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Center Frequency and Bandwith of Bandpass Filter?

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine the center frequency and bandwidth of the bandpass filter.
    VRYp3g4.png

    2. Relevant equations
    center frequency of bandpass filter = ωc occurs when the magnitude of H = 1

    H = V0/Vs

    Center frequency occurs when Im(Z) = 0
    Zc = 1/(jωC) = 1/s
    ZL = jωL
    ZR = R

    I don't know a formula for finding bandwidth.
    3. The attempt at a solution
    Zseries = 1 + 1/s
    Zll = ((1/s)(1 + 1/s))/(2/s + 1)
    Zeq = 1 + Zll
    Simplifying gives...
    Zeq = (1 + 3s + s2)/(2s + s2)

    Vs = IZeq
    Vll = IZll
    V0 = I2(1Ω)
    Vseries = I2Zseries = Vll

    V0Zseries = Vll = IZll

    V0Zseries = (VsZll)/Zeq

    V0/Vseries = Zll/(ZeqZseries)

    Plugging in values from above and simplifying gives...

    V0/Vseries = s/(s + 3s + 1) = H

    s2 = -ω2

    How do I find the magnitude of H? What is a bandwidth?
     

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  2. jcsd
  3. May 15, 2015 #2

    rude man

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    First thing you should do is label the components R1, C1, C2, R2 left to right. Then work with these symbols.

    Then: your H(s) can't be right since it has gain = 1/4 at high frequencies whereas the network H(s) must have zero gain there, thanks to C1.
     
  4. May 15, 2015 #3

    BvU

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    Hello GZ,

    Took me a while too to untangle this. In the mean time Rudy chipped in, but here's my 2 cents as well:

    It would have been easier for me (and probably for others too) if you would have told us what you mean with the various symbols.

    Up to Vseries= ... I agree. Then V0 Zseries=Vll x (1Ω) otherwise the dimensions don't fit. But numerically you're still good.

    V0/Vseries = Zll x (1Ω)/(ZeqZseries) is something you could have gotten more straightforward as

    V0/V|| = 1Ω / Zseries multiplied with

    V||/Vs = Zll / Zeq

    The typo that distracted Rudy demonstrates your derivation is a bit complex (too many Z, too many inversions and then inverting again). Never mind.

    Magnitude of H follows from ##\sqrt{ H^{\displaystyle ^*} H }## as with all complex numbers.

    (At first, LaTeX let me down here, with ##H^{\displaystyle ^*}## I mean the complex conjugate of ##H## )


    Bandwith is ##| \omega_1 - \omega_2 |## where ##H^2(\omega_{1,2}) = H^2_{max}/2##
     
    Last edited: May 15, 2015
  5. May 15, 2015 #4
    So from my value for H = s/(s2 + 3s + 1), I can convert back to using jω:

    H = (jω)/(-ω2 + 3jω + 1)
    H = (jω)/(3jω + (1-ω2))

    The complex conjugate of the denominator is then 3jω - (1-ω2). I can multiply the numerator and denominator by this and simplify to get:
    H = ((ω2-1)(jω) - 3ω2)/(-9ω2 -1)

    √(H*H) = 1 → H*H = 1

    ((ω2-1)(jω) - 3ω2)/(-9ω2 -1) = 1
    2-1)(jω) - 3ω2 = -9ω2 -1
    Simplifying gives the quadratic equation:
    (6 + j)ω2 - jω + 1 = 0
    The quadratic equation gives the result:
    ω = -.0219 -.331j

    I don't think I did that right. Did I determine the complex conjugate incorrectly?

    So for the bandwidth, am I supposed to find dH/dω = 0, and plug that ω back in to find Hmax, then solve Hmax2/2 or am I misunderstanding?
     
  6. May 15, 2015 #5

    rude man

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    You never performed HH*. When you multiply numerator & denominator of H(jw) by the cc of the denominator you wind up with the original H(jw), not the cc
    You compute |H| = √|H(jw) H*(jw)|, which is the magnitude of H(jw) and is a real number.
    You can then set d|H|/dw = 0 to solve for w of |H|. Plug into |H| and you get |Hmax|. Then set |H| = 1/√2 Hmax and solve for w again. This time you get two w. Rest is as BvU stated.
     
    Last edited: May 15, 2015
  7. May 15, 2015 #6
    How do I find the complex conjugate of something in the form of (jω)/(3jω + (1-ω2))? I don't see how to simplify it into an x + iy form.

