# 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 – Hemant Yadav 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 – Hemant Yadav 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. – Hemant Yadav Mar 19 '12 at 23:50 Actually there was a huge difference. Its phase angle waas around -45 degrees! – Hemant Yadav 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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There are many way to confirm with a scope

1) For frequency domain, Use FM sweep gen in X-Y mode. This is for visual confirmation using envelope centered at bottom of screen.using Scope X axis sweep to CH1 and Ch1 out to FM in and CH1 as X axis on scope CH2 as Signal current over R then scope shows frequency response. You can shift X axis with CH1 level and scale FM accordingly.

2)For time domain, pulse the circuit with narrow pulse at low rate eg 100 Hz and measure the ringing resonant frequency.

The gain of the filter or Q factor is ratio of reactive to real impedance, in this case 1257ohm /100 =12.6=Q so there should be a few cycles decaying to confirm resonant values.

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