Suppose we have 2 signals x(t) and y(t) with $$ x(t)=\delta(t)$$ If $x(t)$ is the input to a linear and time invariant system $S$, lets say we get a X(t) as the output:

$$ x(t) \overset{S}{\rightarrow} X(t) $$

If we have $y(t)$ as an input to the same system $S$, then we can predict the output of that system without knowing anything more about the system:

$$ Y(t) = \int_{-\infty}^{\infty}X(\tau)y(t-\tau) d\tau $$

Which are different examples of convolution of 2 signals?


1 Answer 1


Your question is very un-clear, but I think I understand what you're asking. It's already been asked, and answered. You can go straight to the answer, but I'd suggest you read through this first:

Let me make a couple of suggestions:

  1. It is common practice to write time-domain signals such as your $x(t)$ with lower case letters, and frequency-domain signals with Upper case letters:
    • If $x(t)$ is your time-domain signal, then $X(f)$ would be your frequency-domain signal (note you can use any name as the dependent variable, you'll sometimes see $x[n]$, $X[\omega]$, $X[k]$, etc.
      The point is lower-case = time, UPPER-CASE = frequency.
      $X(t)$ and $Y(t)$, while not wrong per-se, just look weird.
  2. When talking about LTI (Linear Time Invariant) systems, it is also common practice to name the input $x$ and output $y$. Don't ask me why, it's just what you'll see over and over in textbooks and articles.
    • For your example, I'd write: $$x(t) \overset{S}\rightarrow y(t)$$

With that out of the way, if you input an impulse $\delta(t)$ to a LTI system, the output is called the impulse response, and is usually (again, most common) written $h(t)$.

$$\delta(t) \overset{S}\rightarrow h(t)$$

This impulse response completely characterizes the LTI system $S$. What this means is that once you know the impulse response, you can predict the output of $S$ to any input, through convolution.

Your statement:

then we can predict the output of that system without knowing anything more about the system

is therefore only half-right: yes, you can predict the output of that system, but you DO know something about that system: its impulse response.

With that in mind, I'd re-write your statement and equation as such:

Consider a Linear Time Invariant system $S$, with impulse response $h(t)$. For an input $x(t)$, $S$ will output $y(t)$ as the result of the convolution of $x(t)$ with $h(t)$ (note you had a slight error in yours; it's ($x(\tau)$, not $x(t)$):

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

  • $\begingroup$ yes you are correct I will edit $\endgroup$
    – Miss Mulan
    Aug 26, 2022 at 20:32

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