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RLC parallel circuit; fourier components

  • Thread starter mathman44
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Homework Statement



In my lab we did an experiment using an RLC parallel circuit, with the L and C in parallel. The voltage source was set at square waves. I am being asked to explain why there are frequencies, beyond the resonance, at which there is a peak voltage drop across the capacitor.

For example, in my circuit the resonance is at 9.382khz, and then there are subsequent peak voltage drops across C at:

1.878 khz, 1.342 khz, 1.043 khz... each resonant peak with a lesser voltage drop than the previous.

I have a feeling this has to do with the fourier components of the square wave, but I can't wrap my head around this. Any hints? Thanks in advance.
 

Answers and Replies

  • #2
vela
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You have the right idea. What is the Fourier series for a square wave?
 
  • #3
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A very large number of sine waves. :s
 
  • #4
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[tex]
V(t) = \frac{4V_0}{\pi}\left(\sin(wt) + \frac{1}{3}\sin(3wt) + \frac{1}{5}\sin(5wt) + \dots \right)
[/tex]

Now what :confused:
 
  • #5
vela
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Actually, I may have misinterpreted your original post. What exactly did you do in this experiment?
 
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Basically, why are there frequencies (of square waves) other than the fundamental frequency that give voltage peaks across a capacitor, in parallel with an inductor, in an RLC circuit?
 
  • #7
vela
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I know, but your statement is ambiguous. Are you varying the frequency of the input signal and seeing the response peak at different frequencies, or do you have a fixed-frequency input signal and see a bunch of peaks in the spectrum?
 
  • #8
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The first.
 
  • #9
vela
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Heh, I was betting on the latter based on what you wrote. Good thing I asked.

Try dividing those frequencies into the resonance frequency of your circuit. Notice anything about the results and how they might relate to the Fourier decomposition of the square wave mmmboh gave above?
 
  • #10
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Great, thanks!
 
  • #11
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The Fourier transform of a square wave includes many sine waves. Think about this:

Imagine the resonant frequency of a glass is 150 Hz (meaning 150 Hz sinusoidal, remember), and one needs 10 dB of 150 Hz to shatter it. You have a square wave generator of 150 Hz, at whatever dB you want. What is the minimum square wave dB needed to shatter the glass?

Well, the 150 Hz square wave's Fourier transform has a certain amount of 150 Hz sinusoidal wave in it, but also quite a few other frequencies. Here is just some of the frequencies it contains, in its Fourier series:

http://mathworld.wolfram.com/FourierSeriesSquareWave.html

So, you would need the square wave to have quite a bit more than 10 dB in and of itself, because much of its energy is in those other frequencies, see?

The Fourier transform is beautiful once understood, though I don't think anyone can intuit just how much of any sinusoidal frequency, if any is in any specific non-sinusoidal wave, like a square wave, which is why it is so hard to grasp.
 

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