## ECE1236H Microwave and Millimeter-Wave Techniques. Lecture: Continuum and other transformers. Taught by Prof. G.V. Eleftheriades

February 12, 2016 ece1236 No comments

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave
Techniques, taught by Prof. G.V. Eleftheriades, covering ch. 5 [1] content.

## Continuum transformer

A non-discrete impedance matching transformation, as sketched in fig. 1, is also possible.

../../figures/ece1236/taperedLinesFig1: fig. 1. Tapered impedance matching.

\label{eqn:continuumAndOtherTransformersCore:820}
\begin{aligned}
\Delta \Gamma
&= \frac{ (Z + \Delta Z) – Z }{(Z + \Delta Z) + Z} \\
&= \frac{\Delta Z}{2 Z}
\end{aligned}

\label{eqn:continuumAndOtherTransformersCore:840}
\Delta Z \rightarrow 0

\label{eqn:continuumAndOtherTransformersCore:860}
\begin{aligned}
d\Gamma
&= \frac{dZ}{2 Z} \\
&= \inv{2} \frac{d (\ln Z)}{dz} \\
&= \frac{Z_0}{Z} \frac{d (Z/Z_0)}{dz} \\
&= \frac{1}{Z} \frac{d Z}{dz}.
\end{aligned}

Hence as we did for multisection transformers, associate $$\Delta \Gamma$$ with $$e^{- 2j \beta z}$$ as sketched in fig. 2.

../../figures/ece1236/taperedLinesFig2: fig. 2. Reflection coefficient over an interval

assuming small reflections (i.e. $$Z(z)$$ is a slowly varying (adiabatic). Then

\label{eqn:continuumAndOtherTransformersCore:920}
\begin{aligned}
\Gamma(\omega)
&= \int_0^L e^{ -2 j \beta z} d\Gamma \\
&= \inv{2}
\int_0^L e^{ -2 j \beta z} \frac{d (\ln Z)}{dz} dz
\end{aligned}

This supplies the means to calculate the reflection coefficient for any impedance curve. As with the step impedance matching process, it is assumed that only first order reflections are of interest.

## Exponential taper

Let
\label{eqn:continuumAndOtherTransformersCore:620}
Z(z) = Z_0 e^{a z}, \qquad 0 < z < L subject to \label{eqn:continuumAndOtherTransformersCore:640} \begin{aligned} Z(0) &= Z_0 \\ Z(L) &= Z_0 e^{a L} = Z_{\textrm{L}}, \end{aligned} which gives $$\label{eqn:continuumAndOtherTransformersCore:660} \ln \frac{Z_{\textrm{L}}}{Z_0} = a L,$$ or $$\label{eqn:continuumAndOtherTransformersCore:680} a = \inv{L} \ln \frac{Z_{\textrm{L}}}{Z_0}$$ Also $$\label{eqn:continuumAndOtherTransformersCore:700} \frac{d}{dz} \ln \frac{Z_{\textrm{L}}}{Z_0} = \frac{d}{dz} (az) = a,$$ Hence \label{eqn:continuumAndOtherTransformersCore:740} \begin{aligned} \Gamma(\omega) &= \inv{2} \int_0^L e^{-2 j \beta z} \frac{d}{dz} \ln \frac{Z_{\textrm{L}}}{Z_0} dz \\ &= \frac{a}{2} \int_0^L e^{-2 j \beta z} dz \\ &= \frac{1}{2L} \ln \frac{Z_{\textrm{L}}}{Z_0} \evalrange{ \frac{e^{-2 j \beta z} }{ -2 j \beta} }{0}{L} \\ &= \frac{1}{2L \beta} \ln \frac{Z_{\textrm{L}}}{Z_0} \frac{ 1 - e^{-2 j \beta L} }{2 j} \\ &= \frac{1}{2} \ln \frac{Z_{\textrm{L}}}{Z_0} e^{-j \beta L} \frac{\sin( \beta L )}{\beta L}, \end{aligned} or $$\label{eqn:continuumAndOtherTransformersCore:940} \Gamma(\omega) = \frac{1}{2} \ln \frac{Z_{\textrm{L}}}{Z_0} e^{-j \beta L} \textrm{sinc}( \beta L ).$$

1. $$\beta$$ is constant with $$Z$$ varying: this is good only for TEM lines.
2. $$\Abs{\Gamma}$$ decreases with increasing length.
3. An electrical length $$\beta L > \pi$$, is required to minimize low frequency mismatch ($$L > \lambda/2$$).

This is sketched in fig. 3.

../../figures/ece1236/taperedLinesFig3: fig. 3. Exponential taper reflection coefficient.

Want:

\label{eqn:continuumAndOtherTransformersCore:760}
\beta L = \pi,

or
\label{eqn:continuumAndOtherTransformersCore:780}
\frac{\omega_c}{v_\phi} L = \pi

where $$\omega_c$$ is the cutoff frequency. This gives

\label{eqn:continuumAndOtherTransformersCore:800}
\omega_c = \frac{\pi v_\phi}{L}.

### Triangular Taper

\label{eqn:continuumAndOtherTransformersCore:960}
Z(z) =
\left\{
\begin{array}{l l}
Z_0 e^{2(z/L)^2 \ln(Z_{\textrm{L}}/Z_0)} & \quad \mbox{$$0 \le z \le L/2$$} \\
Z_0 e^{(4z/L – 2 z^2 – 1) \ln(Z_{\textrm{L}}/Z_0)} & \quad \mbox{$$L/2 \le z \le L$$} \\
\end{array}
\right.

\label{eqn:continuumAndOtherTransformersCore:980}
\frac{d}{dz} \ln (Z/Z_0) =
\left\{
\begin{array}{l l}
{(4z/L^2) \ln(Z_{\textrm{L}}/Z_0)} & \quad \mbox{$$0 \le z \le L/2$$} \\
{(4/L – 4z/L^2) \ln(Z_{\textrm{L}}/Z_0)} & \quad \mbox{$$L/2 \le z \le L$$} \\
\end{array}
\right.

In this case

\label{eqn:continuumAndOtherTransformersCore:1000}
\Gamma(\omega) = \frac{1}{2} e^{-\beta L} \ln \frac{Z_{\textrm{L}}}{Z_0} e^{-j \beta L} \textrm{sinc}^2( \beta L/2 ).

Compared to the exponential taper $$\textrm{sinc}( \beta L )$$ for the $$\beta L > 2 \pi$$ the peaks of $$\Abs{\Gamma}$$ are lower, but the first null occurs at $$\beta L = 2 \pi$$ whereas for the exponential taper it occurs at $$\beta L = \pi$$. This is sketched in fig. 4. The price to pay for this is that the zero is at $$2 \pi$$ so we have to make it twice as long to get the ripple down.

../../figures/ece1236/taperedLinesFig4: fig. 4. Triangular taper impedance curve.

### Klopfenstein Taper

For a given taper length $$L$$, the Klopfenstein taper is optimum in the sense that the reflection coefficient in the passband is minimum. Alternatively, for a given minimum reflection coefficient in the passband, the Klopfenstein taper yields the shortest length $$L$$.

Definition:

\label{eqn:continuumAndOtherTransformersCore:1020}
\ln Z = \inv{2} \ln (Z_0 Z_{\textrm{L}}) + \frac{\Gamma_0}{\cosh A} A^2 \phi(2 z/L -1, A), \qquad 0 \le z \le L,

where

\label{eqn:continuumAndOtherTransformersCore:1040}
\phi(x, A) = \int_0^x \frac{I_1(A\sqrt{1 – y^2})}{A \sqrt{1 – y^2}} dy, \qquad \Abs{x} \le 1.

Here $$I_1(x)$$ is the modified Bessel function. Note that
\label{eqn:continuumAndOtherTransformersCore:1060}
\begin{aligned}
\phi(0, A) &= 0 \\
\phi(x, 0) &= x/2 \\
\phi(1, A) &= \frac{\cosh A – 1}{A^2}
\end{aligned}

The resulting reflection coefficient is

\label{eqn:continuumAndOtherTransformersCore:1080}
\Gamma(\omega)
=
\left\{
\begin{array}{l l}
\Gamma_0 e^{-j \beta L} \frac{\cos\sqrt{(\beta L)^2 – A^2}}{\cosh A} & \quad \mbox{$$\beta L > A$$} \\
\Gamma_0 e^{-j \beta L} \frac{\cos\sqrt{A^2 – (\beta L)^2 }}{\cosh A} & \quad \mbox{$$\beta L < A$$} \\ \end{array} \right., where as usual $$\label{eqn:continuumAndOtherTransformersCore:1100} \Gamma_0 = \frac{Z_{\textrm{L}} - Z_0}{Z_{\textrm{L}} + Z_0} \approx \inv{2} \ln (Z_{\textrm{L}}/Z_0).$$ The passband is defined by $$\beta L \ge A$$ and the maximum ripple in the passband is $$\label{eqn:continuumAndOtherTransformersCore:1120} \Gamma_m = \frac{\Gamma_0}{\cosh A}.$$

### Example

Design a triangular taper, an exponential taper, and a Klopfenstein taper (with $$\Gamma_m = 0.02$$ ) to match a $$50 \Omega$$ load to a $$100 \Omega$$ line.

