Exercise 38

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 (lanyu2@illinois.edu), make sure you name the file using the following notation: “Exn_Last_First.nb”, where n is the Exercise number (in this case, 38), and First and Last would be your first and last name.  So my submission to Lan for this exercise would be titled “Ex38_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:

In the past few classes we introduced the concept of impedance matching.  In our computer lab class we discussed impedance matching of a real load using a λ/4 transformer, and of a complex load by using a λ/4 transformer offset from the laod by a distance d1.  Finally, we discussed the single stub tuning method. However, single stub tuning still requires that we fix a shunt TL to our main TL at a fixed distance d1.  If the load impedance of our TL changes, we have to then move the position of the stub.  Double stub tuning does away with this inconvenience, and allows us to impedance match any load, by choosing two fixed positions, and then simply changing the length of our two stubs.

Double Stub Tuning

Double stub tuning works much the same way as single stub tuning, except, of course, one has two stubs instead of one.  The nice thing about double stub tuning is that one can fix the position of the stubs, and then adjust stub length to impedance match.
Below, write code for calculating the impedance of a double-stub structure, as shown below.  In this system, your inputs will be (Subscript[Z, L]), (Subscript[Z, o]) and (Subscript[l, s1]) and (Subscript[l, s2]).  Assume that your first stub is at position (Subscript[d, 1])=λ/4 and (Subscript[d, 2])=(Subscript[d, 1])+3λ/8=5λ/8.  Both stubs have the same characteristic impedance of the primary TL. You don’t need to show your steps on the Smith Chart, if you don’t want to, but you should print out at least the total admittance/impedance at d2.

p1=Graphics[Table[Circle[{r/(r+1),0},(1/(r+1))],{r,{0,.25,.5,1,2,5}}]];
p2=Graphics[Table[Circle[{1,1/x},Abs[1/x]],{x,{5,2,1,.5,.25,.01,-5,-2,-1,-.5,-.25,-.01}}]];
smith1=Show[p1,p2, Axes->True ,AxesLabel-> {(Subscript[Γ, r]),(Subscript[Γ, i])},PlotRange->{{-1,1},{-1,1}}]

Let’s start by plotting the Smith Chart for the system up to d1.

stub2a[zo_,zLre_,zLim_]:=(
zL=zLre+I*zLim; (*write load impedance in terms of real and imaginary components*)
zLnorm=zL/zo;(*normalized laod impedance*)
gamL=(zL-zo)/(zL+zo); (*load refelction coefficient*)
gamd1=gamL*Exp[-2*I*2*Pi*0.25];  (*determine Γ((Subscript[d, 1])=0.25)*)
p3=Graphics[{PointSize[Large],Red,Point[{Re[gamL],Im[gamL]}]}];(*Plot (Subscript[Γ, L])*)
p4=Graphics[{Red,Circle[{0,0},Abs[gamL],{Arg[gamL],Arg[gamd1]}]}];
p5=Graphics[{PointSize[Large],Red,Point[{Re[gamd1],Im[gamd1]}]}];(*Plot (Subscript[Γ, d1])*)
Show[p1,p2,p3,p4,p5,PlotRange->{{-1,1},{-1,1}}])

Manipulate[stub2a[zo,zLre,zLim],{zo,50},{zLre,50},{zLim,50}]

Now let’s add the admittances of the shorted stub and the TL at d1:

stub3a[zo_,zLre_,zLim_,ls1_]:=(
zL=zLre+I*zLim; (*write load impedance in terms of real and imaginary components*)
zLnorm=zL/zo;(*normalized laod impedance*)
gamL=(zL-zo)/(zL+zo); (*load refelction coefficient*)
gamd1=gamL*Exp[-2*I*2*Pi*0.25];  (*determine Γ((Subscript[d, 1])=0.25)*)
zd1=(1+gamd1)/(1-gamd1);(*determine z((Subscript[d, 1])=0.25)*)
p3=Graphics[{PointSize[Large],Red,Point[{Re[gamL],Im[gamL]}]}];(*Plot (Subscript[Γ, L])*)
p4=Graphics[{Red,Circle[{0,0},Abs[gamL],{Arg[gamL],Arg[gamd1]}]}];
p5=Graphics[{PointSize[Large],Red,Point[{Re[gamd1],Im[gamd1]}]}];(*Plot (Subscript[Γ, d1])*)
z1in=I*Tan[2*Pi*ls1];(*input impedance of shorted stub 1*)
yd1tot=(1/z1in)+(1/zd1);(*total admittance of TL at d1 (including shorted stub)*)
zd1tot=1/yd1tot;(*total impedance of TL at d1 (including shorted stub)*)
gamd1tot=(zd1tot-1)/(zd1tot+1);(*refelction coefficient of TL at d1 (including shorted stub)*)
gamd2=gamd1tot*Exp[-2*I*2*Pi*3/8];  (*determine Γ((Subscript[d, 2])=(1/4 + 3/8)λ)*)
zd2=(1+gamd2)/(1-gamd2);(*determine z((Subscript[d, 2])=0.25+3/8)*)
p6=Graphics[{PointSize[Large],Blue,Point[{Re[gamd1tot],Im[gamd1tot]}]}];(*Plot (Subscript[Γ, d1tot])*)
p7=Graphics[{Blue,Circle[{0,0},Abs[gamd1tot],{Arg[gamd1tot],Arg[gamd2]}]}];
p8=Graphics[{PointSize[Large],Blue,Point[{Re[gamd2],Im[gamd2]}]}];(*Plot (Subscript[Γ, d2])*)
Show[p1,p2,p3,p4,p5,p6,p7,p8,PlotRange->{{-1,1},{-1,1}}])

