# Varying EMF and Fourier series

Discussion in 'Homework Help' started by Niles, Nov 30, 2008.

1. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
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?

Sincerely,
Niles.

2. ### blazedaces Active Member

Jul 24, 2008
130
0
This is complex?

-blazed

3. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
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.

4. ### blazedaces Active Member

Jul 24, 2008
130
0
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

5. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
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?

6. ### blazedaces Active Member

Jul 24, 2008
130
0
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...

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
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.

8. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
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.

9. ### blazedaces Active Member

Jul 24, 2008
130
0
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...

10. ### steveb Senior Member

Jul 3, 2008
2,433
469
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.

11. ### blazedaces Active Member

Jul 24, 2008
130
0
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

12. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
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?