#TODO:补充答案;补充一题多解;补充方波与方波卷积的快速手算.
手工计算卷积是一门手艺活,特别是在还没学傅里叶变换、拉普拉斯变换或\(Z\)变换的时候.比较恶心的就是卷积过程中出现分部积分和反因果信号/序列,以及绝对值符号.计算就是老老实实地算,学习了傅里叶等变换后就会有很多奇技淫巧,会方便和快很多(但再怎么快也快不过计算机用FFT算).
下面放几道题,太懒了就不打答案了,有兴趣的话算一算吧.
连续时间卷积
- \(f_1(t)=\cos (t)·[u(t)-u(t-\frac{\pi}{2})], \ f_2(t)=(1-e^{-at})u(t)\)
- \(f_{1}(t)=\mathrm{e}^{-a t} u(t), \ f_{2}(t)=\sin (t) u(t-2 \pi)\)
- \(f_{1}(t)=2 \mathrm{e}^{-t}[u(t)-u(t-3)], \ f_{2}(t)=4[u(t)-u(t-2)]\)
- \(f_{1}(t)=\left\{\begin{array}{ll}1, & 1 \leqslant t \leqslant 3 \\ 0, & \text { Others }\end{array}\right.,\) \(f_{2}(t)=\left\{\begin{array}{ll}t-1, & 1 \leqslant t \leqslant 3 \\ 0, & \text { Others }\end{array}\right.\)
- \(e^{-|t-1|} * u(t)\)
若\(y(t)=f_1(t)*f_2(t)\),计算\(f_1(2t)*f_2(2t)\).
解: \[ \begin{aligned} y(t) &=f_{1}(t) * f_{2}(t) \\ &=\int_{-\infty}^{+\infty} f_{1}(k) f_{2}(t-k) \mathrm{d} k \end{aligned} \]
\[ f_{1}(a t) * f_{2}(a t)=\int_{-\infty}^{+\infty} f_{1}(a k) f_{2}[a(t-k)] \mathrm{d} k \]
\[ \begin{aligned} y(a t) &=\int_{-\infty}^{+\infty} f_{1}(k) f_{2}(a t-k) \mathrm{d} k \\ &\overset{k=am}{=} \int_{-\infty}^{+\infty} f_{1}(a m) f_{2}(a t-a m) \mathrm{d}(a m) \\ &=a \int_{-\infty}^{+\infty} f_{1}(a m) f_{2}(a t-a m) \mathrm{d}(a m) \\ &=a f_{1}(a t) * f_{2}(a t) \end{aligned} \]
故\(f_1(2t)*f_2(2t)=\frac{1}{2}y(2t)\).
离散时间卷积
- \(x_{1}(n)=2^{n} u(-n), \ x_{2}(n)=u(n)\)
- \(x_{1}(n)=2^{n} u(n),\ x_{2}(n)=3^{n} u(n)\)
- \(x_{1}(n)=(-0.5)^{n} u(n-4), \ x_{2}(n)=4^{n} u(-n+2)\)
- \(x_{1}(n)=u(n)-u(-n),\ x_{2}(n)=\left\{\begin{array}{ll}{4^{n},} & {n<0} \\ {0.5^{n},} & { n \geq 0}\end{array}\right.\)
一题多解
暂时懒得找那道题,待更新.