# 6.5 Standing waves/vswr

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This module covers the idea of voltage standing wave ratio (VSWR).

In making this plot , we have made use of the fact that the propagation constant  can also be expressed as $\frac{2\pi }{}$ , and so for the independent variable, instead of showing $s$ in meters or whatever, we normalize the distance away from the load to the wavelengthof the excitation signal, and hence show distance in wavelengths. What we are showing here is called a standing wave . There are places along the line where the magnitude of the voltage $\left|V(s)\right|$ has a maximum value. This is where ${V}^{+}$ and ${V}^{-}$ are adding up in phase with one another, and places where there is a voltage minimum, where ${V}^{+}$ and ${V}^{-}$ add up out of phase. Since $\left|{V}^{-}\right|=\left|{}_{}\right|\left|{V}^{+}\right|$ , the maximum value of the standing wave pattern is $1+\left|{}_{}\right|$ times $\left|{V}^{+}\right|$ and the minimum is $1-\left|{}_{}\right|$ times $\left|{V}^{+}\right|$ . Note that anywhere on the line, the voltage is still oscillating at $e^{it}$ , and so it is not a constant, it is just that the magnitude of the oscillating signal changes as we move down the line. If we were to put an oscilloscope across theline, we would see an AC signal, oscillating at a frequency  .

A number of considerable interest is the ratio of the maximum voltage amplitude to the minimum voltage amplitude,called the voltage standing wave ratio , or VSWR for short. It is easy to see that:

$\mathrm{VSWR}=\frac{1+\left|\right|}{1-\left|\right|}$
Note that because $\left|{}_{}\right|\in \left[0 , 1\right]$ , $\mathrm{VSWR}\in \left\{1\right\}$ .

Although looks like the standing wave pattern is more or less sinusoidal, if we increase $\left|\right|$ to 0.8, we see that it most definitely is not. There is also a temptation to say that the spacing between minima (ormaxima) of the standing wave pattern is  , the wavelength of the signal, but a closer inspection of either or , shows that in fact the spacing between features is only half a wavelength, or $\frac{}{2}$ . Why is this? Well, $(s)$ goes as $-2s$ and $=\frac{2\pi }{}$ , and so every time $s$ increases by $\frac{}{2}$ , $(s)$ decreases by $2\pi$ and we have come one full cycle on the way $\left|V(s)\right|$ behaves.

Now let's go back to the Crank Diagram . At the position shown, we are at a voltage maximum, and $\frac{Z(s)}{{Z}_{0}}$ just equals the VSWR.
$\frac{Z({s}_{{V}_{\mathrm{max}}})}{{Z}_{0}}=\mathrm{VSWR}=\frac{1+\left|{}_{}\right|}{1-\left|{}_{}\right|}$
Note also that at this particular point, that the voltage and current phasors are in phase with one another (lined up in thesame direction) and hence the impedance must be real or resistive.

We can move further down the line, and now the $V(s)$ phasor starts shrinking, and the $I(s)$ phasor starts to get bigger .

If we move even further down the line, we get to a point wherethe current phasor is now at a maximum value, and the voltage phasor is at a minimum value . We are now at a voltage minimum, the impedance is again real (the voltageand current phasors are lined up with one another, so they must be in phase) and
$Z({s}_{{V}_{\mathrm{min}}})=\frac{1}{\mathrm{VSWR}}=\frac{1-\left|{}_{}\right|}{1+\left|{}_{}\right|}$
The only problem we have here is that except at a voltage minimum or maximum, finding $Z(s)$ from the crank diagram is not very straightforward, since the voltage and current are out of phase, and dividing thetwo vectors becomes somewhat tedious.

#### Questions & Answers

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Source:  OpenStax, Introduction to physical electronics. OpenStax CNX. Sep 17, 2007 Download for free at http://cnx.org/content/col10114/1.4
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