Manipulate[stub3a[zo,zLre,zLim,ls1],{zo,50},{zLre,50},{zLim,50},{ls1,0.001,.5,.001}]

Now, our final step: Add the admittance from the second stub to the admittance of the TL at d2.

stub4a[zo_,zLre_,zLim_,ls1_,ls2_]:=(
zL=zLre+I*zLim; (*write load impedance in terms of real and imaginary components*)
zLnorm=zL/zo;(*normalized laod impedance*)
gamL=(zL-zo)/(zL+zo); (*load refelction coefficient*)
gamd1=gamL*Exp[-2*I*2*Pi*0.25];  (*determine Γ((Subscript[d, 1])=0.25)*)
zd1=(1+gamd1)/(1-gamd1);(*determine z((Subscript[d, 1])=0.25)*)
p3=Graphics[{PointSize[Large],Red,Point[{Re[gamL],Im[gamL]}]}];(*Plot (Subscript[Γ, L])*)
p4=Graphics[{Red,Circle[{0,0},Abs[gamL],{Arg[gamL],Arg[gamd1]}]}];
p5=Graphics[{PointSize[Large],Red,Point[{Re[gamd1],Im[gamd1]}]}];(*Plot (Subscript[Γ, d1])*)
z1in=I*Tan[2*Pi*ls1];(*input impedance of shorted stub 1*)
yd1tot=(1/z1in)+(1/zd1);(*total admittance of TL at d1 (including shorted stub)*)
zd1tot=1/yd1tot;(*total impedance of TL at d1 (including shorted stub)*)
gamd1tot=(zd1tot-1)/(zd1tot+1);(*refelction coefficient of TL at d1 (including shorted stub)*)
gamd2=gamd1tot*Exp[-2*I*2*Pi*3/8];  (*determine Γ((Subscript[d, 2])=(1/4 + 3/8)λ)*)
zd2=(1+gamd2)/(1-gamd2);(*determine z((Subscript[d, 2])=0.25+3/8)*)
p6=Graphics[{PointSize[Large],Blue,Point[{Re[gamd1tot],Im[gamd1tot]}]}];(*Plot (Subscript[Γ, d1tot])*)
p7=Graphics[{Blue,Circle[{0,0},Abs[gamd1tot],{Arg[gamd1tot],Arg[gamd2]}]}];
p8=Graphics[{PointSize[Large],Blue,Point[{Re[gamd2],Im[gamd2]}]}];(*Plot (Subscript[Γ, d2])*)
z2in=I*Tan[2*Pi*ls2];(*input impedance of shorted stub 1*)
yd2tot=(1/z2in)+(1/zd2);(*total admittance of TL at d2 (including shorted stub2)*)
zd2tot=1/yd2tot;(*total impedance of TL at d2 (including shorted stub 2)*)
gamd2tot=(zd2tot-1)/(zd2tot+1);(*Reflection coefficient of TL at d2 (including shorted stubs.  This is what we want to go to "0"*)
p9=Graphics[{PointSize[Large],Green,Point[{Re[gamd2tot],Im[gamd2tot]}]}];(*Plot (Subscript[Γ, d2tot])*)
p10=Graphics[Text[{"(Subscript[Γ, d2tot])=",gamd2tot},{.6,.9}]];
Show[p1,p2,p3,p4,p5,p6,p7,p8,p9,p10,PlotRange->{{-1,1},{-1,1}}])

Manipulate[stub4a[zo,zLre,zLim,ls1,ls2],{zo,50},{zLre,50},{zLim,50},{ls1,0.001,.5,.001},{ls2,0.001,.5,.001}]

1) Using the program above, determine the (Subscript[l, s1]) and (Subscript[l, s2]) you need to impedance match a line with characteristic impedance (Subscript[Z, o])=75 and a load impedance (Subscript[Z, L])=34+j12.  I would try to adjust (Subscript[l, s1]) until the real part of the impedance is 1, then go back and adjust (Subscript[l, s2]) to minimize the imaginary component of impedance.  This will change the real part slightly, so you then have to tweak (Subscript[l, s1])...continue in this manner until you have a real part of 1 and an imaginary part of 0.  It shouldn’t take too long.  Give your answer below:

Manipulate[stub4a[zo,zLre,zLim,ls1,ls2],{zo,50},{zLre,50},{zLim,50},{ls1,0.001,.5,.001},{ls2,0.001,.5,.001}]

I get (Subscript[l, s1])=___λ and (Subscript[l, s2])=___λ

2) Now, again using the program above, determine the (Subscript[l, s1]) and (Subscript[l, s2]) you need to impedance match a line with characteristic impedance (Subscript[Z, o])=50 and a load impedance (Subscript[Z, L])=100-j56.

Manipulate[stub4a[zo,zLre,zLim,ls1,ls2],{zo,50},{zLre,50},{zLim,50},{ls1,0.001,.5,.001},{ls2,0.001,.5,.001}]

I get (Subscript[l, s1])=___λ and (Subscript[l, s2])=___λ