Varying EMF and Fourier series

Thread Starter

Niles

Joined Nov 23, 2008
56
Hi all.

I have found the (complex) EMF of an RCL-circuit to be:

\[
\widetilde\varepsilon (t) = \frac{1}{6} - \sum\limits_{n = 1}^\infty {\frac{{\cos (\omega nt)}}{{n^2 \pi ^2 }}},
\]

where the "~" over the ε indicates that it is complex. I have to find the current of the circuit. How can I do that?

Thanks in advance.

Sincerely,
Niles.
 

Thread Starter

Niles

Joined Nov 23, 2008
56
No, it is not complex - my mistake. The correct EMF (real EMF!) is:

\[
\varepsilon (t) = \rm {Re} [\widetilde\varepsilon (t)] = \frac{1}{6} - \sum\limits_{n = 1}^\infty {\frac{{\cos (\omega nt)}}{{n^2 \pi ^2 }}} = \rm {Re} \left[ {\frac{1}{6} - \sum\limits_{n = 1}^\infty {\frac{{\exp (i\omega nt)}}{{n^2 \pi ^2 }}} } \right]
\]

- and thanks for responding so quickly.
 

blazedaces

Joined Jul 24, 2008
130
O...k.

That doesn't look correct to me, but whatever. Do you have any other information about the circuit? Like if you had the R, L, and C values you could use that to find the thevenin equivalent impedance and then use Ohm's Law (V = I * R) to find the current I, where R is not your resistance, but your equivalent impedance Z.

-blazed
 

Thread Starter

Niles

Joined Nov 23, 2008
56
I got the expression for the EMF from the text in the assignment.

I thought of using your strategy as well, i.e. use that:

\[
\widetilde I = \frac{\widetilde \varepsilon}{Z(\omega_n)} = \frac{\widetilde \varepsilon}{R-in\omega L - \frac{i}{n\omega C}}.
\]

Then I find the real current as the real part of the above expression: But here is my problem: When we look at ε, I have the fraction outside the sum. What value must n in the impedance have there?
 

blazedaces

Joined Jul 24, 2008
130
But here is my problem: When we look at ε, I have the fraction outside the sum. What value must n in the impedance have there?
The EMF is a periodic signal being represented by a fourier series. The fourier series can represent any periodic signal by the sum of a constant plus an infinite number of sines and/or cosines.

I'm not quite sure what you're asking. If you're asking what the value of n in the 1/6 fraction should be... there is no n in the 1/6 constant...? If you're asking what value of n is in the sum from n = 1 to infinity... well that's your answer there too...

Your current will end up also being periodic. Leave the sum and the n's in your final answer. I think that's correct anyway...

-blazed

Edit: Wait a second... why do you have n in your impedance values? That doesn't look right...
 

steveb

Joined Jul 3, 2008
2,431
Hi all.

I have found the (complex) EMF of an RCL-circuit to be:

\[
\widetilde\varepsilon (t) = \frac{1}{6} - \sum\limits_{n = 1}^\infty {\frac{{\cos (\omega nt)}}{{n^2 \pi ^2 }}},
\]

where the "~" over the ε indicates that it is complex. I have to find the current of the circuit. How can I do that?

Thanks in advance.

Sincerely,
Niles.
This is a case where the superposition principle could be used to simplify the analysis. You have an input signal which is a sum of separate components (DC and a fundamental frequency, then all upper harmonics). The output response (in this case current) will be the sum of the responses to each individual component.
 

Thread Starter

Niles

Joined Nov 23, 2008
56
Yes, and that is also why I have divided the EMF with the impendace. But I have to divide by the impedance for each n - i.e. for each harmonic.

But the term 1/6 - what harmonic is that? I would say that it is fundemental, so I would set n=0, but then I will be dividing by zero in the impedance of the capacitor, which is no good.
 

blazedaces

Joined Jul 24, 2008
130
Yes, and that is also why I have divided the EMF with the impendace. But I have to divide by the impedance for each n - i.e. for each harmonic.

But the term 1/6 - what harmonic is that? I would say that it is fundemental, so I would set n=0, but then I will be dividing by zero in the impedance of the capacitor, which is no good.
1/6 is a constant, it's not a harmonic. If you want to think about it this way: it's like 1/6 * cos(0).

It doesn't matter. The impedance should definitely not have any n's in it. I don't know what made you think that. Could you provide a source that claims this? I apologize if I'm mistaken but it just makes absolutely no sense...

-blazed

Edit: The impedance is not a function of \[\omega_n\]... Someone correct me if I'm wrong...
 

steveb

Joined Jul 3, 2008
2,431
Yes, and that is also why I have divided the EMF with the impendace. But I have to divide by the impedance for each n - i.e. for each harmonic.

But the term 1/6 - what harmonic is that? I would say that it is fundemental, so I would set n=0, but then I will be dividing by zero in the impedance of the capacitor, which is no good.
The 1/6 is the DC component, so the impedance is just the pure resistive component R. It is sometimes possible to get infinite DC impedance, or zero too. The impedance has n in it because it is frequency dependent and the only frequency components are multiples of n.
 

blazedaces

Joined Jul 24, 2008
130
The 1/6 is the DC component, so the impedance is just the pure resistive component R. It is sometimes possible to get infinite DC impedance, or zero too. The impedance has n in it because it is frequency dependent and the only frequency components are multiples of n.
Are you trying to argue that if you had a DC voltage input on an RLC circuit that the impedance felt would not be based on your capacitor or inductor?

This entire discussion of harmonics is only a way to help us understand circuits by comparing them to springs... isn't it?

-blazed
 

Thread Starter

Niles

Joined Nov 23, 2008
56
Ok, so now we all agree on that the impedance must contain the n.

Now the question is, if the impedance is zero for the 1/6 or infinite. I personally believe it is infinite, because we have a capacitor, i.e. it has infinite impendance, because it is DC.

But you believe the impedance is just R for DC?
 
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