◎ 本文网络下载链接： 信号与系统2022 春季学期作业-第九次作业 : https://blog.csdn.net/zhuoqingjoking97298/article/details/124319518
§01 基础作业
1.1 信号的采样与恢复
1.1.1 采样信号频谱
设三角形和升余弦信号的底宽均为 τ \tau τ ，用抽样间隔为 T s = τ / 8 T_s = \tau /8 Ts=τ/8 时，分别绘制出采样后信号的频谱。
▲ 图1.1.1 三角形与升余弦信号
1.1.2 矩形信号采样频率
已知矩形信号的底宽 τ = 2.5 m s \tau = 2.5ms τ=2.5ms 。设该信号的频谱第 8 个过零点以外的频谱分量可以忽略，对该型号的抽样最小频率是多少？
▲ 图1.1.2 信号与对应的频谱
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
求下列函数的拉普拉斯变换，并绘制出信号的零极点分布图。缺省情况下，都是求信号的单边拉普拉斯变换。
( 1 ) 1 − e − a t ; ( 2 ) sin t + 2 cos t ; \left( 1 \right)\,\,\,\,1 - e^{ - at} ;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\,\,\,\,\sin t + 2\cos t; (1)1−e−at;(2)sint+2cost; ( 3 ) t ⋅ e − 2 t ; ( 4 ) e − t sin 2 t ; \left( 3 \right)\,\,\,\,t \cdot e^{ - 2t} ;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\,\,\,\,e^{ - t} \sin 2t; (3)t⋅e−2t;(4)e−tsin2t; ( 5 ) ( 1 + 2 t ) ⋅ e − t ; ( 6 ) ( 1 − cos α t ) e − β t ; \left( 5 \right)\,\,\,\,\left( {1 + 2t} \right) \cdot e^{ - t} ;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 6 \right)\,\,\,\,\left( {1 - \cos \alpha t} \right)e^{ - \beta t} ; (5)(1+2t)⋅e−t;(6)(1−cosαt)e−βt; ( 7 ) t 2 + 2 t ; ( 8 ) 2 δ ( t ) − 3 e − 7 t ; \left( 7 \right)\,\,\,\,t^2 + 2t;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 8 \right)\,\,\,\,2\delta \left( t \right) - 3e^{ - 7t} ; (7)t2+2t;(8)2δ(t)−3e−7t; ( 9 ) e − α t sinh ( β t ) ; ( 10 ) cos 2 ( Ω t ) ; \left( 9 \right)\,\,\,\,e^{ - \alpha t} \sinh \left( {\beta t} \right);\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {10} \right)\,\,\,\,\cos ^2 \left( {\Omega t} \right); (9)e−αtsinh(βt);(10)cos2(Ωt); ( 11 ) 1 β − α ( e − α t − e − β t ) ; ( 12 ) e − ( t + a ) cos ω t ; \left( {11} \right)\,\,\,\,{1 \over {\beta - \alpha }}\left( {e^{ - \alpha t} - e^{ - \beta t} } \right)\,;\,\,\,\,\,\left( {12} \right)\,\,\,\,e^{ - \left( {t + a} \right)} \cos \omega t; (11)β−α1(e−αt−e−βt);(12)e−(t+a)cosωt; ( 13 ) t ⋅ e − ( t − 2 ) ⋅ u ( t − 1 ) ; ( 14 ) e − t a f ( t a ) \left( {13} \right)\,\,\,\,t \cdot e^{ - \left( {t - 2} \right)} \cdot u\left( {t - 1} \right);\,\,\,\,\,\,\,\,\,\,\left( {14} \right)\,\,\,\,e^{ - {t \over a}} f\left( { {t \over a}} \right) (13)t⋅e−(t−2)⋅u(t−1);(14)e−atf(at) ( 15 ) e − a t f ( t a ) ; ( 16 ) t ⋅ cos 3 ( 3 t ) ; \left( {15} \right)\,\,\,\,e^{ - at} f\left( { {t \over a}} \right);\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {16} \right)\,\,\,\,t \cdot \cos ^3 \left( {3t} \right); (15)e−atf(at);(16)t⋅cos3(3t); ( 17 ) t 2 ⋅ cos ( 2 t ) ; ( 18 ) 1 t ( 1 − e − a t ) ; \left( {17} \right)\,\,\,\,t^2 \cdot \cos \left( {2t} \right);\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {18} \right)\,\,\,\,{1 \over t}\left( {1 - e^{ - at} } \right); (17)t2⋅cos(2t);(18)t1(1−e−at); ( 19 ) e − 3 t − e − 5 t t ; ( 20 ) sin ( α t ) t ; \left( {19} \right)\,\,\,\,{ {e^{ - 3t} - e^{ - 5t} } \over t};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {20} \right)\,\,\,\,{ {\sin \left( {\alpha t} \right)} \over t}; (19)te−3t−e−5t;(20)tsin(αt);
