Computer Lab 11: Lecture 32

In our computer lab lectures we will work together through concepts and challanges we have discussed over the week in class.  As with your after-class Exercises, you will be expected to turn in your computer lab notebooks before the nect lecture. Please do so by e-mailing your completed *nb 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 Lecture number (in this case, 32), and First and Last would be your first and last name.  So my submission to Lan for this computer lab would be titled “Ex32_Wasserman_Dan.nb”.  You will receive full credit for your computer lab if it turned in before the next lecture (Lecture n+1).  No late submissions will be accepted.

Objective/Overview:

In class and our exercises last week, we starting thinking about how waves propagate on TLs.  First we discussed the impulse response of TLs, and used the impulse response to model the propagation of d’Alembert waveforms on TLs.  Then we extended our discussion to multi-segment TLs, using the same impulse response formalism, but accounting for reflection and transmission at the interface between the two TLs, which ended up being remarkably similar to the behavior of plane waves at the interface between two materials.  Finally we discussed a specific kind of TLs which we dubbed “resonant TLs”.  These are TLs which have either open or shorted ends, and because of this, show resonant behavior at certain frequencies.  In these systems, the allowed wavelengths of the resonant waves depend on the TL length.  Alternatively, we could think of a wave of a given frequency/wavelength as only being resonant on TLs of certain lengths.  Today, we extend this discussion to understand how TLs of arbitrary lengths (but still with either shorted or open ends) behave as a function of frequency.

Admittance and Impedance of Shorted TLs

We know that we can describe the combination of an inductor and a capacitor as having an impedance and admittance.  If our inductor and capacitor are in series, then we write the series impedance as
(Subscript[Z, s])=j(ωL-1/ωC)
and if the inductor and capacitor are in parallel, we write the parallel admittance as
(Subscript[Y, p])=1/(Subscript[Z, p])=j(ωC-1/ωL)
(Subscript[Y, p]) and (Subscript[Z, s]) are equal to 0 when ω=1/(Sqrt[LC])=(Subscript[ω, o]).
Now we know that TL circuits have L and C associated with them, so we should ask whether we can think about Admittance or Impedance of a TL Circuit.
In our last lecture we discussed the resonant modes of open/open, shorted/shorted and open/shorted TLs. In the case of a TL that is shorted at the load and open at the input, we determined that the resonant frequencies are given by
(Subscript[f, n])=v/2l(n+1/2), for n≥0
In this case, we know that at the load, V(z=l)=0, and at the input I(z=0)=0
If we call our input impedance (Subscript[Z, in])=(Subscript[V, in])/(Subscript[I, in]), then we can see that our (Subscript[Z, in])=∞!!
This is just like the input impedance of our parallel LC circuit at resonance!  We then call this resonance, when (Subscript[Z, in])=∞, a Parallel Resonance.
If, on the other hand, the length of the shorted TL is l=nλ/2, then we know that the voltage at the input will always be V(z=0)=0, since V(z=l)=0 from the shorted load and that the TL is a multiple of a half-wavelength.
If we call our input impedance (Subscript[Z, in])=(Subscript[V, in])/(Subscript[I, in]), then we can see that our (Subscript[Z, in])=0, and our input admittance (Subscript[Y, in])=∞!
This is just like the input impedance of our series LC circuit at resonance!  We then call this resonance, when (Subscript[Z, in])=0, a Series Resonance.


So we can see that for the correct choice of TL length, we can achieve either an infinite admittance or impedance.  What about for arbitrary lengths?
So now we are going to make a change in how we label things.  Because our load is the one fixed aspect of the TL, we are going to label the load as z=0, and as we move out from the load, we are going to increase in z.  This doesn’t change the physics of our system, but it does change how we label things.


Recognize that we have had to change the expressions for the Voltage and Current phasors on the line by flipping the sign associated with propagation.

Because we know we have V=0 at what is now z=0, we can apply this boundary condition:
(OverTilde[V])(0)=0=(SuperPlus[V])+(SuperMinus[V]), so (SuperPlus[V])=-(SuperMinus[V])
This means that at the input of a TL with length d, we can write Voltage as:
(OverTilde[V])(d)=(SuperPlus[V])((e^(jβd))-(e^(-jβd)))=(SuperPlus[j2V])Sin(βd), which then makes our current
(OverTilde[I])(d)=((SuperPlus[V])/(Subscript[Z, o]))((e^(jβd))+(e^(-jβd)))=(Subscript[jY, o])2(SuperPlus[V])Cos(βd), where (Subscript[Y, o])=1/(Subscript[Z, o]) is the characteristic admittance of the line.
We can write the input impedance as (Subscript[Z, in])(d)=(Subscript[jZ, o])Tan[βd]
Now we can plot the input impedance of the TL as a function of TL length.  However, it is probably more instructive to plot as a function of β*d, since this is 2πd/λ.  
Let’s assume (Subscript[Z, o])=50Ω and v=c=3×(10^8) m/s.  Note that our impedance is entirely COMPLEX!!  In other words, we have a purely reactive input impedance:

zin1=I*50*Tan[betad]

Plot[Im[zin1],{betad,0,3*Pi},Ticks->{{0,Pi/2,Pi,3Pi/2,2 Pi, 5 Pi/2,3 Pi},{-100,100}}]

A couple of things to note here:

For a shorted TL, the input impedance is always REACTIVE

For 0<d<λ/4, or 0<βd<π/2 Z(d)>0  so our TL is INDUCTIVE.

