信号与系统问题 f1（t）=2[u（t）-u（t-1 ）] f2(t)=sin（πt）[u（t）-u（t-1）]求f1（t）卷积f2（t）

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摘要首先，计算 $f1(t)$ 和 $f2(t)$ 的卷积意味着要解决以下积分：$$(f1f2)(t)=\int_{-\infty}^{\infty}f1(\tau)f2(t-\tau)d\tau$$因为 $f1(t)$ 在 $[0,1]$ 之外的值都等于 $0$，所以上式可以被简化为：$$(f1f2)(t)=\int_{0}^{1}f1(\tau)f2(t-\tau)d\tau$$将 $f1(t)$ 和 $f2(t)$ 的表达式代入上式：$$\begin{aligned}(f1f2)(t) &= \int_{0}^{1}2[u(\tau)-u(\tau-1)]sin(\pi(t-\tau))u(t-\tau)d\tau \&= \int_{0}^{1}2sin(\pi(t-\tau))u(\tau)u(t-\tau)d\tau - \int_{0}^{1}2sin(\pi(t-\tau))u(\tau-1)u(t-\tau)d\tau \&= \int_{0}^{t}2sin(\pi(t-\tau))u(\tau)d\tau - \int_{0}^{t-1}2sin(\pi(t-\tau))u(\tau)d\tau \&= \int_{0}^{t}2sin(\pi(t-\tau))d\tau - \int_{0}^{t-1}2sin(\pi(t-\tau))d\tau \&= -\frac{2}{\pi}cos(\pi t) + \frac{4}{\pi}sin(\pi t) - \left(-\frac{2}{\pi}cos(\pi (t-1)) + \frac{4}{\pi}sin(\pi (t-1))\right) \&= -\frac{2}{\pi}cos(\pi t) + \frac{2}{\pi}cos(\pi (t-1)) + \frac{4}{\pi}sin(\pi t) - \frac{4}{\pi}sin(\pi (t-1))\end{aligned}$$因此，$f1(t)$ 和 $f2(t)$ 的卷积为：$$(f1f2)(t) = -\frac{2}{\pi}cos(\pi t) + \frac{2}{\pi}cos(\pi (t-1)) + \frac{4}{\pi}sin(\pi t) - \fra

咨询记录 · 回答于2023-03-26