Electrical engineering and Fourier series

1. Nov 30, 2008

Niles

1. The problem statement, all variables and given/known data
Hi all.

I have used Fourier series alot in my maths study, and now I am about to use it when analyzing electrical circuits. In my notes from a lecture, I wrote down an example done by the professor. We are looking at a RCL-circuit with an applied EMF given by the periodic square function, which is ε0 from 0 to T/2 and -ε0 from T/2 to T. And from this it is just periodic. We wish to find the current.

Now we have found the complex Fourier series for the electromotive force, and we are able to write it as:

ε(t) = -Im[ε'(t)],

where Im[] is the imaginary part and ε' is some complex Fourier series. He then says that "Hence we can write I(t) = -Im[I'(t)]", and this is where I get confused. Why is it that we know that I(t) = -Im[I'(t)], just because ε(t) = -Im[ε'(t)]?

I hope you guys can help me. I've thought about this most of the weekend, and the notes are very poor. Unfortunately Google doesn't help me either.

Sincerely,
Niles.

Last edited: Nov 30, 2008
2. Nov 30, 2008

tiny-tim

Hi Niles!

if ε is voltage, and I is current, for a fixed circuit, then aren't they scalar multiples of each other, so that any linear or conjugate equation for one will work for the other also?

3. Nov 30, 2008

Niles

Ahh, yeah. I guess you are right.. I never thought of it that way.

Thanks!

4. Dec 1, 2008

turin

I am actually a bit confused what is the meaning of this prime. Are you trying to find the input current? So, the basic concept is the LTI system, which means that you can treat each frequency component in the Fourier series independently and then add them up in the end. Or, to put it another way, you can ignore the summation until you are done, and treat the series index as some arbitrary fixed integer.

Are you calculating the input current to the ciruit? In this case, you would calculate the input impedance of the circuit (or you may call it the equivalent impedance). One important point is that the impedance is complex valued, and the complex phase encodes the decay of the reactive elements (capacitors and inductors that store energy and then release it later). So, the time phase that you would see in a sine or cosine is traded for a complex phase in the phasors (the coefficients in the Fourier series). So, when you divide voltage by impedance, the extra complex phase encodes the time lag of the current w.r.t. the voltage.

5. Dec 2, 2008

Niles

The prime just means that it is the complex valued function, and taking the real part of it gives me the desired current (i.e. the current in the RCL-circuit).

By the way, when we look at a RCL-circuit with a DC-current, is it OK to say the the impedance of the capacitor - and hence the impedance of the whole circuit - is infinite?

6. Dec 2, 2008

turin

If the cap is in series with the rest of the circuit, then yes. In fact, this is a simple trick employed in, for example, RF systems, to set a DC bias independently of the RF signal. If you put a cap between two points in a circuit, then you can set the DC points independently on each side of the cap.