# How to calculate C for an AM demodulator?

I have read this equation from Savant's book that is used to calculate the capacitor for a demodulator:

I've tried to calculate my circuit with this equation, but it didn't work. Then I tried to recalculate this example from the book, but I don't know what is 'w' here in the example:

A 15MHz radio frequency carrier is modulated with a 5kHz signal with a modulation index (m) of 0.5. If the load resistance is 5 kOhms, what capacitor value must be added in parallel to the load to filter the radio frequency signal?

response: 0.013 uF

Can you please teach me how to obtain w so I can calculate C? I tried with 15MHz, and 5kHz, but I don't get the same answer.

$$\\omega\$$ is the angular frequency.

$$\ \omega = 2\pi f \$$

This makes your equation: $$\ C = \frac {1}{2\pi f R_Lm}\$$

$$\ C = \frac {1}{2\pi (5000) (5000) (0.5)} = 0.01273239µF \approx 0.013µF\$$
$$\f\$$ is the frequency of the modulation signal (5kHz,) not the carrier (15MHz.)