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.

Origins of Newton's laws of motion

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

## Sunday, June 23, 2013

### 3.16c Steady state response of the circuit: transmission line, coupling inductor and resonator

 All postings by author previous: 3.16b The resonator alone up: Contents next: 3.16d Transmission line resonators

3.16c Steady state response of the circuit: transmission line, coupling inductor and resonator

Keywords: resonator, steady state, ac circuit analysis, coupling inductor

Topics covered in this posting

• The entire equivalent circuit including resonator, coupling inductor, transmission line impedance, and source is mathematically analyzed.
• The results of the analysis is graphed: the resonator response as a function of coupling inductor value.
• A trick is used to quickly analyze the resonator response at unity coupling.
• The size of the coupling hole in the acoustical circuit as required for unity coupling and efficient coupling of the waves into the resonator is determined.

Contents of this posting

1. The circuit
2. The equations
3. Graphs of the equations
4. Simple calculations of response at unity coupling
5. Coupling hole size for acoustical circuits of Fig. 35 of posting 3.15 for unity coupling

16c. Steady state response of the circuit in Fig. 42 of the posting 3.16a
 Fig. 1. Circuit analyzed in this posting.

1. The circuit

In the previous posting, we have considered only the resonator part of our equivalent circuit, labeled as "resonator" in Fig. 1 at the right. In this posting, we consider the whole circuit in Fig. 1, including the coupling inductor, driving voltage source, and its output resistance. The "output resistance" R0 of the signal source represents a convenient way of accounting for the characteristic impedance of the transmission line and the losses incurred by the resonator in radiating waves back out the transmission line towards the source, as discussed in an earlier posting.

2. The equations

We start with the fact that the current flowing from the wave source equals that flowing through the total impedance of the rest of the circuit as presented to the wave source, IS = VS/Ztotal  where Ztotal = R0 + jωLC + Zresonator and Zresonator = Zres from Eqn. (1) of the previous posting:

.   (1)

 Fig. 2. Graph of Equation (2) with (1) substituted in for IS plotted versus drive frequency. As in the previous posting the parameters used to make this graph were, L = 2H, C = 0.08F, Z0 = R0 = 5Ω, ω0 = 2.5rad/sec, Q0 = 20, R = 100Ω and VS = 30V. The units on the vertical axis are (V), i.e. volts. The graph was made using gnuplot.
 Fig. 3. Graph of Equation (2) with (1) substituted in for IS plotted versus coupling inductance. The component values are the same as in Fig. 2 above. The units on the vertical axis are volts. The drive frequency was adjusted for maximum response at each value of coupling inductance.
 Fig. 4. A similar graph as that shown in Fig. 3. This graph was made using Octave to more exactly adjust the drive frequency at each value of the inductance. We also show the values from Fig. 3 for comparison.

The source voltage VS is known (it is whatever we choose to set the source level at), so that (1) is sufficient to calculate the steady state AC current flowing through the source, through R0, LC and the resonator as a whole.

The voltage generated across the resonator alone indicates the excitation level of the resonator. We call this voltage Vres. The voltage across the resonator is just IS times the impedance of the resonator (or divided by the admittance of the resonator):

.   (2)

3. Graphs of the equations

Fig. 2, at the right, plots the voltage across the resonator Vres as a function of the angular frequency ω of the driving sound source, for three different values of coupling inductors LC (in Henries).

We notice two effects that changing the coupling inductor causes:

• shifts the resonance curve and

To be quantitative concerning the first effect, we see that LC = 8H produces the maximum voltage across the resonator at its peak. This agrees within the expected error of the more precise calculation using Eqn. (5) of the previous posting (see the comment in the last paragraph of Section 2 of the last posting) that yielded LC = 7.82Henries for unity coupling, i.e. maximum power into the resonator. The previous calculation also indicated that the resonant frequency at unity coupling would be 2.79radians/sec which agrees with the green curve in Fig. 2.

As to the second effect, the broadening of the resonance curve in Fig. 2 for smaller coupling inductor LC values agrees with the concept that stronger coupling causes reduced loaded QL's as indicated by the graph in Fig. 4 of posting 3.16a.

