ECE3300 Lecture 12b-5 Smith Chart admittance and open,short circuits

Uploaded by cfurse on 24.09.2009

Now let's talk about how to find the admittance on a Smith Chart.
I'm going to begin with the example
where ZL was 50+j50 on a 50 ohm transmission line, so it was 1+j1,
which we plotted right here on the
Smith Chart. This was ZL. If I want to find YL that is 1 over ZL.
The way we find that is called translating
through the center of the Smith Chart. You take your straight edge
and you draw a line from the
center through your point and you continue it on here.
So it's one straight line, right through there,
right through the center of the Smith Chart. Then you mark your
distance like so and you copy that
distance right here, starting at the center, going to this point right there,
this is YL so you can say it's
translating through the center of the Smith Chart to the point YL.
Let's read the value of YL. YL has a real part of about 0.5 and it has an imaginary part
which is negative. Remember down here we have
the ngative imaginary part. Read it right out here.
It has a negative j0.5. If we wanted to know the
admittance in mhos, we would have to denormalize.
So in order to find YL in mhos we are going to
take YL on the Smith Chart and we're going to multiply it by Y-naught.
Y-naught is 1 over Z -naught so it
is 1 over 50. So YL is going to be equal to 0.5 minus j0.5
then we need to multiply 1 divided by 50 and
that's going to give us the value in mhos. And that's equal to 0.01 minus j0.01 mhos.
That's our YL.
Now let's also talk about some very special cases.
We're going to consider an open circuit and short
circuit because they are very important because.
We use them for building things and also many of
our devices end up acting like high or low impedance close to open circuit or short
circuit. So for an open we know that the impedance of an open circuit is
equal to infinity. So let's find where the real
part becomes very very big. Here's the real part of 1, here's the real
part of 2, 50 and so on. Right here
is where we have the impedance of an open circuit. The impedance of a short
circuit is 0. So where is our real part 0 and our imaginary part 0. Our real part
is 0 every place on the outside of the Smith Chart.
So this circle right there is where the real part of 0.
The imaginary part of 0 right along this axis.
So that point right there is Z of a short circuit. Now how about the
admittance of an open or short?
Remember if we wanted to find the admittance we simply
translate through the center of the Smith Chart.
Let's do the short circuit first. We're just going to translate
that through the center of the Smith Chart right over to here.
So this is the admittance of a short circuit. Let's do the same thing with open
circuit and this is the admittance of an open circuit.
Let's also read the reflection coefficient.
So the reflection coefficient of an open circuit we know that ought to be 1.
Let's check ourselves. Right here is the point where I have the Z of the open
circuit and that's where I'm going to find the reflection coefficient.
Its phase is 0, just as we expect, and it's magnitude if you take these two points,
put them on your piece of paper and mark them down here is magnitude of one.
How about reflection coefficient of the short circuit? We know it should be
minus one. So here's the point where we are going to consider the short circuit.
Let's come over here and measure the phase which is 180 degrees,
and then let's bring this magnitude right there, down here to the center,
and it has a magnitude of 1 and a phase of 180 degrees which is -1.
Let me make a note here that when finding the reflection coefficient use the
impedance points. Do not use the Y points in order to find your reflection coefficient.