Power contained in a periodic signal (complex exponentials)

[Jd303]

Compute the power contained in the periodic signal x(t) = 10.0[cos(160.7πt)]^4

The problem I have is I end up with a constant value for ak for all values of k
-I start by using inverse Euler formula
-Do the appropriate integration
-Then consider k for odd and even values

My working is attached, if anyone is able to show me the correct procedure, or is able to shed some light for me to understand the topic better it would be much appreciated.

 

[rude man]

Power = energy per integer number of periods/time of that number of periods
= ∫x(t)²dt over n periods T/nT.

An expansion I found that might be useful is
cos⁴(x) = (1/8)[3 + 4cos(2x) + cos(4x)].

 

[Jd303]

Thanks for the expression, I have since found that i had a calculation error, but fixing this up leaves me with a result of 0. I have tried this both using exponentials and trigonometric identities both yielding a final answer of 0.

Can anyone point me in the right direction or plot out some steps?

 

[rude man]

Thanks for the expression, I have since found that i had a calculation error, but fixing this up leaves me with a result of 0. I have tried this both using exponentials and trigonometric identities both yielding a final answer of 0.

Can anyone point me in the right direction or plot out some steps?

Perform the integration!

 

[rezeew]

I understand your confusion and would be happy to provide some clarification on this topic.

Firstly, it is important to understand that the power contained in a periodic signal is the average amount of energy per unit time that the signal carries. In mathematical terms, it is defined as the square of the signal's amplitude divided by its period.

In the given signal x(t) = 10.0[cos(160.7πt)]^4, we can see that it is a periodic signal with a period of 1/160.7 seconds. Therefore, the power contained in this signal can be computed as follows:

Power = (Amplitude)^2 / Period
= [10.0]^2 / [1/160.7]
= 1607 watts

Now, let's take a look at your working. You have correctly used the inverse Euler formula and performed the integration. However, your confusion seems to be arising from the fact that you are getting a constant value for ak for all values of k.

This is because when you consider k for odd and even values, you are essentially separating the signal into its even and odd components. In this case, both the even and odd components will have the same amplitude and frequency, resulting in the same value for ak.

To better understand this, let's take a look at the Fourier series representation of the given signal:

x(t) = a0 + ∑[akcos(kωt) + bksin(kωt)]

Here, a0 is the DC component and ak and bk are the coefficients for the cosine and sine terms, respectively. In your working, you have correctly computed a0 as 10.0 and ak as 0 for all values of k.

This is because when you expand the given signal, it can be written as:

x(t) = 10.0[1 + cos(2*80.35πt) + cos(4*80.35πt) + cos(6*80.35πt)]

As you can see, all the terms have the same amplitude of 10.0 and frequency of 80.35π. Hence, the coefficients ak for all values of k will be the same, resulting in a constant value.

In conclusion, the power contained in a periodic signal can be computed by taking the square of its amplitude and dividing it by its period. However, when considering the Fourier series representation, the