Discussion in 'Wireless & RF Design' started by raks_universe, May 4, 2009.

1. ### raks_universe Thread Starter Active Member

Mar 15, 2009
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Does the resonance circuit selects only the resonance frequency(at which Xc=XL)?
If yes,what makes the circuit to accept only the resonant frequency and not the other frequencies(Xc≠XL)?
Is there any compulsion for the circuit to accept only that resonant frequency?

2. ### PRS Well-Known Member

Aug 24, 2008
989
35
I think you're puzzled over the superheterodyne technique? If you build an IF amplifier tuned to 455 kHz, then only signals of 455 kHz (plus the bandwidth of the IF amp -- about 5 or 10 kHz) will pass through this amplifier. Hold this idea in mind...

If you have an antenna capturing signals and send them (after some amplification) into a mixer that mixes a frequency whose sum or difference equals the IF frequency , and if that frequency is 455 kHz, then that signal will pass through the IF amp. The mixing frequency is provided by a local oscillator. This is how radios are channel-selective.

More if you want.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Fortunately the superheterodyne receiver was invented.

I remember long ago playing with AM crystal radio receivers. It was obvious that LC tuned circuits didn't just respond /tune to a single radio station or frequency - you often got interference from adjacent radio frequencies in the band. It was sometimes hard reject the unwanted stations as you tuned across the band.

In fact radio wouldn't work particularly well if the only frequency the tuned circuit responded to was the carrier frequency. There's information in a band of frequencies centered around the carrier frequency or in the sidebands actually transmitted. It's a long story ... given there are several modulation techniques used other than just simple analog AM.

4. ### raks_universe Thread Starter Active Member

Mar 15, 2009
67
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I have doubt in the BASIC concepts of rlc tuning circuit.
Please explain why the circuit always selects only the resonant frequency(at which Xc=XL) and not the other frequencies(Xc≠XL).Is there any compulsion given to the circuit to do so.

Please explain the steps involved in receiving the resonant frequency(step by step) !!!!!!

Mar 6, 2009
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6. ### raks_universe Thread Starter Active Member

Mar 15, 2009
67
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I understand the concept of resonance.

But still i have doubt in "selecting the frequencies",so please let me explain the radio reception.

If i have done any mistake,please mention it.

STATEMENT:
*.Resonance will happen only at the condition Xc=XL
*.For one value of c,there will only one resonant frequency

1.By adjusting the variable capacitor,First set the capacitance value.

2.Now the Xc value will change

3.for this Xc value,there will be a frequency at which Xc=XL

4.Thus resonance occurs.

DOUBTS:
****At the step 3, why the circuit does not selects the other frequency and remains in non resonant condition?****

My doubt is that what about the other frequencies,did the rlc circuit only allows the resonant frequency from the group of frequencies?.

Is there any compulsion for the circuit to be always in resonance?

7. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
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At resonance, a simple LC will have the MOST current flowing through it. It will have the lowest impedance to a signal coming from the "outside world"..an antenna, in the simplest case. At frequencies removed from resonance, the circuit will have greater impedance. The current vs resonance curve is a bell shaped curve that can be either very narrow, in the case of a pure LC circuit with no resistance, or quite broad, if there is significant resistance involved.

In either case, it is the change of impedance of the resonant circuit that gives it the ability to select different frequencies.

It's a lot more meaningful if you can actually build a resonant circuit and look at it with an oscilloscope. Or even better, build a crystal radio! I think the crystal radio is the best teaching tool on the planet....every electronics student should be required by law to build one!

Eric

8. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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t n k:

You evidently have never experienced the thrill of a regenerative receiver! On the verge of oscillation, the simple "regen" approaches INFINITE selectivity! Of course, you have to have the touch of a safecracker to really take advantage of this, but they really are amazing devices!

Incidentally, the main reason for poor performance of most crystal radios is the DIRECT connection of the antenna to the coil...this severely cripples the Q of the circuit. The "loose coupler" took care of this problem by using magnetic coupling to the antenna wire. If you want to play around with some great high performance crystal radio circuits, look at midnightscience.com.

