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My intuition of convolution is that it is just a way to depict multiplication of two signals where each signal is made up of various frequencies and phases.

Since it isn't easy to find the value of $\cos({w1}+{\phi1})+\cos({w2}+{\phi2})$ multiplied by $\cos({w3}+{\phi3})$, we need a simpler method to represent our resulting signal, similar to how we would use phasors when adding signals of the same frequency.

And intuitively, convolutions are seen as sliding windows.

But I am confused about how that logic gets translated into our equation with $t$ and ${\tau}$

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

${\tau}$ is very clearly some dummy variable but beyond that I feel a little confused.

The value $t$ is used to used as time to check out value at a certain instance. And I think using t as a time-shift allows us shift the impulse function to just see it as a moving periodic function function and we are purely changing the origin point of our impulse response.

Because that matches what you are taught as an undergraduate and you draw different regions of overlap.

Is two variables then just to depict:

  1. How much overlap there is between signals since there will be points of the signal that "crossed" the LTI system or haven't reached our filter yet.

  2. What the status of signals are during that overlap?

Would my intuition so far be correct? Which variable is representing this term and how does it relate to the pen and paper method for looking at regions of overlap?

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    $\begingroup$ Sorry, your intuition appears to be wrong on almost all parts. Convolution is NOT about making it easier to multiply signals. Simple example: it doesn't make sense to convolve to sine waves, since the integral doesn't even converge. Since your basic assumption feels incorrect, it's difficult to answer the following questions. $\endgroup$ – Hilmar Apr 2 at 15:12
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    $\begingroup$ As @Hilmar said, “your intuition appears to be wrong on almost all parts.” But intuition is learned, not innate, so may the figures I drew for an answer here might help in understanding convolution. Also, the convolution article in wikipedia is helpful. $\endgroup$ – Ed V Apr 2 at 16:47
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My intuition of convolution is that it is just a way to depict multiplication of two signals where each signal is made up of various frequencies and phases.

Not really, no. I can't think of a case where you'd actually receive two signals and convolve them (correlate them -- yes, but not convolve). This may be why you're not understanding the rest.

Convolution is used to find the output of a linear system* when you know the system's impulse response.

And intuitively, convolutions are seen as sliding windows. ... $\tau$ is very clearly some dummy variable but beyond that I feel a little confused.

$\tau$ is the dummy variable that implements the sliding window in the convolution. For each time $t$, you multiply the signal by the time-reversed impulse response, and you integrate that product. $\tau$ is just the variable over which you do the integration.

* I purposely left the "time-invariant" part of that out -- convolution applies to time-varying systems, too, it's just more complicated.

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    $\begingroup$ I think this is making a little sense now. Convolution is just a means to use a sliding window and answer the question of how an LTI system gets excited by some input. At some point t, Tau acts to slide over the signals, multiplying and getting an impulse response $\endgroup$ – Mr. Johnny Doe Apr 3 at 11:08
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The most practical physical application of convolution can be found in the Linear System Theory, in which the behaviour of a (LTI) system is described by a function, called the impulse response, $h(t)$, and the input signal is described by the function $x(t)$.

The response of this (LTI) system when excited by an arbitrary input $x(t)$ is, then, given by the so called convolution integral: $y(t) = x(t) \star h(t)$ of the input $x(t)$ and the impulse-response $h(t)$, where $y(t)$ gives a mathematical description of the output response of the system.

This convolution integral, helps to avoid (if not more difficult to find) the solution of a linear constant coefficient differential equation to find the output $y(t)$, for a given input $x(t)$.

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    $\begingroup$ (+1) Clear and succinct! $\endgroup$ – Ed V Apr 2 at 16:48
  • $\begingroup$ I've studied LTIs in the past, but more along the lines of "Solve this equation" rather than trying to build an intuition of convolution. And since it is not as simple as multiplying, we just use a sliding window? $\endgroup$ – Mr. Johnny Doe Apr 3 at 11:09

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