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Homework Help: Frequency At Amplitude of Voltage Across Capacitor Decreases

  1. Feb 17, 2010 #1
    Hi, I know the relationship between Vc and the frequency is Vc=I0/(wc)...and if the output amplitude isn't adjusted doesn't this mean that any increase in frequency results in a decrease in voltage? I am asking this because I did a lab and I had to "raise the square wave frequency, but don't adjust the output amplitude of the generator. At some frequency the output wave will begin to look like a sawtooth, and the output amplitude will begin to fall." I had to record the frequency at which this decrease occurs, and I did, but I don't understand why the decrease doesn't start happening at zero. This is an RC circuit by the way.

    Thanks :)
     
  2. jcsd
  3. Feb 17, 2010 #2

    collinsmark

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    Hello mmmboh,

    The reason is that there is a constant in the denominator, of the equation govorning the frequency response. This constant does not vary with [tex] \omega [/tex].

    Suppose you have an RC circuit, with the R and C in series. Call the voltage source Vs. Then the voltage across the capacitor, Vc is (in the s domain), is obtained by a voltage divider,

    [tex] V _C = \frac{\frac{1}{sC}}{R + \frac{1}{sC}}V _S [/tex]

    [tex] = \frac{\frac{1}{sC}}{\frac{sRC}{sC} + \frac{1}{sC}}V _S [/tex]

    [tex] = \frac{\frac{1}{sC}}{\frac{sRC + 1}{sC}} V _S [/tex]

    [tex] = \frac{1}{sRC + 1} V _S [/tex]

    Thus the transfer function,

    [tex] \frac {V _C}{V _S} = \frac{1}{sRC + 1} [/tex]

    To find the frequency response, we substitue [tex] j \omega [/tex] into s.

    [tex] \frac {V _C}{V _S} = \frac{1}{j \omega RC + 1} [/tex]

    So as you can see, if [tex] \omega [/tex] is even somewhat less than 1/RC, the frequency response is still somewhat close to 1. It's only after the angular frequency rises above the neighborhood of 1/RC that the amplitude begins to significantly diminish.
     
  4. Mar 1, 2010 #3
    Sorry this is a late response, but if w is 1/2(RC) then Vc/Vs=1/1.5=0.6...is that really that close to one?
     
  5. Mar 2, 2010 #4

    collinsmark

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    Well, if you want the magnitude, you can't add real and imaginary numbers like that. You need to use the Pythagorean theorem to get the magnitude.

    [tex]
    \left| \frac {V _C}{V _S} \right| = \frac{1}{\sqrt{(\omega RC)^2 + 1}}
    [/tex]

    So if [tex]\omega[/tex] = (1/2)/RC, then |Vc/Vs| = 0.894, which is still relatively close to 1. My point is that you won't see a big drop in the frequency response until [tex]\omega[/tex] rises into the neighborhood of 1/RC (which is the same thing as saying the normal frequency, f [in Hz], rises into the neighborhood of [tex] 1/(2\pi RC) [/tex]). Only at frequencies higher than that does the magnitude start to fall ~proportionally with frequency. Much below that, the magnitude is nearly constant. When [tex]\omega = 1/RC [/tex] the magnitude = [tex] 1/ \sqrt{2} [/tex].
     
    Last edited: Mar 2, 2010
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