Eric

9. ### davebee Well-Known Member

Oct 22, 2008
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Resonant circuits respond to a range of frequencies around the center frequency, which is called the bandwidth of the circuit. If you draw a graph of the response, it looks like a mountain with a center peak. On both sides of the peak, the response curve slopes down towards zero. Given a center frequency, say 455 kHz, the circuit also responds relatively strongly several kHz on both sides. This is needed for radio to allow several kHz of audio frequiencies to pass through the circuit. But as you get farther from the peak, like 20, 40 or more kHz both higher and lower in frequency, the circuit tends to block the signal.

No, there is no compulsion for the circuit to be in resonance. The circuit will only respond to externally applied frequencies. It will respond more if the applied requency is near resonance, but the circuit does not somehow generate its own energy at the resonant frequency. If no energy is applied to the circuit at the center resonant frequency then no energy will come out of the circuit at that frequency (assuming that the resonant circuit doesn't contain non-linear components like semiconductor junctions that are capable of generating new frequencies).

10. ### PRS Well-Known Member

Aug 24, 2008
989
35
I remember playing with a store-bought crystal set when a youngster. Any commercial station whose signal was strong enough would be detected; there was no tuning to it. The strongest signal would dominate, but the others could be heard in the background.

11. ### RAH1379 Well-Known Member

Dec 13, 2005
69
1
in a resonant circuit the capacitor and coil work together, the time it takes them to charge and discharge is what determines the resonant frequency and the bandwidth. Higher frequencies require less time and lower frequencies take more time. If the time it takes the signal to complete one cycle matches with the "window" of the resonant circuit more of it gets through.Also parallel resonant circuits have a low impedance at resonance while series resonant circuits have a high impedance at the resonant frequency.A high Q circuit has very little resistance and so it has a narrow bandwidth which gives its output a tall narrow peak on the oscilloscope, while a resonant circuit with more resistance will have a wider bandwidth and a broader frequency response with a little less peak voltage.

12. ### PRS Well-Known Member

Aug 24, 2008
989
35
Raks Universe, the radio is frequency selective because only a 455 kHz signal will pass through an amplifier tuned to 455 kHz. This is your Intermediate frequency amplifier.

The radio also contains a mixer with a local oscillator just before this IF amp. A mixer adds and subtracts the frequency of the signal created by the local oscillator to the radio signal on the antenna. Only when this math coincides with 455 kHz, does the IF amp pass the signal through to the detector and audio amplifier.

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Thanks KL7AJ / Eric - you're right. I've missed many of the 'fun' things possible in playing with electronics - radio especially. Thanks also for the advice on improved crystal radios. I'll certainly take a look. Cheers!

14. ### BillB3857 Senior Member

Feb 28, 2009
2,402
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????
I was taught that in a series resonant circuit, there was maximum line current at resonance and in a parallel resonant circuit, there was maximum circulating current.

A Google search provided this as a first hit. It is a pretty good explaination.. www.ece.rutgers.edu/~psannuti/ece224/PEEII-Expt-1-07.pdf.

15. ### CDRIVE Senior Member

Jul 1, 2008
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You were taught correctly.

16. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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I use a resonant tank to demonstrate Kirchoff's law for AC! Measure the circulating current, and measure the line current, while sweeping a generator through resonance. The line current dips at the same point the circulating current peaks. Very cool and graphic!

Eric

17. ### PRS Well-Known Member

Aug 24, 2008
989
35
"I was taught that in a series resonant circuit, there was maximum line current at resonance and in a parallel resonant circuit, there was maximum circulating current."

Yes, I was going to say something but I thought someone just forgot the difference like I might do. Age has aged me.

18. ### CDRIVE Senior Member

Jul 1, 2008
2,223
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We can also add a couple of points that may help those struggling to get a grasp on this. I've always found the Parallel Resonant (Tank) circuit the most fascinating of the two because there's more going on there. While the Tank's circulating current is at max @ resonance, it's input current is at minimum and it's input impedance is at max.

Check out these curves that I spiced. I suspect they're similar to what Eric made reference to.

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19. ### CDRIVE Senior Member

Jul 1, 2008
2,223
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I believe you're describing my first encounter with eardrum damage. Fond memories though!

20. ### RAH1379 Well-Known Member

Dec 13, 2005
69
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Yes, i had that backwards. I do that often.