Effective power of sinusoidal signals

Discussion in 'Homework Help' started by faen, Jan 20, 2011.

1. faen Thread Starter New Member

Jan 20, 2011
4
0
Here's the question:

The relevant forumla is supposed to look something like this:

However i tried inserting the numbers and it doesnt work. Probably because my formula doesnt have 3 terms but only 2 terms, while the sinusoidal signals in the question has 3 terms.

So basically i need help in how to find the answer to the question in the first pic.. Any help is greatly appreciated

2. Accipiter New Member

Sep 20, 2009
9
0
If I'm not mistaken....

Contributions to power come only from the voltage-current products of same frequencies. So, find the power contributed from each frequency, and then sum the powers. Remember, to find the power contribution from one frequency (or equivalently for DC which is like zero Hz) you take the RMS voltage for that frequency and the RMS current at that frequency, and then multiply them together. Don't forget to multiply by the power factor also (cosine of the angle difference between voltage and current). RMS for a sinusoid is simply the square of the amplitude divided by √2. RMS for DC is simply the DC value. Also, it would be good to work with only sines or cosines, lest you use an incorrect power factor.

See what you get, and then come back if you need more help.

3. faen Thread Starter New Member

Jan 20, 2011
4
0
I tried like this:

2*10 + (5^2)*(3^2)*cos(30) / 2 = 194

I skipped the last term in the voltage and current time function since they had different frequencies.

It's different from the answer of the paper. But it seems like the DC contribution already exceeds the value of the answer. So i'm not sure yet which way is correct or if i did everything as i was supposed to..

Jul 7, 2009
1,585
141
Effective power is the RMS of the instantaneous power. That is an integral which you need to evaluate (note it's NOT the average power). You can check it with a few lines of some python code to make sure you get the right answer.

5. faen Thread Starter New Member

Jan 20, 2011
4
0
I think i figured it out.. My mistake was not converting the cosine to sine.. Thanks for help guys

6. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I believe the effective power is given by

$P=\frac{1}{T}\int^{T}_{0}p(t)dt$

Irrespective of the definition used I would suggest that solving this by integration of the instantaneous power p(t)=u(t)*i(t) would be an unnecessarily lengthy exercise.

Last edited: Jan 20, 2011