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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 $\tau$

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Royolh
    Apr 19, 2022 at 14:48
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    $\begingroup$ @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. $\endgroup$
    – Matt L.
    Apr 19, 2022 at 15:06
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    $\begingroup$ @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. $\endgroup$
    – Matt L.
    Apr 19, 2022 at 15:12

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