Fig. 3 plots the resonator amplitude versus the coupling inductance assuming that the drive frequency is adjusted at each value of inductance for the maximum response of the resonator, i.e. for the peaks shown in the last graph (Fig. 2). The graph in Fig. 3 was made using an approximation to calculate the optimal frequency for each coupling inductance. We started each calculation assuming a frequency of ω0 and calculated the reactance of the coupling inductor using this XC = ω0LC . We then set XC equal to the negative of the imaginary part of Eqn. (2) of the last posting, assuming that at the optimal frequency, the reactance of the coupling inductor would null out the reactance of the resonator. We then used the quadratic formula to calculate δ and used the approximation in the first part of Eqn. (9) to calculate ω i.e. used ω ≅ ω0 + ½ω0δ . We then used this ω in (1) and (2) to calculate Vres for the particular inductance value.

Fig. 4, at the right, shows a more precise calculation of the same quantity as the previous graph showed. For this graph, we used Octave to find the peak value of Vres for each inductance value and plotted these peak values versus the coupling inductance. We see that this "exact curve" roughly agrees with the approximate plot, although the agreement could be better. The previous two graphs were made with gnuplot.

4. Simple calculation of response at unity coupling

In general, it is messy to calculate the voltage across the resonator at the resonance peak. We would need a formula for the frequency which would involve taking the derivative of the magnitude of (2) (which involves (2) ) with respect to the frequency ω, setting that equal to zero and solving for ω. This involves long, messy formulas.

On the other hand, at resonance and critical coupling (LC = 8H), there is a power method that is very easy. It uses properties of transmission lines and a matched load (i.e. a resonator at critical coupling). We demonstrate this now using the component values used in the above example.

The incident power traveling from the source towards the resonator is:

,   (3)

where 30V is the amplitude we used for VS in making Fig. 2 above. As discussed in reference to Eqn. 3 of posting 3.16a, the incident wave amplitude is half the amplitude VS of the ideal voltage source in our modeling circuit shown above in Fig. 1. (V is the symbol for unit of voltage, i.e. for volt, and W is the symbol for watt, the unit of power.)

At unity coupling, all the incident power is absorbed by the resonator. Since the only dissipative element in the resonator is the resistor R, it is this resistor that dissipates the incident power. The power dissipated by R is given by:

.   (4)

Solving for Vres yields:

,   (5)

which agrees with the peak value of the green curve in the Figs. 3 and 4 above.

5. Coupling hole size for acoustical circuit of Fig. 35 of posting 3.15 for unity coupling

To be more concrete, we now relate the above to actual physical dimensions of the acoustic circuit of posting 3.15. The details of the coupling hole are shown in Fig. 1 of posting 3.16a. We start with the equation for the equivalent inductance (eqn. 2 of 3.16a) of a small hole:

Solving for Vres yields:

,   (6)

where we have taken the limit for holes of very short lengths. Also a is the coupling hole's radius, len is its length, Ah is the hole's cross sectional area, ρ is the density of air (in kg/m3) and K is the equivalence factor we choose between the acoustical circuit and electrical circuit.

We next change this into a dimensionless quantity ω0LC / Z0:

,   (7)

where the equivalent characteristic impedance of an acoustical transmission line is Z0 = Kρc / A, c is the speed of sound, A = πr2 is the cross sectional area of the transmission line and r is the transmission line's radius (assuming it has a circular cross section). Also λ0 = c/f0 = 2πc/ω0 is the wavelength of the sound in the transmission line at resonance. (We used this in the form ω0 = 2πc/λ0 .)

Using the circuit parameters in the caption of Fig. 2 above, we calculate this same normalized coupling inductance for the electrical circuit of our graphs above at the optimal coupling:

.   (8)

Equating (7) and (8) we get:

which implies         .   (9)

To give a concrete example, in the last line we assumed λ0 = 20cm and r = 1cm.

We see that to produce the condition for optimal coupling (unity coupling) we need to restrict the opening to the resonator to a radius of about a third of that of the acoustical transmission line. This means that the cross sectional area of the coupling hole is about one tenth that of the transmission line. In spite of its relatively small size, all the power is absorbed by the resonator.

This exercise is intended to demonstrate the power of a resonator. The higher the Q0 the smaller the coupling should be for 100% absorption of the incident wave. It's as if the resonator reaches out and sucks in the incident wave. Alternately, we can say that the higher the Q0 the stronger the canceling wave will be, because of the fact that a resonator recycles its energy and builds up intense fields inside itself. The canceling (i.e. radiated) wave is proportional to the strength of the internal fields.

 All postings by author previous: 3.16b The resonator alone up: Contents next: 3.16d Transmission line resonators