Exercise 34

As with all of the exercises you will complete in this course, you will be expected to turn your *.nb file in to the course TA before the next lecture.  Please do so by e-mailing your completed exercise file to Lan Yu (ece329.uiuc@gmail.com), make sure you name the file using the following notation: “Exn_Last_First.nb”, where n is the Exercise number (in this case, 34), and First and Last would be your first and last name.  So my submission to Lan for this exercise would be titled “Ex34_Wasserman_Dan.nb”.  You will receive full credit for your exercise if it turned in before the next lecture (Lecture n+1).  No late submissions will be accepted.

Objective/Overview:

Today in class we introduced the Smith Chart.  The Smith Chart often throws people in 329 for a loop, so in our exercise today we are going to look a little deeper into the Smith Chart.  We are then going to solve a “Smith Chart” problem with and without a Smith Chart and compare our results.

Make Your Own Smith Chart

If you remember from class, the Smith Chart is a graphical tool that helps us move back and forth between the complex line impedance (z(d)=r(d)+j*x(d)) and the complex line reflection coefficient (Γ(d)).  In class we worked out the relationship between x and r and (Subscript[Γ, r]) and (Subscript[Γ, i]), the real and imaginary components of the line impedance.



The Figure above shows the basic utility of the concepts of generalized line impedance and generalized reflection coefficients.  Basically, we start with a TL of arbitrary length and with arbitrary load coefficient.  Knowing our load reflection coefficient means that we can replace any portion of our TL with the generalized reflection coefficient Γ(d), and the TL behavior will not change at all.We can take this to the extreme and replace the entire line with an input impedance Z(l), and if we place this into a circuit, it will behave the same as if we had the entire TL in the circuit.

In class we derived the relationship between line impedance and and reflection coefficient.  Let’s do this in Mathematica now.  First we start with the lines of constant r:

p1=Graphics[Table[Circle[{r/(r+1),0},(1/(r+1))],{r,{0,.25,.5,1,2,5}}]]

To that we add the lines of constant x:

p2=Graphics[Table[Circle[{1,1/x},Abs[1/x]],{x,{5,2,1,.5,.25,.01,-5,-2,-1,-.5,-.25,-.01}}]]

All together now!

smith1=Show[p1,p2, Axes->True ,AxesLabel-> {(Subscript[Γ, r]),(Subscript[Γ, i])},PlotRange->{{-1,1},{-1,1}}]

Typical Smith Chart Problem

Example Problem: Let’s say you have a TL with (Subscript[Z, o])=50Ω, and (Subscript[Z, L])=50+j100 Ω.  What is (Subscript[Z, IN]) for the TL if the TL length is l=λ/8?
Let approach this in parts.  

First, we need to find our normalized load impedance and load reflection coefficient

zL=(50+I*100)/zo

gamL=(zL-1)/(zL+1)

Let’s plot this point on our Smith Chart

p3=Graphics[{PointSize[Large],Red,Point[{Re[gamL],Im[gamL]}]}];

Note that we plot the point using the complex Γ, since Mathematica recognizes Cartesian coordinates.

Show[p1,p2, p3,Axes->True ,AxesLabel-> {(Subscript[Γ, r]),(Subscript[Γ, i])},PlotRange->{{-1,1},{-1,1}}]

You should notice that the point in complex space corresponding to (Subscript[Γ, L])={(Subscript[Γ, r]),(Subscript[Γ, i])}  lies at the intersection of the constant r=1 and x=2 lines!!  This makes sense, because the impedence at the load is (Subscript[z, L])=1+2j!!
We can find the impedance ANYWHERE else on the TL in two ways:,
1) Analytically, we recognize that the generalized reflection coefficient on the TL is given by Γ(d)=(Subscript[Γ, L])(e^(-j2βd)).  We can find the reflection coefficient at d=λ/8:

gamin=gamL*Exp[-2*2*I*Pi/8]

p4=Graphics[{PointSize[Large],Blue,Point[{Re[gamin],Im[gamin]}]}];

Show[p1,p2, p3,p4,Axes->True ,AxesLabel-> {(Subscript[Γ, r]),(Subscript[Γ, i])},PlotRange->{{-1,1},{-1,1}}]