For λ/4<d<λ/2 or π/2<βd<π, Z(d)<0, so our TL is CAPACITIVE.

We also see that the input impedance of our TL is periodic in d with period of λ/2.

What is interesting is that because our input impedance is a Tan function, we can give our TL ANY reactive impedance between -∞ and +∞ simply by changing the TL length d.

What is particularly interesting is that we have a shorted TL, but at certain combinations of frequency and TL length, the TL will have an input impedance of ∞, which means that if we put the TL into a circuit, it will look like an OPEN circuit (even though it is shorted at the load!).

Now consider a TL with an open load.  In this case, we can use the same expressions for (OverTilde[V])(z) and (OverTilde[I])(z) as we do above, but now, our boundary condition is that (OverTilde[I])(z=0)=0.  Use this to write and plot the input impedance for an OPEN TL as a function of βd.

zin2=

Plot[Im[zin2],{betad,0,3*Pi},Ticks->{{0,Pi/2,Pi,3Pi/2,2 Pi, 5 Pi/2,3 Pi},{-100,100}}]

Shorted TLs in Circuits

Let’s see what happens when we put our TL in a lumped element circuit.

Example 1

In particular, let’s assume we have a shorted TL of length d=5m in series with a resistor (Subscript[R, L])=100Ω.  Determine the frequency response of the circuit H(ω)=(Subscript[(OverTilde[V]), L])/(OverTilde[F]) as a function of f=ω/2π if (Subscript[Z, o])=50Ω, and v=c.  Let’s work in units of μs, so our v=c=300m/μs and our frequencyis in units of MHz.

We can plot (Subscript[Z, IN]) of the TL as a function of frequency f

zin3=I*50*Tan[2*Pi*f*5/(300)]

Plot[Im[zin3],{f,0,60}]

We can find the response function by writing the voltage over the load resistor using the voltage division expression, assuming (OverTilde[F])=1:
H(ω)=(Subscript[(OverTilde[V]), L])/(OverTilde[F])=(Subscript[R, L])/((Subscript[R, L])+(Subscript[Z, IN]))=100/(100+j*50*Tan(2πfd/v)

h1=100/(100+zin3)

Plot[{Abs[h1],Arg[h1]},{f,0,60},PlotLegends->"Expressions"]

We can see that when f=0, 30, 60, ...MHz our Voltage over the load is maximum, and our input impedance for the TL is 0...these are the Series Resonances of the circuit.  When f=15, 45, ...MHz, all Voltage is dropped over the TL, these are our Parallel Resonances!

Example 2

Now let’s assume we have the same shorted TL of Example 1 length d=5m in series with a resistor (Subscript[R, L])=100Ω.  Now we add an inductor to the circuit.  Determine the frequency response of the circuit H(ω)=(Subscript[(OverTilde[V]), L])/(OverTilde[F]) as a function of f=ω/2π if (Subscript[Z, o])=50Ω, and v=c for the cases where the inductor is in series and parallel to the TL.  Let’s work in units of μs, so our v=c=300m/μs and our frequency is in units of MHz.  Let L be a variable that you can adjust.  Compare the response function to the response function of Example 1, looking in particular at the positions of the series and parallel resonances.

zin4[ind_]:=I*2*Pi*f*ind+I*50*Tan[2*Pi*f*5/(300)]

h4[ind_]:=100/(100+zin4[ind])

Manipulate[Plot[{Abs[h4[ind]],Arg[h4[ind]]},{f,0,60}],{ind,0,1}]

This is pretty cool!  We see that changing the value of the inductor in series can change the series resonance, but not the parallel.
What about for the inductor and resistor in parallel?  We plot for L>0 so that we don’t run into a 1/0 situation...

zin5[ind_]:=1/(1/(I*2*Pi*f*ind)+1/(I*50*Tan[2*Pi*f*5/(300)]))

Manipulate[Plot[{Abs[zin5[ind]],Arg[zin5[ind]]},{f,0,60}],{ind,0.01,1}]

h5[ind_]:=100/(100+zin5[ind])

Manipulate[Plot[{Abs[h5[ind]],Arg[h5[ind]]},{f,0,60}],{ind,0.01,1}]

Here we see that the series resonance is fixed, but the parallel resonance can shift!

Example 3

Now you do the same, but replace the series/parallel inductor with a capacitor. Assume we have the same shorted TL of Example 1. Determine the frequency response of the circuit H(ω)=(Subscript[(OverTilde[V]), L])/(OverTilde[F]) as a function of f=ω/2π if (Subscript[Z, o])=50Ω, and v=c for the cases where the capacitor is in series and parallel to the TL.  Let’s work in units of μs, so our v=c=300m/μs and our frequency is in units of MHz.  Let C be a variable that you can adjust.  Compare the response function to the response function of Examples 1 and 2, looking in particular at the positions of the series and parallel resonances.

zin6[cap_]:=(*In Series*)

h6[cap_]:=

Manipulate[Plot[{Abs[h6[cap]],Arg[h6[cap]]},{f,0,60}],{cap,0.0001,.01}]

zin7[cap_]:=(*In Parallel*)

h7[cap_]:=

Manipulate[Plot[{Abs[h7[cap]],Arg[h7[cap]]},{f,0,60}, PlotRange->{-1,1}],{cap,0.0001,.01}]