- #1

SlideMan

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## 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: [tex]2.5 + 10.01cos(\omega t)[/tex] (10.01V wave shifted up 2.5V)

Output Voltage: [tex]3.46cos(\omega t - 60^\circ)[/tex]

R = 5k

[tex]\omega = 2\pi 100[/tex]

## Homework Equations

[tex] V_O = \frac {V_I}{1 + \frac {1}{j\omega RC}}[/tex]

## 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:

[tex]3.46\angle60 + \frac {3.46}{(628.32)(5k)C}\angle-30 = 10.01\angle0[/tex]

[tex]\frac {3.46}{(628.32)(5k)C}\angle-30 = 8.81\angle-19.9[/tex]

Where'd I go wrong?