• Triangular taper:

\label{eqn:continuumAndOtherTransformersCore:1140}
Z(z) =
\left\{
\begin{array}{l l}
Z_0 e^{ 2(z/L)^2 \ln Z_{\textrm{L}}/Z_0 } & \quad \mbox{$$0 \le z \le L/2$$} \\
Z_0 e^{ (4 z/L – 2 z^2/L^2 – 1)\ln Z_{\textrm{L}}/Z_0 } & \quad \mbox{$$L/2 \ge z \ge L$$}
\end{array}
\right.

The resulting $$\Gamma$$ is

\label{eqn:continuumAndOtherTransformersCore:1160}
\Abs{\Gamma} = \inv{2} \ln (Z_{\textrm{L}}/Z_0) \textrm{sinc}^2\lr{ \beta L/2 }.

• Exponential taper:

\label{eqn:continuumAndOtherTransformersCore:1180}
\begin{aligned}
Z(z) &= Z_0 e^{a z}, \qquad 0 \le z \le L \\
a &= \inv{L} \ln (Z_{\textrm{L}}/Z_0) = \frac{0.693}{L} \\
\Abs{\Gamma} &= \inv{2} \ln (Z_{\textrm{L}}/Z_0) \textrm{sinc}( \beta L )
\end{aligned}

• Klopfenstein taper:
\label{eqn:continuumAndOtherTransformersCore:1200}
\begin{aligned}
Z(z) &= \inv{2} \ln (Z_{\textrm{L}}/Z_0) = 0.346 \\
A &= \cosh^{-1}\lr{ \frac{\Gamma_0}{\Gamma_m}} = \cosh^{-1}\lr{ \frac{0.346}{0.02}} = 3.543 \\
\Abs{\Gamma} &= \Gamma_0 \frac{\cos\sqrt{(\beta L)^2 – A^2}}{\cosh A},
\end{aligned}

The passband $$\beta L > A = 3.543 = 1.13 \pi$$. The impedance $$Z(z)$$ must be evaluated numerically.

To illustrate some of the differences, we are referred to fig. 5.21 [1]. It is noted that

1. The exponential taper has the lowest cutoff frequency $$\beta L = \pi$$. Then is the Klopfenstein taper which is close $$\beta L = 1.13 \pi$$. Last is the triangular with $$\beta L = 2 \pi$$.
2. The Klopfenstein taper has the lowest $$\Abs{\Gamma}$$ in the passband and meets the spec of $$\Gamma_m = 0.02$$. The worst $$\Abs{\Gamma}$$ in the passband is from the exponential taper and the triangular ripple is between the two others.

# References

[1] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.

## ECE1236H Microwave and Millimeter-Wave Techniques. Lecture 7: Multisection quarter-wavelength transformers. Taught by Prof. G.V. Eleftheriades

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave Techniques, taught by Prof. G.V. Eleftheriades, covering ch 5. [3] content.

## Terminology review

\label{eqn:uwavesDeck7MultisectionTransformersCore:20}
Z_{\textrm{in}} = R + j X

\label{eqn:uwavesDeck7MultisectionTransformersCore:40}
Y_{\textrm{in}} = G + j B

• $$Z_{\textrm{in}}$$ : impedance
• $$R$$ : resistance
• $$X$$ : reactance
• $$Y_{\textrm{in}}$$ : admittance
• $$G$$ : conductance
• $$B$$ : susceptance

Apparently this notation goes all the way back to Heavyside!

## Multisection transformers

Using a transformation of the form fig. 1 it is possible to optimize for maximum power delivery, using for example a matching transformation $$Z_{\textrm{in}} = Z_1^2/R_{\textrm{L}} = Z_0$$, or $$Z_1 = \sqrt{R_{\textrm{L}} Z_0}$$. Unfortunately, such a transformation does not allow any control over the bandwidth. This results in a pinched frequency response for which the standard solution is to add more steps as sketched in fig. 2.

../../figures/ece1236/Feb10Fig2: fig. 2. Pinched frequency response.
../../figures/ece1236/deck7Fig1: fig. 3. Single and multiple stage impedance matching.

This can be implemented in electronics, or potentially geometrically as in this sketch of a microwave stripline transformer implementation fig. 3.

../../figures/ece1236/deck7Fig3: fig. 3. Stripline implementation of staged impedance matching.

To find a multistep transformation algebraically can be hard, but it is easy to do on a Smith chart. The rule of thumb is that we want to stay near the center of the chart with each transformation.

There is however, a closed form method of calculating a specific sort of multisection transformation that is algebraically tractable. That method uses a chain of $$\lambda/4$$ transformers to increase the bandwidth as sketched in fig. 4.

../../figures/ece1236/deck7Fig4: fig. 4. Multiple $$\lambda/4$$ transformers.

The total reflection coefficient can be approximated to first order by summing the reflections at each stage (without considering there may be other internal reflections of transmitted field components). Algebraically that is

\label{eqn:uwavesDeck7MultisectionTransformersCore:60}
\Gamma(\Theta) \approx \Gamma_0
+ \Gamma_1 e^{-2 j \Theta} +
+ \Gamma_2 e^{-4 j \Theta} + \cdots
+ \Gamma_N e^{-2 N j \Theta},

where

\label{eqn:uwavesDeck7MultisectionTransformersCore:80}
\Gamma_n = \frac{Z_{n+1} – Z_n}{Z_{n+1} + Z_n}

Why? Consider reflections at the Z_1, Z_2 interface as sketched in fig. 5.

../../figures/ece1236/deck7Fig5: fig. 5. Single stage of multiple $$\lambda/4$$ transformers.

Assuming small reflections, where $$\Abs{\Gamma} \ll 1$$ then $$T = 1 + \Gamma \approx 1$$. Here

\label{eqn:uwavesDeck7MultisectionTransformersCore:100}
\begin{aligned}
\Theta
&= \beta l \\
&= \frac{2 \pi}{\lambda} \frac{\lambda}{4} \\
&= \frac{\pi}{2}.
\end{aligned}

at the design frequency $$\omega_0$$. We assume that $$Z_n$$ are either monotonically increasing if $$R_{\textrm{L}} > Z_0$$, or decreasing if $$R_{\textrm{L}} < Z_0$$.

### Binomial multisection transformers

Let

\label{eqn:uwavesDeck7MultisectionTransformersCore:120}
\Gamma(\Theta) = A \lr{ 1 + e^{-2 j \Theta} }^N

This type of a response is maximally flat, and is plotted in fig. 1.
../../figures/ece1236/multitransformerFig1: fig. 1. Binomial transformer.

The absolute value of the reflection coefficient is

\label{eqn:uwavesDeck7MultisectionTransformersCore:160}
\begin{aligned}
\Abs{\Gamma(\Theta)}
&=
\Abs{A} \lr{ e^{j \Theta} + e^{- j \Theta} }^N \\
&=
2^N \Abs{A} \cos^N\Theta.
\end{aligned}

When $$\Theta = \pi/2$$ this is clearly zero. It’s derivatives are

\label{eqn:uwavesDeck7MultisectionTransformersCore:180}
\begin{aligned}
\frac{d \Abs{\Gamma}}{d\Theta} &= -N \cos^{N-1} \Theta \sin\Theta \\
\frac{d^2 \Abs{\Gamma}}{d\Theta^2} &= -N \cos^{N-1} \Theta \cos\Theta N(N-1) \cos^{N-2} \Theta \sin\Theta \\
\frac{d^3 \Abs{\Gamma}}{d\Theta^3} &= \cdots
\end{aligned}

There is a $$\cos^{N-k}$$ term for all derivatives $$d^k/d\Theta^k$$ where $$k \le N-1$$, so for an N-section transformer

\label{eqn:uwavesDeck7MultisectionTransformersCore:140}
\frac{d^n}{d\Theta^n} \Abs{\Gamma(\Theta)}_{\omega_0} = 0,

for $$n = 1, 2, \cdots, N-1$$. The constant $$A$$ is determined by the limit $$\Theta \rightarrow 0$$, so

