# Homework Help: Calculate an average power of a signal

1. Oct 25, 2015

1. The problem statement, all variables and given/known data
Given the following signal,

$v(t)=2\cos (2\pi f_{0}t)+4\cos (4\pi f_{0}t)+6\cos (6\pi f_{0}t)+8\cos (8\pi f_{0}t)$

1. Calculate the signal's average power in the time domain.
2. Calculate the signal's average power in the frequency domain.

2. Relevant equations
Didn't make a real attempt for a solution as yet, but I try to think of a shorter, easier way other than a straight calculation using the exact definition. If there's a hint according to the even form of the given cosines' amplitudes and frequencies that you could advise me of - it might make my way significantly easier here, I feel.

Thanks in advance for any guidelines suggested!

2. Oct 25, 2015

### Staff: Mentor

The average power is zero.

Oh wait, the average power is very high.

Oh wait, what is missing that keeps us from calculating a power given the voltage only...?

3. Oct 25, 2015

### Staff: Mentor

I think we have to presume that the "signal" is not necessarily electrical, but is a signal in the more general mathematical sense. In such a case, for a periodic function f(t) the power is defined to be (If my memory serves... it's been a while...):

$$P = \frac{1}{T} \int_0^T |f(t)|^2 dt$$

I think it gets more complicated when the signal is not periodic and you have to integrate using limits from -∞ to +∞.

There may be a trick for finding the power of a sum of sinusoids involving summing the squares of the amplitudes and dividing by two... don't quote me on this , as I say, it's been a while...

4. Oct 26, 2015

Ok, so
$P_{av}=2^2+4^2+6^2+8^2=4+16+36+64=120\left [W \right ]$

But why you decided to divide this sum by 2 if $T$ is unknown?

5. Oct 26, 2015

### Staff: Mentor

T is the period of the periodic signal. In this case you can see that the lowest frequency is $2 \pi f_o$, and all the other terms have frequencies that are multiples of this. That means the signal as a whole will repeat every $1/f_o$, so that's its period.

As to why the sum of squares is divided by two, that follows from the power integral for a cosine function. Do the integral for a single cosine; you can use the angular period in place of the time period since it covers the same domain for the function:
$$\frac{1}{2 \pi} \int_0^{2 \pi} |cos(\theta)|^2 d \theta$$

6. Oct 26, 2015

### rude man

For the freq. domain, what did you learn about power for a Fourier series?
For the time domain, it's (1/T)∫T v2(t) dt where T comprises an integer number of FUNDAMENTAL cycles.
The assumption in problems like this is R = 1 Ω.
Main thing is to get the same answer both ways!

7. Oct 26, 2015

### Staff: Mentor

Sorry, I don't get it. How can you ignore the load impedance when calculating the power of a voltage waveform supplied to that load?

8. Oct 26, 2015

### rude man

Yes, but in communications theory for example we talk of the power of a signal, assuming 1 ohm:
http://mathworld.wolfram.com/AveragePower.html
Note that v(t) was not specified to be a voltage, just a "signal".

9. Oct 27, 2015

### rude man

For sine waves, rms = peak/√2 so power = rms2 = peak2/2.