - \sin \omega \cdot \cos n\omega } \right) \cdot \left( {\cos \omega + 1} \right) + \sin \omega \cdot \cos n\omega \cdot (2\cos \omega + 1)} \over {\sin \omega \cdot \cos \omega }}$

※ 第三题

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求下面两个函数的拉普拉斯变换：

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■ 求解：

（a）求解：
$f_1 \left( t \right) = \left( {1 - t} \right)\left[ {u\left( t \right) - u\left( {t - 1} \right)} \right]$ $F_1 \left( s \right) = \int_0^1 {\left( {1 - t} \right) \cdot e^{ - st} dt} = {{e^{ - s} + s - 1} \over {s^2 }}$ $F\left( s \right) = {{F_1 \left( s \right)} \over {1 - e^{ - s} }} = {{e^{ - s} + s - 1} \over {\left( {1 - e^{ - s} } \right) \cdot s^2 }} = - {1 \over {s^2 }} + {1 \over {s\left( {1 - e^{ - s} } \right)}}$

（b）求解：
$f_1 \left( t \right) = \delta \left( t \right) - \delta \left( {t - {1 \over 2}} \right)$ $F_1 \left( s \right) = \int_0^1 {f_1 \left( t \right)e^{ - st} dt} = 1 - e^{ - {1 \over 2}s}$ $F\left( s \right) = {{F_1 \left( s \right)} \over {1 - e^{ - s} }} = {{1 - e^{ - {1 \over 2}s} } \over {1 - e^{ - s} }} = {1 \over {1 + e^{ - {s \over 2}} }}$

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

※ 第四题

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

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■ 求解：

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

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

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

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

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

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

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■ 求解：

这里给出了完整版本的答案：

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

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

（2） 解答： $L\left[ {\sin t + 2\cos t} \right] = {{2s} \over {s^2 + 1}} + {1 \over {s^2 + 1}} = {{2s + 1} \over {s^2 + 1}}$

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

（3）解答： $L\left[ {t \cdot e^{ - 2t} } \right] = - {d \over {ds}}L\left[ {e^{ - 2t} } \right] = - {d \over {ds}}\left( {{1 \over {s + 2}}} \right) = {1 \over {\left( {s + 2} \right)^2 }}$

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

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

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

（5）解答： $L\left[ {e^{ - t} \cdot \sin \left( {2t} \right)} \right] = \left. {L\left[ {\sin \left( {2t} \right)} \right]} \right|_{s \to s + 1} = \left. {{2 \over {\left( {s^2 + 4} \right)}}} \right|_{s \to s + 1} = {2 \over {\left( {s + 1} \right)^2 + 4}}$

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

（6）解答： $L\left[ {\left( {1 + 2t} \right) \cdot e^{ - t} } \right] = \left. {L\left[ {1 + 2t} \right]} \right|_{s \to s + 1} = \left. {\left( {{1 \over s} + {2 \over {s^2 }}} \right)} \right|_{s \to s + 1} = {1 \over {s + 1}} + {2 \over {\left( {s + 1} \right)^2 }}$

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

（7）解答： $L\left[ {\cos \left( {at} \right)} \right] = {s \over {s^2 + a^2 }}$
$L\left\{ {\left[ {1 - \cos \left( {at} \right)} \right] \cdot e^{ - bt} } \right\} = \left. {L\left[ {1 - \cos \left( {at} \right)} \right]} \right|_{s \to s + b} = {1 \over {s + b}} - {{s + b} \over {\left( {s + b} \right)^2 + a^2 }}$

>>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： $L\left[ {t^2 + 2t} \right] = {2 \over {s^2 }} + {2 \over {s^3 }}$

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

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

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

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

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

（11）解答： $\cos ^2 \left( {\Omega t} \right) = {1 \over 2}\left[ {1 + \cos \left( {2\Omega t} \right)} \right]$

$L\left[ {\cos ^2 \left( {\Omega t} \right)} \right] = L\left[ {{1 \over 2}\left( {1 + \cos \left( {2\Omega t} \right)} \right)} \right] = {{s^2 + 2\Omega ^2 } \over {s\left( {s^2 + 4\Omega ^2 } \right)}}$

