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Power at resonance frequency 
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#1
Dec2413, 09:14 AM

P: 61

Hi.
When using spectrum analyzer to measure the response of a coil (a RLC circuit), I see that there is a peak at one frequency (resonance frequency). This is logical because we can model a coil as a RLC circuit. If I change the frequency of the sinusoidal source, the peak reduces its value. My question is: why is there a higher peak at resonance frequency? If what I measure is the power dissipated (by the resistor, of course), why is there a dependence with frequency? If complex power is written as [itex]P = P_{loss}+2j\omega\left(W_mW_e\right)[/itex], where [itex]P_{loss} = \frac{1}{2}\leftI\right^2R[/itex] is the dissipated power, [itex]W_e[/itex] is the electric energy stored in the capacitor and [itex]W_m[/itex] is the magnetic energy stored in the inductor, why the analyzer doesn't show a constant peak value (given by [itex]P_{loss}[/itex]) whatever the frequency of the sinusoidal source? Thank you. 


#2
Dec2413, 10:17 AM

Mentor
P: 11,925

I would guess that the driver power reaches its maximum at the resonance frequency, but it is hard to tell if you don't show the setup and explain where you measure what.



#3
Dec2413, 10:26 AM

P: 61

I rolled two wires around the coil: one coming from the sinusoidal signal generator and another one going to the spectrum analyzer, which also showed "power" (dissipated power, I suppose).
So, I was measuring the voltage drop in the coil (which can be modeled as a RLC circuit). Sweeping frequencies I noted there was a higher peak at resonance frequency. Thanks. 


#4
Dec2413, 02:25 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 7,175

Power at resonance frequency
If the driving frequency is constant, then once the starting transient has decayed the power dissipated in the resistor is (obviously) the same as the power supplied by the signal generator. The phase angle between the the generator voltage and current changes as you go through the resonant frequency. That is why the power (in watts) from the generator is not necessarily the same as the voltamps it is producing. Maybe this is easier to see with a mechanical oscillator like a mass on a spring. If you apply a force an a frequency much lower than resonance, you can ignore the inertia of the mass. The applied force is in phase with the motion, so the positive work you do compressing the spring is equal to the negative work done by the spring when you release it. At a frequency much higher than resonance, you can ignore the force in the spring compared with the inertia force needed to accelerate the mass. The displacement is now 180 degrees out of phase with the force, but the effect is the same: the work you do on the mass averages out to zero, over a complete cycle. Away from resonance, a plot of work done against time may have large positive and negative amplitudes in each cycle (analogous to measuring voltamps) even though the average work done over the cycle is low (analogous to measuring watts). But when you are at the resonant frequency, the displacements are 90 degree out of phase with the force, and you do work on the system over the complete cycle. And of course as the amplitude increases, a constant force does even more work, because the product of force x distance increases. 


#5
Dec2513, 12:06 PM

P: 589

To give you an idea where I am going with this, if you are using coax to extend source & spectrum analyzer down to your inductor, you will see periodic dips in response due the the cable lengths. 


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