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3.13 Coupling strength
In the previous posting, we derived equations for the excitation of a resonator by waves traveling down a waveguide. A flash animation in that posting showed the build up and decay of a resonator due to a finite length (or pulse) of waves traveling down the waveguide. Below we show a snapshot of the graphs from that animation after the action had played out. We include graphs for the three different coupling strength options of that animation.
Envelopes of graphs for various coupling strengths  

The following graphs are screen captures of the graphs in the animation, Fig. 26 of the previous posting, with various coupling strengths. The frequency of the incident wave equals the resonant frequency of the resonator. We have added the envelopes of the curves in yellow. The bottom graph of each image shows the oscillations versus time that occur at the right end of the string. The other graphs are various components of these oscillations versus time (at the same point).
 
↑ Fig. 27a. The undercoupled case. Q_{0} = 30 , Q_{coupling} = 70 .  ↑ Fig. 27b. The critically coupled case. Q_{0} = 30 , Q_{coupling} = 30 . 
← Fig. 27c. The overcoupled case. Q_{0} = 30 , Q_{coupling} = 10 .

Relation between coupling strength and peak heights
Fig. 28. Features of negative going wave for the overcoupled case. 
We next derive the relationship between the Q's and the peak heights (which are shown in Fig. 27 above) of the negativegoing wave. More specifically, we use the features labeled in Fig. 28, at the right. The utility of doing this is that experimentalists can conveniently measure these features of the negativegoing wave and use these measurements to determine the Q's and coupling strength of the resonator they are examining.
To explain some of the labeling of Fig. 28: at the first instant of the pulse, the energy in the resonator has not had time to build up and so is zero. So at that instant, the negativegoing wave is made up entirely of the reflected incident wave. Thus, we label the first peak as "incident wave amplitude". Also, at the instant the incident wave pulse stops, only the radiated wave remains and because (at this instant) it has not yet decayed, the negativegoing wave will be composed of only the radiated wave at the same amplitude it had just before the drive pulse ends, i.e. the steadystate radiated wave amplitude.
The following derivation is based on conservation of power: that the power absorbed by the resonator (due to the damping force inside the resonator) equals the power of the incident wave minus the power of the negativegoing wave:
P_{absorbed} = P_{incident} − P_{negativegoing} . (150)
We will use (150) at the time just before the pulse turns off, the time in Fig. 28 labeled "steadystate negative wave amplitude". At this point, the negativegoing wave has two components: the inverse of the incident wave (i.e. the reflected wave) and the radiated wave. We write this in terms of the potential parameter of the wave, i.e. the vertical force transmitted down the string:
f_{negativegoing} = − f_{incident} + f_{radiated} . (151)
We also use the fact that the power transmitted by each of these wave components is given by (see Eqn(8.24) in this reference):
where Z_{0} is the characteristic impedance of the transmission line (of the string in our case). Equation (152) can be solved for f: f = √2Z_{0}P and applied to the three waves in (151) to yield:
√P_{negativegoing} = − √P_{incident} + √P_{radiated} , (153)
where we have canceled out the "2Z_{0}" which occurs in all the terms. We square (153) and substitute it in for P_{negativegoing} in (150) to have:
where we have also substituted ωU/Q_{0} in for P_{absorbed} from (115) in an earlier posting. We also need to point out that the Q_{0} we use here and for the remainder of this posting includes all losses to the resonator except those due to coupling to the waveguide powering the resonator. Thus if there are other connections to the resonator that each have a power loss and Q associated with them, these are assumed to be rolled into the Q_{0} we use here:
where, in this case, Q_{1} accounts for the losses in the resonator itself and Q_{2}, Q_{3}, etc. account for the losses due to other devices, waveguides, etc that may be sucking signal and power from the resonator, EXCEPT for the coupling losses due to the waveguide powering the resonator which are accounted for by Q_{coupling} .
We solve equation (125) of an earlier posting (i.e. P_{radiated} = ω_{0}U/Q_{coupling} ) for the energy in the resonator U and substitute this in (154) to have:
Finally, dividing through by P_{radiated} yields:
The final squared term can be expanded and partially canceled with the first term on the right to give:
. (157)
Fig. 29. Peak height ratios as a function of the coupling coefficient β = Q_{0}/Q_{coupling} . 
Equation (157) expresses the ratio of Q_{coupling} to Q_{0} in terms of the two peaks of the negativegoing wave, both in terms of powers and in terms of amplitudes. This ratio is called the coupling coefficent and is often labelled β or k.
We can solve (157) for the two ratios to get:
In Fig. 29 we graph the ratios of the powers and of the amplitudes in (158) as a function of the coupling coefficient. The graphed power ratio is the ratio of the power radiated by the resonator over the power incident on the resonator (exciting it). It asymptotically approaches 4 for very large coupling constants. The graphed amplitude ratio is the amplitude of the radiated wave to that of the incident wave. It asymptotically approaches 2 for very large coupling constants.
We see that with β less than 1 (undercoupled), both ratios are less than 1 and appproach zero as the coupling constant goes to zero (very weak coupling). With zero coupling coefficient, the waves on the transmission line are totally ineffective at exciting the resonator.
With β equal to 1 (critically or unit coupling), the ratios equal one. This means that the radiated wave exactly cancels the incident wave reflected off the resonator, and there is no negativegoing wave. The resonator totally absorbs the incident wave. This is the condition that is usually sought after by people designing wave driven resonators. People call this condition "perfect matching" (of the waveguide to the resonator) and say the waveguide is "perfectly terminated".
A dilemma  creating power?
For β greater than 1 (overcoupling) the ratios are greater than one but less than 2 for the amplitude ratio and less than 4 for the power ratio. This brings up an interesting dilemma: how can the resonator driven by the incident power radiate more power than is in the incident wave? Is it somehow magically multiplying the power!
The answer is that the radiated wave is inseparably mixed with the reflected version of the incident wave and this combined negativegoing wave is never larger than the incident wave. This is because the reflected wave is opposite in phase to the radiated wave, so the two partially cancel. Thus, when there is unity coupling, the two exactly cancel, and when there is an over coupling condition, the reflected wave reduces the size of the radiated wave. In extreme overcoupling, when the amplitude of the radiated wave is twice that of the incident (and reflected) wave, the negativegoing wave equals the incident wave in amplitude.
The negativegoing wave
Just to further explain this, we calculate the negativegoing wave next. The negativegoing wave is the sum of the inverse of the incident wave plus the radiated wave:
Fig. 30. Ratio of the negativegoing wave to the incident wave as a function of the coupling coefficient . Note that the ratio goes to zero for unity coupling, i.e. the negativegoing wave is zero when the coupling coefficient β equals one. It is negative for weak coupling, meaning that the incident wave (inverted and reflected by the resonator) dominates. For strong coupling, the radiated wave, which is not inverted dominates. This ratio is called the reflection coefficient by most textbooks. We avoid calling it that here to avoid confusion with our "reflected wave" which we take to be only the incident wave after inversion and 100% reflection at the resonator (discussed in the last posting). 
Substituting f_{radiated} = f_{incident} β , we get:
This can be solved for the ratio of the negativegoing wave to the incident wave:
We graph this ratio as a function of the coupling coefficient at the right in Fig. 30. As Fig. 30 shows, this ratio varies smoothly from 1 to +1 . As discussed above, at unity coupling (coupling coefficient equal to one) the negativegoing wave is zero, because at unity coupling the radiated wave is exactly the negative of the reflected wave so the two perfectly cancel each other.
Superposition of waves
Note that it is the waves themselves that linearly combine (in a linear wave system such as we're considering). Two waves traveling together add together using the laws of phasor addition (both amplitude and phase is important). The powers (as in Equation (158b) above) contained in the waves do not add together in such a simple mannor. Instead, to find the power of the sum of the reflected and the radiated waves, we need to add the waves, and then square the resulting amplitude and divide by 2Z_{0} (see Equation (152) above), provided we are adding the potential parameters, i.e. the vertical forces, transmitted by the waves.

