# 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