Low pass filter

Discussion in 'Homework Help' started by skusku, Sep 5, 2012.

  1. skusku

    Thread Starter Active Member

    Aug 9, 2009
    63
    1
    Hi again guys :)

    Circuit: http://i46.tinypic.com/5owp3l.jpg

    Calculate the cut off freq when a)R3 is not connected
    and b) When R3 is connected
    c)calculate the attenuation in dB when the signal frequency is 4Fc.

    -------------------------------
    a)Fcut-off=1/(2∏(R1+Rs)C1) = 1015.15Hz
    b)Fcut-off=1/(2∏((R1+Rs)//R3)C1)=1137.57Hz

    c)[I choose Vin=1V]as nowhere it states an input voltage. Also i assume calculations for circuit with R3 and no R3 needs to be done:
    1-[NO R3 connected]
    Xc=1/(2∏*4Fc*C1)=3015Ω
    Vout=Vin*Xc/(√((R1+Rs)^2 + Xc^2))=242.53mV
    Att(dB)=20log(Vout/Vin)=-12.3dB

    I then do the same with R3 connected to get the Attenuation. Which gets me -11.38dB.

    Formulas in a) should be correct but I included it for c). Is my cut off calculation correct with the R3 included? And do I understand c) correct?

    Thanks
     
  2. mlog

    Member

    Feb 11, 2012
    276
    36
    I found the same answers as you except for the voltage gain (attenuation) with R3 included at 4 times fc. I got -13.3 db @ 4550 Hz.
     
  3. skusku

    Thread Starter Active Member

    Aug 9, 2009
    63
    1
    Thanks mlog
    So my other calculations calculations are correct?

    I do realise I used the freq value with no R3 included as a mistake.
    However I do not get the -13.3dB .
    With R3 included:

    Xc=1/(2*pi*4*1137.57) = 2690.51 Ohm

    Rparallel=12060//100 000 = 10762 Ohm

    Vout=Vin*Xc/(√((Rparallel)^2 + Xc^2)) = 0.24253 V
    Atenuation = 20log(Vout/Vin) = 20log(0.24253) = -12.3dB which means the same answer as for when R3 is not included. Am I missing something?
     
  4. mlog

    Member

    Feb 11, 2012
    276
    36
    For R3 included, you cannot use the same equation for finding the voltage gain (attenuation). The generalized equation for the voltage gain is:

    A = Z/(R+Z).

    where R=Rs+R1.

    With no R3, the value of Z includes only C1. However, with R3, the value of Z includes both R3 and C1.

    In the 1st case, Z = 1/(jωC). In the 2nd case, Z = R/(1+jωRC). In both cases, R=R3 and C=C1, and j is the imaginary operator.
     
  5. skusku

    Thread Starter Active Member

    Aug 9, 2009
    63
    1
    Mlog im nog getting those values. The j is screwing my calculations around.
     
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