# Impulse Response Formula

How can you determine the impulse response if you know the output of the system?

You should change the input signal with the dirac function with the argument equal to $$t$$ or $$t-\tau$$?

I have this system right here:

$$y(t) = \int_{-\infty}^{t}e^{-(t-\tau)}x(\tau -2)d\tau$$

Sadly there is already a $$\tau$$ here so I'm gonna name the new one $$\tau '$$

$$h(t,\tau ') = \int_{-\infty}^{t} e^{-(t-\tau )} \delta (\tau - \tau ' - 2) d\tau$$

which becomes something like:

$$h(t, \tau ') = e^{-(t-\tau ' -2)}$$

Since we have $$h(t, \tau)$$ which is in function of $$t - \tau '$$, so we can have

$$h(t-\tau ')$$

Is this correct or I have to use $$\delta (t)$$ without $$\tau$$

You need to try to massage the input-output relation into the form of a convolution integral:

$$y(t)=\int_{-\infty}^{\infty}h(t-\tau)x(\tau)d\tau\tag{1}$$

This can be done using variable substitution:

\begin{align}y(t)&=\int_{-\infty}^te^{-(t-\tau)}x(\tau-2)d\tau\\&=\int_{-\infty}^{t-2}e^{-(t-2-\tau)}x(\tau)d\tau\\&=\int_{-\infty}^{\infty}e^{-(t-2-\tau)}u(t-2-\tau)x(\tau)d\tau\tag{2}\end{align}

where $$u(t)$$ is the unit step function. Comparing $$(2)$$ with $$(1)$$, the impulse response of the given system should be obvious.

• So, isn't the impulse response just the response to the Dirac impulse? Is this just a faster way to obtain the impulse response because of how $y(t)$ is made? Can't we determine the impulse response just by putting as input the Dirac impulse? Commented Apr 19, 2022 at 14:48
• @Royolh: Yes, that's right, but you want to be sure that the input-output relation has the form of Eq. (1), otherwise you can't be sure that the system is linear AND time-invariant, and can be described by a (one-dimensional) impulse response. Try plugging a Dirac impulse into the given input-output relation and you should arrive at the same result. Commented Apr 19, 2022 at 15:06
• @Royolh: What you did is basically correct, but you forgot that the result of the integral is zero if $t-\tau'-2<0$. You showed that the impulse response only depends on the difference $t-\tau'$, which shows that the system is not only linear but also time-invariant. What I did in my answer is just an alternative way of solving the problem. Commented Apr 19, 2022 at 15:12