An Energy Analysis of a Simple Ideal Source, Part 1: Zero Average Interference

Introduction

The debate concerning energy events inside a source when reflected energy is incident upon that source is decades old and seemingly without resolution. [1][2] Part 1 of this series of articles will take a close look at what occurs inside a simple ideal source under special case conditions of zero average interference. Part 2 will address destructive interference at the source resistor and Part 3 will cover constructive interference. Perhaps the simple concepts presented in this elementary three-part series will help to "shed some light" on the subject. [3]

Please note that any power referred to in this paper is an AVERAGE POWER. Instantaneous power is beyond the scope of this article, irrelevant to the following discussion, and "of limited utility" according to Eugene Hecht. [4]

Zero Average Interference Example

Please note that it takes only one example to prove an explicit assertion to be false, e.g. "Reflected energy incident upon a source is always re-reflected back toward the load." The following example will prove that explicit assertion to be false.

Everything presented in this first article will be based on the following example chosen as a special case from the infinite number of possibilities. This example is so simple that almost everyone should be able to understand "where the power goes". More complex examples will be presented in Parts 2 and 3.

1. The magnitude of the transient forward power is equal to the magnitude of the steady-state forward power, i.e. the forward power, Pfor, is always equal to 50 watts no matter what the value of the load resistor.

2. After the forward wave is incident upon the load resistor, the reflected power, Pref, from the load is constant as long as the load resistance is unchanged. The voltage reflection coefficient at the load is: rho = (R_L-50)/(R_L+50) so Pref = 50w(rho²) watts, where rho² is the power reflection coefficient.

3. Steady-state is achieved at the time that the reflected energy wave reaches the source because R_S = Z0 = 50 ohms. Thus the reflected wave is not re-reflected at the source.

4. Since the transmission line is 1/8 wavelength (45 degrees) long and the load is purely resistive, the reflected wave incident upon the source resistor will be 2(45) = 90 degrees out of phase with the forward wave at the source resistor. This is the necessary and sufficient condition to produce zero average interference at the source resistor. Although instantaneous interference exists within a cycle, it averages out to zero over each complete cycle.

This last characteristic is interesting because it ensures that interference [4] will not exist anywhere in the example. Therefore, the subject of interference between two RF waves will be postponed until Parts 2 and 3. The complete absence of interference is what makes this simple example so interesting.

Where does the reflected energy go?

Remembering that this special case was chosen specifically to allow us to track the reflected energy through the system, let’s see where the reflected energy goes.

When R_L = 50 ohms, the system is matched and reflected power equals zero. In the matched case, an equal amount of power is dissipated in R_S and R_L, i.e. 50 watts in each resistor and maximum power transfer takes place. Recall that the forward power is 50 watts no matter what the value of the load resistor. This allows us to assert that 50 watts of the dissipation in the source resistor is due to the forward wave and is constant for the above example.

In this special case, as the reflected power is increased above zero by varying the load resistance away from the matched-case 50 ohms, the power dissipated in the source resistor, R_S, is exactly equal to the matched-case 50 watts plus the magnitude of the reflected power. Thus (for this special case) the following equation applies - the power dissipation in the source resistor, R_S, is:

P_Rs = 50w + Pref       Eq. 1

In the following table, random values of R_L are chosen to illustrate the validity of Eq. 1.

Please note that although we have not calculated a single voltage or current, any valid voltage and/or current analysis of the above example will yield identical results in agreement with this energy analysis. Eq. 1, above, will be valid for any voltage or current analysis performed on this particular special-case example.

For this special case, it is obvious that the reflected energy from the load is flowing through the source resistor, R_S, and is being dissipated there. But remember, we chose a special case (resistive R_L and 1/8 wavelength feedline) in order to make that statement true and it is usually not true in the general case. What we have proved false is the assertion that: "Reflected energy is never dissipated in the source." What we have NOT implied or proved is the equally false statement that: "Reflected energy is always dissipated in the source." In the next two parts of this article, we will cover the general cases for ideal class-A voltage (and current) sources. What we will discover is that: "The part of the reflected energy that is not dissipated in the source resistor is redistributed back toward the load as part of the forward wave." The part of the reflected energy that is redistributed back toward the load varies from 0% to 100%, depending upon system configuration.

Notes and References

[1] Bruene, Warren B., W5OLY, "On Measuring Rs", QEX, May 2002

[2] Maxwell, Walter, W2DU, http://www.w2du.com/BrueneRebuttal.pdf

[3] Since the "light" being "shed" comes from the field of optical physics, this seems to be an appropriate play on words.

[4] Hecht, Eugene, Optics, Fourth Edition, (c)Aug. 2001, Addison-Wesley, ISBN 0805385665, Chapter 9: "Interference". This reference lays the groundwork for Parts 2 and 3 of this article.