Discussion in 'Homework Help' started by sloppygiuseppi, Mar 30, 2008.

1. ### sloppygiuseppi Thread Starter New Member

Mar 30, 2008
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If L = 10 micro Henries, C=1Nf, then calculate the value of the resistor when the resonant frequency is 1.4 Mhz

2. ### Mark44 Well-Known Member

Nov 26, 2007
626
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It would take all the fun out of things if we did this homework problem for you. Most of us are willing to help, but that implies that you're doing something to get you part way there.

Do you have any formulas that seem relevant? If so, have you tried to get a value using one or more of them?

3. ### beenthere Retired Moderator

Apr 20, 2004
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Here's a hint - at resonance, Xc = Xl.

4. ### sloppygiuseppi Thread Starter New Member

Mar 30, 2008
6
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i guess it'd take the fun out of it but i have to hand in tomorow and i have so much to do

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
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The equations you need to find the value of R are here. And just for clarification, Xc = Xl at resonance only if the circuit is undamped, i.e., R=0.

6. ### Wendy Moderator

Mar 24, 2008
21,712
3,000
Correct me if I'm wrong, but resonance frequency doesn't care about the resistance, does it? Resistance affects the Q of the circuit.

It's been a while for me.

7. ### beenthere Retired Moderator

Apr 20, 2004
15,808
295
One suspects it's a trick question. The reactances may not be equal at the stated frequency, so the value of R will make that equality while degrading performance (poor Q).

8. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
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No tricks. The natural (undamped) resonant frequency is indeed Fr=1/(2pi*sqrt(LC)), which in this case is 1.59MHz. The series resistor "pulls" the resonant frequency (defined as the frequency where the impedance is highest) to a value lower than the natural resonant frequency. It is possible, using the equations on the page I referenced previously, to solve for the resistance value that causes damped resonance to occur at 1.4MHz.
Have a look at the graph on this page to see the effect of damping on resonant frequency.

9. ### JoeJester AAC Fanatic!

Apr 26, 2005
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I just don't agree with calling the 1.4 MHz the "resonant" frequency.

I agree a 1.4 MHz signal applied to a circuit with resonance at 1.59 MHz would pass a damped 1.4 MHz signal, and you can reduce the damping by lowering the Q of the circuit.

How much loss is acceptable before shifting the circuit's resonance to be close to the frequency of interest?

1.4 MHz is not a resonant frequency for the two components listed. In the sence of resonant frequencies, or 1 / (2pi(LC)^.5), 1.4 isn't in the ball park, unless your willing to accept a much lower signal level or a damped signal.

10. ### Ron H AAC Fanatic!

Apr 14, 2005
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Well, I agree that the Bode plot (attached) of the solution to his problem looks quite different than that of a high-Q tank circuit, but the peak is at 1.4MHz, and the circuit is the one he posted, with the resistor value set to yield maximum impedance at that frequency. If you don't want to call that the resonant frequency, what name would you give it? The page I referenced calls it the damped resonance frequency, and, a I said, 1 / (2pi(LC)^.5) is the undamped resonant frequency.

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11. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Ron,

I agree you'll capture the 1.4 MHz signal. The anti-resonant circuit [parallel] will show a 1.4 MHz peak as the lower Q increases the bandwidth.

Calling the 1.4 MHz a "resonant" frequency with the circuit parameters provided is incorrect. There is only one resonant frequency.

Of course, all the professors in the world can call it whatever they want. Till it follows fr = 1 / (2pi(LC)^.5), it's not the resonant frequency.

12. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
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Here's a little mental experiment: Try banging the tank circuit with a short current pulse. Change the series resistor. Bang it again. Does the ringing frequency change? If it does, would you not say that it resonates at a different frequency? Suppose your tank circuit was a tuning fork, or a pendulum. Does the damping affect their resonant frequency? I maintain that it does.
I think we have a difference in semantics here. I (and the sites I cited) think of resonance as a physical phenomenon. You think of it as a mathematical equation. Perhaps we just have to agree to disagree.