Resonances, waves and fields: 3.16b Looking at the resonator part of the circuit

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3.16b Looking at the resonator part of the circuit

Keywords: resonator, impedance, matching, unity coupling, signal transformer, middle ear

Topics covered in this posting

In this posting we restrict our analysis to the resonator part of the equivalent electric circuit.
An equation for the impedance of the resonator is derived.
This impedance is graphed in a few different ways.
Using the graphs we discuss a possible strategy for coupling to the resonator, using a coupling inductor to efficiently match the waveguide's characteristic impedance to that of the resonator.
The required inductance for matching is algebraically derived, as well as the new resonant frequency, shifted because of the coupling to the transmission line.
The required inductance value and resonant frequency is graphed versus the characteristic impedance of the transmission line.
We discuss the problem of matching a particular range of characteristic impedance values.
The problems of coupling in this range of characteristic impedance values can be overcome by using a transformer instead of coupling inductor. This is related to the mechanical signal transformer inside the human ear.

Contents of this posting

The impedance of the resonator
Getting to unity coupling, i.e. a perfect match
Graphs of the above equations
Signal transformers and the ear

1. The impedance of the resonator

In the previous posting we created an electrical circuit that is equivalent to the acoustical circuit of Fig. 35 of posting 3.15. In this posting we will examine the resonator part of that circuit.
The impedance of the resonator alone (i.e. not including R₀ and L_C) is given by: . (1) The real and imaginary parts of this are graphed at the right in Fig. 1. As we see, both the real and imaginary parts of the resonator's impedance change drastically with the driving frequency ω. The factor "j" is an alternate notation for the imaginary constant i = √−1 . "j" is preferred by electrical engineers who often use "i" to refer to electrical current.	Fig. 1. Graph of Eqn. (1) at the left with R = 100Ω, L = 2H, and C = 0.08F. These values mean that ω₀ = 1/√LC = 2.5 radians/sec and Q₀ = 20.
Fig. 2 at the right shows a different way to plot the impedance, this time as the magnitude and phase of the impedance. The magnitude peaks at the resonant frequency of the resonator while the phase changes from +π/2 to −π/2 when the frequency is swept through the resonance. The phase right at resonance is zero, meaning that the voltage and current through the resonator are in phase, i.e. the resonator acts as a pure resistor with a resistance equal to 100Ω, the value of R. This is to say that at resonance, the effects of the inductor L and capacitance C cancel each other. Incidentally, Figs. 1 and 2 showing the complex impedance versus frequency is another way to graph a resonance, a way that was omitted from various looks of resonance curves . Other parameters are often substituted for the impedance, such as the complex voltage across the resonator or complex current through the resonator for a very similar looking graph of a resonance.	Fig. 2. A graph similar to that in Fig. 1 with the impedance of Eqn. 1 plotted versus the angular frequency. Here the magnitude and phase of the complex impedance is graphed instead of the real and imaginary parts. The same parameters were used as are listed in the caption of Fig. 1.
In Fig. 3 at right we show a polar plot of the impedance graph. The magnitude of the complex impedance is graphed on the radial axis, while the argument (phase) is plotted azimuthally (angularly). As we have noted before in Fig. 13 of the posting various looks of resonance curves on this type of graph, a resonance produces a circle. One sees, in this polar plot, a large phase shift on either side of resonance ... +90 degrees at frequencies less than resonance to −90 degrees after resonance. The phase shift at resonance is 0 degrees. In making the polar plot, the angular frequency was varied from 2 to 3 radians per second. The 2rad/sec point lies vertically just above the origin. The 2.5rad/sec point (the resonator driven at resonance) lies to the right of the origin, while the 3rad/sec point is just below the origin.	Fig. 3. Polar plot of the complex impedance of the resonator.

2. Getting to unity coupling, i.e. a perfect match

Fig. 4. A copy of Fig. 1 above with lines added to show the method of choosing L_C for a perfect match, i.e. for unity coupling to the resonator.

a. Graphical solution

A perfectly matched load for the transmission line means that all the waves launched towards the load will be absorbed and none reflected. In resonator language this means unity coupling.

In Fig. 4 at the right, the vertical dotted line indicates the needed drive frequency (2.78 radians per second) to have the real part of the impedance equal 5Ω (the characteristic impedance of the transmission line that we will be using) and so be perfectly "matched" to the transmission line. Perfect matching also requires that we set the coupling inductor to an impedance ωL_C = 22Ω to cancel the negative reactance of the resonator at ω = 2.78 radians per second. A different coupling inductor will result in undercoupling or overcoupling. The frequency of 2.78 radians per second will be a new resonant frequency when we have perfect match. Changing the inductor's value from that required for perfect match will shift the resonant frequency as we shall see in the next section.

b. Algebraic/computational solution

To algebraically calculate the required impedance and resonant frequency of Fig. 4 we follow the same steps algebraically as the graphical ones above:

adjust the drive frequency so that the real part of the resonator impedance equals the characteristic impedance of the transmission line; we assume the characteristic impedance is real.
adjust the coupling inductance to cancel the imaginary part of the resonator's impedance as the above selected frequency.

