Convolution in Frequency Domain

[f00lishroy]

Homework Statement

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).

Homework 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) ===> \tausinc(\omega\tau/2pi)

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

 

[rude man]

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

How about using u(t-π) and u(t+π) along with sin(t) and forgetting about your rectangle function?

 

[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)

 

[rude man]

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)

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.