Energy in the resonator
We next derive the energy in the resonator as a function of coupling strength. We start with (157) and substitute ωU/Q_{coupling} = ωUβ/Q_{0} in for P_{radiated} as we did for (155) above. This turns (157) into:
We solve this for the energy in the resonator U:
Figure 31 graphs this energy versus the coupling coefficient β. We see that the energy in the resonator peaks at a coupling coefficient equal to one and falls off on either side of that. As we stated above, at unity coupling, the resonator absorbs all of the incident wave, making the negativegoing wave zero. This maximizes the energy in the resonator. At larger coupling coefficients (stronger coupling) the radiated wave increases (as Fig. 29 shows), sucking more amplitude and energy from the resonator. The conclusion of Fig. 31 is that if you are trying to excite a resonator to the maximum with a certain wave, there is an optimum coupling strength, and that is unity coupling.
Ghost wave
The radiated wave at unity coupling can be thought of as an analogue to the virtual particle waves in quantum mechanics. At unity coupling it is a ghost wave or virtual wave because it is hidden to the experimentalist, being canceled out by the reflected wave. On the other hand it does real things to the resonator. It dampens the resonance, lowers the loaded Q, and broadens the resonance curve over what would be the case if the coupling were less than unity coupling. We see this in Fig. 32 below which plots the loaded Q versus the coupling coefficient. Fig. 33 similarly plots the fractional bandwidth versus coupling coefficient.
The equation plotted in Fig. 32 was obtained by solving the definition of the coupling coefficient, Equation (157), β = Q_{0}/Q_{coupling} for Q_{coupling} = Q_{0}/β . This was substitued into (123), i.e. into 1/Q_{L} = 1/Q_{0} + 1/Q_{coupling} to yield:
Note that (164) is only valid when Q_{L} is considerably larger than one. In fact all the Q's are only well defined if their values are considerably larger than one.
The equation plotted in Fig. 33 was obtained by using (118), Δω_{½}/ω_{0} = 1/Q_{L} for the fractional bandwidth, and substituting (164) into this for Q_{L} . The result was:
Similar to the case for (164), this equation is only valid if Q_{0} is considerably larger than one and Δω_{½}/ω_{0} is considerably smaller than one.
Second dilemma  starting a resonance without power absorbed
Above we have seen that initially, before the resonator has time to buildup it resonance, all of the incident wave is returned in the negativegoing wave. Figs. 27 and 28 show this in graphical forms and the text following Fig. 28 discusses this. The question is: if all the incident wave and the power of the wave is rejected by the resonator, how does the resonator get started? Doesn't it take power to start the resonance?
The answer is that: no it doesn't take power at the very first instant to get the resonance started, but as soon as it starts a little, it does take power to continue the build up. But of course after the first instant, power is absorbed from the incident wave, so all it ok. What gets the resonance start is the vertical force that the incident wave applies to the resonator and that force is not dependent of the resonator's response.
This dilemma is similar to the question of how a force can start a ball rolling. If we approach the problem from the point of view of Newton's second law of motion, F = ma then all is ok: the force causes an acceleration and the ball starts moving. But at the first instant of the application of force, the power absorbed by the ball P = Fv = F×0 = 0 where v is the ball's velocity is zero because at the first instant the velocity of the ball is zero. So there is zero power absorbed by the ball, but yet it accelerates just fine.
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