Let's move on to some other Smith Chart
applications. Suppose, somehow, we can obtain a plot of
Vs
V
s
on a line with some unknown load on it. The data might
look like Figure 1. What can we tell from this plot? Well,
Vmax=1.7
V
max
1.7
and
Vmin=0.3
V
min
0.3
which means
VSWR=1.70.3=5.667
VSWR
1.7
0.3
5.667
(1)
and hence
|Γ|=VSWR-1VSWR+1=4.6676.667=0.7
Γ
VSWR
1
VSWR
1
4.667
6.667
0.7
(2)
Since
|rs|=|Γ|
r
s
Γ
, we can plot
rs
r
s
on the Smith Chart, as shown
here. We do this by setting the compass at
a radius of 0.7 and drawing a circle! Now,
Z
L
Z
0
Z
L
Z
0
is
somewhere on this circle. We
just do not know where yet! There is more information to be
gleaned from the VSWR plot however.
Firstly, we note that the plot has a periodicity of about 10
cm. This means that λ the wavelength of the signal on the
line is 20 cm. Why? According to
this equation,
|Vs|
V
s
goes as
cosφs
φ
s
and
φs=
θ
Γ
-2βs
φ
s
θ
Γ
2
β
s
and
β=2πλ
β
2
λ
, thus
|Vs|
V
s
goes as
cos4πsλ
4
s
λ
. Thus each
λ2
λ
2
, we are back to where we started.
Secondly, we note that there is a voltage minima
at about 2.5 cm away from the load. Where on Figure 2 would we expect to find a voltage minima? It
would be where
rs
r
s
has a phase angle of
180
°
180
°
or point "A" shown in here. The voltage minima is
always where the VSWR circle passes through
the real axis on the left hand side. (Conversely a voltage
maxima is where the circle goes through the real axis on the
right hand side.) We don't really care about
Zs
Z
0
Z
s
Z
0
at a voltage minima, what we want is
Zs=0
Z
0
Z
s
0
Z
0
, the normalized load impedance. This should be easy! If
we start at "A" and go
2.520=0.125λ
2.5
20
0.125
λ
towards the load we should end up
at the point corresponding to
Z
L
Z
0
Z
L
Z
0
. The arrow on the mini-Smith Chart says "Wavelengths
towards generator" If we start at A, and want to go towards the
load, we had better go around the opposite
direction from the arrow. (Actually, as you can see on a
real Smith Chart, there are arrows pointing
in both directions, and they are appropriately marked for your
convenience.)
So we start at "A" go
0.125λ
0.125
λ
in a counter-clockwise direction, and mark a new point
"B" which represents our
Z
L
Z
0
Z
L
Z
0
which appears to be about
0.35-0.95ⅈ
0.35-0.95 or so
Figure 4. Thus, the load in this
case (assuming a
50Ω
50
Ω
line impedance) is a resistor, again by co-incidence of about
50Ω
50
Ω
, in series with a capacitor with a negative reactance
of about
47.5Ω
47.5
Ω
. Note that we could have started at the minima at 12.5
cm or even 22.5 cm, and then have rotated
12.520=0.625λ
12.5
20
0.625
λ
or
22.520=1.125λ
22.5
20
1.125
λ
towards the load. Since
λ2=0.5λ
λ
2
0.5
λ
means one complete rotation around the Smith Chart, we
would have ended up at the same spot, with the same
Z
L
Z
0
Z
L
Z
0
that we already have! We could also have started at a
maxima, at say 7.5 cm, marked our starting point on the right hand
side of the Smith chart, and then we would go
0.375λ
0.375
λ
counterclockwise and again, we'd end up at "B".
Now,
here is another example. In this
case the
VSWR=1.50.5=3
VSWR
1.5
0.5
3
, which means
|Γ|=0.5
Γ
0.5
and we get a circle as shown in
Figure 6.
The wavelength
λ=225-10=30cm
λ
2
25
10
30
cm
. The first minima is thus a distance of
1030=0.333λ
10
30
0.333
λ
from the load. So we again start at the minima, "A"
and now rotate as distance
0.333λ
0.333
λ
towards the load.
