作业要求链接： 信号与系统2022春季作业-第九次作业: https://zhuoqing.blog.csdn.net/article/details/124319518
§09 参考答案
1.1 信号的采样与恢复
1.1.1 采样信号频谱
◎ 求解： 对三角形 f 1 ( t ) f_1 \left( t \right) f1(t) 进行抽样之后，得到 f 1 s ( t ) = ∑ n = − ∞ + ∞ f 1 ( n ⋅ τ 8 ) δ ( t − n ⋅ τ 8 ) f_{1s} \left( t \right) = \sum\limits_{n = - \infty }^{ + \infty } {f_1 \left( {n \cdot {\tau \over 8}} \right)\delta \left( {t - n \cdot {\tau \over 8}} \right)} f1s(t)=n=−∞∑+∞f1(n⋅8τ)δ(t−n⋅8τ) 三角信号的频谱 F 1 ( ω ) = E τ 2 S a 2 ( ω τ 4 ) F_1 \left( \omega \right) = { {E\tau } \over 2}Sa^2 \left( { { {\omega \tau } \over 4}} \right) F1(ω)=2EτSa2(4ωτ) 采样后信号的频谱 F 1 s ( ω ) = 4 E ∑ n = − ∞ + ∞ S a 2 [ τ 4 ( ω − 16 n π τ ) ] F_{1s} \left( \omega \right) = 4E\sum\limits_{n = - \infty }^{ + \infty } {Sa^2 \left[ { {\tau \over 4}\left( {\omega - { {16n\pi } \over \tau }} \right)} \right]} F1s(ω)=4En=−∞∑+∞Sa2[4τ(ω−τ16nπ)]
三角形脉冲信号抽样信号与周期频谱如下：
▲ 图1.1.1 三角脉冲抽样信号以及对应的周期频谱
下面是使用Python 绘制出三角脉冲信号的频谱。
▲ 图1.1.2 在E=1的情况下绘制的0~6pi的频谱
from headm import *
Ts = 1
E = 1
f = [E/4, E/2, E*3/4, E, E*3/4, E/2, E/4]
omiga = linspace(0, 6*pi, 500)
fabs = []
fa = array(f)
for o in omiga:
expo = array(exp(-1j*o)**list(range(len(f))))
result = abs(sum(f*expo))
fabs.append(result)
plt.plot(omiga, fabs)
plt.xlabel("omiga")
plt.ylabel("fabs")
plt.grid(True)
plt.tight_layout()
plt.show()
对于升余弦信号，单个升余弦信号的频谱为 F 2 ( ω ) = E τ 2 S a ( ω τ 2 ) 1 − ( ω τ 2 π ) 2 F_2 \left( \omega \right) = {
{E\tau } \over 2}{
{Sa\left( {
{
{\omega \tau } \over 2}} \right)} \over {1 - \left( {
{
{\omega \tau } \over {2\pi }}} \right)^2 }} F2(ω)=2Eτ1−(2πωτ)2Sa(2ωτ) 它对应的抽样信号的频谱为：
F 2 s ( ω ) = 4 E ∑ n = − ∞ + ∞ S a [ τ 2 ( ω − 16 n π τ ) ] 1 − [ τ ( ω − 16 n π τ ) 2 π ] 2 F_{2s} \left( \omega \right) = 4E\sum\limits_{n = - \infty }^{ + \infty } {
{
{Sa\left[ {
{\tau \over 2}\left( {\omega - {
{16n\pi } \over \tau }} \right)} \right]} \over {1 - \left[ {
{
{\tau \left( {\omega - {
{16n\pi } \over \tau }} \right)} \over {2\pi }}} \right]^2 }}} F2s(ω)=4En=−∞∑+∞1−[2πτ(ω−τ16nπ)]2Sa[2τ(ω−τ16nπ)]
▲ 图1.1.3 升余弦信号抽样信号及其周期频谱
如下为Python绘制的采样频谱的幅度谱：
▲ 图1.1.4 利用Python绘制离散抽样信号的频谱
f = (cos(linspace(-pi, pi, 9)) + 1)/2
omiga = linspace(0, 6*pi, 500)
