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I have a rudimentary understanding of Convolution, the Convolution Theorem and why the output z(t) of an LTI system can be found using the convolution of input signal x(t) and the impulse response h(t).

In an old post, I came across a statement "...assume you want to pass signal x(n) through filter y(n). The output of the filter z(n) will be convolution of x(n) and y(n)".

I don't quite understand why that is the case - it is almost as if the "impulse response" can be replaced by the "filter". Can someone please provide an explanation for this?

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  • $\begingroup$ This is about the core of "Linear System Theory" (an EE subdiscipline that might now be renamed "Signals and Systems" by the Oppenheim and Wilsky book). If you have any Linear system, the way it works is convolution with a two-variable kernel that they often call the impulse response. If the Linear system is also Time-Invariant (LTI), then the two-variable kernel becomes a simpler, single-variable kernel, also called the "impulse response". This is in textbooks. It's fundamental to electrical engineering and signal processing. $\endgroup$ Sep 8, 2023 at 21:22
  • $\begingroup$ I'll come back to this within 24 hours and try to write a comprehensive answer. $\endgroup$ Sep 8, 2023 at 21:31
  • $\begingroup$ Thank you. I am an ME, and I have a basic understanding of FT/convolution from a Math POV, and basic understanding of high/low pass filters. Unfortunately, I don't have a strong understanding of filtering, but after some reading, I've taken it on face value that the convolution of an I/P signal with the IR of the filter gives the filtered O/P. I still have fundamental questions like: Are all filtering operations convolutions? If not, when do I use convolution? I will go through the book that you mentioned, and edit and add more questions to this comment if needed, while awaiting your answer. $\endgroup$
    – SNIreaPER
    Sep 9, 2023 at 0:53

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The convolution operation essentially computes a weighted sum of the input signal's values, with the weights determined by the filter. This process allows filters to capture patterns, features, or transformations within the input signal.

Think of what convolution actually does: you have your signal and your impulse response, you flip one of them and place them at $-\infty$. Then you start multiplying and summing as you slide your flipped signal all the way to $+\infty$. Then the question becomes why is this sufficient to compute the output?

Well your filter being LTI is why this is sufficient! Because of linearity any input signal can be expressed as a weighted sum of their responses to individual impulses. Moreover, filters that are time-invariant exhibit the same response to an impulse at any point in time. This means that the impulse response characterizes the filter's behavior for all times.

This is the reason why the convolution operation arises naturally as the consequence of the processing of signals by a LTI system. But it's a mathematical tool that extends to other fields as well (sum of two Random Variables).

For example, consider a brickwall LPF (ideal) which has an impulse response as shown below on the left and a frequency response as shown below on the right:

LPF

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    $\begingroup$ For an LTI system, assuming x(t), h(t) and y(t) as the I/P signal, impulse response and O/P, I understand why y(t) = x(t)*h(t), and how it elegantly allows us to find the response corresponding to an arbitrary input. However, I am having a tough time understanding its relation to filtering. What exactly is it "filtering" out? If I now replace the impulse response with a random filter f(t), and write the O/P as z(t) = x(t)*f(t), what exactly does z(t) represent now? I am trying to understand it in context of Gabor filters, but any explanation with respect to simpler filters is also fine. $\endgroup$
    – SNIreaPER
    Sep 7, 2023 at 17:24
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    $\begingroup$ f(t) is another IR that can accomplish the task of filtering as characterized by the filter REPRESENTED by f(t). Let's say h(t) is IR of a LPF (lowpass) and f(t) is IR of an HPF (highlass), then their frequency responses (FT of the IR) will be functions that are zero or take small values in regions (frequencies) where we want to suppress the input signal and larger values where we want to do little to no attenuation (or perhaps do amplification). Gabor filters, for example, are BPF (bandpass) which allow frequencies in a certain band. IR of such filter is inverse FT of its frequency response! $\endgroup$ Sep 7, 2023 at 22:09
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    $\begingroup$ So the impulse response is just the inverse Fourier Transform of the frequency response. In the frequency domain we can see more clearly how a certain filter behaves. I will add some plots to my answer soon showing what the FRs and IRs of different filters look like... $\endgroup$ Sep 7, 2023 at 22:16
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You can describe the system with an operator acting over an input $x(t)$ transforming it into $z(t)$. If $L$ is the operator, $z(t)=L[x(t)]$.

Remember that the system and the operator are linear and time-invariant, and consider the use of the impulse function to write $x(t)$ in the following way:

$x(t)=\int_\limits{-\infty}^{\infty} x(\zeta)\delta(t-\zeta)d\zeta$.

Then:

$z(t)=L\left[\int_\limits{-\infty}^{\infty} x(\zeta)\delta(t-\zeta)d\zeta\right]$

as $L$ is a linear operator that acts over a function of $t$:

$z(t)=\int_\limits{-\infty}^{\infty} x(\zeta)L[\delta(t-\zeta)]d\zeta$

We can define $h(t,\zeta)=L[\delta(t-\zeta)]$ as the response of the system to the impulse $\delta(t-\zeta)$. This is useful for a time-invariant system (and operator) where $h(t,\zeta)=h(t-\zeta)$ and $z(t)$ can be written as:

$z(t)=\int_\limits{-\infty}^{\infty} x(\zeta)h(t-\zeta)d\zeta$

This equation is known as the convolution integral, then in a shorter form can be written as: $z(t)=x(t)*h(t)$.

This result is interesting because it shows that a linear and time-invariant system can be described just by its impulse response $h(t)$.

If you want more detail about this, I recommend you: Probability, random variables, and random signal principles, P. Peebles, chapter 8 in 2nd ed.

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  • $\begingroup$ I just now returned to this question and this answer is closest to what I would have written. But I would have done it in discrete time, which is easier to see. $\endgroup$ Sep 10, 2023 at 4:04

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