Why unit impulse function is used to find impulse response of an LTI system?

Question

Hello i am working in digital image restoration field, recently i have studied concept of convolution, i studied that to find the impulse response/point-spread function of an LTI system, an unit impulse function is passed from the system and its output is considered as impulse response, My question is that, Why an unit impulse function is used to find the impulse response of any LTI system, what is the mathematical and practical reason behind it.

Answer 1

I'm not really sure what you're asking. A unit impulse is used as the input to find a system's impulse response because, by definition, an LTI system's impulse response is equal to its output for a unit impulse input. If you used any other input, then the output wouldn't be an impulse response.

Answer 2

It is important to realize that due to the LTI property the response to an impulse $\delta(n)$ describes the system completely, because any input signal can be written as a weighted sum of shifted impulses:

$$x(n)=\sum_kx(k)\delta(n-k)\tag{1}$$

Let $\mathcal{T}\{\cdot\}$ denote the system operator, so the response to a shifted impulse is

$$\mathcal{T}\{\delta(n-k)\}=h(n-k)\tag{2}$$

From (1) and (2), due the linearity and the time- (shift-)invariance of the system, the response to any input signal $x(n)$ can be written as a weighted sum of responses to shifted impulses:

$$y(n)=\mathcal{T}\{x(n)\}=\sum_kx(k)\mathcal{T}\{\delta(n-k)\}= \sum_kx(k)h(n-k)$$

This last equality is the convolution sum describing the input-output relation of a discrete LTI system.

Answer 3

The reason an impulse response is used (as a theoretical construct) is because the Fourier transform of the impulse $\delta[n]$ is a constant.

Because convolution in the time domain is simply multiplication in the frequency domain, that means that the frequency response of the output (to an impulse as input) will just be the system's frequency response.