fabs = []
fa = array(f)
for o in omiga:
expo = array(exp(-1j*o)**list(range(len(f))))
result = abs(sum(f*expo))
fabs.append(result)
plt.plot(omiga, fabs)
clipboard.copy(str(fabs))
1.1.2 矩形信号采样频率
◎ 求解： 高度为 E E E ，宽度为 τ \tau τ 矩形脉冲信号的频谱为 F ( ω ) = E τ S a ( ω τ 2 ) = E τ sin ( ω τ 2 ) ω τ 2 = 2 E ω sin ( ω τ 2 ) F\left( \omega \right) = E\tau Sa\left( { { {\omega \tau } \over 2}} \right) = E\tau { {\sin \left( { { {\omega \tau } \over 2}} \right)} \over { { {\omega \tau } \over 2}}} = { {2E} \over \omega }\sin \left( { { {\omega \tau } \over 2}} \right) F(ω)=EτSa(2ωτ)=Eτ2ωτsin(2ωτ)=ω2Esin(2ωτ) 频谱每个过零点对应的频率为 f n = n τ = n ⋅ 1 2.5 m s = 400 ⋅ n ( H z ) f_n = {n \over \tau } = n \cdot {1 \over {2.5ms}} = 400 \cdot n\,\,\left( {Hz} \right) fn=τn=n⋅2.5ms1=400⋅n(Hz) 第八个过零点对应的频率为 f 8 = 8 × 400 = 3200 ( H z ) f_8 = 8 \times 400 = 3200\,\,\left( {Hz} \right) f8=8×400=3200(Hz) 根据抽样定理， 如果需要保存 3200 Hz 对应的频率分量，则对该信号抽样的最小频率为： 2 × 3200 = 6400 ( H z ) 2 \times 3200 = 6400\,\,\left( {Hz} \right) 2×3200=6400(Hz)
▲ 图1.1.5 第八个过零点对应的频率
1.2 拉普拉斯变换与z变换
1.2.1 求信号的拉普拉斯变换
在缺省的情况下，使用 0 − 0_ - 0− 系统单边拉普拉斯变换求解。
◎ 求解：
（1） L [ 1 − e − a t ] = 1 s − 1 s + a = a s ( s + a ) L\left[ {1 - e^{ - at} } \right] = {1 \over s} - {1 \over {s + a}} = {a \over {s\left( {s + a} \right)}} L[1−e−at]=s1−s+a1=s(s+a)a
（2） L [ sin t + 2 cos t ] = 1 s 2 + 1 + 2 s s 2 + 1 = 2 s + 1 s 2 + 1 L\left[ {\sin t + 2\cos t} \right] = {1 \over {s^2 + 1}} + {
{2s} \over {s^2 + 1}} = {
{2s + 1} \over {s^2 + 1}} L[sint+2cost]=s2+11+s2+12s=s2+12s+1
（3） L [ t ⋅ e − 2 t ] = − d d s ( 1 s + 2 ) = 1 ( s + 2 ) 2 L\left[ {t \cdot e^{ - 2t} } \right] = - {d \over {ds}}\left( {
{1 \over {s + 2}}} \right) = {1 \over {\left( {s + 2} \right)^2 }} L[t⋅e−2t]=−dsd(s+21)=(s+2)21
（4） 根据 L [ sin 2 t ] = 2 s 2 + 4 L\left[ {\sin 2t} \right] = {2 \over {s^2 + 4}} L[sin2t]=s2+42 ，根据 s 域平移性质，得到： L [ e − t sin 2 t ] = 2 ( s + 1 ) 2 + 4 L\left[ {e^{ - t} \sin 2t} \right] = {2 \over {\left( {s + 1} \right)^2 + 4}} L[e−tsin2t]=(s+1)2+42
（5） 因为 L [ t n ] = n ! s n + 1 L\left[ {t^n } \right] = {
{n!} \over {s^{n + 1} }} L[tn]=sn+1n! ，所以 L [ 1 + 2 t ] = 1 s + 2 s 2 = s + 2 s 2 L\left[ {1 + 2t} \right] = {1 \over s} + {2 \over {s^2 }} = {
