2nd order low pass fillter

Discussion in 'Homework Help' started by Jaywin, Apr 22, 2008.

  1. Jaywin

    Thread Starter New Member

    Apr 20, 2008
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    For the 2nd order low pass filter
    is the input impedence calculate like this:
    Zin = 1/jwC1 + R2||(R2+1/jwC2)
    if freq-> inf then the Zin is just R1||R2.
    Am i correct? That is wave generated using Pspice.^^
    [​IMG][​IMG]
     
  2. Caveman

    Active Member

    Apr 15, 2008
    471
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    Actually as f->infinity. Zin = R1. Assuming perfect opamp of course.
     
  3. Caveman

    Active Member

    Apr 15, 2008
    471
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    Otherwise, you just analyze the circuit to get the voltage at the node between R1 and R2, Va.

    Zin = (Vin-Va)/R1.

    There might be an easier way, but I don't know it.
     
  4. Jaywin

    Thread Starter New Member

    Apr 20, 2008
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    Zin = (Vin-Va)/R1?

    (Vin-Va)/R1 isn't we wil obtain current? by the way, Va is same as Vc1 right?
    Actually, the problem is to fulfill the specification which our
    Zin >= 10kOhm
    So if i put the R1 = 10kohm is also satisfy the specification right?

    Zin = 1/jwC1 + R2||(R2+1/jwC2) is wrong?
     
  5. Caveman

    Active Member

    Apr 15, 2008
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    Whoops,

    Zin = R * Vin/(Vin-Va)

    But yeah, R1 =10k should solve it, I think.
    I frankly haven't had a chance to check your Zin since it is fairly complex to quickly calculate. I might look at it tonight.
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Is the input impedance your only specification? Don't you also have to meet a frequency response spec?

    Go back to the Spice plot of Zin in your first post, and look carefully at the region between 3000 Hz and 4000 Hz. This may be a trick question.

    If you make R1 = 10k, you also might want to make R2 = 10k, and make C1 and C2 bigger to meet the frequency response spec.

    But if you do that, plot Zin at about 7000 Hz and look carefully at its behavior.

    I think Zin has a second order numerator.
     
  7. Jaywin

    Thread Starter New Member

    Apr 20, 2008
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    The cutoff frequency is given f0 = 1kHz
    If change the R then C also have to change also using f0 = 1/2paiRC
    which RC = 159.15us right?
    Why have to observe certian frequency to know the Zin whether meet the
    >= 10k specificatin?
    Is it just look at the red-circled region if it is >=10k then we meet the specification. Am i right?

    R1=R2=10k
    [​IMG]
    R1= R2=20k
    [​IMG]
     
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Adjust your plot so that the maximum vertical axis magnitude is 25k instead of 300k or 500k.
     
  9. Caveman

    Active Member

    Apr 15, 2008
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    Man, this is nasty, but here goes:

    Zin = R1/(1-H*(1+jwC2*R2)/G)

    In this case H is the transfer function of the 2-pole filter. Look at wikipedia.
    G is the DC gain = (3.5k + 6k)/3.5k.

    That's as far as I got.
     
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