# Convolution in Frequency Domain

1. Dec 10, 2012

### f00lishroy

1. The problem statement, all variables and given/known data
Find the Fourier transform of the following signal JUST by using the FT table and the FT properties

x(t) = sin(t) -pi<=t<=pi
0 otherwise

NOTE: I am using CONVOLVED WITH as a substitute for * (the real convolution operator) because I cannot express multiplication in any other way that I know of).

2. Relevant equations

Frequency Convolution : x1(t)*x2(t) ==> (1/2pi)X1($\omega$) CONVOLVED WITH X2($\omega$)

Fourier Transforms:
sin(ω0t) ===> (pi/j)[δ(ω-ω0)-δ(ω+ω0)]

rect(t/$\tau$) ===> $\tau$sinc($\omega$$\tau$/2pi)

3. The attempt at a solution

I know that x(t) can be expressed as a sine multiplied with a rectangle function

x(t) = sin(t) * rect(t/2pi)

so by the frequency convolution property, X(ω)= (1/2pi)[$\frac{pi}{j}$[δ(ω-1)+δ(ω+1)] CONVOLVED WITH 2pi*sinc(ω)]

My problem is how to do the convolution. Is it easier than it looks? I don't know how the imaginary (j) factors into the convolution. Can someone help explain how to simplify X(ω)? Thank you

2. Dec 11, 2012

### rude man

I get from the problem statement that you're not supposed to doany convolving.

3. Dec 11, 2012

### f00lishroy

My teacher actually said that using the rectangle function was the correct method of doing it, but didn't elaborate. All he said was set x(t) = sin(t)*rect(t/2pi)

4. Dec 11, 2012

### rude man

EDIT:
OK, then, how about the fact that

f(ω)**δ(ω-ω0) = f(ω-ω0) where I use ** to denote convolution. You have delta functions in your transform of sin(t); these are easily eliminated by using this theorem in convolving with the transform of the rectangular time function. The rest is very messy algebra.

Last edited: Dec 11, 2012