2020年春季学期信号与系统课程作业参考答案-第十次作业

第十次作业参考答案  

 

01第一题

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第一小题中的求解除了（14）（15）小题之外，其他的各题都可以在MATLAB中使用MATLAB的符号计算帮助求解，一边检查求解的结果正确性。
使用MATLAB求解第一小题可以参见下面的链接：

• 利用MATLAB帮助求解作业中的Laplace变换和Z变换

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求解：

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

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

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

（4） 由：
$L\left[ {\sin \left( {2t} \right)} \right] = {2 \over {s^4 + 4}}$
再根据s域平移性质，可以得到：

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

（5） 因为
$L\left[ {t^n } \right] = {{n!} \over {s^{n + 1} }}$

所以： $L\left[ {1 + 2t} \right] = {1 \over s} + {2 \over {s^2 }} = {{s + 2} \over {s^2 }}$

再由s域平移性质，可得： $L\left[ {\left( {1 + 2t} \right) \cdot e^{ - t} } \right] = {{s + 3} \over {\left( {s + 1} \right)^2 }}$

（6） 由 $L\left[ {1 - \cos \left( {\alpha t} \right)} \right] = {1 \over s} - {s \over {s^2 + \alpha ^2 }}$

再由s域平移性质，可得： $L\left\{ {\left[ {1 - \cos \left( {\alpha t} \right)} \right]e^{ - \beta t} } \right\} = {1 \over {s + \beta }} - {{s + \beta } \over {\left( {s + \beta } \right)^2 + \alpha ^2 }}$

（7） $L\left[ {t^2 + 2t} \right] = {2 \over {s^3 }} + {2 \over {s^2 }} = {{2s + 2} \over {s^3 }}$

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

（9） 由 $L\left[ {\sinh \left( {\beta t} \right)} \right] = {\beta \over {s^2 - \beta ^2 }}$

再由s域的平移性质，可得： $L\left[ {e^{ - \alpha t} \sinh \left( {\beta t} \right)} \right] = {\beta \over {\left( {s + \alpha } \right)^2 - \beta ^2 }}$

（10） 因为 $\cos ^2 \left( {\Omega t} \right) = {1 \over 2} + {1 \over 2}\cos \left( {2\Omega t} \right)$

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

（11）
$L\left[ {{1 \over {\beta - \alpha }}\left( {e^{ - \alpha t} - e^{ - \beta t} } \right)} \right] = {1 \over {\beta - \alpha }}\left( {{1 \over {s + \alpha }} - {1 \over {s + \beta }}} \right) = {1 \over {\left( {s + \alpha } \right)\left( {s + \beta } \right)}}$

（12） 由于： $L\left[ {e^{ - t} \cos \left( {\omega t} \right)} \right] = {{s + 1} \over {\left( {s + 1} \right)^2 + \omega ^2 }}$

所以： $L\left[ {e^{ - \left( {t + a} \right)} \cdot \cos \left( {\omega t} \right)} \right] = {{\left( {s + 1} \right)e^{ - a} } \over {\left( {s + 1} \right)^2 + \omega ^2 }}$

（13） 因为：
$t \cdot e^{ - \left( {t - 2} \right)} u\left( {t - 1} \right) = e \cdot \left[ {\left( {t - 1} \right) \cdot e^{ - \left( {t - 1} \right)} + e^{ - \left( {t - 1} \right)} } \right] \cdot u\left( {t - 1} \right)$

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

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

（14） 由拉普拉斯变换的s域平移特性： $L\left[ {e^{ - t} f\left( t \right)} \right] = F\left( {s + 1} \right)$
由尺度变换性质，得到：
$L\left[ {e^{ - {t \over a}} f\left( {{t \over a}} \right)} \right] = aF\left( {as + 1} \right)$

（15） 由拉普拉斯变换的尺度特性： $L\left[ {f\left( {{t \over a}} \right)} \right] = aF\left( {as} \right)$

