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3.17 - Circuit analysis of resonant scattering with two output channels
Topics covered in this posting
- This posting deals with a resonator interacting with waves on a transmission line where the resonator hangs off the side of an otherwise continuous uniform transmission line.
- The resonator can be thought of as a scatterer of the waves, reflecting some, absorbing some and allowing some to continue propagating down the transmission line.
- Various aspects of this system are calculated and graphed.
- The system was simulated using SPICE.
- Using these results we draw some conclusions about three dimensional resonant scattering situations.
Contents of this posting
- Reflection and transmission coefficients
- Graph of reflection, transmission and resonant voltages
- SPICE simulation
- Unloaded, loaded and coupling Q's
- Resonant frequency variation
- Power in various wave components
- Another way to calculate β and Q's
- Energy in the resonator
- Various ways to view scattering
- Looking at impedance again
|Snapshot of animation in 3.15 (Fig. 36) to be analyzed in this section|
Fig. 1. This is a snap shot of Fig. 36 of 3.15 which animates resonant scattering of acoustical waves in a situation where there are two output channels. The waves travel from the source, on the left, to the resonator in the center. There they excite the resonator and are scattered both in the backward and forward directions (the two channels).
This animation makes various graphs of the waves. Those in the middle show the acoustical pressure versus position along the waveguide. These graphs change as the waves progress. Those at the bottom are like an oscilloscope picture graphing the acoustical pressures at particular points (the positions of microphones) as a function of time.
This posting analyzes an animation in posting 3.15 which shows the scattering of acoustical waves in a waveguide by a resonator attached in the middle of the waveguide. A snapshot of this animation is shown in Fig. 1 at the right. Previous postings (3.16 through 3.16g) analyzed a slightly different scattering setup shown in Fig. 35 of posting 3.15.
The animation and the analysis in this posting relate to one dimensional (1D) scattering in which both the incident waves and the scattered waves are free to propagate in the forward direction as shown in Fig. 38 of 3.15 . The scatter wave can also propagate in the negative direction as was the case in the previous discussion. The scattering setup for the previous posting is shown in Fig. 37 for comparison. Both types of 1D scattering are closely related to common 3D scattering as shown in Figs. 37 and 38. We analyze the 1D case partially because the math is considerably easier than the 3D cases and this analysis can help us by analogy to better understand the 3D cases. We also analyze the 1D case because it has a direct application in waveguide situations (see Fig. 24 of an earlier posting (3.12) for examples.
2. Reflection and transmission coefficients of the transmission line circuit
|Equivalent electrical circuit|
|Fig. 3. Equivalent circuit with electrical transmission lines (in red) replacing the acoustical ones in Figs. 1 and 38. An LRC resonator replaces the acoustical resonator and a coupling inductor replaces the coupling hole. The transmission lines are are not drawn to scale and generally would be much longer than the physical size of the other components.|
In Fig. 3 we show the electrical circuit which is equivalent to the acoustical circuits in Fig. 1. The acoustical waveguide is replaced by an electrical transmission line, the acoustical resonator by an LRC circuit and the coupling hole by a coupling inductor.
To compute the wave amplitude reflected by the resonator, that absorbed by the resonator, and that which passes on through to the right transmission line, we focus on the point (labeled "M") in the transmission line directly above the coupling inductor LC . The waves launched from this midpoint down the output transmission line are totally absorbed by the matched termination as if launched into a black hole. To the point M, the terminated waveguide acts as a resistor of value R0 which we show in Fig. 4a. In Fig. 4b we have combined the impedances of the resonator, coupling inductor and output transmission line into one load impedance ZL . Because of the resonator, this total impedance will be very frequency dependant.
|Evolution of equivalent circuits for the reflection problem|
|Fig. 4a. Equivalent circuit to Fig. 3 where we have replaced the output transmission line and terminating resistor with the equivalent resistor.||Fig. 4b. Equivalent circuit to Fig. 4a where we have replaced all the circuit elements on the right side of Fig. 4a with a total combined load impedance.|
where R0 is the characteristic impedance of the transmission line which we assume is real, i.e. a lossless line. The total load impedance ZL in Fig. 4b is the impedance of all the elements in the dotted box in Fig. 4a:
and Zres is the impedance of the resonator and is given by:
Equation (3) is graphed in Figs. 1 and 2 of posting 3.16b.
