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Homework Help: Fourier Transform question (kind of Urgent)

  1. Sep 7, 2012 #1
    This is for an assignment, (not sure if its in the right section) but anyway I'm considering the system response to H(w) = 10/(jw + 10)

    when the input is x(t) = 2 + 2*cos(50*t + pi/2)

    so I know that Y(w) = X(w).H(w) but I'm not sure what to do about the '2 + ' in the input.

    I know that '1' transforms into 2piδ(w) frequency domain.

    so does this mean that 2 becomes 4piδ(w) ? times by H(w)...?

    And if so, does this mean it in x(t) is only 4pi at w = 0 ? Because the w = 50 from that input.

    Alternatively I concidered 2*H(0) or 2*H(50), but I see no reason for them.

    My calculation would be as follows:

    % y(t) = 2*something??* + |H(50)|*2.*cos(50*t + pi/2 + <H(50) )
    % where <H(50) = ( 10<0 / 50.99<78.69 )
    % and 360 - 78.69 = 281.31 and plus pi/2 that equals 11.31 deg
    % in rad that is 0.197
    % y(t) = 2 + 0.392*cos(50*t + 0.197)

    Please help asap!

  2. jcsd
  3. Sep 7, 2012 #2

    rude man

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    You shold stay in the Fourier domain until the end.

    You have X(w) = 4πδ(w) + the Fourier transform of 2*cos(50*t + pi/2)
    You might wanty to exploit the fact that cos(x+π/2) = -sin(x)
    You have H(w) = 10/(jw + 10)
    So you have Y(w) = X(w)H(w)
    and finally you have y(t) = F-1(Y(w).
  4. Sep 7, 2012 #3
    What you said makes perfect sense, except that we were told that
    x(t) = Acos(w0t + theta) = Re[Ae^(j.theta).e^(j.w.t)] (I'm assuming from Euler identity)

    and so the response is y(t) = A|H(w0)|cos(w0t + theta + <H(w0) )

    So I don't have to inverse that bit?

    Ah, I thought it might be 4πδ(w), so by what you've said I'd need to inverse transform:
    4πδ(w)* 10/(jw + 10) = Y(w) of the response to '2 +'.

    What would that be? (is w = 50 or 0?)

  5. Sep 7, 2012 #4

    rude man

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    w is not a constant.

    What is the inverse transform integral? I.e. going from Y(w) to y(t)?
  6. Sep 7, 2012 #5
    Good point!

    U'mm, ok so Y(w) being 40piδ(w)

    y(t) = 1/2pi∫-∞→∞ [H(w).Y(w).e^(j.w.t)] dt for -∞=< t =< ∞

    y(t) = 1/2pi∫-∞→∞ [(4piδ(w).e^(j.w.t))*10/(jw + 10)] dt for -∞=< t =< ∞

    so y(t) = 1/2pi∫-∞→∞ [(40piδ(w).e^(j.w.t))/(jw + 10)] dt for -∞=< t =< ∞

    therefore y(t) = (20.e^(jwt))/(10j.w - w^2) ?????????? that doesn't look right
  7. Sep 8, 2012 #6

    rude man

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    No, we said X(w) = 4πδ(w), right? So Y(w) = X(w)H(w) with H(w) = 10/(10 + jw).

    So since y(t) = F-1{Y(w)}, the inverse integral had better be (1/2π)∫-∞+∞X(w)H(w)ejwtdw, don't you think?

    BTW yes you could compute the cosine part of your output the way you said. That is the phasor method which is normally used with continuous sinusoidal inputs. For that matter, you could have handled the "2" input the same way: |H(0)| = 20/(10 + jw) with w = 0 so output = 2. But I thought you were supposed to use the Fourier transform.

    EDIT: I noticed later your integral is almost right. But the integrating variable is w, not t.

    Make the correction to that integral the way I said and now ask yourself the $64,000 question: what is
    -∞+∞f(x)δ(x)dx for any f(x)? Hint: it's a really easy integration!!!
    Check with the sampling theorem if you don't know.
    Last edited: Sep 8, 2012
  8. Sep 8, 2012 #7
    Sorry, I ment X(w)!
    Umm, its just finding the system response to the input, I was using the transform because I thought I had to.

    That cosine part "phasor method" was just in my lecture notes, I'm still not 100% on the basic principle of how it works, hence my confusion in using it here with the '2 +' part of the input.

    So X(w) being 40piδ(w) was what I did correct?

    (I really am struggling with applying the δ function, I'm also trying to find the exponential Fourier series of: x(t) = Ʃ δ(t - kT) for k = -∞ → ∞ But I can't understand how to apply it.

    Thanks again!

    EDIT: I just saw your EDIT and disregard this, I haven't read it enough to properly reply
  9. Sep 8, 2012 #8
    Ok, *composed reply:*
    Well from what I do understand about δ
    -∞+∞f(x)δ(x)dx = f(0)

    EDIT: [ I lost my train of thought and forgot to include the δ amongst my messy working out,
    f(x) = 40pi, δ(x) = δ.exp(jwt)
    so 1/2pi * Int [ f(x)δ(x) ] dx * H(w) = f(0)*H(w) = 40pi/(2pi(jw+10)) = 2 at w = 0

    1/(jw+10) * [20*1] (the '1' being f(0)) I hope you don't see what I embarrassingly originally wrote]

    So with 'w' as the variable and '2' as the input signal:

    y(t) = 20/(10j.w + 10) = 2 at w = 0 because the input of '2' is DC, so to speak?