    Is the complex conjugate (-jω)/(3jω - (1-ω2))?
    If this is the complex conjugate then I get...

    √[(jω)/(3jω + (1-ω2))][(-jω)/(3jω - (1-ω2))] = 1

    [(jω)/(3jω + (1-ω2))][(-jω)/(3jω - (1-ω2))] = 1
    Simplifying gives...
    ω4 + 8ω2 + 1 = 0 which only has imaginary roots, so it's probably also wrong.
     
  8. May 15, 2015 #7

    BvU

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    No need to simplify. Just replace every instance of j by ##-##j
     
  9. May 15, 2015 #8

    BvU

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    That's the idea. (##d|H|\over d\omega## you mean, of course ... :wink:).
     
  10. May 15, 2015 #9
    Okay, so when I replace every instance of j by -j, the complex conjugate is (-jω)/(-3jω + (1-ω2)).

    [(jω)/(3jω + (1-ω2))][(-jω)/(-3jω + (1-ω2))] = 12

    This simplifies to 0 = ω4 + 6ω2 + 1
    so ω = ±j(.414) and ω = ±j(2.414)

    Is this part correct?
     
  11. May 15, 2015 #10

    rude man

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    Why are you setting |H|2 = 1?
     
  12. May 15, 2015 #11
    I thought that a bandpass filter was defined as having its |H(ωc)| = 1. Is this not the case?
     
  13. May 15, 2015 #12

    BvU

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    Check it out ! You already have an expression for ## |H(\omega)|##
     
  14. May 15, 2015 #13

    rude man

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    No it isn't. You need to go d|H|/dw = 0, solve for w, replace in |H|.
     
  15. May 15, 2015 #14
    |H(ω)| = [(jω)/(3jw + (1-ω2))][(-jω)/(-3jw + (1-ω2))]
    |H| = ω2/(7ω2 + ω4 + 1)

    d|H|/dω = 2ω(7ω2 + ω4 + 1)-12(7ω2 + ω4 + 1)-2(14ω + 4ω3) = 0
    Simplifying gives ω = 1. Is this the center frequency?

    |H(1)| = 1/9 Is this the bandwidth?
     
  16. May 15, 2015 #15

    rude man

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    Yes, if you did the math correctly. I did not check it.
    No. |H|(1) = max. gain.
    I did not check your math but your approach is correct up to then.
    Follow what I said in post 5 near the end to get the bandwidth.
     
  17. May 15, 2015 #16
    |H(ω)| = ω2/(7ω2 + ω4 + 1) = (1/√2)Hmax = 1/(9√2)

    Solving for ω gives ω = ±2.355 and ±.425

    Bandwidth = B = ω2 - ω1 = 2.78 or 1.93

    Is this correct?
     
  18. May 15, 2015 #17

    rude man

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    ω cannot be negative, at least not for you now.
    Otherwise, looks right if you did the math right.
    Check #1: does |H| → 0 as ω → 0? Yes.
    Check #2. does |H| → 0 as ω → ∞? Yes.
    Check #3: does √(ω1ω2) = ω0 = 1? √[(2.355)(0.425)] = 1. Yes. (This test rests on what I state below in my last sentence. So you most probably got the mid-band and corner frequencies right.

    EDIT: But it's not sure. What you can say is that if √(ω1ω2) is not = ωmax then you got something wrong. In other words, it's a necessity but not a sufficiency test.)

    So, H is a bandpass, since gain is zero at DC (ω=0) and at arbitrarily high frequencies, and gain > 0 somewhere inbetween.
    But you don't have the right answer for bandwidth. Bandwidth is the width of the passband around the peak, with rolloff frequencies at gain = |Hmax|/√2 on either side of the max. frequency ω0. If you're using log paper (for the x axis), the two points w1 and w2 are equidistant from w0. ω0 is the geometric mean between ω1 and ω2. So take another shot at bandwidth.
     
    Last edited: May 15, 2015
  19. May 15, 2015 #18
    If I know that ω1 and ω2 are correct by calculating that √ω1ω2 = 1, and I know that I should take the positive values of ω1,2, then why isn't the bandwidth = |ω12| = |.425-2.355| = |-1.93| = 1.93?
     
  20. May 15, 2015 #19

    rude man

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    I said ω is always > 0 but you still got the same right answer: bandwidth = 2.355 - 0.425 = 1.93 rad/s.

    BTW look again at my post #17. I edited it a bit.
     
    Last edited: May 15, 2015
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