再由s域平移性质，可得： $L\left[ {e^{ - at} f\left( {{t \over a}} \right)} \right] = aF\left[ {a\left( {s + a} \right)} \right] = aF\left( {as + a^2 } \right)$

（16） 根据： $\cos ^3 \left( {3t} \right) = \cos \left( {3t} \right)\cdot{{1 + \cos \left( {6t} \right)} \over 2} = {1 \over 4}\cos \left( {9t} \right) + {3 \over 4}\cos \left( t \right)$

所以： $L\left[ {\cos ^3 \left( {3t} \right)} \right] = {1 \over 4} \cdot {s \over {s^2 + 81}} + {3 \over 4} \cdot {s \over {s^2 + 9}}$

再根据s域的微分性质，可得： $L\left[ {t \cdot \cos ^3 \left( {3t} \right)} \right] = - {d \over {ds}}\left( {{1 \over 4}{s \over {s^2 + 81}} + {3 \over 4} \cdot {s \over {s^2 + 9}}} \right) = {1 \over 4}\left[ {{{s^2 - 81} \over {\left( {s^2 + 81} \right)^2 }} + {{3s^2 - 27} \over {\left( {s^2 + 9} \right)^2 }}} \right]$
（17） 根据： $L\left[ {\cos \left( {2t} \right)} \right] = {s \over {s^2 + 4}}$

连续两次应用s域微分性质，可得： $L\left[ {t \cdot \cos \left( {2t} \right)} \right] = {{s^2 - 4} \over {\left( {s^2 + 4} \right)^2 }}$

$L\left[ {t^2 \cdot \cos \left( {2t} \right)} \right] = {{2s^2 - 24s} \over {\left( {s^2 + 4} \right)^3 }}$

（18） 根据
$L\left[ {1 - e^{ - \alpha t} } \right] = {1 \over s} - {1 \over {s + \alpha }}$

再由s域的积分性质，可得：
$L\left[ {{1 \over t}\left( {1 - e^{ - \alpha t} } \right)} \right] = \int_s^{ + \infty } {\left( {{1 \over {s_1 }} - {1 \over {s_1 + \alpha }}} \right)ds_1 } = \ln \left( {s + \alpha } \right) - \ln s = - \ln \left( {{s \over {s + \alpha }}} \right)$

（19） 根据 $L\left[ {e^{ - 3t} - e^{ - 5t} } \right] = {1 \over {s + 3}} - {1 \over {s + 5}}$

再由s域的积分性质可得：
$L\left[ {{{e^{ - 3t} - e^{ - 5t} } \over t}} \right] = \int_s^{ + \infty } {\left( {{1 \over {s_1 + 3}} - {1 \over {s_1 + 5}}} \right)ds_1 } = \ln \left( {{{s + 5} \over {s + 3}}} \right)$

（20） 根据： $L\left[ {\sin \left( {\alpha t} \right)} \right] = {\alpha \over {s^2 + \alpha ^2 }}$

再由s域的积分性质可得：
$L\left[ {{{\sin \left( {\alpha t} \right)} \over t}} \right] = \int_s^{ + \infty } {{\alpha \over {s_1^2 + \alpha ^2 }}ds_1 } = \int_s^{ + \infty } {{1 \over {\left( {{{s_1 } \over \alpha }} \right)^2 + 1}}d\left( {{{s_1 } \over \alpha }} \right)} = {\pi \over 2} - \arctan \left( {{s \over \alpha }} \right) = \arctan \left( {{\alpha \over s}} \right)$

02第二题

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求解：

（1）解答：
$X\left( z \right) = \sum\limits_{n = 0}^\infty {\left( {{1 \over 4}} \right)^n z^{ - n} } = {{4z} \over {4z - 1}} = {z \over {z - {1 \over 4}}},\,\,\left| z \right| > {1 \over 4}$

（2）解答：
$X\left( z \right) = \sum\limits_{n = 0}^\infty {\left( { - {1 \over 3}} \right)^n z^{ - n} } = {z \over {z + {1 \over 3}}} = {{3z} \over {3z + 1}},\;\;\;\;\left| z \right| > {1 \over 3}$

