In making
this plot,
we have made use of the fact that the propagation constant
ββ can also be expressed as
2πλ
2
λ
, and so for the independent variable, instead of
showing
ss in meters or whatever,
we normalize the distance away from the load to the wavelength
of 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
|Vs|
V
s
has a maximum value. This is where
V
+
V
+
and
V
-
V
-
are adding up in phase with one another, and places
where there is a voltage minimum, where
V
+
V
+
and
V
-
V
-
add up out of phase. Since
|
V
-
|=|
Γ
ν
||
V
+
|
V
-
Γ
ν
V
+
, the maximum value of the standing wave pattern is
1+|
Γ
ν
|
1
Γ
ν
times
|
V
+
|
V
+
and the minimum is
1−|
Γ
ν
|
1
Γ
ν
times
|
V
+
|
V
+
. Note that anywhere on the line, the voltage is
still oscillating at
eiωt
ω
t
, 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 the
line, we would see an AC signal, oscillating at a frequency
ωω.
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:
VSWR=1+|Γ|1−|Γ|
VSWR
1
Γ
1
Γ
(1)
Note that because
|
Γ
ν
|∈
0
1
Γ
ν
0
1
,
VSWR∈
1
∞
VSWR
1
.
Although Figure 1 looks like the
standing wave pattern is more or less sinusoidal, if we increase
|Γ|
Γ
to 0.8, we see that it most definitely is not. There
is also a temptation to say that the spacing between minima (or
maxima) of the standing wave pattern is
λλ , the wavelength of the
signal, but a closer inspection of either Figure 1
or Figure 2, shows that in fact the spacing between
features is only half a wavelength, or
λ2
λ
2
. Why is this? Well,
φs
φ
s
goes as
-2βs
-2
β
s
and
β=2πλ
β
2
λ
, and so every time ss
increases by
λ2
λ
2
,
φs
φ
s
decreases by
2π
2
and we have come one full cycle on the way
|Vs|
V
s
behaves.
Now let's go back to the
Crank Diagram. At the position shown,
we are at a voltage maximum, and
Zs
Z
0
Z
s
Z
0
just equals the VSWR.
Z
s
V
max
Z
0
=VSWR=1+|
Γ
ν
|1−|
Γ
ν
|
Z
s
V
max
Z
0
VSWR
1
Γ
ν
1
Γ
ν
(2)
Note also that at this particular point, that the voltage and
current phasors are in phase with one another (lined up in the
same direction) and hence the impedance must be
real or resistive.
We can move further down the line, and now the
Vs
V
s
phasor starts shrinking, and the
Is
I
s
phasor starts to get bigger Figure 3.
If we move even further down the line, we get to a point where
the current phasor is now at a maximum value, and the voltage
phasor is at a minimum value
Figure 4. We are now
at a voltage minimum, the impedance is again real (the voltage
and current phasors are lined up with one another, so they must
be in phase) and
Z
s
V
min
=1VSWR=1−|
Γ
ν
|1+|
Γ
ν
|
Z
s
V
min
1
VSWR
1
Γ
ν
1
Γ
ν
(3)
The only problem we have here is that except at a voltage
minimum or maximum, finding
Zs
Z
s
from the crank diagram is not very straightforward,
since the voltage and current are out of phase, and dividing the
two vectors becomes somewhat tedious.
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