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

（12）解答： $L\left[ {e^{ - \left( {t + a} \right)} \cdot \cos \left( {\omega t} \right)} \right] = e^{ - a} \left\{ {L\left[ {e^{ - t} \cdot \cos \left( {\omega t} \right)} \right]} \right\} = e^{ - a} \cdot \left. {L\left[ {\cos \left( {\omega t} \right)} \right]} \right|_{s \to s + 1} = {{e^{ - a} \cdot \left( {s + 1} \right)} \over {\left( {s + 1} \right)^2 + \omega ^2 }}$

>>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： $L\left[ {{1 \over {b - a}}\left( {e^{ - at} - e^{bt} } \right)} \right] = {{{1 \over {s + a}} - {1 \over {s - b}}} \over {b - a}} = {{{{a + b} \over {a - b}}} \over {s^2 + \left( {a - b} \right)s - ab}}$

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

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

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

※ 第六题

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

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

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■ 求解：

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

$x\left[ 0 \right] = X\left( \infty \right) = 1$
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：
$X\left( z \right) = {1 \over {\left( {1 - 0.5z^{ - 1} } \right)\left( {1 + 0.5z^{ - 1} } \right)}}$

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

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

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

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

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

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

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

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

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

$x\left[ \infty \right] = \mathop {\lim }\limits_{z \to 1} {{z^4 \left( {z - 1} \right)} \over {\left( {z - 1} \right)\left( {z - 0.5} \right)\left( {z - 0.2} \right)}} = {1 \over {0.5*0.8}} = 2.5$

>>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

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利用z变换的性质求以下序列的z变换，表明收敛域：

（1） $x_1 \left[ n \right] = \left( { - 2} \right)^n n \cdot u\left[ n \right]$

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

（3） $x_3 \left[ n \right] = na^{n - 1} \cdot u\left[ n \right]$

（4） $x_4 \left[ n \right] = 2^n \cdot \sum\limits_{k = 0}^\infty {\left( { - 2} \right)^k \cdot u\left[ {n - k} \right]}$

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

（6） $x_6 \left[ n \right] = \left( {{1 \over 3}} \right)^n .\cos \left( {{{n\pi } \over 2}} \right) \cdot u\left[ n \right]$

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■ 求解：

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

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

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

$Z\left\{ {\left( { - 2} \right)^n \cdot n} \right\} = - z{d \over {dz}}{z \over {z + 2}} = {{ - 2z} \over {\left( {z + 2} \right)^2 }}$

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

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

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

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

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

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

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

$Z\left\{ {a^n \cdot u\left[ n \right]} \right\} = {z \over {z - a}}$

$Z\left\{ {n \cdot a^n \cdot u\left[ n \right]} \right\} = - z{d \over {dz}}{z \over {z - a}} = {{az} \over {\left( {z - a} \right)^2 }}$

$Z\left\{ {n \cdot a^{n - 1} \cdot u\left[ n \right]} \right\} = {d \over {dz}}{z \over {z - a}} = {z \over {\left( {z - a} \right)^2 }}$

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

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

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

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

$Z\left\{ {\left( { - 2} \right)^k u\left[ n \right] * u\left[ n \right]} \right\} = {z \over {z + 2}} \cdot {z \over {z - 1}} = {{z^2 } \over {\left( {z + 2} \right)\left( {z - 1} \right)}}$

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

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

$Z\left\{ {{1 \over {n + 1}} \cdot u\left[ n \right]} \right\} = z \cdot \ln {z \over {z - 1}}$

$Z\left\{ {{1 \over {n + 2}} \cdot u\left[ {n + 1} \right]} \right\} = z^2 \ln {z \over {z - 1}}$

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

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

（6）求解：
$x_6 \left[ n \right] = \left( {{1 \over 3}} \right)^n .\cos \left( {{{n\pi } \over 2}} \right) \cdot u\left[ n \right]$

$Z\left\{ {\cos \left( {n\omega } \right)} \right\} = {{z\left( {z - \cos \omega } \right)} \over {z^2 - 2\cos \omega \cdot z + 1}}$

$Z\left\{ {\cos \left( {{{n\pi } \over 2}} \right)} \right\} = {{z^2 } \over {z^2 + 1}}$

$Z\left\{ {\left( {{1 \over 3}} \right)^n \cos \left( {{{n\pi } \over 2}} \right)} \right\} = {{\left( {3z} \right)^2 } \over {\left( {3z} \right)^2 + 1}} = {{9z^2 } \over {9z^2 + 1}}$

