A transmission line in TEM mode can carry waves that comprise both E field and H field in two directions. The geometry of the line, its distributed self capacitance and self inductance, and to a lesser extent usually, series resistance and shunt conductance constrain the ratio of E to H for a traveling wave. The E and H fields have a related equivalent V and I, and the ratio V/I is known as the Characteristic Impedance, Z0.
So for example, lets take a line with Z0=50+j0 (yes, it is lossless for the simple mental arithmetic model that follows) connected to a load of 150+j0, and lets say that the load voltage is 150V.
Load current is found using ohms law, IL=150/150=1A.
But, the line only supports waves with E and H fields in the ratio of 50, or wave V/I ratio equal to 50, so how does it propagate the energy that is transferred to the load in this case.
To obtain the resultant E/H or V/I at the line end, there must be two traveling waves interacting, we will call them the ‘forward’ wave (traveling towards the load) and a ‘reflected’ wave traveling towards the source.
The resolution of the physical constraints that each wave V/I equals 50 and load V/I equals 150 is:
When you sum voltage and current components of these two waves, you get VL=100+50=150V, and IL=2+-1=1A which reconciles with the load V and I.
So now we have two waves traveling in opposite directions. Their phase relationship changes along a long line, and at some points, Vfwd is in phase with Vref, so they reinforce and the maximum voltage is 100+50=150V, and at other points they oppose and the minimum voltage is 100-50=50V. The ratio of this maximum to minimum voltage is known as the VSWR and in this example is 3, often written as 3:1.
The ratio of V/I (or Z) varies along the line. In the general case, they are complex quantities, ie they have magnitude and phase.
But lets choose a line that is an electrical quarter wave in length. At the line input, the relationship of the forward and reflected waves is such that V=50V and I=3A, and they are in phase, so Z=(50<theta)/(3<theta)=16.7+j0.
You could connect this to a transmitter that requires a 50+j0 load with a broadband 3:1 transformer, or you could use an ATU to transform the impedance up to 50+j0, albeit with a little loss (<0.5dB). The transmitter will deliver its rated power into its load, a little is lost in the transformer or ATU, and the rest is carried by the lossless line to the load (antenna).
In the real world, the line is not lossless, loads are not nice round numbers to manipulate in your head, but the same principles apply and the problems are solved using the same techniques. The maths might not look pretty, but you can use tools like RF Transmission Line Loss Calculator (TLLC) to perform the calcs.
The real world solution of a quarter wave of RG-58C/U at 7MHz with a
150+j0 load from TLLC is given below. Note the input impedance is not
exactly 16.7+j0 because Z0 is not exactly 50+j0, and the
line is lossy. VSWR is likewise affected. Line loss and efficiency are
calculated under the specific mismatch conditions.
The key thing is that it is possible to explain what happens, including the line loss in the practical line case, without re-re-reflection stuff, no conjugate matching, no conjugate mirrors, no photons, no irradiance, no Mismatch Loss, no Additional Loss due to VSWR, just conventional linear circuit theory and solution of the Telegrapher’s Equation. The power wasn’t stated explicitly, but the efficiency figure reveals that the power at the line output was 91% of that at the line input.
Transmission Line Belden 8262 (RG-58C/U) Code B8262 Data source Belden Frequency 7.000 MHz Length 90.000 deg Zload 150.00+j0.00 Ω Yload 0.006667+j0.000000 S Results
Zo 50.01-j0.84 Ω Velocity Factor, VF -2 0.660, 2.293 Length 90.00 °, 0.250 λ, 7.070 m Line Loss (matched) 0.258 dB Line Loss 0.415 dB Efficiency 90.88 % Zin 17.98-j0.58 Ω Yin 0.055572+j0.001780 S VSWR(50)in 2.78 Γ, ρ<θ, RL, VSWR, MismatchLoss (source end) -4.710e-1-j5.931e-3, 0.471<-179.3°, 6.5 dB, 2.78, 1.09 dB Γ, ρ<θ, RL, VSWR, MismatchLoss (load end) 4.999e-1+j6.295e-3, 0.500<0.7°, 6.0 dB, 3.00, 1.25 dB S11, S21 3.086e-4-j1.630e-2, 5.027e-6-j9.708e-1 Y11, Y21 5.942e-4+j9.973e-6, -3.354e-4+j1.998e-2 NEC NT NT t s t s 5.942e-4 9.973e-6 -3.354e-4 1.998e-2 5.942e-4 9.973e-6 ‘B8262, 90.000 deg, 7.000 MHz k1, k2 1.023e-6, 2.320e-11 C1, C2 4.119e-1, 2.953e-1 γ 3.304e-4+j1.745e-2 Loss model source data frequency range 1.000 MHz – 1000.000 MHz Correlation coefficient (r) 0.999884