How to find impulse response $h(t)$, given an output of LTI system?

Question

A LTI system takes $x(t)$ as input and $y(t)$ as output. Given $$y(t)=\int_{t-5}^{t-1}2x(\tau)d\tau$$, how to find $h(t)$, the impulse response?

Convolution $x(t)*h(t)=\int_{-\infty}^\infty x(\tau)h(t-\tau)d\tau$, so by comparing this to the given output, $h(t)$ is $2\delta(t)$, is this correct?

Answer 1

Just use $\delta(t)$ as the input. By definition you get $h(t)$ as the output: $$y(t)=\int_{t-5}^{t-1}2x(\tau)d\tau$$ $$h(t) =\int_{t-5}^{t-1}2\delta(\tau)d\tau =2\int_{t-5}^{t-1}\delta(\tau)d\tau =\left\{ \begin{array}{r,l} 2,& t-5<0<t-1\\ 0,& otherwise \end{array}\right. =\left\{ \begin{array}{r,l} 2,& 1<t<5\\ 0,& otherwise \end{array}\right.$$ It is a square pulse. You can also write it as a combination of step functions: $$h(t)=2\left[u(t-1)-u(t-5)\right].$$

Answer 2

Hint: Convolution is commutative, so use the variable substitution $\tau = t-\tau'$. Don't forget to change the bounds, too. Can you see what function is integrated against $x$ now?