All postings by author  previous: 3.18 Separating out the radiated and simply reflected components  up: Contents  next: 3.20 Wave properties for transmission lines 
3.19  Applying methods of 3.18 to the circuit of 3.17
Summary: This posting completes the analysis of the two output channel circuit by applying the methods of the previous posting, i.e. 3.18 to the circuit in 3.17 .
Keywords: LRC resonator, radiated waves, transmitted waves, transmission line, absorbed power
Topics covered in this posting (basically the same as in the previous posting but applied to a different circuit)
 A review of the defining system of differential equations for a resonant LRC circuit as excited with waves on a transmission line.
 A look at the numerical solutions for the differential equations.
 Separation of the circuit into two parts, one appropriate for the radiated waves and one appropriate for the simply reflected waves.
 A look at the differential equations for both components and numerical solutions for each.
 A showing that the sum of the radiated and the simply reflected waves equals the total voltage for the initial circuit. This is done algebraically and graphically.
 A discussion of the concept of canceling waves as a method of absorbing power.
Contents of this posting
 The general setup
 Equivalent circuit, differential equations and numerical solution
 Separating out the wave components
 Obtaining the incident and the simply reflected components from V_{1}
 Showing that the sum of the components V_{1} and V_{2} equal the voltage V_{A} of the complete circuit.
 Power absorption via canceling radiation
1. The general setup 

Fig. 1. LRC resonator excited by waves on a transmission line with two output channels (reflection and transmission). The physical size of the components ( i.e. V, R_{0} , L_{C} , L, C and R ) is assumed to be much smaller than the length of the transmission lines and also much smaller than a wavelength of the waves traveling on the transmission lines. This posting is similar to the last posting, except that the transmission line circuit is a little different as shown in fig. 1 at the right (compare with fig. 1 of the last posting which is missing the output transmission line and the rightmost R_{0} shown at the right). This posting is the final mathematical analysis of the scattering circuit shown in previous animations in postings 3.15 and 3.17. Because this posting parallels the previous posting, it is rather minimalistic to avoid repeating a lot of the verbiage of the previous posting. The reader should examine the previous posting for a more detailed discussion of various analytic steps. 
2. Equivalent circuit, differential equations, and numerical solution  

Fig. 2 shows the equivalent circuit of the transmission line circuit of fig. 1 from the point of view of the upper end of the coupling inductor L_{C} , point A in Fig. 2. The defining differential equations of the circuit are: (2) voltage drop across coupling inductor (3) conservation of current at point A (4) voltage drop across source resistor The above equations can be combined to yield the following differential equations: (5) Equations (5)  (7) are interactive and must be solved as a set. Equation (8) uses the solution of eqns. (5)(7) to calculate V_{A} , which is the result we are most interested in. Below is the numerical solution of these equations in response to V_{} being a burst of a sinusoid at the resonant frequency. The resonant frequency was taken from 3.17 fig. 9.

3. Separating out the wave components  

As we did in the previous posting we separate the total voltage at point A into two components. These are:
For more details, see section 3 of the previous posting. Below are the partial circuits, modified from fig. 2 above that are appropriate for this discussion.

4. Obtaining the incident and the simply reflected components from V_{1}  

The math from section 4 of the previous posting (posting 3.18) was derived based on circuit details including point A and everything to the left of point A, i.e. the part of the circuit that is the same between the circuits for the previous posting and this posting. Thus we can simply use the results of section 4 of the previous posting here. The results of that work were: Below we graph the incident and simply reflected waveforms:

5. Showing that the sum of the components V_{1} and V_{2} equal the voltage V_{A} of the complete circuit.  

This section parallels the work in section 5 of the last posting 3.18. We start by substituting into (11) I_{s1} = I_{Lc1} + I_{Ro} = I_{Lc1} + V_{1}/R_{0} and differentiating the result to get: Equation (14) from above is: We subtract (18) from (17) to get: Since the last term in (19) equals V_{2}/L_{C} (19) becomes: Since V_{A} = V_{1} + V_{2} and I_{Lc} = I_{Lc1} − I_{Lc2} (20) becomes: which is the same as equation (2) above, one of the defining equations for the whole circuit. The other defining equations are much easier. Equation (1) defines the resonator's response and nothing to do with our breaking the circuit into pieces in section 3, since in section 3 we replaced the resonator with a unspecified voltage source but presumably governed by (1). Equations (3) and (4) are rather trivial and come from adding similar equations for the two partial circuits similar to what was done just above. This shows conclusively that the components are consistent with the defining equations of the original circuit. As further proof, in Fig. 6 we graphically compare numerical solutions of (V_{1} + V_{2} ) and V_{A} . One curve has been slightly shifted so that one curve doesn't totally cover the other one. They are identical. 
6. Power absorption via canceling radiation 

In the abstract sense, the setup in this posting (see Figs. 1 and 2) can be drawn as shown in Fig. 7 with waves coming in through one "channel" and leaving through two other "channels" (since it is straightforward to separate the positive and negative going waves of the transmission line supplying the power). The power that is absorbed internally by the resonator is the difference between that in the input channel and the output channels: where the power in the negative going channel is given by: and that in the transmitted channel is given by: Without the resonator excited and if the coupling inductor has reasonably large impedance, the simply reflected wave is fairly small and will carry off minimal power. Without the resonator excited, most power will go straight through to the transmitted channel. Suppose we wish to maximize power absorbed by the resonator. According to (22), this requires minimizing the power in the waves in the two output channels. When the resonator becomes excited, most of the negative going wave is due to that radiated by the resonator. So to reduce the power in the negative going output channel we need to reduce the resonator's radiation, which means coupling more weakly to the resonator. At the same time, to reduce the power in the transmitted channel, we need reasonable coupling so that the resonator radiates sufficiently to cancel the transmitted wave. Thus considering the backward direction, radiation should be minimized, while in the forward direction, it is essential to radiate a significant canceling wave. So to maximize power absorbed by the resonator we need to compromise and set the coupling and radiated power at a level that is not too large or too small. This is essentially what the graphs in fig. 11 of posting 3.17 tell us, that there is an optimal coupling to maximize the excitation of the resonator. This chain of logic also tells us that there is no coupling strength that results in all the incident power being absorbed by the resonator. That is, we need some radiated wave to cancel some of the transmitted wave. On the other hand, any radiated wave will result in increased loss of power in the backward direction. Indeed, fig. 11 of posting 3.17 shows us that the best we can do is 50% absorption of power by the resonator. The graphs in Fig. 8a and 8b below show the two wave components in the forward or transmitted output channel. Of particular note is Fig. 8b which is an expanded view of the phasing between the simply reflected and radiated wave components from Fig. 8a. We see that these two components are phased exactly 180degrees out of phase, a phasing that allows them to cancel to the maximum extent possible. Study the captions for more details of these graphs. CONCLUSION: Similar to what we saw in posting 3.18 we see above that the radiated wave is phased to cancel the simply transmitted wave to the maximum extent possible and maximize the power available to the resonator. Although we didn't show this in our graphing, the perfect cancelling phasing only occurs when the source frequency matches the resonator's frequency (with coupling inductor attached). The radiated wave is a canceling wave. 
All postings by author  previous: 3.18 Separating out the radiated and simply reflected components  up: Contents  next: 3.20 Wave properties for transmission lines 