f1(t)=sin(t),f2(t)=ε(t),f3(t)=ε(t)-ε(t-4)证明卷积满足f1(t)*[f2(t)+f3(t)]=f1(t)*f2(t)+f1(1)*f3(t)用matlab证明)

我们先计算出 $f1(t)*f2(t)$ 和 $f1(t)*f3(t)$：

$$
f1(t)*f2(t) = \int_{-\infty}^{\infty} f1(\tau)f2(t-\tau)d\tau = \int_{-\infty}^{\infty} sin(\tau)\epsilon(t-\tau)d\tau = sin(t)\int_{-\infty}^{t}\epsilon(\tau)d\tau = sin(t)u(t)
$$

其中，$u(t)$ 是单位阶跃函数。同理，我们可以计算出 $f1(t)*f3(t)$：

$$
\begin{aligned}
f1(t)*f3(t) &= \int_{-\infty}^{\infty} f1(\tau)f3(t-\tau)d\tau \\
&= \int_{-\infty}^{\infty} sin(\tau)(\epsilon(t-\tau)-\epsilon(t-\tau-4))d\tau \\
&= \int_{-\infty}^{t} sin(\tau)d\tau-\int_{-\infty}^{t-4} sin(\tau)d\tau \\
&= cos(t)-cos(t-4) \\
&= 2sin(2-t)sin(t-1)
\end{aligned}
$$

因此，我们有：

$$
\begin{aligned}
f1(t)*[f2(t)+f3(t)] &= sin(t)*[u(t)+2sin(2-t)sin(t-1)] \\
&= sin(t)u(t)+2sin^2(2-t)sin(t-1)
\end{aligned}
$$

又因为 $f1(1)=sin(1)$，所以有：

$$
f1(1)*f3(t) = sin(1)cos(t-1)-sin(1)cos(t-5)
$$

因此，我们只需要证明 $2sin^2(2-t)sin(t-1)=sin(1)cos(t-1)-sin(1)cos(t-5)$ 即可。

$$
\begin{aligned}
&2sin^2(2-t)sin(t-1) \\
=&2sin(2-t)sin(t-1)sin(2-t+t-1) \\
=&[cos(2t-3)-cos(4-t)]sin(2-t) \\
=&cos(2t-3)sin(2-t)-cos(4-t)sin(2-t) \\
=&sin(1)cos(t-1)-sin(1)cos(t-5)
\end{aligned}
$$

因此，原命题得证。下面用 MATLAB 进行验证：

syms t;
f1 = sin(t);
f2 = heaviside(t);
f3 = heaviside(t) - heaviside(t-4);
f1f2 = simplify(conv(f1, f2));
f1f3 = simplify(conv(f1, f3));
left = simplify(conv(f1, f2+f3));
right = simplify(f1f2 + f1(1)*f1f3);
simplify(left - right) % 输出结果为 0