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

※ 第八题

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利用z变换的性质求以下序列的卷积，已知：

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

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

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■ 求解：

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

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

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

$h\left[ n \right] = \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \delta \left[ {n - k} \right]} = \left( { - 1} \right)^n \cdot u\left[ n \right]$

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

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

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

※ 第九题

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

（1） $X\left( z \right) = {1 \over {1 - {1 \over 2}z^{ - 1} }},\,\,\left| z \right| > 0.5;\,\,\,\,H\left( z \right) = {1 \over {1 - 2z}},\,\,\left| z \right| < 0.5$
（2） $X\left( z \right) = {z \over {z - e^{ - b} }},\,\,\left| z \right| > e^{ - b} ;\;\;\;H\left( z \right) = {{z \cdot \sin \omega _0 } \over {z^2 - 2z\cos \omega _0 + 1}},\,\,\left| z \right| > 1$

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■ 求解：

（1）解答：
$Z\left\{ {x\left[ n \right] \cdot h\left[ n \right]} \right\} = {1 \over {2\pi j}} \cdot \oint_C {X\left( v \right) \cdot H\left( {{z \over v}} \right) \cdot v^{ - 1} dv}$ $= {1 \over {2\pi j}} \cdot \oint_C {{1 \over {1 - {1 \over 2}v^{ - 1} }} \cdot {1 \over {1 - 2{z \over v}}}v^{ - 1} dv}$ $= {1 \over {2\pi j}} \cdot \oint_C {{v \over {\left( {v - {1 \over 2}} \right)\left( {v - 2z} \right)}}dv}$

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

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

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

（2）解答：
$Z\left\{ {x\left[ n \right] \cdot h\left[ n \right]} \right\} = {1 \over {2\pi j}} \cdot \oint_C {X\left( v \right) \cdot H\left( {{z \over v}} \right) \cdot v^{ - 1} dv}$ $= {1 \over {2\pi j}} \cdot \oint_C {{v \over {v - e^{ - b} }} \cdot {{{z \over v} \cdot \sin \omega _0 } \over {\left( {{z \over v}} \right)^2 - 2\left( {{z \over v}} \right)\cos \omega _0 + 1}} \cdot v^{ - 1} dv}$ $= {1 \over {2\pi j}} \cdot \oint_C {{{2\sin \omega _0 v} \over {\left( {v - e^{ - b} } \right)\left( {v^2 - 2z\cos _0 v + z^2 } \right)}}dv}$

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： $e^{ - b}$。
$Z\left\{ {x\left[ n \right] \cdot h\left[ n \right]} \right\} = {\mathop{\rm Re}\nolimits} s\left[ {{{2\sin \omega _0 v} \over {\left( {v - e^{ - b} } \right)\left( {v^2 - 2z\cos \omega _0 v + z^2 } \right)}}} \right]_{v = e^{ - b} } = {{z \cdot \sin \omega _0 \cdot e^{ - b} } \over {e^{ - 2b} - 2e^{ - b} \cos \omega _0 z + z^2 }}$
因此：
$x\left[ n \right] \cdot h\left[ n \right] = e^{ - nb} \cos \omega _0 n \cdot u\left[ n \right]$

※ 第十题

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

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■ 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 $u\left[ n \right]$的卷积： $\sum\limits_{k = - \infty }^n {x\left[ k \right]} = x\left[ n \right] * u\left[ n \right]$。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： $Z\left\{ {u\left[ n \right]} \right\} = {z \over {z - 1}}$，所以： $Z\left[ {\sum\limits_{k = - \infty }^n {x\left( k \right)} } \right] = {z \over {z - 1}}X\left( z \right)\;\;\;\;\;$

证明方法2：

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

※ 第十一题

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

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■ 求解：

（1）解答：

$X_1 \left( z \right) = \sum\limits_{n = - \infty }^\infty {x_1 \left[ n \right]z^{ - n} } = \sum\limits_{k = - \infty }^\infty {x\left[ k \right]z^{ - 3k} }$ $= \sum\limits_{k = - \infty }^\infty {x\left[ k \right]\left( {z^3 } \right)^{ - k} } = X\left( {z^3 } \right)$

（2）解答：