Simple High Pass Filter Problem

[SlideMan]

Homework Statement

Given a simple RC high pass filter, I need to find the value of C that will give the output as shown on a scope capture.

Input Voltage: $$2.5 + 10.01cos(\omega t)$$ (10.01V wave shifted up 2.5V)
Output Voltage: $$3.46cos(\omega t - 60^\circ)$$
R = 5k
$$\omega = 2\pi 100$$

Homework Equations

$$V_O = \frac {V_I}{1 + \frac {1}{j\omega RC}}$$

The Attempt at a Solution

I'm not sure how to take into account the DC component of the input voltage. I know it will be filtered out in the end, but I would think that it would be needed for the computations. Ignoring the DC component and using phasor notation, I come up with a phase angle of -19.9 equal to a phase angle of -30:

$$3.46\angle60 + \frac {3.46}{(628.32)(5k)C}\angle-30 = 10.01\angle0$$

$$\frac {3.46}{(628.32)(5k)C}\angle-30 = 8.81\angle-19.9$$

Where'd I go wrong?

 

[nilesh_pat]

I'm assuming it's the initial step of ignoring the DC component, but I don't know what to do with it.

 

[piercas]

In order to solve this problem, you need to use the full equation for the output voltage of a high pass filter, which includes both the input voltage and the DC component. The equation should be V_O = V_I - \frac {V_I}{1 + \frac {1}{j\omega RC}}. This takes into account the DC component of the input voltage and will give you the correct phase angle of -19.9 degrees. From there, you can solve for C using the given values for R, \omega, and the desired output voltage.