（3）解答：
$X\left( z \right) = \sum\limits_{n = - \infty }^0 {\left( {{1 \over 2}} \right)^n z^{ - n} } = {1 \over {1 - 2z}},\,\,\,\,\left| z \right| < {1 \over 2}$

（4）解答：
$X\left( z \right) = \sum\limits_{n = - \infty }^0 {\left( { - {1 \over 2}} \right)^n z^{ - n} } = {1 \over {1 + 2z}},\,\,\,\,\,\left| z \right| < {1 \over 2}$

（5）解答：
$X\left( z \right) = \sum\limits_{n = - \infty }^{ - 1} { - \left( {{1 \over 6}} \right)^{n - 1} z^{ - n} u\left[ { - n - 1} \right]} = {{36z} \over {6z - 1}},\,\,\,\,\left| z \right| < {1 \over 6}$

（6）解答：
$X\left( z \right) = {1 \over {z - 1}},\,\,\,\,\left| z \right| > 1$

（7）解答：
$X\left( z \right) = \sum\limits_{n = 0}^3 {\left( {{1 \over {10}}} \right)^n \cdot z^{ - n} = {{1 - \left( {{1 \over {10z}}} \right)^3 } \over {1 - {1 \over {10z}}}} = {{\left( {10z} \right)^3 - 1} \over {\left( {10z} \right)^2 \left( {10z - 1} \right)}},\,\,\,\,\,\left| z \right| > 0}$

（8）解答：
$X\left( z \right) = {z \over {z - {1 \over 5}}} + {z \over {z - {1 \over 6}}} = {{z\left( {60z - 11} \right)} \over {\left( {5z - 1} \right) \cdot \left( {6z - 1} \right)}},\,\,\,\,\left| z \right| > {1 \over 5}$

（9）解答：
$X\left( z \right) = 1 - {1 \over 2}z^{ - 8} = {{2z^8 - 1} \over {2z^8 }},\,\,\,\,\left| z \right| > 0$

03第三题

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求双边序列 $x\left[ n \right] = \left( {{1 \over a}} \right)^{\left| n \right|} ,\,\,\,\,\left| a \right| > 1$
的z变换，并表明收敛于及绘制出零极点图。

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求解：

将双边序列分成右边和左边序列，分别求出各自的Z变换。
$x_R \left[ n \right] = \left( {{1 \over a}} \right)^n ,\,\,\,\,\,n \ge 0,\,\,\,\,\,\,x_L \left[ n \right] = \left( {{1 \over a}} \right)^{ - n} = a^n ,\,\,\,\,\,n < 0$

$Z\left\{ {x_R \left[ n \right]} \right\} = {z \over {z - {1 \over a}}},\,\,\,\,\left| z \right| > {1 \over a}$

$Z\left\{ {x_L \left[ n \right]} \right\} = \sum\limits_{n = - \infty }^{ - 1} {a^n z^{ - n} } = \sum\limits_{n = 1}^\infty {z^n a^{ - n} }$ $= \sum\limits_{n = 0}^\infty {z^n a^{ - n} } - 1 = {1 \over {1 - za^{ - 1} }} - 1 = {{ - z} \over {z - a}}$

$Z\left\{ {x\left[ n \right]} \right\} = Z\left\{ {x_R \left[ n \right] + x_L \left[ n \right]} \right\}$ $= {z \over {z - {1 \over a}}} + {{ - z} \over {z - a}} = {{\left( {1 - a^2 } \right)z} \over {az^2 - \left( {1 + a^2 } \right)z + a}}$

04第四题

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求解：

（1）解答

$L^{ - 1} \left( {{1 \over {s + 2}}} \right) = e^{ - 2t} ,\,\,\,\,t \ge 0$

（2）解答：
$L^{ - 1} \left( {{4 \over {2s + 5}}} \right) = L^{ - 1} \left( {{2 \over {s + {5 \over 2}}}} \right) = 2 \cdot e^{ - {5 \over 2}t} \,\,\,\,t \ge 1$