其中，如果包括函数 f ( t ) f\left( t \right) f(t) ，那么假设它对应的Laplace变换为 F ( s ) F\left( s \right) F(s) 。
（1）必做题
上述个信号中标号为偶数的小题。
其它信号的Laplace变换都是选做题。
1.2.2 求信号的z变换
求下面个序列信号的 z z z 变换 X ( z ) X\left( z \right) X(z) ，并表明收敛域，会出 X ( z ) X\left( z \right) X(z) 的零极点图。
( 1 ) ( 1 4 ) n u ( t ) ; ( 2 ) ( − 1 3 ) n u ( t ) ; \left( 1 \right)\,\,\,\,\left( { {1 \over 4}} \right)^n u\left( t \right);\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\,\,\,\,\left( { - {1 \over 3}} \right)^n u\left( t \right); (1)(41)nu(t);(2)(−31)nu(t); ( 3 ) ( 1 2 ) n u ( t ) ; ( 4 ) ( − 1 2 ) n u [ − n ] ; \left( 3 \right)\,\,\,\,\left( { {1 \over 2}} \right)^n u\left( t \right);\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\,\,\,\,\left( { - {1 \over 2}} \right)^n u\left[ { - n} \right]; (3)(21)nu(t);(4)(−21)nu[−n]; ( 5 ) − ( 1 6 ) n − 1 u [ − n − 1 ] ; ( 6 ) u [ n − 1 ] ; \left( 5 \right)\,\,\,\, - \left( { {1 \over 6}} \right)^{n - 1} u\left[ { - n - 1} \right];\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 6 \right)\,\,\,\,\,u\left[ {n - 1} \right]; (5)−(61)n−1u[−n−1];(6)u[n−1]; ( 7 ) ( 1 10 ) n { u [ n ] − u [ n − 3 ] } ; \left( 7 \right)\,\,\,\,\left( { {1 \over {10}}} \right)^n \left\{ {u\left[ n \right] - u\left[ {n - 3} \right]} \right\}; (7)(101)n{ u[n]−u[n−3]}; ( 8 ) ( 1 5 ) n u [ n ] + ( 1 6 ) n u [ n ] ; \left( 8 \right)\,\,\,\,\left( { {1 \over 5}} \right)^n u\left[ n \right] + \left( { {1 \over 6}} \right)^n u\left[ n \right]; (8)(51)nu[n]+(61)nu[n]; ( 9 ) δ [ n ] − 1 2 δ [ n − 8 ] \left( 9 \right)\,\,\,\,\delta \left[ n \right] - {1 \over 2}\delta \left[ {n - 8} \right] (9)δ[n]−21δ[n−8]
（1）必做题
上述各个信号中标号为偶数的小题。
其它信号的z变换都是“选做题”。
1.2.3 双边序列的z变换
求是小编序列 x [ n ] x\left[ n \right] x[n] 的 z z z 变换，并标明收敛域。 绘制出 z z z 变换的零极点图。
x [ n ] = ( 1 a ) ∣ n ∣ , ∣ a ∣ > 1 x\left[ n \right] = \left( { {1 \over a}} \right)^{\left| n \right|} ,\,\,\left| a \right| > 1 x[n]=(a1)∣n∣,∣a∣>1
§02 实验作业
2.1 MATLAB中的Laplace，ZT的变换
在MATLAB中练习使用 laplace 、 ilaplace 、 ztrans 、 iztrans 命令完成Laplace变换和z变换以及反变换。并对于前面系统中的相应结果进行验证。
2.1.1 MATLAB中求解LT代码
（1）Laplace变换
测试Laplace变换例子：
（1） f f ( t ) = t 2 f\left( t \right) = t^2 f(t)=t2 ；
（2） f ( t ) = e − 3 t f\left( t \right) = e^{ - 3t} f(t)=e−3t
>> syms t
>> laplace(t^2)
ans = 2/s^3
>> laplace(exp(4*t))
ans = 1/(s-4
（2）Laplace反变换
测量Laplace反变换例子：
( 1 ) 1 ( s + 1 ) 2 ; ( 2 ) 1 s 2 + 4 ; \left( 1 \right)\,\,\,\,{1 \over {\left( {s + 1} \right)^2 }};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\,\,\,\,{1 \over {s^2 + 4}}; (1)(s+1)21;(2)s2+41;
>> syms s
>> ilaplace(1/(1+s)^2)
ans = t*exp(-t)
>> ilaplace(1/(s^2+4))
ans = 1/4*4^(1/2)*sin(4^(1/2)*t)
2.1.2 MATLAB中求解z变换代码
下面是测试的代码：
ztrans(f)
>> f=n^4
>> ztrans(f)
>ans = z*(z^3+11*z^2+11*z+1)/(z-1)^5
iztrans(f)
>> iztrans(z/(z-2))
ans = 2^n
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