\label{eqn:uwavesDeck7MultisectionTransformersCore:200}
\Gamma(0) = 2^N A = \frac{Z_{\textrm{L}} – Z_0}{Z_{\textrm{L}} + Z_0},

because the various $$\Theta$$ sections become DC wires when the segment length goes to zero. This gives

\label{eqn:uwavesDeck7MultisectionTransformersCore:220}
\boxed{
A = 2^{-N} \frac{Z_{\textrm{L}} – Z_0}{Z_{\textrm{L}} + Z_0}.
}

The reflection coefficient can now be expanded using the binomial theorem

\label{eqn:uwavesDeck7MultisectionTransformersCore:240}
\begin{aligned}
\Gamma(\Theta)
&= A \lr{ 1 + e^{ 2 j \Theta } }^N \\
&= \sum_{k = 0}^N \binom{N}{k} e^{ -2 j k \Theta}
\end{aligned}

Recall that

\label{eqn:uwavesDeck7MultisectionTransformersCore:260}
\binom{N}{k} = \frac{N!}{k! (N-k)!},

providing a symmetric set of values

\label{eqn:uwavesDeck7MultisectionTransformersCore:280}
\begin{aligned}
\binom{N}{1} &= \binom{N}{N} = 1 \\
\binom{N}{1} &= \binom{N}{N-1} = N \\
\binom{N}{k} &= \binom{N}{N-k}.
\end{aligned}

Equating \ref{eqn:uwavesDeck7MultisectionTransformersCore:240} with \ref{eqn:uwavesDeck7MultisectionTransformersCore:60} we have

\label{eqn:uwavesDeck7MultisectionTransformersCore:300}
\boxed{
\Gamma_k = A \binom{N}{k}.
}

### Approximation for $$Z_k$$

From [1] (4.6.4), a log series expansion valid for all $$z$$ is

\label{eqn:uwavesDeck7MultisectionTransformersCore:320}
\ln z = \sum_{k = 0}^\infty \inv{2 k + 1} \lr{ \frac{ z – 1 }{z + 1} }^{2k + 1},

so for $$x$$ near unity a first order approximation of a logarithm is

\label{eqn:uwavesDeck7MultisectionTransformersCore:340}
\ln x \approx 2 \frac{x -1}{x+1}.

Assuming that $$Z_{k+1}/Z_k$$ is near unity we have

\label{eqn:uwavesDeck7MultisectionTransformersCore:360}
\begin{aligned}
\inv{2} \ln \frac{Z_{k+1}}{Z_k}
&\approx
\frac{ \frac{Z_{k+1}}{Z_k} – 1 }{\frac{Z_{k+1}}{Z_k} + 1} \\
&=
\frac{ Z_{k+1} – Z_k }{Z_{k+1} + Z_k} \\
&=
\Gamma_k.
\end{aligned}

Using this approximation, we get

\label{eqn:uwavesDeck7MultisectionTransformersCore:380}
\begin{aligned}
\ln \frac{Z_{k+1}}{Z_k}
&\approx
2 \Gamma_k \\
&= 2 A \binom{N}{k} \\
&= 2 \lr{2^{-N}} \binom{N}{k} \frac{Z_{\textrm{L}} – Z_0}{Z_{\textrm{L}} + Z_0} \\
&\approx
2^{-N} \binom{N}{k} \ln \frac{Z_{\textrm{L}}}{Z_0},
\end{aligned}

I asked what business do we have in assuming that $$Z_{\textrm{L}}/Z_0$$ is near unity? The answer was that it isn’t but surprisingly it works out well enough despite that. As an example, consider $$Z_0 = 100 \Omega$$ and $$R_{\textrm{L}} = 50 \Omega$$. The exact expression

\label{eqn:uwavesDeck7MultisectionTransformersCore:880}
\begin{aligned}
\frac{Z_{\textrm{L}} – Z_0}{Z_{\textrm{L}} + Z_0}
&= \frac{100-50}{100+50} \\
&= -0.333,
\end{aligned}

whereas
\label{eqn:uwavesDeck7MultisectionTransformersCore:900}
\inv{2} \ln \frac{Z_{\textrm{L}}}{Z_0} = -0.3466,

which is pretty close after all.

Regardless of whether or not that last approximation is used, one can proceed iteratively to $$Z_{k+1}$$ starting with $$k = 0$$.

### Bandwidth

To evaluate the bandwidth, let $$\Gamma_{\mathrm{m}}$$ be the maximum tolerable reflection coefficient over the passband, as sketched in fig. 6.

../../figures/ece1236/deck7Fig6: fig. 6. Max tolerable reflection.

That is

\label{eqn:uwavesDeck7MultisectionTransformersCore:400}
\begin{aligned}
\Gamma_m
&= 2^N \Abs{A} \Abs{\cos \Theta_m }^N \\
&= 2^N \Abs{A} \cos^N \Theta_m,
\end{aligned}

for $$\Theta_m < \pi/2$$. Then $$\label{eqn:uwavesDeck7MultisectionTransformersCore:420} \Theta_m = \arccos\lr{ \inv{2} \lr{ \frac{\Gamma_{\mathrm{m}}}{\Abs{A}}}^{1/N} }$$ The relative width of the interval is \label{eqn:uwavesDeck7MultisectionTransformersCore:440} \begin{aligned} \frac{\Delta f_{\mathrm{max}}}{f_0} &= \frac{\Delta \Theta_{\mathrm{max}}}{\Theta_0} \\ &= \frac{2 (\Theta_0 - \Theta_{\mathrm{max}}}{\Theta_0} \\ &= 2 - \frac{2 \Theta_{\mathrm{max}}}{\Theta_0} \\ &= 2 - \frac{4 \Theta_{\mathrm{max}}}{\pi} \\ &= 2 - \frac{4 }{\pi} \arccos\lr{ \inv{2} \lr{ \frac{\Gamma_{\mathrm{max}}}{\Abs{A}}}^{1/N} }. \end{aligned}

### Example

Design a 3-section binomial transformer to match $$R_{\textrm{L}} = 50 \Omega$$ to a line $$Z_0 = 100 \Omega$$. Calculate the BW based on a maximum $$\Gamma_{\textrm{m}} = 0.05$$.

### Solution

The scaling factor
\label{eqn:uwavesDeck7MultisectionTransformersCore:460}
\begin{aligned}
A
&= 2^{-N} \frac{Z_{\textrm{L}} – L_0}{Z_{\textrm{L}} + Z_0} \\
&\approx
\inv{2^{N+1}} \ln \frac{Z_{\textrm{L}}}{Z_0} \\
&= -0.0433
\end{aligned}

Now use

\label{eqn:uwavesDeck7MultisectionTransformersCore:940}
\ln \frac{Z_{n+1}}{Z_n}
\approx 2^{-N} \binom{N}{n} \ln \frac{R_{\textrm{L}}}{Z_0},

starting from

• $$n = 0$$.

\label{eqn:uwavesDeck7MultisectionTransformersCore:480}
\ln \frac{Z_{1}}{Z_0} \approx 2^{-3} \binom{3}{0} \ln \frac{R_{\textrm{L}}}{Z_0},

or
\label{eqn:uwavesDeck7MultisectionTransformersCore:500}
\begin{aligned}
\ln Z_{1}
&= \ln Z_0 + 2^{-3} \binom{3}{0} \ln \frac{R_{\textrm{L}}}{Z_0} \\
&= \ln 100 + 2^{-3} (1) \ln 0.5 \\
&= 4.518,
\end{aligned}

so
\label{eqn:uwavesDeck7MultisectionTransformersCore:520}
Z_1 = 91.7 \Omega

• $$n = 1$$

\label{eqn:uwavesDeck7MultisectionTransformersCore:540}
\ln Z_{2}
= \ln Z_1 + 2^{-3} \binom{3}{1} \ln \frac{50}{100} = 4.26

so
\label{eqn:uwavesDeck7MultisectionTransformersCore:560}
Z_2 = 70.7 \Omega

• $$n = 2$$

\label{eqn:uwavesDeck7MultisectionTransformersCore:580}
\ln Z_{3} = \ln Z_2 + 2^{-3} \binom{3}{2} \ln \frac{50}{100} = 4.0,

so

\label{eqn:uwavesDeck7MultisectionTransformersCore:600}
Z_3 = 54.5 \Omega.

With the fractional BW for $$\Gamma_m = 0.05$$, where $$10 \log_{10} \Abs{\Gamma_m}^2 = -26$$ dB}.