From our plot, it is clear that our Γ(d=λ/8) sits on the intersection of the constant r=1 and x=-2 lines.  We can check this analytically by calculating the normalized line impedance at d=λ/8 using z(d)=((1+Γ(d))/(1-Γ(d)))

zin=(1+gamin)/(1-gamin)

It works!!
Now let’s try this graphically

2) Graphically.  We know that we are starting at (Subscript[z, L])=1+2j.  To find (Subscript[z, IN]) graphically we need to take the circle of radius |(Subscript[Γ, L])| and rotate from our iintial position by 2βd=2βλ/8=2*2πλ/8λ=π/2 radians in the Clockwise direction.
Let’s start by plotting the |(Subscript[Γ, L])| circle:

p5=Graphics[{Red,Circle[{0,0},Abs[gamL]]}];

Show[p1,p2, p3,p5,Axes->True ,AxesLabel-> {(Subscript[Γ, r]),(Subscript[Γ, i])},PlotRange->{{-1,1},{-1,1}}]

In order to plot the arc from ((Subscript[Γ, L]),(Subscript[z, L])) to ((Subscript[Γ, IN]),(Subscript[z, IN])), we can still use the Circle command, we just need to specify the angles, in radians of our arc.  We know that the initial angle is nothing more than the argument of (Subscript[Γ, L]), to rotate by 2βd we just just subtract  2βd from (Subscript[Γ, L])  :

p6=Graphics[{Blue,Circle[{0,0},Abs[gamL],{Arg[gamL],Arg[gamL]-2*2*Pi/8}]}];

Show[p1,p2, p3,p5,p6,Axes->True ,AxesLabel-> {(Subscript[Γ, r]),(Subscript[Γ, i])},PlotRange->{{-1,1},{-1,1}}]

You see that the arc ends at the intersection of the constant r=1 and x=-2 lines, which is exactly where we ended up from our analytical solution!!

Your Turn

Now you try the same for a TL, except this time, have the (Subscript[Z, L]), (Subscript[Z, o]) and TL length be inputs.  Input the TL length in units of wavelength.  So if your TL has a length λ/5, then your input should be 1/5.  Your result should be a Smith Chart showing the position of ((Subscript[Γ, L]),(Subscript[z, L])) and ((Subscript[Γ, IN]),(Subscript[z, IN])), as well as the |(Subscript[Γ, L])| circle and the arc connecting ((Subscript[Γ, L]),(Subscript[z, L])) and ((Subscript[Γ, IN]),(Subscript[z, IN])) on the |(Subscript[Γ, L])| circle.  Have your plot also label your points with their Γ and Z values.

smith2[zo1_,zL1_,tlL_]:=(zLnorm=;gamL1=;
p7=Graphics[{PointSize[Large],Red,Point[{Re[gamL1],Im[gamL1]}]}];
p7l1=Graphics[Text[{"(Subscript[Γ, L])=",gamL1},{Re[gamL1],Im[gamL1]+.1}]];
p7l2=Graphics[Text[{"(Subscript[Z, L])=",zL1},{Re[gamL1],Im[gamL1]+.25}]];
gamin1=;
zin1=;
p8=Graphics[{PointSize[Large],Blue,Point[{Re[gamin1],Im[gamin1]}]}];
p8l1=Graphics[Text[{"(Subscript[Γ, IN])=",gamin1},{Re[gamin1],Im[gamin1]-.1}]];
p8l2=Graphics[Text[{"(Subscript[Z, IN])=",zin1*zo1},{Re[gamin1],Im[gamin1]-.25}]];
p9=Graphics[{Red,Circle[{0,0},Abs[gamL1]]}];p10=Graphics[{Blue,Circle[{0,0},Abs[gamL1],{Arg[gamL1],Arg[gamin1]}]}];
Show[p1,p2,p7,p7l1,p7l2,p8,p8l1,p8l2,p9,p10,Axes->True ,AxesLabel-> {(Subscript[Γ, r]),(Subscript[Γ, i])},PlotRange->{{-1,1},{-1,1}}])

smith2[]

This little program you just wrote should be useful as you continue to learn about Smith Charts and also as you do Smith Chart problems for class!  In fact, we will rely on this code for our next Computer Lab Lecture...