So, what do we do for
Z
L
Z
L
? A quick glance at a transmission line problem shows that at
the load we have a resistor and an inductor in parallel. This
was done on purpose, to show you one of the powerful aspects of
the Smith Chart. Based on what you know from circuit theory you
would calculate the load impedance by using the formula for two
impedances in parallel
Z
L
=iωLRiωL+R
Z
L
ω
L
R
ω
L
R
which will be somewhat messy to calculate.

Let's remember the formula for what the Smith
Chart represents in terms of the phasor
rs
r
s
.

Z
L
Z
0
=1+rs1−rs
Z
L
Z
0
1
r
s
1
r
s

(1)Let's invert this expression

Z
L
Z
0
=1
Z
L
1
Z
0
=
Y
L
Y
0
=1−rs1+rs
Z
L
Z
0
1
Z
L
1
Z
0
Y
L
Y
0
1
r
s
1
r
s

(2) Equation 3 says that is we want to get an

admittance instead of an impedance, all we have to
do is substitute

−rs
r
s
for

rs
r
s
on the Smith Chart plane!

Y
0
=1
Z
0
=150=0.02
Y
0
1
Z
0
1
50
0.02

(3)in our case. We have two elements in parallel for
the load (

Y
L
=Y+iB
Y
L
Y
B
), so we can easily add their admittances, normalize them to

Y
0
Y
0
, put them on the Smith Chart, go

180
°
180
°
around (same thing as letting

−rs=rs
r
s
r
s
) and read off

Z
L
Z
0
Z
L
Z
0
. For a

200Ω
200
Ω
resistor,

GG, the
condunctance equals

1200=0.005
1
200
0.005
.

Y
0
=0.02
Y
0
0.02
so

G
Y
0
=0.25
G
Y
0
0.25
. The generator is operating at a frequency of

200MHz
200
MHz
, so

ω=2πf=1.25×109s-1
ω
2
f
1.25
10
9
s
-1
and the inductor has a value of

160nH
160
nH
, so

iωL=200i
ω
L
200
and

B=1iωL=-0.005i
B
1
ω
L
-0.005
and

B
Y
0
=-0.25i
B
Y
0
-0.25
.

We plot this on the Smith
Chart by first finding the real part = 0.25 circle, and
then we go down onto the lower half of the chart since that is
where all the negative reactive parts are, and we find the curve
which represents
-0.25i
-0.25
and where they intersect, we put a dot, and mark the
location as
Y
L
Y
0
Y
L
Y
0
. Now to find
Z
L
Z
0
Z
L
Z
0
, we simply reflect half way around to the opposite
side of the chart, which happens to be about
Y
L
Y
0
=2+2i
Y
L
Y
0
22
, and we mark that as well. Note that we can take the
length of the line from the center of the Smith Chart to our
Z
L
Z
0
Z
L
Z
0
and move it down to the
|Γ|
Γ
scale and find that the reflection coefficient has a
magnitude of about 0.6. On a real Smith Chart, there is also a
phase angle scale on the outside of the circle (where our
distance scale is) which you can use to read off the phase angle
of the reflection coefficient as well. Putting that scale on
the "mini Smith Chart" would clog things up too much, but the
phase angle of ΓΓ is about
3.0
°
3.0
°
.

Now the wavelength of the signal on the line is
given as

λ=
ν
p
f=2.8×108200×106=1m
λ
ν
p
f
2.8
10
8
200
10
6
1
m

(4)
The input to the line is located

21.5cm
21.5
cm
or

0.215λ
0.215
λ
away from the load. Thus, we start at

Z
L
Z
0
Z
L
Z
0
, and rotate around on a circle of constant radius a
distance

0.215λ
0.215
λ
towards the generator. To do this, we extend a line out from
our

Z
L
Z
0
Z
L
Z
0
point to the scale and read a relative distance of

0.208λ
0.208
λ
. We add

0.215λ
0.215
λ
to this, and get

0.423λ
0.423
λ
Thus, if we rotate around the Smith Chart, on our
circle of constant radius Since, after all, all we are doing is
following

rs
r
s
as it rotates around from the load to the input to the line.
When we get to

0.423λ
0.423
λ
, we stop, draw a line out from the center, and where
it intercepts the circle, we read off

Z
L
Z
0
Z
L
Z
0
from the grid lines on the

Smith
Chart. We find that

Z
in
Z
0
=0.3+-0.5i
Z
in
Z
0
0.3-0.5

(5)
Thus,

Z
in
=15+-25i
Z
in
15-25
ohms

Figure 3. Or, the impedance at the
input to the line looks like a

15Ω
15
Ω
resistor in series with a capacitor whose reactance

iX=-25i
X
-25
, or, since

X
cap
=1iωC
X
cap
1
ω
C
, we find that,

C=12π200×200×106=31.8pF
C
1
2
200
200
10
6
31.8
pF

(6)To find

V
in
V
in
, there is no avoiding doing some complex math:

V
in
=15+-25i50+15+-25i10
V
in
15-25
50
15-25
10

(7)
Which, we write in polar notation, divide, figure the voltage
and then return to rectangular notation.

V
in
=29.1∠5969.6∠-2110
V
in
29.159
69.6-21
10

(8)
V
in
=0.418∠-38×10=4.18∠-38=3.30+-2.58i
V
in
0.418-38
10
4.18-38
3.30-2.58

(9)
If at this point we needed to find the actual voltage phasor

V
+
V
+
we would have to use the equation

V
in
=
V
+
eiβL+Γ
V
+
e−(iβL)=
V
+
eiβL+|Γ|
V
+
ei(
θ
r
−βL)
V
in
V
+
β
L
Γ
V
+
β
L
V
+
β
L
Γ
V
+
θ
r
β
L

(10)Where

β=2πλ
β
2
λ
is the propagation constant for the line as mentioned
in the

last
chapter, and

LL is the
length of the line.

For this example,
βL=2πλ0.215λ=1.35radians
β
L
2
λ
0.215
λ
1.35
radians
and
θ
Γ
=Γ=0.52radians
θ
Γ
Γ
0.52
radians
. Thus we have:

V
in
=
V
+
ei1.35+0.52
V
+
ei(0.52−1.35)
V
in
V
+
1.35
0.52
V
+
0.52
1.35

(11)
Which then gives us:

V
+
=
V
in
ei1.35+0.52ei(0.52−1.35)
V
+
V
in
1.35
0.52
0.52
1.35

(12)
When you expand the exponentials, add and combine in rectangular
coordinates, change to polar, and divide, you will get a phasor
value for

V
+
V
+
. If you do it correctly, you will find that

V
+
=5.04∠-71.59
V
+
5.04-71.59
Many times we don't care about
V
+
V
+
itself, but are more interested in how much power is being
delivered to the load. Note that power delivered to the input
of the line is also the amount of power which is delivered to
the load! Finding
I
in
I
in
is easy, it's just
V
in
Z
in
V
in
Z
in
. All we have to do is change
Z
in
Z
in
to polar form.

Z
in
=15+-25i=29.1∠59
Z
in
15-25
29.159

(13)
I
in
=
V
in
Z
in
=4.18∠3829.1∠59=0.144∠21
I
in
V
in
Z
in
4.1838
29.159
0.14421

(14)
Comments:"Accessible versions of this collection are available at Bookshare. DAISY and BRF provided."