The forward wave complex voltage amplitude is described by the complex algebraic term V+. The modulus, or size, of this amplitude is written |V+|.
For sinusoidal signals, the rms voltage in the forward wave is |V+| (1/sqrt[2]). It conveys power in the positive x direction, and the power flow is |V+||V+|/2Zo watts.
Similarly, the power flow in the negative x direction, that is from load to generator, is |V-||V-|/2Zo watts. The ratio V-/V+ is the "complex reflection coefficient," gamma. The ratio of return power to forward power is therefore the (modulus of gamma) squared.
The "return loss" is the number of dB by which the reflected power is lower than the forward power. A return loss of 3 dB means that half the power is reflected, and half absorbed in the load. A return loss of 20 dB means that only 1% of power is returned, and 99% is absorbed in the load.
Remember, power has to go somewhere. If it isn't absorbed it must be reflected. The difference between forward and reflected power flow is the power delivered to the load.
We have the following: total line voltage V = V+ + V-, is the total voltage on the line and it is the sum of the forward and backward wave complex voltage amplitudes, taking account of any phase shifts between the forward and backward waves. This is why the amplitudes have to be expressed as COMPLEX amplitudes, and we have to add up two complex numbers. In the Argand diagram this is done vectorially; you can think of the forward and backward wave voltage amplitudes as being phasors.
Similarly, the total line current I = I+ + I- is the sum of the complex amplitudes of forward and backward wave current.
We also have, as discussed in the lectures, for the forward wave, V+ = Zo I+ but for the backward wave, since current is measured against the wave propagation direction, V- = -Zo I-.
A little algebra shows V+ = (V+ + V-)/2 + (V+ - V-)/2
This process is known as "matrixing" and it allows us to write V+ in terms of the total voltage and current on the line. You can quickly see by substitution that V+ = (V+ZoI)/2.
Similarly you can show that V- = (V-ZoI)/2.
We can now determine the power delivered to the load. From the results above, the power into the load is (|V+||V+| - |V-||V-|)/(2Zo), ie the difference between forward and backward wave powers. Since there may be phase shifts between the current and voltage, one has to take these into account in forming the complex amplitudes V+ZoI and V-ZoI.
Exercise for the reader. For a load complex impedance ZL = R + jX, which sets the voltage and current on the line as V = ZL I , find the power delivered to the load impedance and show that it is equal to |I|.|I|.R/2 as expected.
The load impedance is 75 + j60 and Zo is 50, so (ZL+Zo) is 125+j60 and (ZL-Zo) is 25+j60. So the reflection coefficient gamma = (ZL-Zo)/(ZL+Zo) works out at 0.3498 + j0.3121 to 4 significant figures . The modulus squared of gamma is then 0.2198 and so 21.98% of the forward wave power is reflected. [Note, these calculations are simple with a calculator which allows complex number operations. I get on very well with my HP 32S which is a standard rpn calculator, particularly well-adapted to the complex number multiplications and divisions, and conversions between polar and rectangular forms.]
The forward wave power flow is 25*25/50 = 12.5 watts, so the reflected power is 0.2198 times 12.5 or 2.747 watts. The difference appears in the 75 ohms of the load. Thus there is 9.753 watts into the 75 ohms which represents a total load rms current of 0.3606 amps. The modulus of gamma is sqrt(0.2198) or 0.4688 and so the voltage rms in the reflected wave is 25 times 0.4688 or 11.72 volts.
The peak voltage in the forward wave is sqrt(2) times 25 or 35.36 volts; the peak voltage in the backward wave is sqrt(2) times 11.72 or 16.57 volts. The sum of these is 51.93 volts which is the largest voltage between the conductors anywhere along the line. The difference is 18.79 volts and the VSWR is 51.93/18.79 = 2.764.
The size of the rms current in the reflected wave is (rms voltage)/(line impedance), or 11.72/50 = 0.2344 amps. If the generator is connected at a VSWR maximum, the voltages add in phase but the currents add "out of phase." Therefore they subtract. The total rms current in the generator source internal impedance (50 ohms) is therefore 0.5 - 0.2344 amps. We note the forward wave current is 0.5 amps rms, 25 volts divided by 50 ohms characteristic impedance. The source impedance current of 0.2656 amps is squared and multiplied by the 50 ohms source impedance to give 3.527 watts of power dissipated in the generator's internal resistance.
The power in the load was 9.753 watts, so the overall transmission efficiency is (9.753)/(9.753+3.527) or 73.44%
If we had connected the generator at a VSWR minimum the currents would have added and the voltages subtracted. The total current in the 50 ohms internal generator impedance would have been 0.7344 amps, and that would be dissipating 26.97 watts in this impedance. The efficiency would have dropped to (9.753)/(9.753+26.97) or 26.56%.
The moral of this tale is that the length of cable connecting the generator to the load makes a critical engineering difference to the efficiency of power transmission, in cases where there is a mismatch at the load end of the cable.
We shall see later on how to match a load impedance to the characteristic impedance of a transmission line by adding lossless tuning stubs. However, this will not necessarily result in the highest efficiency of power transmission from generator to load; such efficiency would only be 50% for a matched load of 50 ohms in our example, and we have seen that we can do better than that.