A system at resonance, produces higher than input power at its output, at resonance frequency (At most of the times, considering the system design).

I want to know, whether this extra energy is justified considering the systems stability?

How does this extra energy at oscillations influence the output in terms of its magnitude and phase characteristics ?

What happens if the systems stays in resonance for too long time ? and also the case if the systems resonance is short lived ?

PS: just to make it simple, I want to analyse for open loop systems.

Thanks.

 

There is no extra energy. Where did you come up with this quaint notion? You're welcome to show us an example, if you have one. What is actually happening is that energy is passed back and forth among the reactive components. Any losses must be covered by the availability of input power. A stable oscillator, which requires an active component with gain, will keep oscillating until the cows come home.

An unstable system will have an output that increases to the limit of the supply at which time it will become non-linear. It might or might not destroy itself. The classic example is a microphone placed close to a speaker that produces an amplified version of the microphone input. Most amplifiers limit the output to prevent this kind of destruction.

 

A system at resonance, produces higher than input power at its output, at resonance frequency (At most of the times, considering the system design).

Wow... Energy crisis SOLVED!

 

What I think you mean is that the store resonant energy is greater than the energy required to keep the system oscillating, not that it puts out more energy than you put in (which, of course, is impossible perpetual motion).
Not sure I understand your questions though.
The resonant oscillations will built up from the applied stimulus until the energy dissipated in the oscillations equals the applied energy.
If the resonant system has low losses (high Q) then the oscillations can build up until the system fails (see Tacoma Narrows Bridge).

 

What I think you mean is that the store resonant energy is greater than the energy required to keep the system oscillating, not that it puts out more energy than you put in (which, of course, is impossible perpetual motion).
Not sure I understand your questions though.
The resonant oscillations will built up from the applied stimulus until the energy dissipated in the oscillations equals the applied energy.
If the resonant system has low losses (high Q) then the oscillations can build up until the system fails (see Tacoma Narrows Bridge).

Thanks for the reply. What I meant was the magnitude bode plot shoots above the 0dB and my efficiency calculation at resonant frequency shows above 100%, which made me think about this particular question. But I know my question is wrongly framed. I know it seemed like I am questioning the law of conservation of energy. But I am still a pupil in this field I cant go that far

Can you please tell me what it means when the magnitude plot shoots over the 0dB at resonant frequency ? under two circumstances 1. It takes a long time to fall back below 0dB and 2. It falls back immediately.

I have attached a magnitude plot. you can see the system oscillates at around 2MHz, and the gain at resonant frequency is well below 0dB, but at certain load it shoots up aboe 0dB, why is that?

 

It shoots above the 0dB point because at the resonant point the voltage increases due to the stored energy in the resonant system.
But the net energy out always equals the net energy in.
Understand that dB here is voltage dB, not power, i.e. the reference input impedance is not equal to the output impedance of the resonant circuit as would be required for dB to be measuring power.

 

To add to what @crutschow said when you have a voltage peak across a reactive element, the current is out of phase with the voltage and is probably at or near a trough. Power is the product of instantaneous voltage and current and is pretty small when either voltage or current is at a peak.

 

It shoots above the 0dB point because at the resonant point the voltage increases due to the stored energy in the resonant system.
But the net energy out always equals the net energy in.
Understand that dB here is voltage dB, not power, i.e. the reference input impedance is not equal to the output impedance of the resonant circuit as would be required for dB to be measuring power.

So even if the system output voltage shoots above 0dB at resonance, there wont be any problem? to the stability of the system?

 

So even if the system output voltage shoots above 0dB at resonance, there wont be any problem? to the stability of the system?

The only way there could be a problem is if the output increases without an upper bound. Oscillation is not the same thing as unstable.