\label{eqn:uwavesDeck7MultisectionTransformersCore:920}
\begin{aligned}
\frac{\Delta f}{f_0}
&\approx
2 – \frac{4}{\pi} \arccos\lr{ \inv{2} \lr{ \frac{\Gamma_m}{\Abs{A}} }^{1/N} } \\
&=
2 – \frac{4}{\pi} \arccos\lr{ \inv{2} \lr{ \frac{0.05}{0.0433} }^{1/3} } \\
&= 0.7
\end{aligned}

At $$2$$ GHz, $$BW = 0.7$$ (70%), or $$1.4$$ GHz, which is the range $$[2.3,2.7]$$ GHz, whereas a single $$\lambda/4$$ transformer $$Z_T = \sqrt{ (100)(50) } = 70.7 \Omega$$ yields a BW of just $$0.36$$ GHz (18%).

# References

[1] DLMF. NIST Digital Library of Mathematical Functions. https://dlmf.nist.gov/, Release 1.0.10 of 2015-08-07. URL https://dlmf.nist.gov/. Online companion to Olver:2010:NHMF.

[2] F. W. J. Olver, D. W. Lozier, R. F. Boisvert, and C. W. Clark, editors. NIST Handbook of Mathematical Functions. Cambridge University Press, New York, NY, 2010. Print companion to NIST:DLMF.

[3] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.

## Average power for circuit elements

February 9, 2016 ece1236 No comments , , , , , , ,

In [2] section 2.2 is a comparison of field energy expressions with their circuit equivalents. It’s clearly been too long since I’ve worked with circuits, because I’d forgotten all the circuit energy expressions:

\label{eqn:averagePowerCircuitElements:20}
\begin{aligned}
W_{\textrm{R}} &= \frac{R}{2} \Abs{I}^2 \\
W_{\textrm{C}} &= \frac{C}{4} \Abs{V}^2 \\
W_{\textrm{L}} &= \frac{L}{4} \Abs{I}^2 \\
W_{\textrm{G}} &= \frac{G}{2} \Abs{V}^2 \\
\end{aligned}

Here’s a recap of where these come from

### Energy lost to resistance

Given
\label{eqn:averagePowerCircuitElements:40}
v(t) = R i(t)

the average power lost to a resistor is

\label{eqn:averagePowerCircuitElements:60}
\begin{aligned}
p_{\textrm{R}}
&= \inv{T} \int_0^T v(t) i(t) dt \\
&= \inv{T} \int_0^T \textrm{Re}( V e^{j \omega t} ) \Real( I e^{j \omega t} ) dt \\
&= \inv{4 T} \int_0^T
\lr{V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{I e^{j \omega t} + I^\conj e^{-j \omega t} }
dt \\
&= \inv{4 T} \int_0^T
\lr{
V I e^{2 j \omega t} + V^\conj I^\conj e^{-2 j \omega t}
+ V I^\conj + V^\conj I
}
dt \\
&= \inv{2} \textrm{Re}( V I^\conj ) \\
&= \inv{2} \textrm{Re}( I R I^\conj ) \\
&= \frac{R}{2} \Abs{I}^2.
\end{aligned}

Here it is assumed that the averaging is done over some integer multiple of the period, which kills off all the exponentials.

### Energy stored in a capacitor

I tried the same sort of analysis for a capacitor in phasor form, but everything cancelled out. Referring to [1], the approach used to figure this out is to operate first strictly in the time domain. Specifically, for the capacitor where $$i = C dv/dt$$ the power supplied up to a time $$t$$ is

\label{eqn:averagePowerCircuitElements:80}
\begin{aligned}
p_{\textrm{C}}(t)
&= \int_{-\infty}^t C \frac{dv}{dt} v(t) dt \\
&= \inv{2} C v^2(t).
\end{aligned}

The $$v^2(t)$$ term can now be expanded in terms of phasors and averaged for

\label{eqn:averagePowerCircuitElements:100}
\begin{aligned}
\overline{{p}}_{\textrm{C}}
&= \frac{C}{2T} \int_0^T \inv{4}
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} } dt \\
&= \frac{C}{2T} \int_0^T \inv{4}
2 \Abs{V}^2 dt \\
&= \frac{C}{4} \Abs{V}^2.
\end{aligned}

### Energy stored in an inductor

The inductor energy is found the same way, with

\label{eqn:averagePowerCircuitElements:120}
\begin{aligned}
p_{\textrm{L}}(t)
&= \int_{-\infty}^t L \frac{di}{dt} i(t) dt \\
&= \inv{2} L i^2(t),
\end{aligned}

which leads to

\label{eqn:averagePowerCircuitElements:140}
\overline{{p}}_{\textrm{L}}
= \frac{L}{4} \Abs{I}^2.

### Energy lost due to conductance

Finally, we have conductance. In phasor space that is defined by

\label{eqn:averagePowerCircuitElements:160}
G = \frac{I}{V} = \inv{R},

so power lost due to conductance follows from power lost due to resistance. In the average we have

\label{eqn:averagePowerCircuitElements:180}
\begin{aligned}
p_{\textrm{G}}
&= \inv{2 G} \Abs{I}^2 \\
&= \inv{2 G} \Abs{V G}^2 \\
&= \frac{G}{2} \Abs{V}^2
\end{aligned}

# References

[1] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.

[2] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.

## Survey: Are there any additional comments you would like to add about your graduate student experience at this university?

February 8, 2016 Incoherent ramblings No comments

I definitely enjoying the formal study of the courses I am taking, however, I find University courses inferior to self study in many ways. There are two specific benefits to a University course:

1) Having a knowledgeable instructor at hand to answer questions.
2) Having that same knowledge base available to set the study curriculum, since that instruction can direct attention to the most important aspects of the study.

The fixed formal lecture format is not terribly effective, at least for the scientific topics that I have been studying. In this day and age, when the technology to record lectures is so pervasive and freely available, there is really no excuse for the formal and fixed lecture style still found in the classroom. In an ideal learning environment, there would be time available between concepts to work through problems and gain complete understanding of each idea before going on to the next. Such an ideal learning environment would interleave practical work, text study and lectures. Testing should be used as a metric for whether or not the material is fully understood before continuing to the next aspect of study (or returning to areas of deficiency). In University classes testing is designed to produce a grade, not understanding, something that is completely backwards.

I am also surprised to see that grading for many graduate courses is still primarily based on exams. Having spent twenty years working before coming back to school it is a rude shock to come back to such an artificial metric for success. In nowhere other but school is it considered reasonable to make the metric for success based on how quickly a student can rush through a test, with no references at hand, attempting to construct answers that are optimized for maximal marks. In industry, the kind of mistakes that are inevitable in an examination context would lead to millions of dollars of service and product maintenance costs, and get people fired.

It is asinine that there is no feedback mechanism in place to review and retest of failed or partially successful final exam material. This seems to highlight the fact that university courses do not seem to be designed with learning as the primary goal. It is not clear what that goal is.

## ECE1236H Microwave and Millimeter-Wave Techniques. Lecture 3: Smith charts and impedance transformations. Taught by Prof. G.V. Eleftheriades

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave Techniques, taught by Prof. G.V. Eleftheriades, covering [2] chap. 2 content.

## Short circuited line

A short circuited line, also called a shorted stub, is sketched in fig. 1.

../../figures/ece1236/deck5smithChartsAndImpedenceTxFig1: fig. 1. Short circuited line.

With
\label{eqn:uwavesDeck5SmithChartCore:20}
Z_{\textrm{L}} = 0,

the input impedance is

\label{eqn:uwavesDeck5SmithChartCore:40}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{ Z_{\textrm{L}} + j Z_0 \tan(\beta l) }{ Z_0 + j Z_{\textrm{L}} \tan(\beta l)} \\
&= j Z_0 \tan(\beta l)
\end{aligned}

For short line sections $$\beta l \ll \pi/2$$, or $$l \ll \lambda/4$$, the input impedance is approximately

\label{eqn:uwavesDeck5SmithChartCore:80}
\begin{aligned}
Z_{\textrm{in}}
&= j Z_0 \tan(\beta l) \\
&\approx j Z_0 \sin(\beta l) \\
&\approx j Z_0 \beta l
\end{aligned}

Introducing an equivalent inductance defined by $$Z_{\textrm{in}} = j \omega L_{\mathrm{eq}}$$, we have

\label{eqn:uwavesDeck5SmithChartCore:100}
\begin{aligned}
L_{\mathrm{eq}}
&=
\frac{Z_0}{\omega} \beta l \\
&=
\frac{Z_0}{\omega} \frac{\omega}{v_\phi} l \\
&=
\frac{Z_0 l}{v_\phi}.
\end{aligned}

The inductance per unit length of the line is $$C = Z_0/v_\phi$$. An application for this result is that instead of using inductors, shorted stubs can be used in high frequency applications.

This is also the case for short sections of high impedance line.

