How Do You Calculate the Magnitude of a Power Spectral Density Function?

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Homework Statement


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For now I am trying to find the power spectral density only.

The Attempt at a Solution


  • From the given graph, I got g(t) = -1 + 2rect(t/2), with a perdiod of 4s
  • PSD = Sg(f) = lim(T [itex]\rightarrow\infty[/itex] ) [itex]\frac{1}{2T}[/itex] || G(f) || [itex]^{2}[/itex]
  • My Fourier transform of g(t), G(f), I got to = - [itex]\delta[/itex] (f) + 4sinc(2f)
where " [itex]\delta[/itex] (f)" is the unit impulse function.
I want to continue but I am not sure this is correct? If it is correct, I need to get the magnitude and square it. I don't know why I am not too comfortable with magnitude. So in this case i have || - [itex]\delta[/itex] (f) + 4sinc(2f) || which I know ≠ ||- [itex]\delta[/itex] (f) || + || 4sinc(2f) ||. So what do I do? How do I find magnitude?

EDIT: Can this post be moved to the electrical engineering section? I feel it is more likely to get a reply there.
 
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on Phys.org
You know from Fourier series that the psd must be a sequence of delta functions situated at odd harmonics of the fundamental frequency (0.25 Hz). Compute the height of the delta functions as you were taught.

You can see by inspection what the dc component is and also the average power.

For the autocorrelation function I will give you a hint: it's a triangular function in tau space. You can obtain this function graphically if you shift the square wave against itself and compute the product for each shift position.
 
rude man said:
You know from Fourier series that the psd must be a sequence of delta functions situated at odd harmonics of the fundamental frequency (0.25 Hz). Compute the height of the delta functions as you were taught.

You can see by inspection what the dc component is and also the average power.

For the autocorrelation function I will give you a hint: it's a triangular function in tau space. You can obtain this function graphically if you shift the square wave against itself and compute the product for each shift position.

Thank you for your help. I was able to do it. For anyone that wants to know how, here is what you have to use:

PSD = ∑ |Cn|[itex]^{2}[/itex] δ(f-nf[itex]_{0}[/itex])
Cn =(1/T[itex]_{0}[/itex]) G(f)|[itex]^{for f = nf_{0}}[/itex]
Extract a single period from g(t) to get a new g(t) = -rect(t/4) + 2rect(t/2)

Then solve for PSD .
 
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