    Appreciate your help so much!
    Last edited: Sep 8, 2012
  10. Sep 8, 2012 #9
    As I said, I hope you didn't see my un-edited last reply before I changed it, it would be enough to demoralise anyone from giving further advice ever again.
    ∫-∞+∞f(x)δ(x)dx = f(0) was a helpful relation, though I was wondering if: f(x) = 40pi, δ(x) = δ.exp(jwt) is the right way of 'putting it'?

    Could I lastly ask, the fourier series of:
    x(t) = Ʃ δ(t - kT) for k = -∞ → ∞

    I'm assuming they mean that 'T' is the period, so would x(t) be like: a δ on the multiples of the period, like -T, 0, T, 2T... → ∞ ?

    so integrating that Fourier Series would Cn = 1/T ∫ 0→T [1* exp(-jKwt)] dt ?

    Last edited: Sep 8, 2012
  11. Sep 8, 2012 #10

    rude man

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    I don't understand this question, toneboy.

    You did get the right answer for y(t) = 2 going the Fourier route and yes, that's a dc input (w = 0 → zero frequency = dc!).

    If you want to practice your Fourier transform, doing the cosine input is a really good exercise! I just went thru it myself with another op & it had me in a spin for a bit. But, as I said, the conventional way is to use phasors. BTW the term 'phasor' is really not applicable for dc voltages. You just set w = 0 everywhere and get your answer pronto.

    Yes, T is the time spacing between delta functions. This is a very important function in sample-data theory. The idea is every delta function δ(t - kT), k = 1,2, ... with the appropriate coefficient, represents one sample of a sampled continuous ("analog") function, like think voltage from a microphone. The series thus represents discretization, and suggests digitization, of the analog signal.

    Finding the Fourier series is quite straight-forward. But why do you have "1" in your integrand for finding the complex Cn coefficients? (I assume you're referring to the exponential form of the series). Your function is not "1", it's δ!

    Use the exponential form of the series and compute the Cn complex coefficients. The period is indeed T.

    Cheers also.
  12. Sep 8, 2012 #11

    rude man

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  13. Sep 8, 2012 #12
    I thought '1' because I thought that was the magnitude of δ at T was '1'.
    would it be fair to say:

    Cn = 1/T ∫ 0→T [δ(t-kT)* exp(-jKwt)] dt

    = 1/-jkwT*[δ(T-kT)*exp(-jkwT) - δ(0-kT)*1]

    =1/jkwT * δ(0-kT) ?

    (thanks for the fourier link btw)
  14. Sep 9, 2012 #13

    rude man

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    The value of δ(t-T) at t = T is infinity!

    It's the integral of δ(t-T) from -∞ to +∞ that's equal to 1.

    Well, first, you don't want to integrate from 0 to T because T is where the function is infinite & you need to include your entire function x(t) when integrating over one period T. You might be missing half of infinity! Let's say instead you integrate from t = T-ε to T+ε where ε is an arbitrarily small number; that would include your entire function over one period T, right? Then, realize that you can integrate δ(t-T) from -∞ to +∞ and get the same answer!

    In what follows I assume Cn are the coefficients of the exponential form of the Fourier series.

    I don't think your formula for Cn is correct. For one thing, n is missing on the right-hand side. One correct form for Cn is:

    Cn = (1/T)∫ over one period T of δ(t - kT)e-j2nt/Tdt

    By that convention then,

    x(t) = Ʃ from m = -∞ to +∞ of Cmejmw0t
    where w0 = 2π/T.

    Now, think about solving that integral, bearing in mind what we said about solving an integral with δ(t-T) in it - how simple that is!
  15. Sep 9, 2012 #14

    rude man

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    I notice now that you've confused n and k in your Cn expression, otherwise it's actually quite spot-on.
    So it's δ(t - kT) but it's exp(-jnwt) but realize that your w is really 2π/T.
    Also of course your evaluation of the integral is wrong. What did we say about ∫f(x)δ(x-x0)dx?

    Bed time for me, be back in about 9 hrs. I'm at GMT-7.
  16. Sep 9, 2012 #15
    Do you have like 1 more min! ? please!
  17. Sep 9, 2012 #16
    Ah, I thought you'd be in a different time zone. I had no idea it was so late. (locally)

    So Cn = 1/T ∫ T-ε→T+ε [δ(t-kT)* exp(-jn2pi.t/T)] dt

    = 1/T*[exp(-jn.k.2pi)] (that is, only f(x0) remain at t = Tk and the 'T' in the exponent cancel out)

    IF you're still there, is that right?

    EDIT: forget it, go to bed lol, I don't want to keep you up, I need too long to think about it, thanks
    Last edited: Sep 9, 2012
  18. Sep 9, 2012 #17

    rude man

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    Right you are!
  19. Sep 9, 2012 #18
    You're a genius, a legend and a saviour...not to mention a gentlemen and a scholor.

    Thanks for your help, rest well.
  20. Sep 9, 2012 #19

    rude man

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    BTW note that exp(j2πk) = 1 for integer k!
  21. Sep 9, 2012 #20
    H'mm, I'm yet to reason why that is...

    EDIT: ofcourse, Euler Identity!
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