{s + 2} \over {s^2 }} L[1+2t]=s1+s22=s2s+2 再由 s 域的平移性质，得： L [ ( 1 + 2 t ) ⋅ e − t ] = s + 3 ( s + 1 ) 2 L\left[ {\left( {1 + 2t} \right) \cdot e^{ - t} } \right] = {
{s + 3} \over {\left( {s + 1} \right)^2 }} L[(1+2t)⋅e−t]=(s+1)2s+3
（6） 先求得 L [ 1 − cos α t ] = 1 s − s s 2 + α 2 L\left[ {1 - \cos \alpha t} \right] = {1 \over s} - {s \over {s^2 + \alpha ^2 }} L[1−cosαt]=s1−s2+α2s 再根据 s 域平移性质，得到： L [ ( 1 − cos a t ) e − β t ] = 1 s + β − s + β ( s + β ) 2 + α 2 L\left[ {\left( {1 - \cos at} \right)e^{ - \beta t} } \right] = {1 \over {s + \beta }} - {
{s + \beta } \over {\left( {s + \beta } \right)^2 + \alpha ^2 }} L[(1−cosat)e−βt]=s+β1−(s+β)2+α2s+β
（7） L [ t 2 + 2 t ] = 2 s 3 + 2 s 2 = 2 ( s + 1 ) s 3 L\left[ {t^2 + 2t} \right] = {2 \over {s^3 }} + {2 \over {s^2 }} = {
{2\left( {s + 1} \right)} \over {s^3 }} L[t2+2t]=s32+s22=s32(s+1)
（8） L [ 2 δ ( t ) − 3 e − 7 t ] = 2 − 3 s + 7 = 2 s + 11 s + 7 L\left[ {2\delta \left( t \right) - 3e^{ - 7t} } \right] = 2 - {3 \over {s + 7}} = {
{2s + 11} \over {s + 7}} L[2δ(t)−3e−7t]=2−s+73=s+72s+11
（9） 根据 L [ sinh β t ] = β s 2 − β 2 L\left[ {\sinh \beta t} \right] = {\beta \over {s^2 - \beta ^2 }} L[sinhβt]=s2−β2β ，再由 s 域平移性质，得： L [ e − α t sinh β t ] = β ( s + α ) 2 − β 2 L\left[ {e^{ - \alpha t} \sinh \beta t} \right] = {\beta \over {\left( {s + \alpha } \right)^2 - \beta ^2 }} L[e−αtsinhβt]=(s+α)2−β2β
（10） 因为 cos 2 Ω t = 1 2 + 1 2 cos 2 Ω t \cos ^2 \Omega t = {1 \over 2} + {1 \over 2}\cos 2\Omega t cos2Ωt=21+21cos2Ωt ，所以： L [ cos 2 Ω t ] = 1 2 s + 1 2 ⋅ s s 2 + 4 Ω 2 = 1 2 ( 1 s + s s 2 + 4 Ω 2 ) L\left[ {\cos ^2 \Omega t} \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) L[cos2Ωt]=2s1+21⋅s2+4Ω2s=21(s1+s2+4Ω2s)
（11） L [ 1 β − α ( e − α t − e − β t ) ] = 1 β − α ( 1 s + α − 1 s + β ) = 1 ( s + α ) ( s + β ) 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)}} L[β−α1(e−αt−e−βt)]=β−α1(s+α1−s+β1)=(s+α)(s+β)1
（12） 由于 L [ e − t cos ω t ] = s + 1 ( s + 1 ) 2 + ω 2 L\left[ {e^{ - t} \cos \omega t} \right] = {
