Power Spectrum Problem

Discussion in 'Homework Help' started by Stereoblind, Mar 24, 2015.

  1. Stereoblind

    Thread Starter New Member

    Mar 10, 2015
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    I'm looking through a past paper for my course just now and I have come across a problem about power spectrums. I've looked through my course notes and the recommended textbook and attacked Google but can't seem to find any answer to my question :(

    I've attached the problem as an image, if anyone can help me out I'd be so grateful!
     
  2. t_n_k

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    Mar 6, 2009
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    Presumably the half power bandwidth is defined by

    \{\frac{\sin\(\pi f {10}^{-2}\)}{\pi f {10}^{-2}}\}^2=0.5

    Solve for the value of f that satisfies the relationship.
     
    Last edited: Mar 24, 2015
  3. Stereoblind

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    Mar 10, 2015
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    I'd be at a loss at how to solve that because we have the f as an argument of the sin function but also have it underneath?
    We've been given an answer sheet that says the answer is approximately 161.7Hz. Even if I punch that in I'm not getting a half :/ .
     
  4. t_n_k

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    I would plot the function and look for the value when it drops to around 0.5 G(0). Then try to tweak the value by any method that works. Are there tables for the sinc function? That's the best I can offer.
     
  5. WBahn

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    Mar 31, 2012
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    Another case of an author not giving a damn about units.

    Since the argument to the sine function HAS to be dimensionless, the value of

    <br />
\pi f 10^{-2}<br />

    has to have units of radians. This will not be the case if 'f' has units of Hz. To sort it out, we would need to know where this power spectrum came from and since it is just a given in the problem there's no way to do that.
     
  6. Stereoblind

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    Mar 10, 2015
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    Thanks for your help guys. I've now resorted to emailing my lecturer asking for clarification on what to do - I will keep you posted :)
     
  7. Stereoblind

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    Mar 10, 2015
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    "You cannot find the solution for this problem analytically (except part c), hence something like this will not come at the exam/test.

    To solve it, observe that G(f) is maximum when f=0 and the max is G(f=0)=10. Half power bandwidth is the frequency f* when G(f*)=G(f=0)/2=5. You can now replace this value in the equation and obtain:
    5=10{sin(pi f 0.01)/(pi f 0.01)}^2 . This is an equation with one unknown that you should solve for f. The obtained f is the 3-dB bandwidth. As I said this cannot be done analytically, but you can use Matlab or any numerical method to solve it.

    Part b) is similar, only we are looking now for f** such that G(f**)=G(f=0)/4."
     
  8. WBahn

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    It's pretty clear that the lecturer hasn't tried to actually solve it yet -- which is not too surprising and is not even a condemnation of them.

    Did you tell the lecturer what you are getting for an answer and that it doesn't match the answer given. That will usually get them to verify the answer themselves.
     
  9. t_n_k

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    Which is what post #2 succinctly proposes.
     
  10. t_n_k

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    @Stereoblind : you searched for information on Google about this particular problem. I suspect with many of these mathematical "perspectives" of signal processing, one is likely to lose sight of the origin & usefulness of such mathematical expressions. If you Google energy spectral density or power spectrum of a rectangular pulse, you might find some further enlightenment.
    Adopting a bit of 'Peaseology', it might also be worth asking the (hopefully pertinent) question - "What's all this power spectrum stuff, Anyhow?"
     
    Last edited: Mar 25, 2015
  11. Stereoblind

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    Mar 10, 2015
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    Very true, thanks for your help.

    I will chase the lecturer up further to ask if he has actually set about solving the problem and I will see what he says.

    Sorry, I did not mean to imply that I had been searching for this particular problem. I had indeed been searching for power spectrums and energy spectral densities and resorted to posting here when looking at these did not help me answer the tutorial problem. You're right though :) far too often people only look for the answer to their question rather than an understanding of their question!
     
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