## Open circuited line

An open circuited line is sketched in fig. 2.

../../figures/ece1236/deck5smithChartsAndImpedenceTxFig2: fig. 2. Open circuited line.

This time with $$Z_{\textrm{L}} \rightarrow \infty$$ we have

\label{eqn:uwavesDeck5SmithChartCore:120}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{ Z_{\textrm{L}} + j Z_0 \tan(\beta l) }{ Z_0 + j Z_{\textrm{L}} \tan(\beta l)} \\
&= -j Z_0 \cot(\beta l).
\end{aligned}

This time we have an equivalent capacitance. For short sections with $$\beta l \ll \pi/2$$

\label{eqn:uwavesDeck5SmithChartCore:140}
Z_{\textrm{in}}
\approx
-j \frac{Z_0}{\beta l}

Introducing an equivalent capacitance defined by $$Z_{\textrm{in}} = 1/(j \omega C_{\mathrm{eq}})$$, we have

\label{eqn:uwavesDeck5SmithChartCore:160}
\begin{aligned}
C_{\mathrm{eq}}
&=
\frac{ \beta l}{\omega Z_0} \\
&=
\frac{ \omega/v_\phi l}{\omega Z_0} \\
&=
\frac{ l}{v_\phi Z_0}
\end{aligned}

The capacitance per unit length of the line is $$C = 1/(Z_0 v_\phi)$$.

This is also the case for short sections of low impedance line.

## Half wavelength transformer.

A half wavelength transmission line equivalent circuit is sketched in fig. 3.

../../figures/ece1236/deck5smithChartsAndImpedenceTxFig3: fig. 3. Half wavelength transmission line.

With $$l = \lambda/2$$

\label{eqn:uwavesDeck5SmithChartCore:180}
\begin{aligned}
\beta l
&= \frac{2 \pi}{\lambda} \frac{\lambda}{2} \\
&= \pi.
\end{aligned}

Since $$\tan \pi = 0$$, the input impedance is

\label{eqn:uwavesDeck5SmithChartCore:200}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{ Z_{\textrm{L}} + j Z_0 \tan(\beta l) }{ Z_0 + j Z_{\textrm{L}} \tan(\beta l)} \\
&= Z_{\textrm{L}}.
\end{aligned}

## Quarter wavelength transformer.

A quarter wavelength transmission line equivalent circuit is sketched in fig. 4.

../../figures/ece1236/deck5smithChartsAndImpedenceTxFig4: fig. 4. Quarter wavelength transmission line.

With $$l = \lambda/4$$

\label{eqn:uwavesDeck5SmithChartCore:220}
\begin{aligned}
\beta l
&= \frac{2 \pi}{\lambda} \frac{\lambda}{4} \\
&= \frac{\pi}{2}.
\end{aligned}

We have $$\tan \beta l \rightarrow \infty$$, so the input impedance is

\label{eqn:uwavesDeck5SmithChartCore:240}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{ Z_{\textrm{L}} + j Z_0 \tan(\beta l) }{ Z_0 + j Z_{\textrm{L}} \tan(\beta l)} \\
&= \frac{Z_0^2}{Z_{\textrm{L}}}.
\end{aligned}

This relation

\label{eqn:uwavesDeck5SmithChartCore:280}
\boxed{
Z_{\textrm{in}}
= \frac{Z_0^2}{Z_{\textrm{L}}},
}

is called the impedance inverter.

• A large impedance is transformed into a small one and vice-versa.
• A short becomes an open and vice-versa.
• A capacitive load becomes inductive and vice-versa.
• If $$Z_{\textrm{L}}$$ is a series resonant circuit then $$Z_{\textrm{in}}$$ becomes parallel resonant.

See [1] for an explanation of the term series resonant.

### Matching with a $$\lambda/4$$ transformer.

Matching for a quarter wavelength transmission line equivalent circuit is sketched in fig. 5.

../../figures/ece1236/deck5smithChartsAndImpedenceTxFig5: fig. 5. Quarter wavelength transmission line matching.

For maximum power transfer

\label{eqn:uwavesDeck5SmithChartCore:300}
Z_{\textrm{in}} = \frac{Z_0^2}{R_{\textrm{L}}} = R_{\textrm{G}},

so

\label{eqn:uwavesDeck5SmithChartCore:320}
Z_0 = \sqrt{ R_{\textrm{G}} R_{\textrm{L}} }.

We have

\label{eqn:uwavesDeck5SmithChartCore:340}
\Abs{\Gamma_{\textrm{L}}} = \frac{ R_{\textrm{L}} – Z_0 }{R_{\textrm{L}} + Z_0} \ne 0,

and still maximum power is transferred.

## Smith chart

A Smith chart is a graphical tool for making the transformation $$\Gamma \leftrightarrow Z_{\textrm{in}}$$. Given

\label{eqn:uwavesDeck5SmithChartCore:360}
Z_{\textrm{in}} = Z_0 \frac{ 1 + \Gamma }{ 1 – \Gamma },

where $$\Gamma = \Gamma_{\textrm{L}} e^{- 2 j \beta l }$$, we begin by normalizing the input impedance, using an overbar to denote that normalization

\label{eqn:uwavesDeck5SmithChartCore:380}
Z_{\textrm{in}} \rightarrow \overline{{Z}}_{\textrm{in}} = \frac{Z_{\textrm{in}}}{Z_0},

so

\label{eqn:uwavesDeck5SmithChartCore:400}
\begin{aligned}
\overline{{Z}}_{\textrm{in}}
&= \frac{ 1 + \Gamma }{ 1 – \Gamma } \\
&= \frac{ (1 + \Gamma_r) + j \Gamma_i }{ (1 – \Gamma_r) – j \Gamma_i } \\
&= \frac{ \lr{ (1 + \Gamma_r) + j \Gamma_i}\lr{(1 – \Gamma_r) + j \Gamma_i} }{ (1 – \Gamma_r)^2 + \Gamma_i^2 } \\
&= \frac{ (1 – \Gamma_r^2 – \Gamma_i^2) + j \Gamma_i (1 – \Gamma_r + 1 + \Gamma_r ) }{ (1 – \Gamma_r)^2 + \Gamma_i^2 } \\
&= \frac{ (1 – \Abs{\Gamma}^2) + 2 j \Gamma_i }{ (1 – \Gamma_r)^2 + \Gamma_i^2 }.
\end{aligned}

If we let $$\overline{{Z}}_{\textrm{in}} = \overline{{\Gamma}}_{\textrm{L}} + j \overline{{X}}_{\textrm{L}}$$, and equate real and imaginary parts we have

\label{eqn:uwavesDeck5SmithChartCore:420}
\begin{aligned}
\overline{{\Gamma}}_{\textrm{L}} &= \frac{ 1 – \Abs{\Gamma}^2 }{ (1 – \Gamma_r)^2 + \Gamma_i^2 } \\
\overline{{X}}_{\textrm{L}} &= \frac{2 \Gamma_i }{ (1 – \Gamma_r)^2 + \Gamma_i^2 }
\end{aligned}

It is left as an exercise to demonstrate that these can be rearranged into

\label{eqn:uwavesDeck5SmithChartCore:440}
\begin{aligned}
\lr{ \Gamma_r – \frac{\overline{{\Gamma}}_{\textrm{L}}}{1 + \overline{{\Gamma}}_{\textrm{L}} } }^2 + \Gamma_i^2 &= \lr{ \inv{1 + \overline{{\Gamma}}_{\textrm{L}} }}^2 \\
\lr{ \Gamma_r – 1 }^2 + \lr{ \Gamma_i – \inv{\overline{{X}}_{\textrm{L}} } }^2 &= \inv{\overline{{X}}_{\textrm{L}}^2},
\end{aligned}

which trace out circles in the $$\Gamma_r, \Gamma_i$$ plane, one for the real part of $$\Gamma$$ and one for the imaginary part. This provides a graphical way for implementing the impedance transformation.

### Real impedance circle

The circle for the real part is centered at

\label{eqn:uwavesDeck5SmithChartCore:580}
\lr{ \frac{\overline{{\Gamma}}_{\textrm{L}}}{1 + \overline{{\Gamma}}_{\textrm{L}} }, 0 },

with radius
\label{eqn:uwavesDeck5SmithChartCore:600}
\inv{1 + \overline{{\Gamma}}_{\textrm{L}} }.

