Homework Help: Frequency At Amplitude of Voltage Across Capacitor Decreases

1. Feb 17, 2010

mmmboh

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. Feb 17, 2010

collinsmark

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 $$\omega$$.

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,

$$V _C = \frac{\frac{1}{sC}}{R + \frac{1}{sC}}V _S$$

$$= \frac{\frac{1}{sC}}{\frac{sRC}{sC} + \frac{1}{sC}}V _S$$

$$= \frac{\frac{1}{sC}}{\frac{sRC + 1}{sC}} V _S$$

$$= \frac{1}{sRC + 1} V _S$$

Thus the transfer function,

$$\frac {V _C}{V _S} = \frac{1}{sRC + 1}$$

To find the frequency response, we substitue $$j \omega$$ into s.

$$\frac {V _C}{V _S} = \frac{1}{j \omega RC + 1}$$

So as you can see, if $$\omega$$ 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.

3. Mar 1, 2010

mmmboh

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?

4. Mar 2, 2010

collinsmark

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.

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

So if $$\omega$$ = (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 $$\omega$$ 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 $$1/(2\pi RC)$$). Only at frequencies higher than that does the magnitude start to fall ~proportionally with frequency. Much below that, the magnitude is nearly constant. When $$\omega = 1/RC$$ the magnitude = $$1/ \sqrt{2}$$.

Last edited: Mar 2, 2010