The amplitude of the waves reflected back towards the source (i.e. the negative going waves in the input transmission line - See Fig. 3) is given by:
V− = Γ Vinc . (4)
The amplitude of the waves launched down the output transmission line is given by:
Vout = T Vinc . (5)
The voltage amplitude Vres applied to the resonator is given by the voltage divider equation:
where we have also used (5) in the last step.
We loaded the above equations into gnuplot and produced the graph shown in Fig. 5 below.
We should point out that most of the work in this posting is a steady state analysis, good if a continuous stream of sinusoidal oscillations come from the wave source. This is not exactly the case protrayed in the animation where the exciting waves only last for a finite period of time, i.e. only give a burst of sinusoidal excitation. At the same time in the animation discussed here, towards the end of the burst, just before the source stops, the wave behavior should be approximately steady state. Fig. 6d below shows a SPICE simulation of the time dependent behavior of this circuit that accounts for the burst behavior accurately. Also, the analysis in a future posting 3.19 will further discuss the time dependent burst behavior of this circuit.
3. Graphs of reflected, transmitted, and resonator voltage amplitudes
|Graphs of Equations (1) and (6) above|
Fig. 5. Graphs of the magnitudes of the complex reflection coefficient, the transmission coefficient, and the voltage amplitude of the resonator, Vres (divided by the incident amplitude).
All three are plotted for five different values of the coupling inductor LC as labeled (all the transmission coefficient curves (in blue) and reflection coefficient curves (in green) are not labeled but are clear from the position of their peaks as lined up with those of the resonator voltage Vres - in red).
The values used for the other components are R = 100Ω, C = 0.08F, L = 2, and Z0 = R0 = 5Ω. These component values mean that ω0 = 2.5rad/sec and Q0 = 20.
Looking at the above graphs (Fig. 5), we see that:
- The smaller the coupling inductor, the broader the resonance curve is (which means stronger coupling and lower loaded Q).
- With the coupling inductor equal to 5H and at resonance (2.95rad/sec):
If we look at the other curves (for different values of LC we see that at the resonance of each curve lower values of coupling inductance result in:
- the reflection coefficient is at its peak of about 0.55 .
- The voltage amplitude on the resonator reaches its maximum value (3.2 times the incident amplitude), and
- The transmission coefficient is at its minimum value of about 0.45 .
- the peak resonance frequency increasing,
- the peak excitation amplitude of the resonator increasing and then decreasing,
- the peak transmission coefficient decreasing,
- the peak reflection coefficient increasing.
- The effect on the transmission and reflection coefficients can be attributed to the coupling inductor and resonator having a lower and lower impedance with decreasing LC values and effectively "shorting out" the transmission line.
- An optimal LC of approximately 5H results in the maximum resonator excitation. We shall see in section 5 (fig. 7c) that the optimal LC is actually 5.5H for unity coupling to the resonator, i.e. for β = 1.0 .
Checking out the rough analysis made in posting 3.15, section 3 we see that:
In item 1 of posting 3.15, we predicted that the voltage would not be appreciably increased at the resonator site because of a reflected wave. Looking at fig. 5 with LC = 5H, we see that the reflection coefficient is about 0.55 meaning that there is an appreciable reflected wave at unity coupling. If we calculate the reflection coefficient using eqn. (1) above, we see it is almost entirely real (very little imaginary component) so that the total wave presented to the coupling inductor is enhanced by about 55%.
On the other hand, if we consider only that part of the wave due to the source voltage (with the resonator shorted out) discussed in the next posting, we get a reflection coefficient of 15% (mostly imaginary) so that the total exciting voltage at the coupling inductor is actually decreased by 1% over the naked incident wave. For the first circuit, we get an enhancement of 1.91 . The ratio of the two cases is 0.52 so our assertion in posting 3.15 is correct.
- In item 2 we asserted that for a given Vres the radiated voltage will be approximately ½ that for the previously analyzed circuit (the circuit in which the right most transmission line and R0 was absent). A more careful analysis of the two circuits involved indicates that in the second case the radiated amplitude would be 52% of that in the first case.
- Putting the two together, 0.52×0.52 = 0.27 so our guess of one quarter is not bad.
4. Confirming our results with SPICE
As a way of confirming our results, we turn to SPICE, a standard electrical engineering simulation tool. Below we show the results of SPICE which confirm the graph in Fig. 5.
|Confirming results of SPICE simulation|
|Fig. 6a. As a check, the circuit above was simulated in a standard spice program (a standard electrical engineering tool for analyzing circuits). (a tutorial) The simulated circuit is shown at the left. Below we see the results of the SPICE simulation for the voltage across the resonator as a function of time. We used the same component and parameter values as given in the caption of Fig. 5 above.|
Fig. 6b. A frequency sweep of the circuit showing results similar to that shown in Fig. 5 for LC = 5H. The differences from Fig. 5's graph are that:
The magnitudes agree with those shown in Fig. 5.