{s + 1} \over {\left( {s + 1} \right)^2 + \omega ^2 }} L[e−tcosωt]=(s+1)2+ω2s+1 ，所以： L [ e − ( t + a ) cos ω t ] = s + 1 ( s + 1 ) 2 + ω 2 ⋅ e − a L\left[ {e^{ - \left( {t + a} \right)} \cos \omega t} \right] = {
{s + 1} \over {\left( {s + 1} \right)^2 + \omega ^2 }} \cdot e^{ - a} L[e−(t+a)cosωt]=(s+1)2+ω2s+1⋅e−a
（13） 因为 t ⋅ e − ( t − 2 ) ⋅ u ( t − 1 ) = e ⋅ [ ( t − 1 ) e − ( t − 1 ) + e − ( t − 1 ) ] ⋅ u ( t − 1 ) t \cdot e^{ - \left( {t - 2} \right)} \cdot u\left( {t - 1} \right) = e \cdot \left[ {\left( {t - 1} \right)e^{ - \left( {t - 1} \right)} + e^{ - \left( {t - 1} \right)} } \right] \cdot u\left( {t - 1} \right) t⋅e−(t−2)⋅u(t−1)=e⋅[(t−1)e−(t−1)+e−(t−1)]⋅u(t−1) 其中 L [ ( t − 1 ) e − ( t − 1 ) u ( t − 1 ) ] = e − s ( s + 1 ) 2 , L [ e − ( t − 1 ) u ( t − 1 ) ] = e − s s + 1 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}} L[(t−1)e−(t−1)u(t−1)]=(s+1)2e−s,L[e−(t−1)u(t−1)]=s+1e−s 所以 L [ t ⋅ e − t ( t − 2 ) u ( t − 1 ) ] = e [ 1 ( s + 1 ) 2 + 1 s + 1 ] ⋅ e − s = ( s + 2 ) e − ( s − 1 ) ( s + 1 ) 2 L\left[ {t \cdot e^{ - t\left( {t - 2} \right)} u\left( {t - 1} \right)} \right] = e\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 }} L[t⋅e−t(t−2)u(t−1)]=e[(s+1)21+s+11]⋅e−s=(s+1)2(s+2)e−(s−1)
（14） 根据 s 域 平移性质： L [ e − t f ( t ) ] = F ( s + 1 ) L\left[ {e^{ - t} f\left( t \right)} \right] = F\left( {s + 1} \right) L[e−tf(t)]=F(s+1) 再由LT尺度特性： L [ e − t a f ( t a ) ] = a F ( a s + 1 ) L\left[ {e^{ - {t \over a}} f\left( {
{t \over a}} \right)} \right] = aF\left( {as + 1} \right) L[e−atf(at)]=aF(as+1)
（15） 根据 L [ f ( t a ) ] = a F ( a s ) L\left[ {f\left( {
{t \over a}} \right)} \right] = aF\left( {as} \right) L[f(at)]=aF(as) ， 再有 s 域平移性质， 得： L [ e − a t f ( t a ) ] = a F [ a ( s + a ) ] = a F ( a s + a 2 ) 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) L[e−atf(at)]=aF[a(s+a)]=aF(as+a2)
（16） 先根据三角恒等式 cos 3 3 t = cos 3 t ⋅ 1 + cos 6 t 2 = 1 2 cos 3 t + 1 2 [ 1 2 ( cos 9 t + cos 3 t ) ] = 1 4 cos 9 t + 3 4 cos 3 t \cos ^3 3t = \cos 3t \cdot {
{1 + \cos 6t} \over 2} = {1 \over 2}\cos 3t + {1 \over 2}\left[ {
