Fork me on GitHub
Math for the people, by the people.

User login

Convolution over finite interval

Primary tabs

Convolution over finite interval

Hi there,

I want to calculate a convolution over a finite interval, that is I have

int_{0}^{t} f(t-t')g(t')dt'

which I want to compute using the convolution theorem. However this is not an exact convolution since the integral does not run from -infty to +infty. Of course I figured out that

int_{0}^{t} f(t-t')g(t')dt'=int_{-infty}^{+infty} Rect(t'/t+1/2)f(t-t')g(t')dt'

where Rect(x) is the rectangle function valuing 1 between -1/2 and 1/2 and 0 elsewhere.

Is I suppose that
FT(int_{0}^{t} f(t-t')g(t')dt')= sinc(w)*F(w)G(w)

where FT means fourier transform, F=FT(f), G=FT(g)and * means convolution.

Is this correct ? (I might be missing a complex exponential from the translation of the rect function)

Do you know a better way to compute FT(int_{0}^{t} f(t-t')g(t')dt') ?

Thank you !

Hi there,

Sure I agree with you but this substitution must have an impact ont he convolution. Here is a better way I found to do this :

int_{0}^{t} f(t-t')g(t')dt'= int_{-infty}^{infty}O(t-t')f(t-t')O(t')g(t')dt'
where O(x) is the step function (O(x)=0 when x<0).
Thus using now the convolution theorem you get that

FT(int_{0}^{t} f(t-t')g(t')dt')=FT(O(t)f(t)) FT(O(t)g(t))

However I now have the following question : is there a formula that gives FT(O(t)g(t)) for an arbitray function g ? By that I mean do you know a formula that gives FT(O(t)g(t)) as a function of FT(g) and FT(O) (this one is such a dirac delta with some constants).

I'm not sure if it is different from what you did, but why not substitute h for g where h(t')=g(t') for 0<t'<t and h(t')=0 elsewhere?

Subscribe to Comments for "Convolution over finite interval"