Time Response - 31.5 - Ex 31.7

[ELECTRICGREMLIN]

Hi,

I have a question about Example 31.7 of the the "Electrical Engineering Reference Manual for the Power, Electrical and Electronics, and Computer PE exams" 8th edition.

Basically you're given a voltage waveform that is v(t)=636cos(2513t+/_60degrees) in the time domain is equal to 450/_60 degrees. How did they do that?

Thanks

Electric Gremlin

 

[Flyer_PE]

The amplitude in the frequency domain is usually an rms voltage.

636/sqrt(2) = 450.

The angle is just the angle portion of the argument.

The frequency for that particular waveform appears to be 400 Hz (2*pi*f=2513 gives f=400 Hz). The frequency doesn't show up in the polar notation.