{1 \over 2}\left( {\cos 9t + \cos 3t} \right)} \right] = {1 \over 4}\cos 9t + {3 \over 4}\cos 3t cos33t=cos3t⋅21+cos6t=21cos3t+21[21(cos9t+cos3t)]=41cos9t+43cos3t 所以 L [ cos 3 3 t ] = 1 4 ⋅ s s 2 + 81 + 3 4 ⋅ s s 2 + 9 L\left[ {\cos ^3 3t} \right] = {1 \over 4} \cdot {s \over {s^2 + 81}} + {3 \over 4} \cdot {s \over {s^2 + 9}} L[cos33t]=41⋅s2+81s+43⋅s2+9s 再由 s 域分为性质，得： L [ t ⋅ cos 3 3 t ] = − d d s ( 1 4 s s 2 + 81 + 3 4 s s 2 + 9 ) = 1 4 [ s 2 − 81 ( s 2 + 81 ) 2 + 3 s 2 − 27 ( s 2 + 9 ) 2 ] L\left[ {t \cdot \cos ^3 3t} \right] = - {d \over {ds}}\left( {
{1 \over 4}{s \over {s^2 + 81}} + {3 \over 4}{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] L[t⋅cos33t]=−dsd(41s2+81s+43s2+9s)=41[(s2+81)2s2−81+(s2+9)23s2−27]
（17） 根据 L ( cos 2 t ) = s s 2 + 4 L\left( {\cos 2t} \right) = {s \over {s^2 + 4}} L(cos2t)=s2+4s ，连续应用 s 域微分性质两次，得： L [ t ⋅ cos 2 t ] = s 2 − 4 ( s 2 + 4 ) 2 L\left[ {t \cdot \cos 2t} \right] = {
{s^2 - 4} \over {\left( {s^2 + 4} \right)^2 }} L[t⋅cos2t]=(s2+4)2s2−4 L [ t 2 ⋅ cos 2 t ] = 2 s 2 − 24 s ( s 2 + 4 ) 3 L\left[ {t^2 \cdot \cos 2t} \right] = {
{2s^2 - 24s} \over {\left( {s^2 + 4} \right)^3 }} L[t2⋅cos2t]=(s2+4)32s2−24s
（18） 根据 L [ 1 − e − α t ] = 1 s − 1 s + α L\left[ {1 - e^{ - \alpha t} } \right] = {1 \over s} - {1 \over {s + \alpha }} L[1−e−αt]=s1−s+α1 由 s 域的积分性质，得： L [ 1 t ( 1 − e − α t ) ] = ∫ s ∞ ( 1 s 1 − 1 s 1 + α ) d s 1 = ln ( s + α ) − ln s = − ln s s + α 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 {s \over {s + \alpha }} L[t1(1−e−αt)]=∫s∞(s11−s1+α1)ds1=ln(s+α)−lns=−lns+αs
（19） 由 L [ e − 3 t − e − 5 t ] = 1 s + 3 − 1 s + 5 L\left[ {e^{ - 3t} - e^{ - 5t} } \right] = {1 \over {s + 3}} - {1 \over {s + 5}} L[e−3t−e−5t]=s+31−s+51 再由 s 域积分性质，得： L [ e − 3 t − e − 5 t t ] = ∫ s ∞ ( 1 s 1 + 3 − 1 s 1 + 5 ) d s 1 = ln s + 5 s + 3 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 {
{s + 5} \over {s + 3}} L[te−3t−e−5t]=∫s∞(s1+31−s1+51)ds1=lns+3s+5
（20） 根据 L [ sin α t ] = α s 2 + α 2 L\left[ {\sin \alpha t} \right] = {\alpha \over {s^2 + \alpha ^2 }} L[sinαt]=s2+α2α 由 s 域的积分性质，得： L [ sin α t t ] = ∫ s ∞ α s 1 2 + α 2 d s 1 = ∫ s ∞ 1 ( s 1 α ) 2 + 1 d ( s 1 α ) = π 2 − arctan ( s α ) = arctan ( α s ) L\left[ {
{
{\sin \alpha t} \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) L[tsinαt]=∫s∞s12+α2αds1=∫s∞(αs1)2+11d(αs1)=2π−arctan(αs)=arctan(sα)
