Calculate an average power of a signal

AI Thread Summary
The discussion revolves around calculating the average power of a given signal composed of multiple cosine terms. Participants emphasize the need to consider the signal's periodic nature, with the average power defined as the integral of the square of the signal over one period. The total power is calculated by summing the squares of the individual amplitudes, resulting in 120 W, while the division by two relates to the root mean square (RMS) value of cosine functions. There is also a debate about the relevance of load impedance in power calculations, with some suggesting a standard assumption of 1 ohm. The conversation highlights the importance of consistency in results obtained from both time and frequency domain analyses.
Adir_Sh
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Homework Statement


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.

Homework 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!

Adir.
 
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Adir_Sh said:

Homework Statement


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.

Homework 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!

Adir.

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...? :smile:
 
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 :nb), as I say, it's been a while...
 
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gneill said:
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 :nb), as I say, it's been a while...

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?
 
Adir_Sh said:
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?
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:
$$\frac{1}{2 \pi} \int_0^{2 \pi} |cos(\theta)|^2 d \theta$$
 
Adir_Sh said:

Homework Statement


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.

Homework 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!

Adir.
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! :smile:
 
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?
 
berkeman said:
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?
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".
 
Adir_Sh said:
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?
For sine waves, rms = peak/√2 so power = rms2 = peak2/2.
 
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