Fourier Transform Problem

Discussion in 'Homework Help' started by wtrow, Mar 15, 2010.

  1. wtrow

    Thread Starter New Member

    Oct 21, 2009
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    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!
     
  2. Nanophotonics

    Active Member

    Apr 2, 2009
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    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.
     
  3. wtrow

    Thread Starter New Member

    Oct 21, 2009
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    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.
     
  4. t_n_k

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    Mar 6, 2009
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    Last edited: Mar 15, 2010
  5. steveb

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    Jul 3, 2008
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    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.
     
  6. t_n_k

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  7. wtrow

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    Oct 21, 2009
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    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....
     
  8. t_n_k

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    So were you able to determine the value of 'a' that satisfied the energy out = 50% energy in condition?
     
  9. wtrow

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    Oct 21, 2009
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    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.
     
  10. t_n_k

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    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?
     
    Last edited: Mar 16, 2010
  11. t_n_k

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    I have a=1/(2*Pi)

    Let's see what others get .....
     
  12. t_n_k

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    In case you are interested I've attached my attempt at solving part B of the question. Not sure about the method.
     
    wtrow likes this.
  13. wtrow

    Thread Starter New Member

    Oct 21, 2009
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    I followed that work and it all makes sense now. Thank you very much!
     
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