RC Bandpass transfer function

Discussion in 'Homework Help' started by fireface, Apr 21, 2014.

  1. fireface

    Thread Starter New Member

    Apr 21, 2014
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    I'm having trouble getting the right transfer function for this using the OC and SC method for the time constants and getting the poles.

    Am I doing this right ...
    Open circuit
    1/s1 = (R1+R2)*C1
    1/s2 = C2*R2
    1/s3 = (R2+R3+R4)*C3
    1/s4 = C4*R4

    Short circuit
    1/s5 = R1*C1
    1/s6 = (R1||R2 +R3)*C2
    1/s7 = R3*C3
    1/s8 = R3||R4 *C4

    Where do I go after this?
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Is this in terms of jw or s?
     
  3. fireface

    Thread Starter New Member

    Apr 21, 2014
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    's'... i think. sorry, been a long time

    s = jw?

    Did I calculate the poles?

    When I run it through multisim with the given values I get a pole at 500 rads/s. In my calcs I'm getting two SC poles at 1000 rads/sec and one OC at 333 rads/s. Is the combining effect giving me a corner frequency at 500 or am I just making things up?
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Replace the capacitors with their impedance [Zc=1/(sC)=1/(jwC)]. Then use node voltage or mesh current method to get Vout in terms of Vin, so you will have Vout=(something)*Vin. Then carry Vin to the left side of the equal sign so you will have:
    Vout/Vin=(something)=Transfer Function=H(s)=H(jw)

    Done!
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Are you trying to do an analysis based on a dominant-pole assumption?
     
  6. fireface

    Thread Starter New Member

    Apr 21, 2014
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    0
    ^Yes

    Indeed approximating the poles and cutoff frequency. So the dominant pole is the smallest for high pass and the opposite for low pass?
     
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