All these circles pass through the point $$(1,0)$$, since
\label{eqn:uwavesDeck5SmithChartCore:620}
\begin{aligned}
\frac{\overline{{\Gamma}}_{\textrm{L}}}{1 + \overline{{\Gamma}}_{\textrm{L}} } + \inv{1 + \overline{{\Gamma}}_{\textrm{L}} }
&=
\frac{1 + \overline{{\Gamma}}_{\textrm{L}}}{1 + \overline{{\Gamma}}_{\textrm{L}} } \\
&= 1.
\end{aligned}

For reactive loads where $$\overline{{\Gamma}}_{\textrm{L}} = 0$$, we have $$\Gamma_r^2 + \Gamma_i^2 = 1$$, a circle through the origin with unit radius.

For matched loads where $$\overline{{\Gamma}}_{\textrm{L}} = 1$$ the circle is centered at $$(1/2, 0)$$, with radius $$1/2$$.

### Imaginary impedance circle

The circle obtained by equating imaginary parts are constant reactance circles with center

\label{eqn:uwavesDeck5SmithChartCore:640}
\lr{ 1, \inv{\overline{{X}}_{\textrm{L}} } },

with radius

\label{eqn:uwavesDeck5SmithChartCore:660}
\inv{\overline{{X}}_{\textrm{L}}}.

These circles also pass through the point $$(1,0)$$. These circles are orthogonal to the constant resistance circles. Some of the features of a Smith chart are sketched in fig. 7.

../../figures/ece1236/deck5smithChartsAndImpedenceTxFig7: fig. 7. Hand sketched Smith chart.

A matlab produced blank Smith chart can be found in fig. 1.

../../figures/ece1236/smithchartFig1: fig. 1. Blank Smith chart.

### Example: Perform a transformation along a lossless line.

../../figures/ece1236/deck5smithChartsAndImpedenceTxFig8: fig. 8. Impedance transformation along lossless line.

Given

\label{eqn:uwavesDeck5SmithChartCore:700}
\overline{{Z}} = \frac{1 + \Gamma}{1 – \Gamma},

\label{eqn:uwavesDeck5SmithChartCore:720}
\Gamma = \Gamma_{\textrm{L}} e^{-2 j \beta l},

and

\label{eqn:uwavesDeck5SmithChartCore:740}
\Gamma_{\textrm{L}} = \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}} }

The total reflection coefficient is

\label{eqn:uwavesDeck5SmithChartCore:760}
\Gamma = \Abs{\Gamma_{\textrm{L}}} e^{j (\Theta_{\textrm{L}} – 2 \beta l) }

If $$\Gamma_{\textrm{L}} = \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}} }$$ is plotted on the Smith chart, then in order to move towards the generator, a subtraction from $$\Theta_{\textrm{L}}$$ of $$2 \beta l$$ is required.

Some worked examples that demonstrate this can be found in fig. 1, fig. 2, and fig. 3.

../../figures/ece1236/smithChartSlidesFig1: fig. 1. Mapping an impedance value onto a Smith chart.
../../figures/ece1236/smithChartSlidesFig2: fig. 2. Moving on the Smith chart towards the generator.
../../figures/ece1236/smithChartSlidesFig3: fig. 3. Moving on the Smith chart.

### Single stub tuning.

Referring to fig. 4, the procedure for single stub tuning is

../../figures/ece1236/smithChartSlidesFig4: fig. 4. Single stub tuning example

1. Plot the load on the Smith Chart.
2. Trace the constant VSWR circle. (blue).
3. Move toward the generator until the constant resistance=1 circle is reached (red). This determines the distance $$d$$.
4. Now the input impedance is of the form $$Z_{\textrm{A}} = 1 + j X$$.
5. We now have to use the stub to cancel out the $$j X$$ and make $$Z_{\textrm{in}} = 1$$ (matched).
6. This can be done on the Smith Chart. If $$X>0$$ then we need a capacitive stub (open). If $$X<0$$ then we need an inductive stub (shorted).
7. Say we need a capacitive stub (open): Start from the position of the open. Now the constant VSWR circle is the exterior unit
circle. Move toward the generator until you hit negative $$X$$. This determines the length of the stub $$l$$.

Notes:

1. In step (3) there are two points where the R=1 circle is intersected . Usually we chose the shortest one
2. By adding multiples of half-wavelength lengths to either $$d$$ or $$l$$ an infinite number of solutions can be constructed.

## Question: Find the Smith chart circle equations

Prove \ref{eqn:uwavesDeck5SmithChartCore:440}.

## Answer

We can write

\label{eqn:uwavesDeck5SmithChartCore:460}
(1 – \Gamma_r)^2 + \Gamma_i^2 = \frac{2 \Gamma_i }{\overline{{X}}_{\textrm{L}}},

or

\label{eqn:uwavesDeck5SmithChartCore:480}
(1 – \Gamma_r)^2 + \lr{ \Gamma_i – \inv{\overline{{X}}_{\textrm{L}}} }^2 = \inv{\lr{\overline{{X}}_{\textrm{L}}}^2},

which is one of the circular equations. For the other, putting the $$\Gamma_r, \Gamma_i$$ terms in the numerator, we have

\label{eqn:uwavesDeck5SmithChartCore:500}
\begin{aligned}
\frac{1 – \Gamma_r^2 – \Gamma_i^2 }{\overline{{\Gamma}}_{\textrm{L}}}
&=
(1 – \Gamma_r)^2 + \Gamma_i^2 \\
&=
1 – 2 \Gamma_r + \Gamma_r^2 + \Gamma_i^2,
\end{aligned}

or
\label{eqn:uwavesDeck5SmithChartCore:520}
\Gamma_r^2 \lr{ 1 + \inv{\overline{{\Gamma}}_{\textrm{L}}} } – 2 \Gamma_r + \Gamma_i^2 \lr{ 1 + \inv{\overline{{\Gamma}}_{\textrm{L}}} }
=
\inv{\overline{{\Gamma}}_{\textrm{L}}} – 1.

Dividing through by $$1 + \ifrac{1}{\overline{{\Gamma}}_{\textrm{L}}} = (\overline{{\Gamma}}_{\textrm{L}} + 1)/\overline{{\Gamma}}_{\textrm{L}}$$, we have

\label{eqn:uwavesDeck5SmithChartCore:540}
\begin{aligned}
\Gamma_r^2 – 2 \Gamma_r \frac{ \overline{{\Gamma}}_{\textrm{L}} }{\overline{{\Gamma}}_{\textrm{L}} + 1} + \Gamma_i^2
&=
\frac{1 – \overline{{\Gamma}}_{\textrm{L}}}{\overline{{\Gamma}}_{\textrm{L}}} \frac{ \overline{{\Gamma}}_{\textrm{L}} }{\overline{{\Gamma}}_{\textrm{L}} + 1} \\
&=
\frac{1 – \overline{{\Gamma}}_{\textrm{L}}}{ \overline{{\Gamma}}_{\textrm{L}} + 1},
\end{aligned}

or
\label{eqn:uwavesDeck5SmithChartCore:560}
\begin{aligned}
\lr{ \Gamma_r – \frac{ \overline{{\Gamma}}_{\textrm{L}} }{\overline{{\Gamma}}_{\textrm{L}} + 1} }^2 + \Gamma_i^2
&=
\frac{1 – \overline{{\Gamma}}_{\textrm{L}}}{ \overline{{\Gamma}}_{\textrm{L}} + 1} + \lr{ \frac{ \overline{{\Gamma}}_{\textrm{L}} }{\overline{{\Gamma}}_{\textrm{L}} + 1} }^2 \\
&=
\frac{ 1 – \overline{{\Gamma}}_{\textrm{L}}^2 + \overline{{\Gamma}}_{\textrm{L}}^2 }{\lr{\overline{{\Gamma}}_{\textrm{L}} + 1}^2} \\
&=
\frac{ 1 }{\lr{\overline{{\Gamma}}_{\textrm{L}} + 1}^2}.
\end{aligned}

# References

[1] EETech Media. Simple Series Resonance, 2016. URL http://www.allaboutcircuits.com/textbook/alternating-current/chpt-6/simple-series-resonance/. [Online; accessed 04-Feb-2016].

[2] David M Pozar. Microwave engineering. John Wiley \& Sons, 2009.

## ECE1236H Microwave and Millimeter-Wave Techniques: Transmission lines. Taught by Prof. G.V. Eleftheriades

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave
Techniques, taught by Prof. G.V. Eleftheriades, covering [1] chap. 2 content.

## Requirements

A transmission line requires two conductors as sketched in fig. 1, which shows a 2-wire line such a telephone line, a coaxial cable as found in cable TV distribution, and a microstrip line as found in cell phone RF interconnects.

../../figures/ece1236/deck4TxlineFig1: fig. 1. Transmission line examples.

A two-wire line becomes a transmission line when the wavelength of operation becomes comparable to the size of the line (or higher spectral component for pulses). In general a transmission line much support (TEM) transverse electromagnetic modes.

