Calculate an average power of a signal

In summary: Same idea here.In summary, the signal described is a periodic function with a period of 1/f₀. The average power in the time domain can be calculated using the integral formula (1/T)∫T v²(t) dt, while in the frequency domain it can be calculated by summing the squares of the amplitudes of the individual sinusoidal components and dividing by two.
  • #1
Adir_Sh
22
0

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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  • #2
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:
 
  • #3
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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  • #4
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?
 
  • #5
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:
$$\frac{1}{2 \pi} \int_0^{2 \pi} |cos(\theta)|^2 d \theta$$
 
  • #6
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:
 
  • #7
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
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".
 
  • #9
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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What is the definition of average power of a signal?

The average power of a signal is the average value of the signal's instantaneous power over a period of time. It is a measure of the signal's strength and is typically expressed in watts (W) or decibels (dB).

How do you calculate the average power of a signal?

The average power of a signal can be calculated by taking the integral of the squared signal over a specific time period, and then dividing the result by the length of the time period. This can also be expressed as the mean of the squared values of the signal.

What is the difference between average power and RMS power?

Average power and RMS (root mean square) power are both measures of a signal's strength, but they use different methods of calculation. While average power takes the mean of the squared values of the signal, RMS power takes the square root of the mean of the squared values. In other words, RMS power gives more weight to the higher amplitude values of the signal.

Why is it important to calculate the average power of a signal?

Calculating the average power of a signal is important for understanding the overall strength and behavior of the signal. It can also be useful for comparing different signals or for determining the power requirements of a system that uses the signal.

What factors can affect the accuracy of the calculated average power of a signal?

The accuracy of the calculated average power of a signal can be affected by the length of the time period over which the power is measured, the sampling rate used, and any noise or interference present in the signal. It is important to carefully consider these factors when calculating and interpreting the average power of a signal.

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