Fourier Transform Problem

Thread Starter

wtrow

Joined Oct 21, 2009
18
I don't know if this forum covers signals and systems, but here it is.

The signal x(t) = exp(-at)u(t) is input into a system with impulse response h(t) = sin(2t)/(pi*t)
A) Find the Fourier Transform Y(w) of the output
B) For what value of a does the energy in the output signal equal one-half the input signal energy?

So for part A) - I take the convolution to get y(t), yes? Which is y(t) = ∫exp(-at)u(t)*sin(2(t-τ))/(pi(t-τ))dτ, which is an integral I can't even begin to solve for. But if and when I solve for y(t), I take the fourier transform of it to get Y(w), right? (Any help with that integral or transform would be greatly appreciated). And I don't even know how I would go about solving B). Thanks!
 

Nanophotonics

Joined Apr 2, 2009
383
A) Finding the Fourier Transform Y(w);
Instead of convolving, you can find the F.T of x(t) and h(t) individually, and their product will give you Y(w). It's a lot easier to do multiplication in the frequency domain than convolution in the time domain.

Y(w) = X(w).H(w)

B) No clue yet.

You're in the right forum. You'll soon see the backup coming.
 

Thread Starter

wtrow

Joined Oct 21, 2009
18
That helps a little. Doing it that way I get:

X(w) = ∫exp(-at)u(t)*exp(-jwt)dt = (since its a step function, change bottom limit from -∞ to 0) ∫exp(-at)exp(-jwt)dt = 1/(a+jw)

H(w) = ∫sin(2t)/(∏t)*exp(-jwt)dt = I cannot solve this intergral to save my life.
 

steveb

Joined Jul 3, 2008
2,436
That helps a little. Doing it that way I get:

X(w) = ∫exp(-at)u(t)*exp(-jwt)dt = (since its a step function, change bottom limit from -∞ to 0) ∫exp(-at)exp(-jwt)dt = 1/(a+jw)

H(w) = ∫sin(2t)/(∏t)*exp(-jwt)dt = I cannot solve this intergral to save my life.
You could definitely solve that if your life depended on it. Remember Euler's formula lets you express the complex exponential in terms of sines and cosines. Similarly, the sine function can be written in terms of complex exponentials. If you still don't know how to complete it from there, just look up the integral in a table.

However, since your life probably does not depend on it, it is simpler to just use a transform table. I've attached the one I use, in case you don't have one yet.
 

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Thread Starter

wtrow

Joined Oct 21, 2009
18
Thanks for all of your answers! They're very helpful. Using that transform table I found one for a sinc function like t_n_k mentioned, giving my final answer for part A) to be Y(w) = 1/(a+jw) if abs(w)<2, 0 if abs(w)>2

Look good?

Now for part B....
 

Thread Starter

wtrow

Joined Oct 21, 2009
18
Unfortunately, I can't find a way to. Looking at Y(w) closer I think that it can be written as rect(4w)/(a+jw) right? But in order to find the value of a that makes the output 50% of the input, I was going to set it up as x(t)=y(t)/2 and solve for a, but I do not have Y(w). Can I get this by doing the reverse Fourier transform? I think its 1/2π∫Y(w)exp(jwt)....not sure if the j is negative or not.
 

t_n_k

Joined Mar 6, 2009
5,455
Because the 50% condition relates to energy rather than simple amplitude then one has to think in terms of the cumulative square of the amplitude terms across the signal spectrum. As an analogy think in terms of power vs voltage (or current) across a resistor element.

You may be able to solve this using spectral density function - I'm not sure.

I think (??) you can show the 'energy' of the input is of the form 1/(2a). There may be a constant multiplier involved, but the dependence on 'a' is the important consideration. What Sinc filter cut-off frequency would reduce that value by 50% is the burning question. Perhaps it would be better to state this - for a known Sinc filter bandwidth what value of 'a' satisfies the 50% condition?
 
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