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Frequency response H(w) problem for rlc circuit

  1. Apr 12, 2014 #1
    1. The problem statement, all variables and given/known data
    for the circuit below, compute frequency response H(w) using method of complex impedence. then if x(t) is input signal (we have equation but, say it is A*sin(wt), find y(t).


    2. Relevant equations

    transfer function H(w)=vout/vin

    3. The attempt at a solution
    R || L = R*jwL/(R+jwL)
    total impedence = (jwRL/(R+jwL))+ R + 1/jwC

    H(w)=vo/in = (R+1/jwC) / (answer above)

    then I'm confused about the 2nd part. if we have x(t), can I use this H(w) above just multiply it by Vin and that will get me Vout?
    so H(w) * x(t)= y(t). sound right?
  2. jcsd
  3. Apr 13, 2014 #2

    rude man

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    First part is correct.
    Second part, you are mixing time and frequency expressions in same equation. Cannot do that.

    x(t) is time domain so you have to change to frequency domain: x(t) = Asin(wt) → A in freq. domain. So output in frequency domain is AH(w).

    Now, do you know how to go back to time domain to get y(t)?
  4. Apr 13, 2014 #3
    If I use the inverse fourier transform to convert freq-> time domain, I would have to do something like H(w) to h(t) first, so [itex]\frac{1}{2\pi}[/itex] -inf to inf∫H(w)*ejwt dw, but the problem is I can't do this with hand calculations, like I have to.
  5. Apr 13, 2014 #4

    rude man

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    You don't use the Fourier transform.
    This transform is much simpler: Asin(wt + ø) transforms to Ae, not F{Asin(wt + ø)}. It is restricted to steady-state, sinusoidal quantities only.

    And similarly with the inverse. You should be able to figure out the inverse based on the forward I just gave you.
  6. Apr 13, 2014 #5
    just to give a bit more information, the eq for x(t) is a DTMF signal, so A1*sin(w1t) + A2*sin(w2t) where A1=A2=10 and w1 and w2 also given. So I remember the transform (for circuits I guess) for A*sin(wt) is A, so X(w) would then be 20? I multiply H(w) by this value, how would I find the inverse h(t) that I need to? would I just need to work backword maybe, like L instead of jwL?
    Last edited: Apr 13, 2014
  7. Apr 13, 2014 #6
    sorry double post (deleted)
  8. Apr 13, 2014 #7

    rude man

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    With two signals x1 and x2 of differing frequencies the transfer function H(jw)will also be different, and you have to deal with one signal at a time, get the output Y(jw) for each, then transform each back to the time domain and add.

    H(jw) you just developed:

    total impedence = (jwRL/(R+jwL))+ R + 1/jwC
    H(jw)=vo/in = (R+1/jwC) / (answer above)

    So H(jw) is still a complex number: H(jw1) = H(jw)|w=w1 for x1 and H(jw)|w=w2 for x2. How'd you get "20"??
  9. Apr 13, 2014 #8

    rude man

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    P.S. you might want to get comfy with a + jb = √(a2 + b2) e
    where ø = tan-1(b/a). Retain the signs of a and b when coming up with ø.
  10. Apr 13, 2014 #9
    sorry i am getting confused. I meant x(t)=A1*sin(w1t)+A2*sin(w2t) (a sum of 2 sinusoids, different frequency w1 and w2 but I don't think its 2 different signals). I got 20 because A1+A2, I thought I just add up the amplitudes for the sines like you said (since theres no phase shift).
  11. Apr 14, 2014 #10

    rude man

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    x(t) = x1(t) + x2(t)
    x1(t) = A1 sin(w1t)
    x2(t) = A2 sin(w2t)
    Therefore, x(t) comprises two signals of differing frequencies: x1(t) and x2(t). You cannot just add their amplitudes.

    Do you know how to use Excel? Pick two different frequencies, assign column A to t, column B to w1*t, column C to w2*t, column D to sin(w1t), column E to sin(wt2), and finally column F to sin(w1t) + sin(w2t). Pick w1 = 1, w2 = 1.3, t increments of 0.05, and use at least 500 rows. Then graph column A on the x axis and D, E and F on the y axis. You will see what x1(t), x2(t) and x(t) look like and why you can't assign an amplitude to x(t).

    I will say again: you must treat each signal separately. x1(t) → A1 → A1|H(jw1)|e → y1(t). y1(t) will look like B1sin(w1t + ø).
    & same for x2(t)
    Then output y(t) = y1(t) + y2(t).
  12. Apr 14, 2014 #11
    yes I'm sorry.. this was explained in my book but I didn't know to look for it. it shows using the fourier trasnform for sin but when they inverse it (I guess Y(w)-> y(t)) it becomes a simple function, A *| H(jw0)| * sin(w0t + ang(H(jw0))). the x(t)=A1*sin(w1t)+A2*sin(w2t) would become y(t)=A1*|H(jw0)|*sin(w0t + ang(H(jw0))) + A2*|H(jw1)|*sin(w1t + ang(H(jw1))). so I was able to figure this out by using the values for magnitude/phase of H(jw) at those 2 frequencies. thanks for the help.
  13. Apr 14, 2014 #12

    rude man

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    Good going! But you should still not invoke the Fourier transform even though of course it works.

    E.g. the Fourier transform for Asin(w0t) is A(jπ)[δ(ω + ω0) - δ(ω - ω0],

    j = √(-1) and δ is the Dirac delta function.

    That's a lot messier than simply A! And if your voltage had been Asin(ωt + ø) it would have been even messier:

    F{Asin(ωt + ø)} = πA{cosø[δ(ω - ω0) + δ(ω + ω0] + jAsinø[δ(ω + ω0) - δ(ω - ω0]}

    as opposed to just Ae.
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