The convolution integral is defined as:
$$y(t) = (h * x)(t) = \int^{+\infty}_{-\infty} h(\tau). x(t-\tau)\ d\tau$$
where $h(t)$ and $x(t)$ are functions in terms of time.
Why is $y$ in terms of '$t$' instead of in terms of $\tau$? Isn't convolution in terms of the time shift? Does $\tau$ represent time shift?
Please help, as my friends and I are having trouble understanding the definition of the convolution integral.
On
Note: this answer just addresses the mathematical formula used in convolution without discussing the meaning behind it.
When you calculate a definite integral with respect to $\tau$, it is not a part of the resulting expression. When you look at that integral, imagine that $t$ is a fixed value, so the integral gives a number. For a different fixed value of $t$, the function $x$ is shifted a different amount, so the integral gives a different number (probably). Finally, imagine that you're interested in all the values of the integral, as $t$ varies: now you have a function of $t$.
On
When I took DSP as an elective in undergrad, I was taught this definition of convolution, along with a "flip and slide" technique for computing the integral. But the whole "$t-\tau$" business never made much sense. The other day I was thinking and finally came up with a geometric interpretation for what the convolution integral actually means.
Recall that the convolution of two discrete-time signals $x[n]$, $y[n]$ where $n\in\mathbb Z$ would be computed as $$w[n]:=(x\star y)[n] = \sum_{k=-\infty}^{\infty}x[k]y[n-k]. $$ This is really just the Cauchy product of two sequences. For example, if we restrict $n$ to be nonnegative and consider the expression: $$\left(\sum_{n=0}^\infty x[n]z^n\right)\left(\sum_{n=0}^\infty y[n]z^n\right),$$ by collecting like powers of $z$ (in the same manner as if you were to multiply two polynomials), we would find that the coefficient of $z^n$ is $$\sum_{k=0}^n x[k]y[n-k].$$ If we index $x$ by $k$ and $y$ by $l$, then in the Cartesian plane, we see that the above sum has exactly $n+1$ summands for a given value of $n$. These are the nonnegative integers $(k,l)$ that satisfy $k+l=n$. When we sum over $n$, we aren't summing over the $x$ or $y$ axes - we are summing over the level sets $$L_n:=\{(k,l) : k+l = n\}$$ to get $$w[n]z^n = \left(\sum_{u,s=0}^\infty x[u]y[s]\chi_{L_n}(u,s)\right) z^n $$ where $\chi$ is the characteristic (or indicator) function in $(\mathbb N\cup\{0\})^2$. Again, summing over those values which add up to $n$. The same idea works applies for continuous-time convolution. If we have two continuous-time signals $x(t)$, $y(t)$, $t\in(-\infty,\infty)$ then the convolution is of course as $$w(t):=(x\star y)(t) = \int_{-\infty}^\infty x(s)y(t-s)\ \mathsf ds. $$
The reason I write $s$ instead of $\tau$ is the same reason I wrote $k$ above in the discrete convolution; seeing $n$ and $k$ together in a sum suggests that they are both integer values, much as seeing $t$ and $s$ together in a sum would suggest they are real values. Now, plot the Cartesian plane again, but label values on the $y$-axis by say, $u$. Then for a given value of $t$, we again want the values $x(s)$ and $y(u)$ which satisfy $s+u=t$, that is, the level set $$L_t\{(s,u)\in\mathbb R^2 : s+u =t \}. $$ Similarly, the convolution integral can be written as $$w(t) = \iint\limits_{L_t}xy\ \mathsf d(s\times u) = \int_{-\infty}^\infty\int_{-\infty}^\infty L_t(s,u)x(s)y(u)\ \mathsf d u\ \mathsf d s. $$
This won't necessarily help to compute a convolution. But hopefully it lends some motivation toward the concept.
In this case $y$ is strictly a function of $t$ because the integral is done with respect to $\tau$. Maybe it is a bit easier to understand when talking about just a normal definite integral:
$$\int_a^b f(x) dx = F(b) - F(a) = K$$
$K$ here is any real number and $F(x)$ is the antiderivative. Notice that what you get back is strictly a number, no $x$'s after you finish the integration with $x$. Apply this logic to see why all the $\tau$'s disappear and only $t$'s remain.