# Finding Resonant Frequency in LC circuit using oscilloscope data

I designed a RLC circuit. I chose L and C such that resonant frequency become 2KHz. L=100mH, C=0.25*10^-6F and I took R to be 100 ohms. Then I generated Bode plot for range 100Hz to 10KHz. Ideally I should look for a phase angle of zero in oscilloscope out for resonant frequency, but there is none.

Now how should I find out experimental resonant frequency? Should I pick up a phase angle nearest to zero and look for corresponding frequency?

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Can you post the schematic of the filter? RLC filters can be in different ways. And what do you mean with "there is none"? –  clabacchio Mar 19 '12 at 7:10
It is not allowing me to post pictures as my rep is under 10. I have uploaded the picture here i42.tinypic.com/25f547m.png –  Windows on ARM Mar 19 '12 at 7:19

Your assumptions are wrong: the resonance frequency, give the values of L and C, is

$$f= \frac{\omega}{2 \pi} = \frac{1}{2 \pi}\sqrt{\frac{1}{LC}} = \frac{1}{6.28 \cdot 0.1 \cdot 250 \cdot 10^{-9}} \simeq 1 \, kHz$$

If you want it to resonate at 2K, and keeping the value of the inductance, it helds

$$C = \frac{1}{(2 \pi f)^{2}L} = \frac{1}{(6.28)^2 \cdot 4 \cdot 10^{6}*0.1} = 0.4 \, \mu F \simeq 63,4 \, nF$$

To see it on the scope, you have to measure and superimpose the voltage and the current; fortunately, since you have a series circuit, the current will cause a voltage drop on the resistor, so you can just take that drop as Vout. Also, at the resonance frequency, you will have the maximum gain.

Check this simulation for demonstration.

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You are missing a factor of 1/(2 x Pi) in your frequency formula - as shown it gives the answer in radians per second. So your answer is high by a factor of ~6.28. f = 1006 Hz. –  Russell McMahon Mar 19 '12 at 15:00
Hi! Thanks for the response. Attached is my Oscilloscope Out i41.tinypic.com/2r4ol91.jpg –  Windows on ARM Mar 19 '12 at 22:01
@HemantYadav but it says 7 kHz...did you find the resonance finally? And did you change the component values? –  clabacchio Mar 19 '12 at 22:15
It says around 2KHz. Yes, I found out the value. It came out to be 1908 Hz. I took the csv dump of the bode plot data, and looked for maximum gain. That was -11.xxxx, and its corresponding frequency was 1908 Hz. Funny thing is phase angle at maximum gain was not zero as expected. –  Windows on ARM Mar 19 '12 at 23:50
Actually there was a huge difference. Its phase angle waas around -45 degrees! –  Windows on ARM Mar 21 '12 at 19:57
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f = 1/ [2 x Pi sqrt(LC)] =~ 1007 Hz

Excellent Resonant circuits.

Discusses the circuit below, which essentially directly matches your problem.
.. and much else:

Useful

http://www.tina.com/English/tina/course/28resonant/resonant.htm

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