■ 下面给出MATLAB下计算结果
( 1 ) 1 − e − a t \left( 1 \right)\,\,1 - e^{ - at} (1)1−e−at
laplace(1-exp(-a*t))’
ans=1/s-1/(a+s)
( 2 ) sin t + 2 cos t \left( 2 \right)\,\,\sin t + 2\cos t (2)sint+2cost
laplace(sin(t)-2*cos(t))’
ans=1/(s2+1)-(2*s)/(s2+1)
( 3 ) t ⋅ e − 2 t \left( 3 \right)\,\,t \cdot e^{ - 2t} (3)t⋅e−2t
laplace(texp(-2t))’
ans=1/(s+2)^2
( 4 ) e − t ⋅ sin ( 2 t ) \left( 4 \right)\,\,e^{ - t} \cdot \sin \left( {2t} \right) (4)e−t⋅sin(2t)
laplace(exp(-t)sin(2t))’
ans=2/((s+1)^2+4)
( 5 ) ( 1 + 2 t ) ⋅ e − t \left( 5 \right)\,\,\left( {1 + 2t} \right) \cdot e^{ - t} (5)(1+2t)⋅e−t
laplace((1+2*t)exp(-t))’
ans=(2(s/2+3/2))/(s+1)^2
( 6 ) [ 1 − cos ( α t ) ] e − β t \left( 6 \right)\,\,\left[ {1 - \cos \left( {\alpha t} \right)} \right]e^{ - \beta t} (6)[1−cos(αt)]e−βt
laplace((1-cos(a*t))exp(-bt))’
ans=1/(b+s)-(b+s)/((b+s)2+a2)
( 7 ) t 2 + 2 t \left( 7 \right)\,\,\,t^2 + 2t (7)t2+2t
laplace(tt+2t)’
ans=2/s2+2/s3
( 8 ) 2 δ ( t ) − 3 e − 7 t \left( 8 \right)\,\,2\delta \left( t \right) - 3e^{ - 7t} (8)2δ(t)−3e−7t
laplace(2dirac(t)-3exp(-7*t))’
ans=2-3/(s+7)
( 9 ) e − α t sinh ( β t ) \left( 9 \right)\,\,\,e^{ - \alpha t} \sinh \left( {\beta t} \right) (9)e−αtsinh(βt)
laplace(exp(-a*t)*sinh(t))’
ans=1/((a+s)^2-1)
( 10 ) cos 2 ( Ω t ) \left( {10} \right)\,\,\cos ^2 \left( {\Omega t} \right) (10)cos2(Ωt)
laplace(cos(omigat).^2)’
ans=(2omiga2+s2)/(s*(4*omiga2+s2))
( 11 ) 1 β − α ( e − α t − e − β t ) \left( {11} \right)\,\,{1 \over {\beta - \alpha }}\left( {e^{ - \alpha t} - e^{ - \beta t} } \right) (11)β−α1(e−αt−e−βt)
laplace((exp(-at)-exp(-bt))/(b-a))’
ans=-(1/(a+s)-1/(b+s))/(a-b)
( 12 ) e − ( t + a ) cos ( ω t ) \left( {12} \right)\,\,\,e^{ - \left( {t + a} \right)} \cos \left( {\omega t} \right) (12)e−(t+a)cos(ωt)
laplace(exp(-(t+a))cos(omigat))’
ans=(s+1)/(exp(a)omiga2+exp(a)*s2+2exp(a)*s+exp(a))
( 13 ) t ⋅ e − ( t − 2 ) u ( t − 1 ) \left( {13} \right)\,\,\,t \cdot e^{ - \left( {t - 2} \right)} u\left( {t - 1} \right) (13)t⋅e−(t−2)u(t−1)
laplace(t*exp(-(t-2))*heaviside(t-1))’
ans=(exp(-s)*exp(1))/(s+1)+(exp(-s)*exp(1))/(s+1)^2
