6.7 The smith chart

Introduction to physical Page 1 / 1

Introduction of Smith Chart.

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

Complex impedence with real part = +1

According to The Bilinear Transform , the first thing we should do is add 1 to $Z s Z_{0}$ . This gives us the line $2 X$ .

Adding 1

Now, we take the inverse of this, which will give us a circle, of diameter 1/2 . Now, according to The Bilinear Transform we take this circle and multiply by -2 .

Inverting

Multiplying by -2

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

Z s Z_{0}

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

∞ to

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

r s

plane.

Adding 1 once again

Let's do the same thing for

Z s Z_{0} 0.5 X

and

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 . 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 thesame things again:

add 1
invert
multiply by -2
add 1

once again. As you can see in , , , and 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 circleswhich are smaller in diameter than the +1 circle. All circles pass through the +1 point on the

r s

plane and are tangent to one another.

Two more examples

Add +1 to each

Inverting

Multiply by -2

The final result

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

Z s Z_{0} X

, the imaginary axis. We will put all of the transform steps together on . 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. Onceagain, you should work your way through the various steps to make sure you have a good understanding as to how this procedureis supposed to happen. Note that even the imaginary axis on the

Z s Z_{0}

plane gets transformed into a circle when we go over onto the

r s

plane.

Another transform

Transforming

X

to the

r s

plane.

The other line we should worry about is

Z s Z_{0}

∞ X . Now

∞ 1 ∞ , and

-2

∞ 0.0 1 1 , and so the line

1 X

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

r s

plane. Even points at

∞ on the

Z s Z_{0}

plane end up on the

r s

plane, and are easily accessible!

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

Other constant real part lines

Adding other constant real part line to the

Z s Z_{0}

plane.

Family of circles

Family of

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 theimaginary part of

Z s Z_{0}

. Let's look at

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

90^{°}

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

90^{°}

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

region of the complex plane .

A line of constant imaginary part

After inverting

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 reflectit back up into the

region of the complex plane .

Mulitply by -2

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

And add 1

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

Z s Z_{0} 0

circles, as beyond that is the region corresponding to negative real part, which we would not expect to encounter inmost 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.

The complete transformation

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

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

Z s Z_{0}

plane transforms into the real axis on the

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 $Z s Z_{0}$ and mapped the whole thing into a circle with diameter 1! Let's put the two of them side by side. (Although we can'tshow the whole $Z s Z_{0}$ plane of course.) These are shown here , where we show how each line on $Z s Z_{0}$ maps into a (curved) line on the $r s$ plane. Note also, that for every point on the $Z s Z_{0}$ plane ("A" and "B") there is a corresponding point on the $r s$ plane. Pick a couple more points, "C" and "D" and locate them either on the $Z s Z_{0}$ plane, or the $r s$ plane, and then find the corresponding point on the other plane.

The mapping

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

Z s Z_{0}

1

(a small square on

Z s Z_{0}

maps into a major fraction of the

r s

plane whereas all the rest of the

Z s Z_{0}

plane, all the way out to infinity in three directions (

∞ ,

∞ , and

∞ ) map into the rest of the

r s

circle .

Mapping

Mapping 1,

1

Mapping the rest

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

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Source: OpenStax, Introduction to physical electronics. OpenStax CNX. Sep 17, 2007 Download for free at http://cnx.org/content/col10114/1.4

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