# question about RLC parallel in lab

Discussion in 'Homework Help' started by pr.ee, Nov 20, 2014.

1. ### pr.ee Thread Starter New Member

Nov 20, 2014
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hello i have one question about RLC parallel circuits in laboratory
i have attached the picture of my circuits and i have a question related to it.I gave it a 4volt(peak to peak) square wave but the output was sinusoidal! I really dont know the reason why it happens.please help me with this

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2. ### MrAl Well-Known Member

Jun 17, 2014
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It is a resonant circuit.

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3. ### WBahn Moderator

Mar 31, 2012
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Have you studied frequency response? Plot the frequency response of this circuit (which is acting as a filter) and then plot the frequency content of your square wave (which will be a Fourier series). You should find that the output is dominated by the response to whatever component of the square wave is closest to the peak of the frequency response of the filter. The other components are there, they are just heavily attenuated.

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Nov 20, 2014
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thank you

5. ### MrAl Well-Known Member

Jun 17, 2014
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Hi again,

Sorry i was in a bit of a hurry last post, but the main reason for this response is because the L and C act as a bandpass filter, and a bandpass filter likes to pass the frequencies that are within the bandwidth of the filter more than other frequencies, so you end up seeing mainly the frequency that is closest to the center frequency.

A square wave can be said to be equivalent to a whole bunch of sine waves mixed together, so when you use a bandpass filter that only passes perhaps one frequency, you see only that sine frequency. WBahn was saying something similar to this too.

For your given filter, the angular center frequency is w=25kHz which means the center frequency is about 3979Hz so you should see the highest amplitude sine when you use a square wave of that frequency because the fundamental of the square wave is the highest of all the other frequencies. The third harmonic is next but because your filter is so sharp (L is much bigger than C) you wont see much of that in the output.
If you adjust the frequency slightly off center, you should see the amplitude of the output go down because it will attenuate the fundamental and since the bandwidth is only about 284Hz the fundamental will be attenuated by a factor of about 0.7 if you adjust either way by about 142Hz. Any farther away from center and the amplitude will go lower and lower.

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6. ### MikeML AAC Fanatic!

Oct 2, 2009
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First, sweep the frequency using a sinosoidal source to find the resonance. Then switch over to a square wave source at the resonant frequency to see what happens...

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Nov 20, 2014
7
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hi dear MrAl
it's ok

8. ### pr.ee Thread Starter New Member

Nov 20, 2014
7
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thank you MikeML

9. ### MrAl Well-Known Member

Jun 17, 2014
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Hi,

I agree with Mike about that very good idea about sweeping the frequency from low to high and look for the highest response.
Normally you use a sine wave to do this but you can get away with using the square wave too as long as it is 50 percent duty cycle.
If you look for the maximum response this way you should see that it is at the resonant frequency of the circuit.

Also, if you make the L lower and C higher you'll get a wider response so you can check the third harmonic too if you like. The harmonic amplitudes are 1/n so the third will be 1/3 of max, the fifth will be 1/5 of max, etc. You'll see mostly odd harmonics because they make up the square wave.

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10. ### pr.ee Thread Starter New Member

Nov 20, 2014
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Actually I did that in the lab and i got those results . You are right but i didnt know about bandpass filter
Sure i would check other harmonies too.thank u