Fig. 6c. This graph is basically the same as that in Fig. 6b except that here LC = 12H.
Fig. 6d below shows the time dependent behavior of this circuit to a burst of sinusoidal excitation by the voltage source. The sinusoidal source turns on at 0seconds and off at 48seconds. In green we see the response of the resonator, i.e. Vres. The resonator shows a building up of the oscillations then a decay. In dark blue we see the voltage on the total load (basically the transmitted voltage).
For this time dependent graph the frequency of the exciting source was set at 2.7radians/second (0.43Hz) and the coupling inductor LC at 12Henries.
5. Unloaded, loaded and coupling Q's
The unloaded Q is the Q when the resonator part of the circuit is disconnected from the rest of the circuit, as shown in Fig. 7a. The unloaded Q is calculated in Eqn. (32) of 3.16d:
The coupling Q and loaded Q can be calculated in a manner similar to that done for Eqn. (6) of 3.16a. Basically we use the formula QL = ω0U/P where P includes all the power dissipated during free decay when the excitation source is turned off:
where Q0 is as given by (7a) and the coupling Q is given by:
|Graph of equation (8a)|
|Fig. 7c. Log-log plot of the coupling coefficient β versus the coupling inductor LC in Henries. The drive frequency ω was adjusted as per equation (16) below for each LC value so as to be right at resonance. (As we shall see in the next section, the resonant frequency changes with LC .) The other constants in (8a) were chosen as per those in fig. 5 above.|
The last expression in (7b) is just a reaffirmation of the universal rule for Q addition (see Eqn. (124) of posting 3.11).
Solving for the coupling constant β:
We plot (8a), i.e. β versus LC at the right in fig. 7c for the other constants chosen as in the graph in fig. 5 above.
We can also solve (8a) for LC as a function of β:
6. Resonant frequency varying with the coupling
|Circuit for frequency calculation|
|Fig. 8. Modified version of Fig. 7b from above with the addition of an ideal AC current source to provide excitation without damping.|
The resonant frequency shifts from the unloaded case as we go to strong coupling (by decreasing the coupling inductance, LC ).
To calculate the resonant frequency we use the circuit shown in Fig. 7b above and imagine an ideal AC current source between ground and the top of the resonator as shown in Fig. 8 point "A". The voltage across the resonator will peak when the impedance is maximum, or the admittance is minimum (see ref for more on this.) The admittance to ground from point A is:
The real part of (10) only slowly varies, but the imaginary part changes drastically at resonance, going through zero right at resonance. Thus to find the resonant frequency, we take the imaginary part of (10), set it equal to zero and start to solve for ω :
To get rid of the denominators we multiply through by (ωL)[(½R0)2 + (ωLC)2] :
We multiply out the first term and collect terms of similar powers in ω to get:
where the constants a, b and c are given by:
|Resonant frequency shifts with coupling|
|Fig. 9. Angular resonant frequency versus coupling inductance LC as given by Eqn (16) ... in green. The five red points represent the frequencies of the peak values of the resonant curves in Fig. 5 above. We see that these points agree with the plotted curve of Eqn. (16).|
Equation (13) is a quadratic equation in ω2 and is easily solvable as:
We ignore the minus part of the ± sign (which creates a negative ω2 ) and take the square root of (15) again for the final answer:
7. Power in the various wave components
Because we are assuming our transmission lines to be lossless and thus their characteristic impedance real, we can calculate the power transported by the various wave components. Using the above equations, we write:
Looking at Fig. 10 we see that at resonance, the transmitted power reaches a minimum while the power absorbed by the resonator reaches a maximum. The reflected power reaches a maximum near the resonance. All three of these are as one might expect, that at resonance, the resonator plays its maximum role at absorbing and reflecting power from the transmitted beam.
We also see that for the choice of LC = 8H, at resonance the transmitted power equals the absorbed power, i.e. the power that is not reflected is split equally between the output transmission line and the resonator. Unexpectedly, the reflected power seems to peak at a frequency slightly larger than the resonant frequency as shown by the other two power peaks. The peak frequency of the transmitted and absorbed curves agrees with the center red dot of Fig. 9.
