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Simple filter circuit: Phase shift

  1. Jan 19, 2007 #1
    Hello all,

    I'm currently taking an upper division circuits class as a MechE student after taking a 2 year break from the topic.

    We were given a problem to find the transfer fuctinon T(s) of a high and low pass filter, and then plot Vout(t)'s phase and magnitude vs. the freq.

    My question is: How to a represent a 2V p-p AC Voltage source in the frequency domain (s)? I think i understand a sin(wt) type transform, but dont i need to transform sin(wt + phi) to get the phase w.r.t. the input source?

    Any hints, clues, or advice is needed.
  2. jcsd
  3. Jan 19, 2007 #2


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    Staff: Mentor

    When plotting a transfer function, you plot the ratio of |Vo|/|Vi| for magnitude, and the phase difference for the phase. So when Vo=Vi, you have 0dB of gain. If you have a low-pass filter, at low frequencies you'll typically have 0dB of gain (or whatever the DC gain of your circuit it), and then it will roll off to lower gains (-10dB, -20dB, etc.) according to the transfer function polynomial.

    So your question about the 2Vpp signal source and how to plot it in the frequency domain doesn't apply directly to the transfer function. Even if your exciting source is a swept 2Vpp signal generator, you still need to do the division Vo/Vi and take the magnitude and phase to plot the T(s).
  4. Jan 19, 2007 #3
    Ok, so now I'm starting to see the part the transfer function plays.

    I have now derived the T(s) for the circuit. Which is probablly pretty classic as far as circuits courses go.

    Vo/Vi = T(s) = 1/(1 + jwCR).

    In my class notes I have the prof setting 1/CR = w0 so that,

    Vo/Vi = T(s) = 1/(1 + j(w/w0))

    I'm not sure what point that serves...

    From this how do I create dB/Phase vs. freq plots? Do I just find the mag and angle of T(s) using complex algebra?

  5. Jan 19, 2007 #4


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    Staff: Mentor

    Yes, split the T(s) up into its magnitude and phase components (you know the trick of multiplying by the complex conjugate of the denominator?), and then plot them on your frequency graph (log frequency on the horizontal, and linear dB and phase on the vertical).

    The reason that your prof said to set w0 = 1/RC is that for a single-pole lowpass filter like this T(s), you will get the flat 0dB gain line from low frequencies that tips over and starts down at -20dB/decade. The frequency it tips over and starts down is called w0, and is defined as the half-power point for the T(s). Since T(s) is a voltage transfer function as you have written it, you get half power when the voltage is down by a factor of [tex]\frac{1}{\sqrt{2}}[/tex]
  6. Jan 19, 2007 #5


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    Staff: Mentor

    BTW, it's a good exercise to prove that w0 is 1/RC for this transfer function. Try doing it after you have finished the regular problem.
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