（3）解答：
${2 \over {s\left( {2s + 3} \right)}} = {{{2 \over 3}} \over s} + {{ - {2 \over 3}} \over {s + {3 \over 2}}}$

$L^{ - 1} \left( {{{{2 \over 3}} \over s} + {{ - {2 \over 3}} \over {s + {3 \over 2}}}} \right) = {2 \over 3}u\left( t \right) - {2 \over 3}e^{ - {3 \over 2}t} .u\left( t \right)$

（4）解答：
${1 \over {s\left( {s^2 + 3} \right)}} = {{{1 \over 3}} \over s} + {{ - {1 \over 3}s} \over {s^2 + 3}}$

$L^{ - 1} \left( {{{{1 \over 3}} \over s} - {{{1 \over 3}s} \over {s^2 + 3}}} \right) = {1 \over 3}u\left( t \right) - {1 \over 3}\cos \sqrt 3 t \cdot u\left( t \right)$

（5）解答：

${3 \over {\left( {s + 4} \right)\left( {s + 3} \right)}} = {3 \over {s + 3}} - {3 \over {s + 4}}$

$L^{ - 1} \left[ {{3 \over {s + 3}} - {3 \over {s + 4}}} \right] = 3e^{ - 3t} - 3e^{ - 4t} ,\,\,\,\,t \ge 0$

（6）解答：
${{3s} \over {\left( {s + 4} \right)\left( {s + 3} \right)}} = {{ - 9} \over {s + 3}} + {{12} \over {s + 4}}$

$L^{ - 1} \left( {{{ - 9} \over {s + 3}} + {{12} \over {s + 4}}} \right) = - 9e^{ - 3t} + 12e^{ - 4t} ,\,\,\,\,t \ge 0$

（7）解答：
$L^{ - 1} \left( {{1 \over {s^2 + 1}} + 1} \right) = \sin t + \delta \left( t \right),\,\,\,\,t \ge 0$

（8）解答：
${1 \over {s^2 - 3s + 2}} = {{ - 1} \over {s - 1}} + {1 \over {s - 2}}$

$L^{ - 1} \left( {{{ - 1} \over {s - 1}} + {1 \over {s - 2}}} \right) = - e^t + e^{2t} ,\,\,\,\,t \ge 0$

（9）解答：
${1 \over {s\left( {RCs + 1} \right)}} = {1 \over s} - {1 \over {s + {1 \over {RC}}}}$

$L^{ - 1} \left[ {{1 \over s} - {1 \over {s + {1 \over {RC}}}}} \right] = 1 - e^{ - {1 \over {RC}}t} ,\,\,\,\,t \ge 0$

（10）解答：
${{1 - RCs} \over {s\left( {1 + RCs} \right)}} = {1 \over s} - {2 \over {s + {1 \over {RC}}}}$

$L^{ - 1} \left( {{1 \over s} - {2 \over {s + {1 \over {RC}}}}} \right) = 1 - 2e^{ - {t \over {RC}}} ,\,\,\,\,t \ge 0$

（11）解答：
${\omega \over {\left( {s^2 + \omega ^2 } \right)}} \cdot {1 \over {\left( {RCs + 1} \right)}}$

${\omega \over {s^2 + \omega ^2 }} \cdot {{{1 \over {RC}}} \over {s + {1 \over {RC}}}} = {{ms + n} \over {s^2 + \omega ^2 }} \cdot {l \over {s + {1 \over {RC}}}}$

${{ms^2 + {m \over {RC}}s + ns + {n \over {RC}} + ls^2 + l\omega ^2 } \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}} = {{{\omega \over {RC}}} \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}}$

${{\left( {m + l} \right)s^2 + \left( {n + {m \over {RC}}} \right)s + l\omega ^2 } \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}} = {{{\omega \over {RC}}} \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}}$

