Instructions on how to use the Smith Chart for simple calculations such as converting from admittance to impedance.
So, what do we do for
? A quick glance at a
transmission line problem shows that at
the load we have a resistor and an inductor in parallel. Thiswas done on purpose, to show you one of the powerful aspects of
the Smith Chart. Based on what you know from circuit theory youwould calculate the load impedance by using the formula for two
impedances in parallel
which will be somewhat messy to calculate.
Let's remember the formula for what the Smith
Chart represents in terms of the phasor
.
Let's invert this expression
says that is we want to get an
admittance instead of an impedance, all we have to
do is substitute
for
on the Smith Chart plane!
in our case. We have two elements in parallel for
the load (
), so we can easily add their admittances, normalize them to
, put them on the Smith Chart, go
around (same thing as letting
) and read off
. For a
resistor,
, the
condunctance equals
.
so
. The generator is operating at a frequency of
, so
and the inductor has a value of
, so
and
and
.
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 iswhere all the negative reactive parts are, and we find the curve
which represents
and where they intersect, we put a dot, and mark the
location as
. Now to find
, we simply reflect half way around to the opposite
side of the chart, which happens to be about
, 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
and move it down to the
scale and find that the reflection coefficient has a
magnitude of about 0.6. On a real Smith Chart, there is also aphase angle scale on the outside of the circle (where our
distance scale is) which you can use to read off the phase angleof the reflection coefficient as well. Putting that scale on
the "mini Smith Chart" would clog things up too much, but thephase angle of
is about
.
Now the wavelength of the signal on the line is
given as
The input to the line is located
or
away from the load. Thus, we start at
, and rotate around on a circle of constant radius a
distance
towards the generator. To do this, we extend a line out from
our
point to the scale and read a relative distance of
. We add
to this, and get
Thus, if we rotate around the Smith Chart, on our
circle of constant radius Since, after all, all we are doing isfollowing
as it rotates around from the load to the input to the line.
When we get to
, we stop, draw a line out from the center, and where
it intercepts the circle, we read off
from the grid lines on the
Smith
Chart . We find that
Thus,
ohms
. Or, the impedance at the
input to the line looks like a
resistor in series with a capacitor whose reactance
, or, since
, we find that,
To find
, there is no avoiding doing some complex math:
Which, we write in polar notation, divide, figure the voltage
and then return to rectangular notation.
If at this point we needed to find the actual voltage phasor
we would have to use the equation
Where
is the propagation constant for the line as mentioned
in the
last
chapter , and
is the
length of the line.
For this example,
and
. Thus we have:
Which then gives us:
When you expand the exponentials, add and combine in rectangular
coordinates, change to polar, and divide, you will get a phasorvalue for
. If you do it correctly, you will find that
Many times we don't care about
itself, but are more interested in how much power is being
delivered to the load. Note that power delivered to the inputof the line is also the amount of power which is delivered to
the load! Finding
is easy, it's just
. All we have to do is change
to polar form.