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What is my bandwidth?

  1. Jul 1, 2012 #1
    1. The problem statement, all variables and given/known data

    My message signal is the Fourier Series (x(t)) of a bipolar pulse. This signal passes through the ideal low pass filter, h(f), which has a cutoff frequency of 4Khz. What the baseband bandwidth of the filtered signal?

    2. Relevant equations


    ƩAn*cos(2*pi*n*f0*t), An= [4*(-1)^((n-1)/2)]/(pi*n)

    In other words: x(t)=(4/pi)*[cos(2*pi*f0*t)-(1/3)*cos(2*pi*3*f0*t)+(1/5)*cos(2*pi*5*f0*t)]

    3. The attempt at a solution
    I don't know if I should apply the Fourier Transform to the Fourier Series because this one is x(t) and my filter is h(f). What should I do?
     

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  3. Jul 1, 2012 #2

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    Welcome to PF, marina87! :smile:

    Your Fourier Series gives you the frequency spectrum.
    For instance at ##f_0## the amplitude is ##{4 \over \pi}##, and at ##3f_0## the amplitude is ##{4 \over \pi}{1 \over 3}##.

    If you apply a low pass filter, all the higher frequencies are cut off.
    To find the baseband bandwidth, you need to know what ##f_0## is.
    Do you?
     
  4. Jul 1, 2012 #3
    I see so if my f0 is 1KHz then my bandwidth will be 3KHz.
    But if I modulate this signal with a carrier signal cos(2*pi*fc*t) I still need to use the Fourier Transform , right? In other words I still need to apply the Fourier Transform to the series or wait because I filtered the signal is not a Fourier series anymore? Am I correct?
     
  5. Jul 1, 2012 #4

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    So what is the highest frequency represented in your fourier series that is less than your cut-off frequency of 4 kHz?
     
  6. Jul 1, 2012 #5

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    Yep. :)

    There's no real need to apply a Fourier transform.
    You can read off the frequency spectrum straight from the Fourier series.

    But if you would apply a Fourier transform to you Fourier series, you'll find a transform that shows you spikes at exactly the aforementioned frequencies.


    Btw, the filtered series is still a Fourier series.
    It's just that the higher order terms all have amplitude zero.
     
  7. Jul 1, 2012 #6
    3Khz but can you help me with my new analysis. I just need someone to tell me if I am in the right track.

    After filtered this signal its not a Fourier Series anymore is just a "regular signal that add to cosines functions. Then I can apply the Fourier Transform right?
     
  8. Jul 1, 2012 #7
    Sorry I am confusing you. Let me explain a little more the problem.
    First I have to filter my signal x(t) (The fourier Series) with and ideal low pass filter which has a cut off frequency 4KHz. Afther that this "new signal x1(t)" is multiply by a cos(wct) to obtain a DSB-SC modulation. I am thinking apply the Fourier transform to the modulated signal to obtain the AM expression.
     
  9. Jul 1, 2012 #8

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    Hmm, the new signal x1(t) is just the sum of 2 cosine terms.
    You can simply multiply them by your new cosine term, yielding your DSB-SC modulation.

    If you want to find the frequency spectrum, the easiest way is to replace the cosine products by cosine sums, using the relevant goniometric identity.

    See for instance the section "Product-to-sum and sum-to-product identities" in http://en.wikipedia.org/wiki/List_of_trigonometric_identities
     
  10. Jul 1, 2012 #9
    That is what I was thinking but ones I have the product of the cosines then I can use the properties of the Fourier Transform to obtain the spectrum?
     
  11. Jul 1, 2012 #10

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    Which properties are you thinking of?
     
  12. Jul 1, 2012 #11
    x(t)*cos(2pif0t) ----> 0.5*(X(f+f0)+X(f-f0))

    If you use the Eulers formula of a cos(x)= 0.5*(e^(jx) + e^(-jx)) and then multiply by a a function x(t) then you can apply the Frequency Shift property
     
  13. Jul 1, 2012 #12

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    Yes, that looks usable.

    You do realize that:
    $$\cos \theta \cos \varphi = {\cos(\theta - \varphi) + \cos(\theta + \varphi) \over 2}$$
    Note that this looks a lot like your property.
     
  14. Jul 1, 2012 #13
    Thanks for the help.
     
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