Now let's see how we can use The Bilinear Transform to get the
co-ordinates on the
Zs
Z
0
Z
s
Z
0
plane transferred over onto the
rs
r
s
plane. The
Bilinear Transform tells us how to take
*any*
Zs
Z
0
Z
s
Z
0
and generate an
rs
r
s
from it. Let's start with an easy one. We will assume
that
Zs
Z
0
=1+iX
Z
s
Z
0
1
X
, which is a vertical line, which passes through 1, and
can take on whatever imaginary part it wants Figure 1.

According to The Bilinear Transform, the first thing
we should do is add 1 to
Zs
Z
0
Z
s
Z
0
. This gives us the line
2+iX
2
X
Figure 2.

Now, we take the inverse of this, which will give us a circle,
of diameter 1/2

Figure 3. Now, according to

The Bilinear Transform we
take this circle and multiply by -2

Figure 4.

And finally, we take the circle and add +1 to it: as shown

here. There, we are done with the
transform. The vertical line on the

Zs
Z
0
Z
s
Z
0
plane that represents an impedance with a real part of
+1 and an imaginary part with any value from

−(i∞)
to

i∞
has been reduced to a circle with diameter 1, passing
through 0 and 1 on the complex

rs
r
s
plane.

Let's do the same thing for

Zs
Z
0
=0.5+iX
Z
s
Z
0
0.5
X
and

Zs
Z
0
=2+iX
Z
s
Z
0
2
X
. We'll call these lines A and B respectively, and just
add these to the sketches we already have

Figure 6.
Follow along with

The
Bilinear Transform, and see if you can figure out where
each of these sketches comes from. We will simply be doing the
same things again:

• add 1; • invert; • multiply by -2; • add 1
once again. As you can see in

Figure 7,

Figure 8,

Figure 9, and

Figure 10 we get more circles. For lines inside the +1
real part, we end up with a circle that is

*larger* than the +1 circle, and for lines
which have a real part greater than +1, we end up with circles
which are smaller in diameter than the +1 circle. All circles
pass through the +1 point on the

rs
r
s
plane and are tangent to one another.

There are two special lines we should worry about. One is

Zs
Z
0
=iX
Z
s
Z
0
X
, the imaginary axis. We will put all of the transform
steps together on

Figure 11. We start on the axis,
shift over one, get a circle with unity diameter when we invert,
grow by two and flip around the imaginary axis when we multiply
by -2, and then hop one to the right when +1 is added. Once
again, you should work your way through the various steps to
make sure you have a good understanding as to how this procedure
is supposed to happen. Note that even the imaginary axis on the

Zs
Z
0
Z
s
Z
0
plane gets transformed into a circle when we go over
onto the

rs
r
s
plane.

The other line we should worry about is

Zs
Z
0
=∞+iX
Z
s
Z
0
X
. Now

∞+1=∞
1
, and

-2∞=0.0+1=1
-2
0.0
1
1
, and so the line

1+iX
1
X
gets mapped into a point at 1 when we do our
transformation onto the

rs
r
s
plane. Even points at

∞
on the

Zs
Z
0
Z
s
Z
0
plane end up on the

rs
r
s
plane, and are easily accessible!

OK, Figure 12 is a plot of the
Zs
Z
0
Z
s
Z
0
plane. The lines shown represent the real part of
Zs
Z
0
Z
s
Z
0
that we want to transform. We run them all through
The Bilinear
Transform, to get them onto the
rs
r
s
plane. Now we have a whole family of circles, the
biggest of which has a diameter of 2 (which corresponds to the
imaginary axis) and the smallest of which has a diameter of 0
(which corresponds to points at ∞)
Figure 13. The circles all fit within one another,
and since a +1 was added to every transform as the final bit of
manipulation, all of the circles pass through the point +1,
0i
0
. Circles with smaller diameters correspond to larger
values of real
Zs
Z
0
Z
s
Z
0
, while the larger circles correspond to the lesser
values of
Zs
Z
0
Z
s
Z
0
.

Well, we're half way there. Now all we have to do is find the
transform for the co-ordinate lines which correspond to the
imaginary part of

Zs
Z
0
Z
s
Z
0
. Let's look at

Zs
Z
0
=R+i1
Z
s
Z
0
R
1
. When we add +1 to this, nothing happens! The line
just slides over 1 unit, and looks just the same

Figure 14. Now we take its inverse. This will gives us
a circle, but since the line we are inverting lies at an angle
of

90
°
90
°
with respect to the real axis, the major diameter of
the circle will lie at an angle of

−
90
°
90
°
when we go through the inversion process. This gives
us a circle which is lying in the

−i
region of the complex plane

Figure 15.

The next thing we do is to take this circle and multiply by -2.
This will make the circle twice as large, but will also reflect
it back up into the

i
region of the complex plane

Figure 16.

And, finally, we add 1 to it, which causes the circle to hop
one over to the right

Figure 17.

We can do the same thing to other lines of constant imaginary
part and we can then add more circles. (Or partial circles, for
it makes no sense to go beyond the

ℜZs
Z
0
=0
Z
s
Z
0
0
circles, as beyond that is the region corresponding to
negative real part, which we would not expect to encounter in
most transmission lines.) Take at least one of the other
circles drawn

here and see if you
can get it to end up in about the right place.

There is one line of interest which we have a take a little care
with. That is the real axis,

Zs
Z
0
=0+iX
Z
s
Z
0
0
X
. This line is a distance 0 away from the origin, and
so when we invert it, we get a circle with

∞ diameter. That's OK though,
because that is just a straight line. So, the real axis of the

Zs
Z
0
Z
s
Z
0
plane transforms into the real axis on the

rs
r
s
plane.

We have done a most wondrous thing! (Although you
may not realize it yet.) We have taken the
*entire* half plane of complex impedance
Zs
Z
0
Z
s
Z
0
and mapped the whole thing into a circle with diameter
1! Let's put the two of them side by side. (Although we can't
show the whole
Zs
Z
0
Z
s
Z
0
plane of course.) These are shown here, where we show how each line on
Zs
Z
0
Z
s
Z
0
maps into a (curved) line on the
rs
r
s
plane. Note also, that for every point on the
Zs
Z
0
Z
s
Z
0
plane ("A" and "B") there is a corresponding point on
the
rs
r
s
plane. Pick a couple more points, "C" and "D" and
locate them either on the
Zs
Z
0
Z
s
Z
0
plane, or the
rs
r
s
plane, and then find the corresponding point on the
other plane.

Note that the mapping is not very uniform. All of the region
where either the real or imaginary part of

Zs
Z
0
Z
s
Z
0
is

1
1
(a small square on

Zs
Z
0
Z
s
Z
0
maps into a major fraction of the

rs
r
s
plane

Figure 20 whereas all the rest of the

Zs
Z
0
Z
s
Z
0
plane, all the way out to infinity in three directions
(

∞
,

i∞
, and

−(i∞)
) map into the rest of the

rs
r
s
circle

Figure 21.

This graph or transformation is called a

Smith
Chart, after the Bell Labs worker who first thought it
up. It is a most useful and powerful graphical solution to the
transmission line problem. In

Introduction to Using the Smith Chart we
will spend a little time seeing how and why it can be so useful.

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