Why does convolution give the output of a passing a signal through a filter?

Question

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?

Answer 1

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:

Answer 2

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.