( 14 ) e − t a f ( t a ) \left( {14} \right)\,\,\,e^{ - {t \over a}} f\left( { {t \over a}} \right) (14)e−atf(at)
设已知： L T [ f ( t ) ] = F ( s ) LT\left[ {f\left( t \right)} \right] = F\left( s \right) LT[f(t)]=F(s)
( 15 ) e − a t f ( t a ) \left( {15} \right)\,\,\,e^{ - at} f\left( { {t \over a}} \right) (15)e−atf(at)
设已知： L T [ f ( t ) ] = F ( s ) LT\left[ {f\left( t \right)} \right] = F\left( s \right) LT[f(t)]=F(s)
( 16 ) t ⋅ cos 3 ( 3 t ) \left( {16} \right)\,\,\,t \cdot \cos ^3 \left( {3t} \right) (16)t⋅cos3(3t)
laplace(tcos(3t)^3)’
ans=(3s2)/(2*(s2+9)2)-1/(4*(s2+81))-3/(4(s2+9))+s2/(2*(s2+81)2)
( 17 ) t 2 cos ( 2 t ) \left( {17} \right)\,\,\,t^2 \cos \left( {2t} \right) (17)t2cos(2t)
laplace(ttcos(2t))’
ans=(8s3)/(s2+4)3-(6*s)/(s2+4)^2
( 18 ) 1 t ( 1 − e − a t ) \left( {18} \right)\,\,\,{1 \over t}\left( {1 - e^{ - at} } \right) (18)t1(1−e−at)
laplace((1-exp(-a*t))/t)’
ans=log(a+s)-log(s)
( 19 ) e − 3 t − e − 5 t t \left( {19} \right)\,\,\,{ {e^{ - 3t} - e^{ - 5t} } \over t} (19)te−3t−e−5t
laplace((exp(-3t)-exp(-5t))/t)’
ans=log(s+5)-log(s+3)
( 20 ) sin ( α t ) t \left( {20} \right)\,\,\,{ {\sin \left( {\alpha t} \right)} \over t} (20)tsin(αt)
laplace(sin(a*t)/t)’
ans=atan(a/s)
1.2.2 求信号的z变换
◎ 求解：
（1） X ( z ) = ∑ n = 0 + ∞ ( 1 4 ) n z − n = 4 z 4 z − 1 = z z − 1 4 , ∣ z ∣ > 1 4 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} X(z)=n=0∑+∞(41)nz−n=4z−14z=z−41z,∣z∣>41
（2） X ( z ) = ∑ n = 0 ∞ ( − 1 3 ) n z − n = z z + 1 3 = z 3 z + 1 , ∣ z ∣ > 1 3 X\left( z \right) = \sum\limits_{n = 0}^\infty {\left( { - {1 \over 3}} \right)^n z^{ - n} } = {z \over {z + {1 \over 3}}} = {z \over {3z + 1}},\,\,\,\,\left| z \right| > {1 \over 3} X(z)=n=0∑∞(−31)nz−n=z+31z=3z+1z,∣z∣>31
（3） X ( z ) = ∑ n = − ∞ 0 ( 1 2 ) n z − n = 1 1 − 2 z , ∣ z ∣ < 1 2 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} X(z)=n=−∞∑0(21)nz−n=1−2z1,∣z∣<21
（4） X ( z ) = ∑ n = − ∞ 0 ( − 1 2 ) n z − n = 1 1 + 2 z , ∣ z ∣ < 1 2 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} X(z)=n=−∞∑0(−21)nz−n=1+2z1,∣z∣<21
（5） X ( z ) = ∑ n = − ∞ − 1 − ( 1 6 ) n − 1 z − n ⋅ u [ − n − 1 ] = 36 z 6 z − 1 , ∣ z ∣ < 1 6 X\left( z \right) = \sum\limits_{n = - \infty }^{ - 1} { - \left( {
