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Power of error and Parseval's theorem

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Emil09

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I'm approximating a square wave using a Fourier series. I need to determine the power of error by using Parseval's theorem. The problem is that I don't understand the connection between power of error and Parseval's theorem. Can anyone explain this?

BTW, here is a graph of what I'm doing (in Maple).

9171-2s864ac.png
 

Hayato

Member
You know that ∫|y(t)|² = ∫|Y(f)|².

So, if you calculate the square wave power on time domain you are going to have the actual power from the signal.

As the Fourier transform of the signal gives you infinite coefficients (an and bn) you won't be able to use them all, so you need to limit the coefficients, let's say, to the 10th harmonic. 10 << ∞, so you are going to have an error.

So, the |error power| is the |actual power - calculated coefficients power|.
 

Emil09

New Member
You know that ∫|y(t)|² = ∫|Y(f)|².

So, if you calculate the square wave power on time domain you are going to have the actual power from the signal.

As the Fourier transform of the signal gives you infinite coefficients (an and bn) you won't be able to use them all, so you need to limit the coefficients, let's say, to the 10th harmonic. 10 << ∞, so you are going to have an error.

So, the |error power| is the |actual power - calculated coefficients power|.

Thanks for the reply. Using Parseval's THM I made the following calculation in Maple:
9198-2e1vqe0.png

This should represent the average power but it seems too small...is this actually the power in the error?
Here is the original function which I made the Fourier series expansion of:
9199-10421k0.png
 

Hayato

Member
Ok.

Let's take a square wave, of 0.5Vp with 0.5 offset. (It will look like your rect pulse). With f = 0.5 Hz. Duty = 50%.

The normalized average power is 1/2 * 1^2 = 1/2 W.

Ok.

If we make the fourier transform of a periodic signal. We'll find that |Cn|=|F(f)|*δ(f-nf0 ) ∀ -∞<n<∞. where f0 is the first harmonic, and the F(f) is the fourier integral of the aperiodic signal.

We know that the F{0.5*rect(t)} = 0.5*sinc(f)

if f0 = 0.5 Hz, then we are going to have harmonics at 0.5 Hz, 1.5 Hz, 2.5 Hz, 3.5 Hz, 4.5 Hz... and so on. And a DC component.

So to get the Fourier coefficient at a given frequency we do this calc:
F(nf0) = 0.5*sinc(nf0)

And we are going to have these: (2* because we have negatives and positives n)
0 Hz = 0.5V
2*0.5 Hz = 0.6366 V
2*1.5 Hz = 0.2122 V
2*2.5 Hz = 0.1273 V
2*3.5 Hz = 0.0909 V
2*4.5 Hz = 0.0707 V

Since each coefficient is a sine wave amplitude (except for the DC), we can calculate the avg. power using the P = A²/2 relation:
DCpower = 0.5² = 0.25W
1st = 0.6366²/2 = 0.2026W
3rd = 0.2122²/2 = 0.0225W
5th = 0.1273²/2 = 0.0081W
7th = 0.0909²/2 = 0.0041W
9th = 0.0707²/2 = 0.0024W

Power = ∑ of those = 0.4897 W

Could you see the Parseval there? First I calculated the squarewave power, so that would be ∫f²(t) dt.
Then I calculated part of the discreete integral of the frequency domain function: 1/2 * ∑ F(nf0)² [-∞<n<∞]

But we know that ∫g(t)² = ∫G(f)², and we didn't find that, because:
0.5W ≠ 0.4897W, so we have an error there.
 

Emil09

New Member
So here is what I've got, complete with the original question and my answers. Do you think it's logical?

Problem:
9212-2mnktqr.png


Answer:
9213-35kmt6u.png

9214-2ur5w0w.png

9215-1zojdon.png
 

Hayato

Member
You are missing C0 in 1.6.
And you are comparing fourier series (geometric) to fourier series (exponencial), how can you find an error with that?

You have to compare the "real" waveform to the fourier approximation.
 
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