We actually still have some options open to
us. One of thenicest, at least in terms of getting some insight,
is call a crank diagram. Note that this equation is a complex
equation, which requires us to take a the ratio of two complex
numbers;
1+
Γ
ν
e−(j2βs)
1
Γ
ν
2
β
s
and
1−
Γ
ν
e−(j2βs)
1
Γ
ν
2
β
s
.
Let's plot these two quantities on the complex
plane, starting at
s=0
s
0
(the load end of the line). We can represent
Γ
ν
Γ
ν
, the reflection coefficient, by its magnitude and its
phase,

Γ
ν

Γ
ν
and
φ
Γ
φ
Γ
. For the numerator we plot a 1, and then add the
complex vector ΓΓ which has a
length
Γ
Γ
and sits at an angle
φ
Γ
φ
Γ
with respect to the real axis Figure 1.
The denominator is just the same thing, except the
ΓΓ vector points in the
opposite direction Figure 2.
The top vector is proportional to
Vs
V
s
and the bottom vector is proportional to
Is
I
s
Figure 3. Of course, for
s=0
s
0
we are at the load so
Vs=0=
V
L
V
s
0
V
L
and
Is=0=
I
L
I
s
0
I
L
.
As we move down the line, the two
"
ΓΓ" vectors rotate around at
a rate of
2βs
2
β
s
Figure 4. As they rotate, one vector gets
longer and the other gets shorter, and then the opposite
occurs. In any event, to get
Zs
Z
s
we have to divide the first vector by the second. In
general, this is not easy to do, but there are
some places where it is not too bad. One of
these is when
2βs=−
θ
Γ
2
β
s
θ
Γ
Figure 5.
At this point, the voltage vector has rotated around so that it
is just lying on the real axis. Obviously its length is now
1+Γ
1
Γ
. By the same token, the current vector is also lying
on the real axis, and has a length
1−Γ
1
Γ
. Dividing one by the other, and multiplying by
Z
0
Z
0
gives us
Zs
Z
s
at this point.
Zs=
Z
0
1+
Γ
ν
1−
Γ
ν

Z
s
Z
0
1
Γ
ν
1
Γ
ν
(1)
Where is this point, and does it have any special meaning? For
this, we need to go back to our expression for
Vs
V
s
in
this
equation.
Vs=
V
+
ejβs(1+
Γ
ν
e2jβs)=
V
+
ejβs(1+
Γ
ν
ej(
θ
Γ
−2βs))=
V
+
ejβs(1+
Γ
ν
ejφs)
V
s
V
+
β
s
1
Γ
ν
2
β
s
V
+
β
s
1
Γ
ν
θ
Γ
2
β
s
V
+
β
s
1
Γ
ν
φ
s
(2)
where we have substituted

Γ
ν
ejθ
Γ
ν
θ
for the phasor
Γ
ν
Γ
ν
and then defined a new angle
φs=
θ
Γ
−2βs
φ
s
θ
Γ
2
β
s
.
Now let's find the magnitude of
Vs
V
s
. To do this we need to square the real and imaginary
parts, add them, and then take the square root.
Vs=
V
+
(1+
Γ
ν
ejφs)=
V
+
1+
Γ
ν
cosφs2+
Γ
ν
2sin2φs
V
s
V
+
1
Γ
ν
φ
s
V
+
1
Γ
ν
φ
s
2
Γ
ν
2
φ
s
2
(3)so,
Vs=
V
+
1+2
Γ
ν
cosφs+
Γ
ν
2cos2φs+
Γ
ν
2sin2φs
V
s
V
+
1
2
Γ
ν
φ
s
Γ
ν
2
φ
s
2
Γ
ν
2
φ
s
2
(4)which, since
sin2·+cos2·=1
·
2
·
2
1
Vs=
V
+
1+
Γ
ν
2+2
Γ
ν
cosφs
V
s
V
+
1
Γ
ν
2
2
Γ
ν
φ
s
(5)
Remember,
φs
φ
s
is an angle which changes with
ss. In particular,
φs=
θ
Γ
−2βs
φ
s
θ
Γ
2
β
s
. Thus, as we move down the line
Vs
V
s
will oscillate as
cosφs
φ
s
oscillates. A typical plot for
Vs
V
s
(for

Γ
ν
=0.5
Γ
ν
0.5
and
θ
Γ
=
45
°
θ
Γ
45
°
) is shown
here.
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