## Time harmonic solutions on transmission lines

In fig. 2, an electronic representation of a transmission line circuit is sketched.

../../figures/ece1236/deck4TxlineFig2: fig. 2. Transmission line equivalent circuit.

In this circuit all the elements have per-unit length units. With $$I = C dV/dt \sim j \omega C V$$, $$v = I R$$, and $$V = L dI/dt \sim j \omega L I$$, the KVL equation is

\label{eqn:uwaves4TransmissionLines:20}
V(z) – V(z + \Delta z) = I(z) \Delta z \lr{ R + j \omega L },

or in the $$\Delta z \rightarrow 0$$ limit

\label{eqn:uwaves4TransmissionLines:40}
\PD{z}{V} = -I(z) \lr{ R + j \omega L }.

The KCL equation at the interior node is

\label{eqn:uwaves4TransmissionLines:60}
-I(z) + I(z + \Delta z) + \lr{ j \omega C + G} V(z + \Delta z) = 0,

or
\label{eqn:uwaves4TransmissionLines:80}
\PD{z}{I} = -V(z) \lr{ j \omega C + G}.

This pair of equations is known as the telegrapher’s equations

\label{eqn:uwaves4TransmissionLines:100}
\boxed{
\begin{aligned}
\PD{z}{V} &= -I(z) \lr{ R + j \omega L } \\
\PD{z}{I} &= -V(z) \lr{ j \omega C + G}.
\end{aligned}
}

The second derivatives are

\label{eqn:uwaves4TransmissionLines:120}
\begin{aligned}
\PDSq{z}{V} &= -\PD{z}{I} \lr{ R + j \omega L } \\
\PDSq{z}{I} &= -\PD{z}{V} \lr{ j \omega C + G},
\end{aligned}

which allow the $$V, I$$ to be decoupled
\label{eqn:uwaves4TransmissionLines:140}
\boxed{
\begin{aligned}
\PDSq{z}{V} &= V(z) \lr{ j \omega C + G} \lr{ R + j \omega L } \\
\PDSq{z}{I} &= I(z) \lr{ R + j \omega L } \lr{ j \omega C + G},
\end{aligned}
}

With a complex propagation constant

\label{eqn:uwaves4TransmissionLines:160}
\begin{aligned}
\gamma
&= \alpha + j \beta \\
&= \sqrt{ \lr{ j \omega C + G} \lr{ R + j \omega L } } \\
&=
\sqrt{ R G – \omega^2 L C + j \omega ( L G + R C ) },
\end{aligned}

the decouple equations have the structure of a wave equation for a lossy line in the frequency domain

\label{eqn:uwaves4TransmissionLines:180}
\boxed{
\begin{aligned}
\PDSq{z}{V} – \gamma^2 V &= 0 \\
\PDSq{z}{I} – \gamma^2 I &= 0.
\end{aligned}
}

We write the solutions to these equations as

\label{eqn:uwaves4TransmissionLines:200}
\begin{aligned}
V(z) &= V_0^{+} e^{-\gamma z} + V_0^{-} e^{+\gamma z} \\
I(z) &= I_0^{+} e^{-\gamma z} – I_0^{-} e^{+\gamma z} \\
\end{aligned}

Only one of $$V$$ or $$I$$ is required since they are dependent through \ref{eqn:uwaves4TransmissionLines:100}, as can be seen by taking derivatives

\label{eqn:uwaves4TransmissionLines:220}
\begin{aligned}
\PD{z}{V}
&= \gamma \lr{ -V_0^{+} e^{-\gamma z} + V_0^{-} e^{+\gamma z} } \\
&=
-I(z) \lr{ R + j \omega L },
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:240}
I(z)
=
\frac{\gamma}{ R + j \omega L } \lr{ V_0^{+} e^{-\gamma z} – V_0^{-} e^{+\gamma z} }.

Introducing the characteristic impedance $$Z_0$$ of the line

\label{eqn:uwaves4TransmissionLines:260}
\begin{aligned}
Z_0
&= \frac{R + j \omega L}{\gamma} \\
&= \sqrt{ \frac{R + j \omega L}{G + j \omega C} },
\end{aligned}

we have

\label{eqn:uwaves4TransmissionLines:280}
\begin{aligned}
I(z)
&=
\inv{Z_0} \lr{ V_0^{+} e^{-\gamma z} – V_0^{-} e^{+\gamma z} } \\
&=
I_0^{+} e^{-\gamma z} – I_0^{-} e^{+\gamma z},
\end{aligned}

where

\label{eqn:uwaves4TransmissionLines:300}
\begin{aligned}
I_0^{+} &= \frac{V_0^{+}}{Z_0} \\
I_0^{-} &= \frac{V_0^{-}}{Z_0}.
\end{aligned}

## Mapping TL geometry to per unit length $$C$$ and $$L$$ elements

From electrostatics and magnetostatics the per unit length induction and capacitance constants for a co-axial cable can be calculated. For the cylindrical configuration sketched in fig. 3

../../figures/ece1236/deck4TxlineFig3: fig. 3. Coaxial cable.

From Gauss’ law the total charge can be calculated assuming that the ends of the cable can be neglected

\label{eqn:uwaves4TransmissionLines:520}
\begin{aligned}
Q
&= \int \spacegrad \cdot \BD dV \\
&= \oint \BD \cdot d\BA \\
&= \epsilon_0 \epsilon_r E ( 2 \pi r ) l,
\end{aligned}

This provides the radial electric field magnitude, in terms of the total charge

\label{eqn:uwaves4TransmissionLines:320}
E =
\frac{Q/l}{\epsilon_0 \epsilon_r ( 2 \pi r ) },

which must be a radial field as sketched in fig. 4.

../../figures/ece1236/deck4TxlineFig4: fig. 4. Radial electric field for coaxial cable.

The potential difference from the inner transmission surface to the outer is

\label{eqn:uwaves4TransmissionLines:340}
\begin{aligned}
V
&= \int_a^b E dr \\
&=
\frac{Q/l}{2 \pi \epsilon_0 \epsilon_r }
\int_a^b \frac{dr}{r} \\
&=
\frac{Q/l}{2 \pi \epsilon_0 \epsilon_r } \ln \frac{b}{a}.
\end{aligned}

Therefore the capacitance per unit length is

\label{eqn:uwaves4TransmissionLines:360}
C = \frac{Q/l}{V} = \frac{2 \pi \epsilon_0 \epsilon_r }{ \ln \frac{b}{a} } .

The inductance per unit length can be calculated form Ampere’s law

\label{eqn:uwaves4TransmissionLines:380}
\begin{aligned}
\int \lr{ \spacegrad \cross \BH } \cdot d\BS
&=
\int \BJ \cdot d\BS + \PD{t}{} \int \BD \cdot d\Bl \\
&=
\int \BJ \cdot d\BS \\
&=
I \\
&=
\oint \BH \cdot d\Bl \\
&=
H ( 2 \pi r ) \\
&=
\frac{B}{\mu_0} ( 2 \pi r )
\end{aligned}

The flux is

\label{eqn:uwaves4TransmissionLines:400}
\begin{aligned}
\Phi
&= \int \BB \cdot d\BA \\
&= \frac{\mu_0 I}{ 2 \pi } \int_A \inv{r} d dr \\
&= \frac{\mu_0 I}{ 2 \pi } \int_a^b \inv{r} l d dr \\
&= \frac{\mu_0 I l}{ 2 \pi } \ln \frac{b}{a}.
\end{aligned}

The inductance per unit length is

\label{eqn:uwaves4TransmissionLines:420}
L = \frac{\Phi/l}{I} = \frac{\mu_0}{ 2 \pi } \ln \frac{b}{a}.

For a lossless line where $$R = G = 0$$, we have $$\gamma = \sqrt{ (j \omega L)(j \omega C)} = j \omega \sqrt{L C}$$,
so the phase velocity for a (lossless) coaxial cable is

\label{eqn:uwaves4TransmissionLines:440}
\begin{aligned}
v_\phi
&= \frac{\omega}{\beta} \\
&= \frac{\omega}{\textrm{Im}(\gamma)} \\
&= \frac{\omega}{\omega \sqrt{LC})} \\
&= \frac{1}{\sqrt{LC})}.
\end{aligned}

This gives

\label{eqn:uwaves4TransmissionLines:460}
\begin{aligned}
v_\phi^2
&= \inv{ L } \inv{C} \\
&=
\frac{ 2 \pi }{ \mu_0 \ln \frac{b}{a} }
\frac
{\ln \frac{b}{a}}
{2 \pi \epsilon_0 \epsilon_r } \\
&=
\frac{1 }{ \mu_0 \epsilon_0 \epsilon_r } \\
&=
\frac{1 }{ \mu_0 \epsilon }.
\end{aligned}

So

\label{eqn:uwaves4TransmissionLines:480}
v_\phi = \inv{\sqrt{\epsilon \mu_0}},

which is the speed of light in the medium ($$\epsilon_r$$) that fills the co-axial cable.