KaTeX parse error: Undefined control sequence: \matrix at position 10: \left\{ {\̲m̲a̲t̲r̲i̲x̲{ {m + l = 0} ${n + {m \over {RC}} = 0}$ ${{n \over {RC}} + l\omega ^2 = {\omega \over {RC}}}$

KaTeX parse error: Undefined control sequence: \matrix at position 10: \left\{ {\̲m̲a̲t̲r̲i̲x̲{ {m = {{ - R… ${n = {\omega \over {1 + \left( {RC\omega } \right)^2 }}}$ ${l = {{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}}$

${\omega \over {s^2 + \omega ^2 }} \cdot {1 \over {RCs + 1}} = {{{{ - RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}s + {\omega \over {1 + \left( {RC\omega } \right)^2 }}} \over {s^2 + \omega ^2 }} + {{{{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}} \over {s + {1 \over {RC}}}}$

${{ - RC} \over {1 + \left( {RC\omega } \right)^2 }}\cos \left( {\omega t} \right) + {1 \over {1 + \left( {RC\omega } \right)^2 }}\sin \left( {\omega t} \right) + {{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}e^{ - {t \over {RC}}} ,\,\,\,\,\,t \ge 0$

（12）解答：
${{4s + 5} \over {s^2 + 5s + 6}} = {{4s + 5} \over {\left( {s + 2} \right)\left( {s + 3} \right)}} = {{ - 3} \over {s + 2}} + {7 \over {s + 3}}$

$L^{ - 1} \left( {{{4s + 5} \over {s^2 + 5s + 6}}} \right) = - e^{ - 2t} + 7e^{ - 3t} ,\,\,\,\,t \ge 0$

（13）解答：
${{100\left( {s + 50} \right)} \over {s^2 + 201s + 200}} = {{{{4900} \over {199}}} \over {s + 1}} + {{{{15000} \over {199}}} \over {s + 200}}$

$L^{ - 1} \left( {{{100\left( {s + 50} \right)} \over {s^2 + 201s + 200}}} \right) = {{4900} \over {199}}e^{ - t} + {{15000} \over {199}}e^{ - 200t} ,\,\,\,\,\,t \ge 0$

（14）解答：
${{s + 3} \over {\left( {s + 1} \right)^3 \cdot \left( {s + 2} \right)}} = {A \over {\left( {s + 1} \right)^3 }} + {B \over {\left( {s + 1} \right)^2 }} + {C \over {s + 1}} + {D \over {s + 2}}$

$D = \left. {{{s + 3} \over {\left( {s + 1} \right)^3 }}} \right|_{s = - 2} = - 1$

$A = \left. {{{s + 3} \over {s + 2}}} \right|_{s = - 1} = 2$

$B = \left. {{d \over {ds}}\left( {{{s + 3} \over {s + 2}}} \right)} \right|_{s = - 1} = \left. {{1 \over {s + 2}} - {{s + 3} \over {\left( {s + 2} \right)^2 }}} \right|_{s = - 1} = - 1$

$C = \left. {{1 \over 2} \cdot {{d^2 } \over {ds^2 }}\left( {{{s + 3} \over {s + 2}}} \right)} \right|_{s = - 1} = \left. {{1 \over 2}{{2\left( {s + 3} \right)} \over {\left( {s + 2} \right)^3 }} - {2 \over {\left( {s + 2} \right)^2 }}} \right|_{s = - 1} = 1$

$L^{ - 1} \left( {{{s + 3} \over {\left( {s + 1} \right)^3 \left( {s + 2} \right)}}} \right) = t^2 e^{ - t} - te^{ - t} + e^{ - t} - e^{ - 2t} ,\,\,\,\,\,t \ge 0$

05第五题

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求下列函数的拉普拉斯逆变换：
$F\left( s \right) = \ln \left( {{s \over {s + 9}}} \right)$

