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Why do we only use impulse response for convolution for getting output of LTI systems. why we don't use unit step or any other signal for this purpose?

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  • $\begingroup$ You can use any signal at the input, there's nothing stopping you from seeing the output of a sine, but if you meant for finding out the response of the system, then try making a simple filter and feeding an impulse at the input; what do you think you'll get at the output? Now suppose you don't know what filter you have, just a black box, how can you find out its response? You could use a step function, but then you'd have to differentiate, and that adds extra calculus, possible noise, etc. $\endgroup$ – a concerned citizen Feb 28 at 17:58
  • $\begingroup$ " Now suppose you don't know what filter you have, just a black box, how can you find out its response?" please kindly elaborate your this comment $\endgroup$ – engr Feb 29 at 12:18
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    $\begingroup$ The answer below gives all you need in the very first proposition. The step is just the integrated impulse, which means the output changes to an integral, so in order to find out the transfer function, you'll have to differentiate. This is what I meant above with the extra calculus and noise. $\endgroup$ – a concerned citizen Feb 29 at 12:53
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Because the impulse response completely characterizes an LTI system. The reason is an arbitrary input $x(t)$ can be written as an infinite sum of time shifted and scaled Delta functions:

$$x(t) = \int_{-\infty}^{+\infty}x(\tau)\delta(t-\tau)d\tau$$

or alternatively, using the convolution operator, $$x(t) = x(t)*\delta(t)$$

Properties of linearity and time invariance lead to the following sequence of input-output pairs:

  • $\delta(t) \longrightarrow h(t)$
  • Time invariance: $\delta(t-\tau) \longrightarrow h(t-\tau)$
  • Linearity: $x(\tau)\delta(t-\tau) \longrightarrow x(\tau)h(t-\tau)$
  • Linearity: $\int_{-\infty}^{+\infty}x(\tau)\delta(t-\tau)d\tau \longrightarrow \int_{-\infty}^{+\infty}x(\tau)h(t-\tau)d\tau$

The last input-output pair can be written as $$x(t) \longrightarrow y(t) = x(t)*h(t)$$

This way, any output can be computed using the input $x(t)$, the impulse response $h(t)$, and the convolution integral (or sum, in discrete time).

@a concerned citizen gave you some other hints regarding your question.

On the contrary, if you use the unit step function, you get

$$\int_{-\infty}^{+\infty}x(\tau)u(t-\tau)d\tau = \int_{-\infty}^t x(\tau)d\tau$$

or alternatively, using the convolution operator, $$\int_{-\infty}^t x(t) = x(t)*u(t)$$

Once again, just for comparison, you get the following sequence of input-output pairs:

  • $u(t) = \int_{-\infty}^t \delta(\tau)d\tau \longrightarrow s(t) = \int_{-\infty}^t h(\tau)d\tau$
  • Time invariance: $u(t-\tau) \longrightarrow s(t-\tau)$
  • Linearity: $x(\tau)u(t-\tau) \longrightarrow x(\tau)s(t-\tau)$
  • Linearity: $\int_{-\infty}^{+\infty}x(\tau)u(t-\tau)d\tau \longrightarrow \int_{-\infty}^{+\infty}x(\tau)s(t-\tau)d\tau$

where $s(t)$ is the step response, the system output when the unit step is presented at the input. The last input-output pair can be written as $$\int_{-\infty}^t x(\tau)d\tau \longrightarrow x(t)*s(t) = x(t)*u(t)*h(t)$$

The latter can be written as $$(x(t)*h(t))*u(t) = y(t)*u(t) = \int_{-\infty}^t y(\tau)d\tau$$

So after all, what you get at the output is the integral of the output you would get using the delta decomposition of an input signal. That is, your output must be differentiated to obtain the actual system output for a given $x(t)$.

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  • $\begingroup$ "Because the impulse response completely characterizes an LTI system. The reason is an arbitrary input x(t) can be written as an infinite sum of time shifted and scaled Delta functions" can't we use unit-step instead of delta in this case? what will happen if we use unit step ? $\endgroup$ – engr Feb 29 at 12:25
  • $\begingroup$ Of course we can use the unit step function. If we know the system response to a unit step function (that is, the step response), we can write any input as an infinite sum of time shifted and scaled unit step functions, and following similar argumentation, we end up in similar results. $\endgroup$ – GKH Feb 29 at 16:39
  • $\begingroup$ "Of course we can use the unit step function." But why impulse is normally and mostly used??If we can even use unit step as you said,what is the issue using unit step because of which people opt for impulse instead of unit step? $\endgroup$ – engr Feb 29 at 16:44
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    $\begingroup$ I have modified my answer to include what happens if you use the step function. $\endgroup$ – GKH Feb 29 at 18:36
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    $\begingroup$ I think you should check and correct the second part of your answer concerning the step response. E.g., the first equation in that part of the answer is probably not what you meant. $\endgroup$ – Matt L. Mar 2 at 7:06
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Because the impulse function is the neutral element for convolution? So it is a consequence of how convolution is defined. So why doesn't one define it differently? The Fourier transform, by and large, is an invertible transform. It transforms a constant 1 into a dirac pulse and vice versa. Dirac pulses are sort-of ugly since they are distributions rather than proper functions, but this makes for symmetric and reasonably powerful domains with regard to Fourier transforms, allowing to represent undampened perpetual continuous sine/cosine functions as an idealisation.

Once you move to other definitions that integrate or differentiate in one direction of the transform, either the transform domain or the time domain become vastly less useful.

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If you try to pass a specific frequency through a system, the system will either stop that signal or attenuate/amplify it or it may even delay the signal(phase shift). The output determines how the system responds to that particular signal. 

What about another input signal with some arbitrary amplitude and frequency? The answer lies in the frequency response of the system.  Then the next question is what is frequency response? Frequency response is nothing but how the system responds to each individual frequency, provided the input amplitude is unity and zero phases (all signals aligned to the zero of the timeline),

To obtain the frequency response of a system what we need to do is to pass through the system is, a signal with all frequency components from $-\infty$ to $+\infty$ with unity amplitude and zero phases (which input signal is having such a property? this is your first answer.)

what the system will actually do is, it will act as a filter and pass only those frequency components which it can process, which is in effect gives you the frequency response. 

Getting frequency response with impulse input

when you try to push any signal with an arbitrary spectrum,  the output will look like the multiplication of the input spectrum of the signal with the frequency response of the system, think of it as a product of point to point value multiplication for each frequency value. 

Response for an arbitrary signal

As you already know multiplication in the frequency domain is convolution in the time domain. ( Hope your second doubt got cleared.)

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Unit impulse is a nice pedagogic construct in order to understand how we can measure and execute a LTI system.

The stimuli of choice when characterizing real-world near-LTI systems is often not a unit impulse. It may be a MLS sequence, a sine sweep or some other signal containing a range of frequencies and more energy than an impulse.

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