Editing Standing wave ratio (section)

==The standing wave pattern==
Using [[phasor|complex]] notation for the voltage amplitudes, for a signal at frequency {{mvar|f}}, the actual (real) voltages V{{sub|actual}} as a function of time {{mvar|t}} are understood to relate to the complex voltages according to:

:<math> V_\mathsf{actual} = \mathcal{R_e} (e^{i 2 \pi f t} V) ~.</math> 

Thus taking the real part of the complex quantity inside the parenthesis, the actual voltage consists of a [[sine wave]] at frequency {{mvar|f}} with a peak amplitude equal to the complex magnitude of {{mvar|V}}, and with a phase given by the phase of the complex {{mvar|V}}. Then with the position along a transmission line given by {{mvar|x}}, with the line ending in a load located at {{mvar|x}}{{sub|o}}, the complex amplitudes of the forward and reverse waves would be written as:

:<math>\begin{align}
  V_\mathsf{fwd}(x) &= e^{-i k(x-x_\mathsf{o})} A \\
  V_\mathsf{rev}(x) &= \Gamma e^{i k(x-x_\mathsf{o})} A
\end{align}</math>

for some complex amplitude {{mvar|A}} (corresponding to the forward wave at {{mvar|x}}{{sub|o}} that some treatments use phasors where the time dependence is according to <math>e^{ -i 2 \pi f t }</math> and spatial dependence (for a wave in the {{mvar|+x}} direction) of <math>\ e^{+i k(x - x_\mathsf{o})} ~.</math> Either convention obtains the same result for {{mvar|V}}{{sub|actual}}.

According to the [[superposition principle]] the net voltage present at any point {{mvar|x}} on the transmission line is equal to the sum of the voltages due to the forward and reflected waves:

:<math>\begin{align}
  V_\mathsf{net}(x) &= V_\mathsf{fwd}(x) + V_\mathsf{rev}(x) \\
                  &= e^{-i k(x - x_\mathsf{o})} \left( 1 + \Gamma e^{i 2k(x - x_\mathsf{o})}\right ) A
\end{align}</math>

Since we are interested in the variations of the ''magnitude'' of {{mvar|V}}{{sub|net}} along the line (as a function of {{mvar|x}}), we shall solve instead for the squared magnitude of that quantity, which simplifies the mathematics. To obtain the squared magnitude we multiply the above quantity by its complex conjugate:

:<math>\begin{align}
  |V_\mathsf{net}(x)|^2 &= V_\mathsf{net}(x)  V^*_\mathsf{net}(x) \\
                 &= e^{-i k \left(x - x_\mathsf{o}\right)} \left(1 + \Gamma e^{i 2 k \left(x - x_\mathsf{o}\right)}\right) A \,  e^{+i k\left(x - x_\mathsf{o}\right)} \left(1 + \Gamma^* e^{-i 2 k \left(x - x_\mathsf{o}\right)}\right) A^* \\
                 &= \left[\ 1 + |\Gamma|^2 + 2\ \operatorname\mathcal{R_e} \left(\Gamma  e^{i 2 k \left(x - x_\mathsf{o}\right)} \right)\ \right] |A|^2
\end{align}</math>

Depending on the phase of the third term, the maximum and minimum values of {{mvar|V}}{{sub|net}} (the square root of the quantity in the equations) are <math>\ \left(1 + |\Gamma|\right) |A|\ </math> and <math>\ \left(1 - |\Gamma|\right) |A|\ ,</math> respectively, for a standing wave ratio of:


:<math> \boldsymbol\mathsf{SWR} =  \frac{|V_\mathsf{max}|}{|V_\mathsf{min}|} = \frac{1 + |\Gamma|}{1 - |\Gamma|}</math>


:<math> \left|\Gamma\right| = \frac{\ \boldsymbol\mathsf{SWR} - 1\ }{\boldsymbol\mathsf{SWR} + 1}</math>


as earlier asserted. Along the line, the above expression for <math>\ |V_\mathsf{net}(x)|^2\ </math> is seen to oscillate sinusoidally between <math>\ |V_\mathsf{min}|^2\ </math> and <math>\ |V_\mathsf{max}|^2\ </math> with a period of {{sfrac| 2{{mvar|π}} |2{{mvar|k}} }}&nbsp;. This is ''half'' of the guided wavelength {{nobr|{{mvar|λ}} {{=}} {{sfrac| 2{{mvar|π}} | {{mvar|k}} }} }} for the frequency {{mvar|f}}&nbsp;. That can be seen as due to interference between two waves of that frequency which are travelling in ''opposite'' directions.

For example, at a frequency {{nobr| {{mvar|f}} {{=}} 20 MHz }} (free space wavelength of 15&nbsp;m) in a transmission line whose [[velocity factor]] is 0.67&nbsp;, the guided wavelength (distance between voltage peaks of the forward wave alone) would be {{nobr| {{mvar|λ}} {{=}} 10 m .}} At instances when the forward wave at {{nobr| {{mvar|x}} {{=}} 0 }} is at zero phase (peak voltage) then at {{nobr| {{mvar|x}} {{=}} 10 m }} it would also be at zero phase, but at {{nobr| {{mvar|x}} {{=}} 5 m }} it would be at 180° phase (peak ''negative'' voltage). On the other hand, the magnitude of the voltage due to a standing wave produced by its addition to a reflected wave, would have a wavelength between peaks of only {{nobr| {{sfrac|1|2}}{{mvar|λ}} {{=}} 5 m .}} Depending on the location of the load and phase of reflection, there might be a peak in the magnitude of {{mvar|V}}{{sub|net}} at {{nobr| {{mvar|x}} {{=}} 1.3 m .}} Then there would be another peak found where {{nobr| {{!}}{{mvar|V}}{{sub|net}}{{!}} {{=}} {{mvar|V}}{{sub|max}} }} at {{nobr| {{mvar|x}} {{=}} 6.3 m ,}} whereas it would find minima of the standing wave {{nobr|  |{{mvar|V}}<sub>net</sub>| {{=}} {{mvar|V}}<sub>min</sub> }} at {{nobr| {{mvar|x}} {{=}} 3.8 m,}} 8.8&nbsp;m, etc.