{1 \over 6}} \right)^{n - 1} z^{ - n} \cdot u\left[ { - n - 1} \right]} = {
{36z} \over {6z - 1}},\,\,\,\left| z \right| < {1 \over 6} X(z)=n=−∞∑−1−(61)n−1z−n⋅u[−n−1]=6z−136z,∣z∣<61
（6） X ( z ) = 1 z − 1 , ∣ z ∣ > 1 X\left( z \right) = {1 \over {z - 1}},\,\,\,\left| z \right| > 1 X(z)=z−11,∣z∣>1
（7） X ( z ) = ∑ n = 0 3 ( 1 10 ) n ⋅ z − n = 1 − ( 1 10 z ) 3 1 − 1 10 z = ( 10 z ) 3 − 1 ( 10 z ) 2 ( 10 z − 1 ) , ∣ z ∣ > 0 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 X(z)=n=0∑3(101)n⋅z−n=1−10z11−(10z1)3=(10z)2(10z−1)(10z)3−1,∣z∣>0
（8） X ( z ) = z z − 1 5 + z z − 1 6 = z ( 60 z − 11 ) ( 5 z − 1 ) ( 6 z − 1 ) , ∣ z ∣ > 1 5 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)\left( {6z - 1} \right)}},\,\,\,\left| z \right| > {1 \over 5} X(z)=z−51z+z−61z=(5z−1)(6z−1)z(60z−11),∣z∣>51
（9） X ( z ) = 1 − 1 2 z − 8 = 2 z 8 − 1 2 z 8 , ∣ z ∣ > 0 X\left( z \right) = 1 - {1 \over 2}z^{ - 8} = {
{2z^8 - 1} \over {2z^8 }},\,\,\,\left| z \right| > 0 X(z)=1−21z−8=2z82z8−1,∣z∣>0
1.2.3 双边序列的z变换
◎ 求解： 将双边序列分成右边序列和左边序列，分别求出各自的 z 变换。
x R [ n ] = ( 1 a ) n , n ≥ 0 , x L [ n ] = ( 1 a ) − n = a n , n < 0 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 xR[n]=(a1)n,n≥0,xL[n]=(a1)−n=an,n<0 Z { x R [ n ] } = z z − 1 a , ∣ z ∣ > 1 a Z\left\{ {x_R \left[ n \right]} \right\} = {z \over {z - {1 \over a}}},\,\,\,\left| z \right| > {1 \over a} Z{
xR[n]}=z−a1z,∣z∣>a1 Z { x L [ n ] } = ∑ n = − ∞ − 1 a n z − n = ∑ n = 1 ∞ z n a − n = 1 1 − z a − 1 − 1 = − z z − a , ∣ z ∣ < 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} } = {1 \over {1 - za^{ - 1} }} - 1 = {
{ - z} \over {z - a}},\,\,\left| z \right| < a Z{
xL[n]}=n=−∞∑−1anz−n=n=1∑∞zna−n=1−za−11−1=z−a−z,∣z∣<a Z { x [ n ] } = Z { x R [ n ] } + Z { x L [ n ] } = z z − 1 a + − z z − a = ( 1 − a 2 ) z a z 2 − ( 1 + a 2 ) z + a Z\left\{ {x\left[ n \right]} \right\} = Z\left\{ {x_R \left[ n \right]} \right\} + Z\left\{ {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}} Z{
x[n]}=Z{
xR[n]}+Z{
xL[n]}=z−a1z+z−a−z=az2−(1+a2)z+a(1−a2)z 收敛域为 1 a < ∣ z ∣ < a {1 \over a} < \left| z \right| < a a1<∣z∣<a 。
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