# Energy gain in a resonant system

1. Jun 30, 2010

### maldata

Something is bothering me about passive systems in resonance. Consider the Bode plot of a second-order system, for example:

http://en.wikipedia.org/wiki/File:Harmonic_oscillator_gain.svg

Let's say we excite this system near its resonance, so the system has a gain > 1. I don't understand how this can be the case for a passive system like an RLC circuit. If the output amplitude is greater than the input amplitude, that energy has to come from somewhere.

What is it that I'm misunderstanding here? It seems like in a passive circuit, the output amplitude should, at most, be the same as the input amplitude. So how is there a gain of greater than 1?

Thanks.

2. Jul 1, 2010

### Studiot

Remember the power (and therefore energy) is not determined by one quantity alone.

Your graph is for the amplitude of a single quantity.

Look at what happens to the other quantities in one of the energy or power equations, particularly the impedance.

3. Jul 1, 2010

### maldata

Hi Studiot, thanks for the reply.

I'm still not sure how that matters, though. In that plot, the gain at resonance is 100. So, say this is an underdamped RLC circuit. We put in a sinusoidal voltage with amplitude 1 at the resonant frequency, and we get out a sinusoidal voltage with amplitude 100. Regardless of what any of the other quantities are, we've just amplified a signal passively. I don't see how that's possible.

Other passive circuits make sense to me... an RC circuit or an overdamped RLC circuit never have gain greater than 1. So how can an underdamped RLC circuit do this?

Obviously there's still some concept I'm missing. Thoughts?

Thanks!

4. Jul 1, 2010

### Studiot

Think of a (perfect) transformer.

24 volts in 240 volts out.

Does it amplify the voltage?
Well yes.

Does it amplify the energy or power
Well no, because the current available is less, by the same factor (1/10 in this case)

Furthermore this may be viewed as an impedance transformation, from low impedance to high impedance, hence the lower current.

Your resonant circuit is the same. You will get out a higher voltage, but at higher impedance, so the minute you apply it to a load you will not be able to draw the same level of current as the original signal could support.

It is difficult to be more specific without specific values, so work a few out to see.

5. Jul 1, 2010

### maldata

Ah. There's the analogy I was looking for... that seems a little more intuitive to me. Thanks for the insight!

6. Jul 1, 2010

### Studiot

In fact with resonant systems the impedance change usually so great that we have to use an amplifier to prevent the rest of the circuit loading (damping) the resonance completely.
We often place the resonant circuit in the feedback path of the amplifier for this purrpose, alternatively we buffer the resonant circuit with an amplifier. Thus we can inject the extra energy to maintain the amplification at lower impedance.