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求解：

本题应用到Laplace变换的一个性质，如下：
$F\left( s \right) = \int_{0_ - }^\infty {f\left( t \right)e^{ - st} dt}$ $F'\left( s \right) = {d \over {ds}}\int_{0_ - }^\infty {f\left( t \right)e^{ - st} dt} = \int_{0_ - }^\infty { - t \cdot f\left( t \right)e^{ - st} dt}$ $t \cdot f\left( t \right) = - L^{ - 1} \left[ {{{dF\left( s \right)} \over {ds}}} \right]$

$f\left( t \right) = L^{ - 1} \left[ {\ln \left( {{s \over {s + 9}}} \right)} \right]$

$- t \cdot f\left( t \right) = L^{ - 1} \left[ {{d \over {ds}}\ln \left( {{s \over {s + 9}}} \right)} \right] = L^{ - 1} \left( { - {1 \over {s + 9}} + {1 \over s}} \right)\, = 1 - e^{ - 9t} ,\,\,\,\,t \ge 0$

$f\left( t \right) = {{ - 1} \over t} + {1 \over t}e^{ - 9t} ,\,\,\,\,t \ge 0$

如果使用MATLAB进行求解，结果如下：

>>ilaplace(log(s/(s+9)))
ans=(exp(-9*t)-1)/t

06第六题

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求解：

（1） $x\left[ n \right] = \delta \left[ n \right]$
（2） $x\left[ n \right] = \delta \left[ {n + 3} \right]$
（3） $x\left[ n \right] = \delta \left[ {n - 1} \right]$
（4） $x\left[ n \right] = - 2\delta \left[ {n - 2} \right] + \delta \left[ n \right] + 2\delta \left[ {n + 1} \right]$
（5） $x\left[ n \right] = a^n u\left[ n \right]$
（6） $x\left[ n \right] = - a^n u\left[ { - n - 1} \right]$

07第七题

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求解：

（1）求解：
$X\left( z \right) = {z \over {z + 0.5}}$

由于 $\left| z \right| > 5$，所以序列是右边序列，因此序列为： $x\left[ n \right] = \left( { - 0.5} \right)^n u\left[ n \right]$

（2）求解：
${{X\left( z \right)} \over z} = {{z - 0.5} \over {\left( {z + {1 \over 4}} \right) \cdot \left( {z + {1 \over 2}} \right)}} = {{ - 3} \over {z + {1 \over 4}}} + {4 \over {z + {1 \over 2}}}$
由于 $\left| z \right| > 5$，所以序列是右边序列，因此序列为： $x\left[ n \right] = \left[ { - 3\left( {{{ - 1} \over 4}} \right)^n + 4\left( {{{ - 1} \over 2}} \right)^n } \right] \cdot u\left[ n \right]$

$= \left( { - {1 \over 4}} \right)^n \left( { - 3 + 2^{n + 2} } \right) \cdot u\left[ n \right]$

（3）求解：
$X\left( z \right) = {{1 - {1 \over 2}z^{ - 1} } \over {1 - {1 \over 4}z^{ - 2} }} = {z \over {z + {1 \over 2}}}$ $\left| z \right| > {1 \over 2}\;\;\;$ $x\left[ n \right] = \left( { - {1 \over 2}} \right)^n \cdot u\left[ n \right]$

（4）求解：
${{X\left( z \right)} \over z} = {{z - a} \over {z\left( {1 - az} \right)}} = {{ - {1 \over a}\left( {z - a} \right)} \over {z\left( {z - {1 \over a}} \right)}} = {{ - a} \over z} + {{a - {1 \over a}} \over {z - {1 \over a}}}$ $\left| z \right| > \left| {{1 \over a}} \right|,$ $x\left[ n \right] = - a\delta \left[ n \right] + \left( {a - {1 \over a}} \right) \cdot \left( {{1 \over a}} \right)^n \cdot u\left[ n \right]$

08第八题

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利用三种方法求解下面 $X\left( z \right)$的你变换 $x\left[ n \right]$。

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

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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

09第九题

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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)

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