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I wonder why when losses are present in a oscillator, the width of the resonance peak is broadened. More precisely: why, when losses are present, can the amplitude reach nearly the maximal one (the one at resonance) over a larger range of frequency than for the lossless case where this range is zero?

I'm looking for an physical/intuitive reason, I know the equations but I cannot find an explanation. I think the explanation should be in the time domain rather than in the frequency domain, using the fact that the lossy case impulse response is a more like decaying wave than a pure sine wave.

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To think about it in the time domain, consider starting the system and watching it evolve:

No loss: the response is a pure frequency $A \cos {\omega t}$

With loss, there’s exponential damping $A e^{-t/\tau} \cos {\omega t}$. Note that’s not a single frequency: due to the (varying) drop from one cycle to the next, the early peaks are a little closer together, effectively at a higher frequency. The frequency response is broadened by the decay.

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I would try this: Don't think of it as of broader resonance, but as less pronounced behavior of the system. Oscillators with less damping just have their resonance more precisely located.

Study this plot in Wikipedia. Clearly it's not like that losses could make things better around th resonant frequency, they just make it evenly worse and therefore growth of less steep is observed around resonance frequency.

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  • $\begingroup$ You mean that since losses decrease the maximal amplitude at the resonance frequency it's a natural consequence that the peak is broader ? $\endgroup$ – David May 27 '15 at 12:57
  • $\begingroup$ Almost. Let's say relatively broader. Width doesn't change rapidly but heigth does. Therefore the peak appears broader. $\endgroup$ – Victor Pira May 27 '15 at 14:02
  • $\begingroup$ Ok. So now I can ask why nature does it this way rather than also shrinking the width so the FWHM stay constant ? Your answer help but that's not (for me at least) a physical reason, even though it's a start. $\endgroup$ – David May 27 '15 at 14:08
  • $\begingroup$ Well, I can't speak for the nature :-), but I really should recommend you not the point of view "losses broadens" but "due to losses the focus on one frequency is decreased". And its of course due to energy dissipation. $\endgroup$ – Victor Pira May 28 '15 at 7:59
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One possible explanation could be based on the fact that for low loss, high $Q$ circuits, the sensitivity of the impedance to the deviation of the frequency from the resonant frequency, is greater than for high loss, low $Q$ circuits.

As a result, relative changes in the voltages and currents in low loss circuits are greater as well and, correspondingly, the peaks of the resonant curves, associated with low loss circuits, are sharper. Or, we can say, that the peaks, associated with the resonant curves of high loss circuits, are broader.

Let's take two RLC resonant circuits, high $Q$ and low $Q$, with identical reactive components, $L$ and $C$, but different resistive components: low, $r$, for the high $Q$ circuit and high, $R$, for the low $Q$ circuit.

At a resonant frequency, the impedances of both resonant circuits, high $Q$ and low $Q$, are purely resistive.

When the frequency deviates from the resonant frequency, the same reactive component, capacitive or inductive (depending on the direction of the frequency change), is added to the impedances of both circuits, but, due to a lower resistance in the high $Q$ circuit, this reactive component would become more dominant in the high $Q$ circuit than in the low $Q$ circuit, leading to a more dramatic change in the impedance of the high $Q$ circuit (including both the magnitude and the phase).

enter image description here

Let's say that both circuits are driven by the same AC voltage source $V$.

At the resonant frequency, $f_0$, the impedances of the two circuits will be defined by their resistances, $Z_1=r$ and $Z_2=R$. Correspondingly, the currents in the two circuits will be $I_1=V/r$ and $I_2=V/R$.

If the frequency increases by $\Delta f$, the impedance of the inductors will exceed the impedance of the capacitors in both circuits, say, by $X$. The magnitudes of the new impedances for the high $Q$ and low $Q$ circuits will become $Z_1'=\sqrt {r^2+X^2}$ and $Z_2'=\sqrt {R^2+X^2}$, respectively.

A relative increase in the magnitude of a non-resonant impedance over a resonant impedance will be greater for the high $Q$ circuit than for the low $Q$ circuits: $$\frac {|Z_1'|}{|Z_1|} > \frac {|Z_2'|}{|Z_2|} \, ,$$ since $$\sqrt {1+ \left(\frac {X} {r} \right)^2} > \sqrt {1+ \left(\frac {X} {R} \right)^2} \, .$$

Correspondingly, a relative decrease of a non-resonant current over a resonant current will be greater for the high $Q$ circuit, leading to a sharper resonant curve.

The same effect could be demonstrated for parallel RLC circuits, except that in that case larger resistance gives larger $Q$.

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If we take an organ pipe oscillator as an example, the reason why becomes clear, as follows.

First we imagine the pipe oscillating without losses. It has standing waves inside it with a velocity zero exactly at the closed end and a pressure zero exactly at the open end and so its resonance wavelength is exactly related to the pipe length.

Now we introduce losses into the system. Here we will use radiation resistance i.e., energy losses due to radiation of sound waves out the open end of the pipe and into the air surrounding it. The oscillator is now damped.

For the sound wave residing inside the pipe to lose energy out the open end of the pipe requires that the open end of the pipe not have a pressure zero, else pressure waves would not be exiting it. this means that the sound wave will protrude slightly out the open end of the pipe in the process of "communicating" with the surrounding air.

This means the resonating length of the air mass inside the pipe now includes a parcel of air outside the pipe which it has to shake in order to move energy out the pipe. This extra mass makes the pipe "look" longer and thereby lowers the undamped natural frequency of the pipe, and makes it play "flat".

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You are right by saying: "the explanation should be in the time domain rather than in the frequency domain". Take, for example an ideal SHO at rest and apply a resonance force at $t=0$. In other words, at $t=0$ you start to pump the amplitude, the driving frequency is still the same. The amplitude grows as $t$ - linearly. Then at $t=\tau$ you stop pumping and leave the oscillator free. For any finite time of observation $T$ the system spectrum will contain different frequencies just due to presence of a variable amplitude within $0<t<T$. Only when $T\gg\tau$ and $T\gg 2\pi/\omega_0$ the other frequencies will disappear. Pumping (gain) is some sort of an opposite case to decaying (loss); the common feature being time variations of the oscillator amplitude in the both cases.

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The basic harmonic oscillator is an idealistic model and assumes linearity in the differential equations that describe it. Any broadening of the peak for this model is due to the Q of the system: how much damping force is present relative to the inertial and restoring forces. The higher the Q, the narrower the peak.

But in actual, physical systems that approximate the model of a harmonic oscillator, there is always non-linearity present on top of the ideal linear harmonic. The non-linearity can also lead to broadening of the peak by harmonic distortion. In the time domain distortion is seen as a sine wave that is not quite a perfect sine wave.

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