convolutionintegrationsignal processing
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.
Intuitively speaking when you are given two signals/or functions $f$ and $g$. You time reverese one of the signals, it doesnt matter which one, and shift it by a value of $t$ then you simply multiply and then sum the area under the intersection.
If you consider a function say function of $x$, then time reversal means inserting $-x$ wherever you see $x$ in this function.
Example:
Question1: Assume you have a function $f(x)$ that is $1$ if $x\in[0,1]$, and $0$ elsewhere then how should you plot $f(-x)$?
Question2: Assume you have $g(x)=f(x)$ is there any intersecting area between $f(x)$ and $g(x)$?
Question3: Now shift $g(-x)$ by $0.5$, that is to find $g(-x+0.5)$. How does it look like when you plot it?
Question4: Where does the intersecting region lie in this case $x\in?$ what is the are of the intersecting region? Answer: below white area$=0.5$ at $x=0.5$ shift.
Question5: If you select the shifting parameter not $0.5$ but all reals in $[0,2]$ what function should you get at the output? check $x=0.5$ and see $f*g(x)$ is $0.5$ as found at step 4
EDIT: You define convolution integral in $[0,t]$ for bounded signals. The integral limits depend on where your signal is non-zero.
If you have two signals as you suggested $f(t)=e^{at}$ and $g(t)=e^{bt}$ then the first question: what is the relation between $a$ and $b$? are they positive? where is the function defined? For example when $a$ and $b$ are some positive terms then we have the following integral
$$h(t)=\int f(\tau)g(t-\tau)d\tau= \int e^{a\tau}e^{b(t-\tau)}d\tau=e^{bt}\int e^{(a-b)\tau}d\tau=\Bigg]_{\tau\in\Omega}\frac{e^{(a-b)\tau}}{a-b}$$
clearly $\Omega=\mathbb{R}$ is not possible because the integral does not converge.
Write each of the signals as
$$e^{-k \tau} \theta(\tau)$$
where $k$ is either of $a$ or $b$, and $\theta(\tau)$ is the Heaviside step function, zero when $\tau < 0$ and $1$ when $\tau > 0$. The convolution integral may then be written as
$$\int_{-\infty}^{\infty} d\tau \, e^{-a \tau} \theta(\tau) \, e^{-b (t-\tau)} \theta(t-\tau)$$
Now, the product of the two Heavisides in the integral is zero outside the interval $[0,t]$. Therefore, we may write the convolution integral as
$$\int_0^t d\tau \, e^{-a \tau} \, e^{-b (t-\tau)} = e^{-b t} \int_0^t d\tau \, e^{-(a-b) \tau} $$
which is
$$\frac{1}{a-b} e^{-b t} \left (1-e^{-(a-b) t} \right ) = \frac{e^{-b t}-e^{-a t}}{a-b}$$
Best Answer
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.