There are all sorts of resonances around us, in the world, in our culture, and in our technology. A tidal resonance causes the 55 foot tides in the Bay of Fundy. Mechanical and acoustical resonances and their control are at the center of practically every musical instrument that ever existed. Even our voices and speech are based on controlling the resonances in our throat and mouth. Technology is also a heavy user of resonance. All clocks, radios, televisions, and gps navigating systems use electronic resonators at their very core. Doctors use magnetic resonance imaging or MRI to sense the resonances in atomic nuclei to map the insides of their patients. In spite of the great diversity of resonators, they all share many common properties. In this blog, we will delve into their various aspects. It is hoped that this will serve both the students and professionals who would like to understand more about resonators. I hope all will enjoy the animations.

For a list of all topics discussed, scroll down to the very bottom of the blog, or click here.

Origins of Newton's laws of motion

Non-mathematical introduction to relativity

Three types of waves: traveling waves, standing waves and rotating waves new

History of mechanical clocks with animations
Understanding a mechanical clock with animations
includes pendulum, balance wheel, and quartz clocks

Water waves, Fourier analysis

Wednesday, July 18, 2012

3.13 Coupling strength

All postings by author previous: 3.12 Coupling of waves to a resonator up: Contents next: Waveguide excited resonator with circulator

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).

  • The incident and reflected curves have simple off-on-off envelopes, indicating that the incident wave is a simple pulse and the "reflected" wave is defined to be the inverted reflection of the incident pulse.
  • The radiated wave comes from the resonator and has an amplitude proportional to the oscillation strength of the resonator. As you might expect, the radiated wave builds up with time. Note that stronger coupling leads to a larger amplitude of the radiated wave. In the extreme overcoupled case, the radiated wave has twice the amplitude of the incident wave. Since it negatively adds to (i.e. subtracts from) the reflected wave, the complete negative-going wave (in the extremely overcoupled case) equals the incident wave in magnitude, meaning that the incident wave is effectively 100% reflected. See the second animation in this posting for more on this (click "total reflection + ").
  • The negative-going wave is simply the sum of the reflected wave and radiated wave. Note that the phase of the radiated wave is 180 degrees out of phase with the reflected wave, so that the two subtract (negatively interfere with each other).
  • To repeat, the total wave is the sum of the incident wave and negative-going wave. It is the only wave an observer of the real setup would see. On the other hand, it is possible to separate out the incident wave from the negative-going wave and this is routinely done for microwave and optical waves.
↑ Fig. 27a. The undercoupled case. Q0 = 30 , Qcoupling = 70 . ↑ Fig. 27b. The critically coupled case. Q0 = 30 , Qcoupling = 30 .
← Fig. 27c. The overcoupled case. Q0 = 30 , Qcoupling = 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 negative-going 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 negative-going 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 negative-going 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 negative-going wave will be composed of only the radiated wave at the same amplitude it had just before the drive pulse ends, i.e. the steady-state 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 negative-going wave:

Pabsorbed = Pincident − Pnegative-going     .    (150)

We will use (150) at the time just before the pulse turns off, the time in Fig. 28 labeled "steady-state negative wave amplitude". At this point, the negative-going 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:

fnegative-going =  − fincident +  fradiated     .    (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):

    ,    (152)

where Z0 is the characteristic impedance of the transmission line (of the string in our case). Equation (152) can be solved for f:     f = √2Z0P and applied to the three waves in (151) to yield:

Pnegative-going  =   − √Pincident  +   √Pradiated     ,    (153)

where we have canceled out the "2Z0" which occurs in all the terms. We square (153) and substitute it in for Pnegative-going in (150) to have:

    ,    (154)

where we have also substituted ωU/Q0 in for Pabsorbed from (115) in an earlier posting. We also need to point out that the Q0 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 Q0 we use here:

    ,    (154a)

where, in this case, Q1 accounts for the losses in the resonator itself and Q2, Q3, 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 Qcoupling .

We solve equation (125) of an earlier posting (i.e. Pradiated = ω0U/Qcoupling ) for the energy in the resonator U and substitute this in (154) to have:

    .    (155)

Finally, dividing through by Pradiated yields:

    .    (156)

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 = Q0/Qcoupling  .

Equation (157) expresses the ratio of Qcoupling to Q0 in terms of the two peaks of the negative-going 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:

        and              .    (158)

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 negative-going 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 negative-going 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 negative-going wave equals the incident wave in amplitude.

The negative-going wave

Just to further explain this, we calculate the negative-going wave next. The negative-going wave is the sum of the inverse of the incident wave plus the radiated wave:

    .    (159)

Fig. 30. Ratio of the negative-going 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 negative-going 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 fradiated = fincident β , we get:

    .    (160)

This can be solved for the ratio of the negative-going wave to the incident wave:

    .    (161)

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 negative-going 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 2Z0 (see Equation (152) above), provided we are adding the potential parameters, i.e. the vertical forces, transmitted by the waves.

Fig. 31. Resonator energy as a function of the coupling coefficient  β = Q0/Qcoupling  . This graph assumes that Pincident = 1, Q0 = 20, and ω = 1 .

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/Qcoupling = ωUβ/Q0  in for Pradiated  as we did for (155) above. This turns (157) into:

    .    (162)

We solve this for the energy in the resonator U:

    .    (163)

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 negative-going 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 = Q0/Qcoupling for Qcoupling = Q0/β . This was substitued into (123), i.e. into 1/QL = 1/Q0 + 1/Qcoupling  to yield:

    .    (164)

Note that (164) is only valid when QL 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), Δω½/ω0 = 1/QL  for the fractional bandwidth, and substituting (164) into this for QL . The result was:

    .    (165)

Similar to the case for (164), this equation is only valid if Q0 is considerably larger than one and Δω½/ω0 is considerably smaller than one.

Fig. 32. The loaded Q, QL, as a function of the coupling coefficient β. For this graph we have assumed an unloaded Q of Q0 = 100. Fig. 33. Fractional bandwidth Δω½/ω0 as a function of the coupling coefficient β . For this graph we have assumed a Q0 of 100.

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 negative-going 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.

All postings by author previous: 3.12 Coupling of waves to a resonator up: Contents next: Waveguide excited resonator with circulator