In Fig. 11 below, we repeat the graphs in Fig. 10 with three different values of LC . We see that larger LC values reduce the coupling to the transmission line and reduce the effect that the resonator has on the transmitted or reflected powers. Reducing the coupling also reduces the power absorbed by the resonator. Increasing the coupling has the reverse effect, by increasing the effect that the resonator has on the transmission line, increasing the reflected power and decreasing the transmitted power. Increasing the coupling also increases the power absorbed by the resonator.
|Repeat of Fig. 10 for various LC values|
|Fig. 11. The curves here are the transmitted power (in blue), power absorbed by the resonator (in red) and reflected power (in green). All powers have been divided by the incident power. Equations (17b), (17c) and (17d) were used to make the curves for various values of LC as indicated on the graph itself. The component values listed in the caption of Fig. 5 above were used in making these curves.|
8. Another way to calculate β and Q's
With the exception of section 4 (the SPICE simulation) the above work is directed towards the steady state response of the waveguide and resonator. This section will relate measurements made in burst mode to Q's, coupling coefficient β and stored energy, similar to that done in posting 3.13 near Eqns. (150)-(158) and (163). It allows one to use measurements made in the burst mode just before and just after the end of the burst while the resonator is still at its steady state energy level.
We start with conservation of energy or power for steady state:
where Pinc is the wave power leaving the source traveling towards the resonator, P− is the negative going wave power traveling from the resonator back towards the source and Ptrans is the (transmitted) positive going wave power traveling away from the resonator to the right of the resonator. This last wave is made up of the incident wave plus that radiated by the resonator in the positive direction. U is the energy stored in the resonator and ω is the frequency of the source which is assumed to be tuned to the resonant frequency of the resonator.
We write a similar equation for the coupling Q which accounts for the power radiated by the resonator just after the burst of the incident wave has finished (i.e. passed the resonator).
In this setup, waves are launched by the resonator in both forward and backward directions. In the backward direction these radiated waves are the only waves since the configuration offers no (or very small) reflecting obstacles in the center except for the resonator. In the forward direction the radiated waves are mixed with the incident waves, except for the time after the burst has passed. The "2" in (18b) accounts for the total radiated wave being split equally between the two directions.
We divide (18b) into (18a):
The middle term equals negative ½. The last term can be manipulated by breaking the transmitted wave into its components and realizing that a wave's power is proportional to amplitude squared as in (17a) above:
where V− = Vrad but 2Prad = 2P− . (Remember that the output current from the radiating resonator is split into two parts by the transmission lines going in two direction, while both left and right radiated voltages are equal and also equal that dumped on the transmission line by the resonator.) Also, right at resonance Vrad will be exactly 180degrees out of phase with Vinc and so will subtract from Vinc .
Substituting (18d) into (18c) we get:
|Normalized reflected power vs coupling strength|
Fig. 12. Graph of Equation (18g) showing the negative going wave power versus β for the two output channels case. This is similar to fig. 29 of posting 3.13 for the one output channel case.
This is similar to equation (157) of posting 3.13.
We can also solve (18e) for P−/Pinc in terms of β:
and to finish:
Equation (18g) is similar to (158) of posting 3.13. We graph (18g) on the right in fig. 12. From the graph we see that for weak coupling, the negative going wave is small. This makes sense because a weakly coupled resonator (β small) would be a minor presence in the waveguide and the waves would go through to the right waveguide unimpaired. For strong coupling (β large) the resonator becomes a major obstacle for the waves, resulting in large reflection and the negative going wave approaches the size of the incident wave (100% reflection in the limit as β approaches infinity).
|Energy in resonator vs coupling strength|
Fig. 13. Graph of Equation (19b) showing resonator energy versus β for the two output channel case. This graph is similar to fig. 31 of posting 3.13 for the one output channel case. In the making of this graph we assumed Pinc = 1W, Q0 = 20, and ω = 1rad/sec. We did not use the same parameters as used in the earlier graphs. We are interested in showing the general shape of the graph for Eqn. (19b).
9. Energy in the resonator
Similar to that done in posting 3.13, we can use:
to solve for U :
and calculate the energy in the resonator as a function of coupling strength similar to (163) of posting 3.13.
The graph in fig. 13 makes sense because it shows maximum resonator response for unity coupling. For small β there is little excitation available for the resonator and for large β the loaded Q is small meaning that energy is not effectively recycled and the energy is less. The peak energy is half that seen in fig. 31 of posting 3.13 because of the lack of the doubling of the excitation at the coupling aperture that occurs in fig. 31.