To begin, we separate out the real and imaginary parts of Eqn. (1):

, (2)

where detuning factor, δ , is given by:

, (3)

and the reactance of the L or C inside the resonator, X_res , is given by:

. (4)

We can use the quadratic formula to solve (3) for drive angular frequency, ω , in terms of the detuning factor δ :

. (5)

To find the detuning needed for unity coupling, we set the real part of the resonator impedance Re{Z_res} = R₀ = Z₀ , i.e. the characteristic impedance, assumed real, of the transmission line to be used:

, (6a)

and solved for δ to give the detuning factor required for unity coupling:

, (6b)

where everything on the right side of the equation is known.

The detuning faction, δ, from (6b) can be substituted into (5) to yield the resonant frequency of the circuit with unity coupling.

To calculate the coupling inductance required for unity coupling we set the magnitude of the second term of (2), the imaginary part, equal to X_Lc = ωL_C:

. (7a)

and solve for the L_C required for unity coupling:

, (7b)

where everything on the right side of the equation is known.

Alternatively we could approximate (3) for δ:

, (8)

which is good if the loaded Q is large. Equation (8) can be solved for ω in terms of δ:

, (9)

which could then be substituted into (7b) to yield an approximate coupling inductance value for unity coupling.

If we use (9) and then (7b) with the parameter value given in the caption of Fig. 1, we calculate the approximate values of ω = 2.77rad/secs and L_C = 7.86H.

An exact solution is had by:

using (6b) to calculate the required detuning factor, δ ,
then using that δ in (5) to calculate the new resonant frequency,
and finally using (7b) to calculate the required coupling inductor, L_C .

With this procedure, we get:

ω = 2.79rad/sec, a difference of 0.5% from our approximate value, and
an optimal coupling inductance of L_C = 7.82Henries, also with a 0.5% difference from the approximate value above.

Graphs of the equations just above are shown below.

3. Graphs of the above equations

At the right we plot the approximate frequency Eqn. (9) and the exact frequency (5) versus impedance of the driving transmission line R₀ . These frequencies have been normalized by dividing by ω₀ = 2.5rad/sec. The resistance in the resonator R is set at 100Ω which makes Q₀ = 20. Also L = 2H and C = 0.08F. On the graph we see that the approximation for the frequency shift is close to the exact value. The graph is for a Q₀ of 20. At lower values of Q₀ the approximation would be worse and at higher values it would be better.

At every value of transmission line impedance, the coupling inductance is adjusted to achieve unity coupling (perfect match between transmission line and LRC resonator).

Also we see that there are no values for the transmission line impedance greater than the resistance of the resonator. This means that this method of matching the resonator and transmission line using a coupling inductor will not work for these values. We must use another means, such as a transformer. Indeed if we look at the human ear where we have such a situation, nature uses a quite complex mechanical transformer consisting of tiny bones in the middle ear to match impedances, because an easy fix such as a small coupling hole would not work in this situation. More on this below.

Fig. 5. Resonant frequency calculated two ways plotted against the characteristic impedance of the transmission line. The frequencies, as plotted, have been divided by ω₀ = 1/√LC .

At the right we plot the approximate and exact coupling inductance values required for perfect match (unity coupling) plotted versus transmission line impedance. The two curves practically lie on top of each other. The curve goes up and then down because of the humped nature of the resonator reactance (the imaginary part of the resonator impedance) as shown in Fig. 1 above. As stated above at transmission line impedances greater than the resistance of the resonator, there is no solution using a single inductor (or single capacitor) to achieve a perfect match.

Fig. 6. Coupling inductance needed for unity coupling plotted versus the characteristic impedance of the transmission line powering the resonator.

4. Signal transformers and the ear

How does one match a resonator or load in the case that the transmission line impedances are greater than the resistance of the resonator? The standard electrical engineering solution is to use a signal transformer, 2nd ref. This is a magnetic component, similar to an inductor, that employs two coils. It is similar to a power transformer used to change the AC voltage in the power grid. A signal transformer is usually much, much smaller than a power transformer and is used to change the impedance of an AC source. When used to match impedances it is placed at the site of the load and its turns ratio is set to change the impedance of an incident wave on a transmission line from the transmission line's impedance to the load's impedance. Signal transformers were widely used in the days of vacuum tubes to match the high output impedance of tubes to the low impedance of speakers in an old sound system. Today, because of their size and cost, most circuit designers shy away from them if possible. Still they are used at microwave frequencies for difficult matching problems.

Interestingly, the human ear (and the ears of most animals), while not resonant, uses a mechanical version of a transformer to match the impedance of air borne sound waves to the very high impedance of the fluid filled inner ear to overcome this problem and allow efficient transfer of the acoustic energy into the inner ear. This mechanical transformer is located in the middle ear and consists of three tiny bones acting as a lever connecting one diaphragm (the ear drum) to another diaphragm of a different size covering the entrance of the inner ear. The diaphragms and bones act similarly in function to the lever shown in Fig. 26 of an earlier posting.