Convolution Functions

[dknguyen]

I'm doing some stuff on matched filters and something is starting to really confuse me here about all the time shiftings and reversals of functions as they go into the convolution integral. It is what is regarded as the actual argument of a function:

ie. f(t) = g(a-t)

Now what would f(b-t) be in terms of g? ie. f(b-t) = g(?). THere's too many t's floating around for me to keep track of but that's the way all the integrals and equations are churned out since the two t's are not the same thing.

 

[MrAl]

Hi there,


I am not entirely sure i understand your question 100 percent, but one thing you can do that is very informative is to sit down and start with a simple filter like a resistor and capacitor (low pass filter) and a simple forcing function like the unit step function, or whatever you choose really.

Start by finding the time domain representation for the filter (impulse response), then knowing that the output of the filter is the convolution of the input function and the filter function, try to arrange the convolution integral until you get the right answer for the output. As you know, the output will be an increasing exponential that starts at Vout=0 and ends at Vout=1.

If you like after that you can change the forcing function to a ramp (whatever slope you want) and see that the output looks right, which you can check with the circuit simulator of your choice.

It also helps to draw a graph of this process.

Also, for real physical systems the integral is often changed to:
y(t)=Integral[0 to +inf] x(t-z)h(z) dz

where it may be a little easier to spot that the variable of integration is z, not t.



Here's a discrete form that makes it a little easier to see too:
y(n)=Sum[k=-inf to +inf] x(k)h(n-k)

where we see we are summing an awful lot just for one n.
It's also very illustrative to convert the low pass filter impulse response into a discrete form (as well as the step or ramp function) and then go through the summation for each n. The result is the discrete form of the output and all this is very easy to graph.

 

[dknguyen]

I figured it out but it wasn't my real problem.

I was basically asking that if you have
g(t) = f(a-t)

Then what would g(t-b) equal to in terms of f(?). THe major problem was just so many t's floating around and they were all slightly different versions of it. It was just the way function operations get superimposed onto each other. Everywhere you see a 't' in the function f, you replace it with 't-b' so you get f[a-(t-b)] = f(a-t+b). THe wrong alternative is you subtract b from the argument of f(a-t) which would give you f[(a-t)-b]

My real problem was I forgot that when convoluting there tends to be a lot of different boundaries of integration that needed to be accounted for. That was a pain.

 

[MrAl]

Hi again,


Ok i see what you mean now.

In cases like that, it is usually a good idea to temporarily switch variable names:
g(t-b)

becomes for example:
g(T-b)

Now when we already have:
g(t)=f(a-t)

in order to get g(T-b) from g(t), we have to set t equal to T-b:
t=T-b

Now when we rewrite g(t)=f(a-t) we have to carefully replace every t with T-b, and so we get:
g(T-b)=f(a-(T-b))

(and the careful part is where we recognize that we are subtracting the entire T-b from a, not just T
because originally we were subtracting t)

and this of course comes out to:
g(T-b)=f(a-T+b)

and then finally swapping back to the old variable name we get:
g(t-b)=f(a-t+b)

It really does help a lot to temporarily switch variable names. Also, using T is not as good of an idea for replacing t
as say Z would be, where it is not the same letter at all.