10. Various ways to view scattering
In most of the analyses of scattering in this posting and previous postings, we have focused on classical circuit analysis, used along with a little understanding of waves on transmission lines. While it is mathematically correct, this analysis tends to hide the dynamics of the waves involved. Below we list several ways to view the wave interactions occurring in these circuits. Each point of view is useful and adds to our understanding of the physics involved.
Ways to view the processes in our circuits:
- Waves interacting with various impedance elements along a transmission line. The impedance elements can be calculated using standard, non-wave based circuit analysis. This is the view that most of our analyses used.
- Waves scattering off physical elements, such as our resonator. In a limited sense this is the same as item 1 but the "scattering" concept opens up our mind to the idea of an incident wave component dynamically scattering off an obstacle. It uses various components of the wave field to help our understanding of the process. It also open us up to the analogies with two and three dimensional scattering situations.
- In the next posting we explore the idea of a canceling wave or radiated wave that is emitted by the scattering resonator and "steals" energy from the incident beam. In that posting we examine the math behind a "radiated wave", a concept that we have used before qualitatively (see Fig. 35 of posting 3.14 for example), but in the next posting we add the math.
11. Looking at impedance again
In the case where a wave is launched down a transmission line or waveguide towards a load (as is shown in Fig. 4b above) some of the wave will be reflected by the load and some will be absorbed. It is well known that to maximize the absorption, one sets the load impedance equal to the complex conjugate of the characteristic impedance of the transmission line or waveguide. We are usually interested in maximizing absorption by the resonator alone. But in the case of this posting, where the resonator is hanging off to the side of the transmission line, what is the rule for maximizing the power absorbed by the resonator? Above we have talked about maximizing power absorbed by the resonator, but here we shall analyze it from an impedance point of view.
We will first analyze the case shown below in Fig. 14a. The average power absorbed by the load resistor R in that circuit is given by:
where VA is the voltage amplitude at point A shown in Fig. 14a and VA is related to V0 by the voltage divider equation:
The incident power is given by Pinc = V02 / 4R0 .
In Fig. 14b below we graph the power absorbed (divided by the incident power) by the resistor R as a function of R/R0 . We see that at max we get half the incident power absorbed by R, which happens when the resistance of R is half the characteristic impedance of the transmission line, i.e. when R = ½R0 .
The next question that comes to mind is "is it realistic to assume that the resonator plus coupling induction will present a totally real impedance?" To answer this question we do an analysis of the circuit of Fig. 14a with R replaced by LC plus the resonator shown in Fig. 15a below.
The impedance of the resonator plus LC is given by:
We graph the real and imaginary parts of (21a) as a function of the angular frequency ω in Fig. 15b below.
|Equivalent circuit of the resonator plus coupling inductor attached to the side of a transmission line||Graph of the real and imaginary impedance of the resonator plus LC as a function of angular frequency|
Fig. 15a. Equivalent circuit of a transmission line with resonator plus LC hanging off the side of the transmission line. The characteristic impedance of the transmission line is R0 .
Fig. 15b. Graph of the real and imaginary parts of the impedance of the coupling inductor and resonator as a function of angular frequency. For component values we used those from Fig. 5 above plus LC = 5.5H as Fig. 7c indicates is appropriate for unity coupling for the other component values. Fig. 9 indicates that for these component values resonance will be achieved at an angular frequency of 2.91rad/sec.
The units of the vertical scale is ohms, and that of the horizontal scale is radians per second.
We see from this graph that at an angular frequency ω = 2.91rad/sec (the expected resonant frequency), that the imaginary component of the impedance is zero. At that same frequency, the real component is approximately 2.5 or 2.6Ω. We were expecting exactly half of R0 or 2.5Ω.
To further understand the above, we plot the impedance of the circuit to ground from the point between the resonator and LC with V0 set to zero. This is the impedance from the "point of view of the resonator". We plot this impedance versus frequency ω in Fig. 15c below. The equation used to generate this plot is:
where the resonator impedance is given by:
Fig. 15d shows the power absorbed by the resonator as a function of frequency. This power is divided by the incident power. We see at resonance, near 2.91rad/sec, half the incident power is absorbed by the resonator. While this result is very similar to Fig. 11 above, we include it to show the use of the equations developed in this section.
The equations used to make the graph of Fig. 15d are:
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