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Homework Help: Transfer functions of active filters with Amplification

  1. Nov 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Derive the transfer function for both circuits [tex]\frac{V_{out}}{V_{in}}[/tex] sketch Bode plots for each circuit (amplitude and phase)

    2. Relevant equations

    3. The attempt at a solution
    We can treat this as a potential divider using the impedances of the resister and caps.
    using the Equation for the first circuit (Low pass active)[tex]\frac{V_{out}}{V_{in}}=\frac{Z_c}{Z_R+Z_C}[/tex][tex]Z_R=R[/tex]
    sustituting for ZC and ZR


    Multiply though by [tex]j{\omega}C[/tex] gives[tex]\frac{V_{out}}{V_{in}}=\frac{1}{j{\omega}RC+1}[/tex]
    substituting for RC with wC

    I am not sure how to implement the gain function, I think it is just a case of multiplying by
    as this is negative feed back [tex]\frac{V_{out}}{V_{in}}=\frac{-G}{j\frac{\omega}{\omega_C}+1}[/tex]
    Am I on the right lines.
    As for the BODE plot when the filter is at cutoff frequance the phase shift will be -45 deg at -3dB with a roll of of 20dB per dec?

    All help will be apreciated

    Attached Files:

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    • Low.JPG
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  2. jcsd
  3. Nov 12, 2016 #2


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    Staff: Mentor

    Your method looks okay for the first circuit. You should specify that the R in your ωC is actually R2.

    Take care with your Bode and phase plots. The 3 dB point will be 3 dB down from the maximum gain (low frequency gain). The phase shift will be 45 degrees from the low frequency phase (and it's a negative gain circuit so near DC the phase shift is....?).
  4. Nov 12, 2016 #3


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    Gold Member


    ZR2,C1 = ZR2 || ZC1 = R2*(1/sC1) / (R2+1/sC1) = R2 / (sC1R2+1)

    ZR1 = R1

    Vout/Vin = -ZR2,C1 / ZR1 = -R2 / ( sC1R2R1 + R1 )

    If you are not familiar with Laplace, just substitute s = jω as for sinusoidal signal ( Bode ).
    Last edited: Nov 12, 2016
  5. Nov 13, 2016 #4
    At DC the phase shift is 0 deg 0dB? and wc -45 deg -180 deg at about -40dB?
  6. Nov 13, 2016 #5


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    Staff: Mentor

    At DC the circuit looks like a standard inverting amplifier. It has a gain set by the resistors R1 and R2:

    ##\frac{V_{out}}{V_{in}} = - \frac{R_2}{R_1}##

    Note carefully the negative sign!
  7. Nov 13, 2016 #6
    Circuit 2 also a low frequency pass (integrator)
    the gain is[tex]\frac{V_{out}}{V_{in}}=\frac{-Z}{R_1}[/tex]
    [tex]Z=\frac{\frac{1}{j{\omega}C}{R_2}} {\frac{1}{j{\omega}C}+{R_2}}[/tex]
    multiplying throgh by jwC.
    the gain is[tex]\frac{V_{out}}{V_{in}}=\frac{-Z}{R_1}=\frac{1}{R_1}*-R_2*\frac{1}{1+j{\omega}CR_2}=\frac{-R_2}{R_1}\frac{1}{1+j{\omega}CR_2}[/tex]
    and as [tex]G=\frac{R_2}{R_1}[/tex]
    then [tex]\frac{-G}{1+j{\omega}CR_2}[/tex]
    This gives the same first order TF as the circuit before I believe it wil has the same BODE plot too?

    As there are no values give how can I sketch a BODE plot for the Gain
  8. Nov 13, 2016 #7


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    Staff: Mentor

    For the second circuit, I'd find the expression for the DC gain first. The feedback path is divided (R2 and R3), so the gain won't be so simple as a ratio of R2/R1.

    A sketch of a bode plot doesn't need to have accurate values. Just label significant features (like the corner frequency) with the appropriate expression from your derivations. That said you can always pick some typical part values and use the result to find the general shape as a basis for your sketch.
  9. Nov 13, 2016 #8
    There is no R3 in the second circuit?
    I may be getting mixed up somewhere.

    Is the TF for the first circuit ok?
    Where have I gone wrong with the second?

    thanks for all your help
  10. Nov 13, 2016 #9


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    Staff: Mentor

    Your second image:

    Definitely an R3 in there!

    Your first circuit TF work looks okay.
  11. Nov 13, 2016 #10
    I see the confusion.
    fig.1(a) and fig.1(b) the above fig is what i am referring to as the first circuit (fig.1(a)) and the second (fig.1(b))
    I should have used the correct idents from the beginning sorry.

    So in the DC gain for fig.1(a) should be [tex]G=\frac{R_2}{R_3}[/tex]
    This is wrong mabe a typo in post " #1"

    and the DC gain for fig.1(b) is[tex]G=\frac{R_2}{R_1}[/tex]
  12. Nov 13, 2016 #11


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    Staff: Mentor

    Ah. Well that's a horse of a different color then :smile:
    You want to check that. Have you analyzed the circuit? Note that since no current can flow into the op-amp inputs, the junction of the R2~R3 divider must be equal to Vin (essentially no input current so no potential drop across R1).
    The gain should be negative for fig.1(b).
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