Simplifying equations involving Dirac Delta (Analog Signal Processing)

[nameless912]

Homework Statement

Signal f(t) = (5+rect(t/4))cos(60pi*t) is mixed with signal cos(60pi*t) to produce signal y(t). Subsequently, y(t) is low-pass filtered with a system having frequency response H(w) = 4rect(w/(4*pi)) to produce q(t). Sketch F(ω), Y(ω), and Q(ω) and determine q(t).

I'm specifically having trouble with taking the Fourier transform of f(t) in order to sketch F(w) and also to move on with the rest of the problem.

Homework Equations

f(t) = (5+rect(t/4))cos(60pi*t)
mixed_signal = cos(60pi*t)

The Attempt at a Solution

I attempted to take the Fourier series as follows:
f(t) = (5+rect(t/4))cos(60pi*t)
= 5cos(60pi*t) + rect(t/4)cos(60pi*t)
I used the transformation for cos(w0t) and rect(t/4) to come up with
F(w) = 5pi(delta(w-60pi) + delta(w+60pi)) + sinc(2w) * pi(delta(w-60pi) + delta(w+60pi))
(note that asterisk in the last line indicates convolution not multiplication)

But now I can't figure out how to convolve sinc(2w) and delta(w-60pi) since they have different coefficients in front of w. I know that f(t) * delta(t-t0) = f(t-t0), but that formula doesn't seem to apply here...

I thought that maybe in order to simplify it I could break sinc(2w) into sin(w)cos(w)/w but that seems like it'll leave me with an insanely complex solution to try to do the inverse Fourier transform on later in the problem. Does anyone have any ideas?

 

[rude man]

What is the problem convolving sinc(2w) and δ(w - w0)?

Write out the convolution integral for these two functions of w. (Pick the appropriate one; there are two versions).

Then, big hint: δ(-x) = δ(x) and you have your convolution expression in w.