This is \underline{not} a coincidence. In any two-wire homogeneously filled transmission line, the phase velocity is equal to the speed of light in the unbounded medium that fills the line.

The characteristic impedance (again assuming the lossless $$R = G = 0$$ case) is

\label{eqn:uwaves4TransmissionLines:500}
\begin{aligned}
Z_0
&= \sqrt{ \frac{R + j \omega L}{G + j \omega C} } \\
&= \sqrt{ \frac{j \omega L}{j \omega C} } \\
&= \sqrt{ \frac{L}{C} } \\
&= \sqrt{
\frac{\mu_0}{ 2 \pi } \ln \frac{b}{a}
\frac{ \ln \frac{b}{a} }{2 \pi \epsilon_0 \epsilon_r }
} \\
&=
\sqrt{ \frac{\mu_0}{\epsilon} } \frac{ \ln \frac{b}{a} }{ 2 \pi }.
\end{aligned}

Note that $$\eta = \sqrt{\mu_0/\epsilon_0} = 120 \pi \Omega$$ is the intrinsic impedance of free space. The values $$a, b$$ in \ref{eqn:uwaves4TransmissionLines:500} can be used to tune the characteristic impedance of the transmission line.

## Lossless line.

The lossless lossless case where $$R = G = 0$$ was considered above. The results were

\label{eqn:uwaves4TransmissionLines:540}
\gamma = j \omega \sqrt{ L C },

so $$\alpha = 0$$ and $$\beta = \omega \sqrt{LC}$$, and the phase velocity was

\label{eqn:uwaves4TransmissionLines:560}
v_\phi = \inv{\sqrt{LC}},

the characteristic impedance is

\label{eqn:uwaves4TransmissionLines:580}
Z_0 = \sqrt{\frac{L}{C}},

and the signals are
\label{eqn:uwaves4TransmissionLines:600}
\begin{aligned}
V(z) &= V_0^{+} e^{-j \beta z} + V_0^{-} e^{j \beta z} \\
I(z) &= \inv{Z_0} \lr{ V_0^{+} e^{-j \beta z} – V_0^{-} e^{j \beta z} }
\end{aligned}

In the time domain for an infinite line, we have

\label{eqn:uwaves4TransmissionLines:620}
\begin{aligned}
v(z, t)
&= \textrm{Re}\lr{ V(z) e^{j \omega t} } \\
&= V_0^{+} \textrm{Re}\lr{ e^{-j \beta z} e^{j \omega t} } \\
&= V_0^{+} \cos( \omega t – \beta z ).
\end{aligned}

In this case the shape and amplitude of the waveform are preserved as sketched in fig. 7.

../../figures/ece1236/deck4TxlineFig7: fig. 7. Lossless line signal preservation.

## Low loss line.

Assume $$R \ll \omega L$$ and $$G \ll \omega C$$. In this case we have

\label{eqn:uwaves4TransmissionLines:640}
\begin{aligned}
\gamma
&= \sqrt{ (R + j \omega L) ( G + j \omega C ) } \\
&=
j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j\omega L} }
\lr{ 1 + \frac{G}{j\omega C} }
} \\
&\approx
j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{2 j\omega L} }
\lr{ 1 + \frac{G}{2 j\omega C} } \\
&\approx
j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{2 j\omega L} + \frac{G}{2 j\omega C} } \\
&=
j \omega \sqrt{L C}
+ j \omega \frac{R \sqrt{C/L}}{2 j\omega}
+ j \omega \frac{G \sqrt{L/C}}{2 j\omega} \\
&=
j \omega \sqrt{L C}
+
\inv{2} \lr{
R \sqrt{\frac{C}{L}}
+
G \sqrt{\frac{L}{C}}
},
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:660}
\begin{aligned}
\alpha &=
\inv{2} \lr{
R \sqrt{\frac{C}{L}}
+
G \sqrt{\frac{L}{C}}
} \\
\beta &= \omega \sqrt{L C}.
\end{aligned}

Observe that this value for $$\beta$$ is the same as the lossless case to first order. We also have

\label{eqn:uwaves4TransmissionLines:680}
Z_0
= \sqrt{ \frac{R + j \omega L}{G + j \omega C} }
\approx
\sqrt{ \frac{L}{C} },

also the same as the lossless case. We must also have $$v_\phi = 1/\sqrt{L C}$$. To consider a time domain signal note that

\label{eqn:uwaves4TransmissionLines:700}
\begin{aligned}
V(z)
&= V_0^{+} e^{-\gamma z} \\
&= V_0^{+} e^{-\alpha z} e^{-j \beta z},
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:720}
\begin{aligned}
v(z, t)
&= \textrm{Re} \lr{ V(z) e^{j \omega t} } \\
&= \textrm{Re} \lr{ V_0^{+} e^{-\alpha z} e^{-j \beta z} e^{j \omega t} } \\
&= V_0^{+} e^{-\alpha z} \cos( \omega t – \beta z ).
\end{aligned}

The phase factor can be written

\label{eqn:uwaves4TransmissionLines:740}
\omega t – \beta z
=
\omega \lr{ t – \frac{\beta}{\omega} z }
\omega \lr{ t – z/v_\phi },

so the signal still moves with the phase velocity $$v_\phi = 1/\sqrt{LC}$$, but in a diminishing envelope as sketched in fig. 8.

../../figures/ece1236/deck4TxlineFig8: fig. 8. Time domain envelope for loss loss line.

Notes

• The shape is preserved but the amplitude has an exponential attenuation along the line.
• In this case, since $$\beta(\omega)$$ is a linear function to first order, we have no dispersion. All of the Fourier components of a pulse travel with the same phase velocity since $$v_\phi = \omega/\beta$$ is constant. i.e. $$v(z, t) = e^{-\alpha z} f( t – z/v_\phi )$$. We should expect dispersion when the $$R/\omega L$$ and $$G/\omega C$$ start becoming more significant.

## Distortionless line.

Motivated by the early telegraphy days, when low loss materials were not available. Therefore lines with a constant attenuation and constant phase velocity (i.e. no dispersion) were required in order to eliminate distortion of the signals. This can be achieved by setting

\label{eqn:uwaves4TransmissionLines:760}
\frac{R}{L} = \frac{G}{C}.

When that is done we have
\label{eqn:uwaves4TransmissionLines:780}
\begin{aligned}
\gamma
&= \sqrt{ (R + j \omega L) ( G + j \omega C ) } \\
&= j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j \omega L} }
\lr{ 1 + \frac{G}{j \omega C} }
} \\
&= j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j \omega L} }
\lr{ 1 + \frac{R}{j \omega L} }
} \\
&= j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{j \omega L} } \\
&= R \sqrt{\frac{C}{L} }
+ j \omega \sqrt{L C} \\
&= \sqrt{R G }
+ j \omega \sqrt{L C}.
\end{aligned}

We have

\label{eqn:uwaves4TransmissionLines:800}
\begin{aligned}
\alpha &= \sqrt{R G } \\
\beta &= \omega \sqrt{L C}.
\end{aligned}

The phase velocity is the same as that of the lossless and low-loss lines

\label{eqn:uwaves4TransmissionLines:820}
v_\phi = \frac{\omega}{\beta} = \inv{\sqrt{L C}}.

## Terminated lossless line.

Consider the load configuration sketched in fig. 9.

../../figures/ece1236/deck4TxlineFig9: fig. 9. Terminated line.

Recall that

\label{eqn:uwaves4TransmissionLines:840}
\begin{aligned}
V(z) &= V_0^{+} e^{-j \beta z} + V_0^{-} e^{+j \beta z} \\
I(z) &= \frac{V_0^{+}}{Z_0} e^{-j \beta z} – \frac{V_0^{-}}{Z_0} e^{+j \beta z} \\
\end{aligned}

At the load ($$z = 0$$), we have

\label{eqn:uwaves4TransmissionLines:860}
\begin{aligned}
V(0) &= V_0^{+} + V_0^{-} \\
I(0) &= \inv{Z_0} \lr{ V_0^{+} – V_0^{-} }
\end{aligned}

So

\label{eqn:uwaves4TransmissionLines:880}
\begin{aligned}
Z_{\textrm{L}}
&= \frac{V(0)}{I(0)} \\
&= Z_0 \frac{ V_0^{+} + V_0^{-} }{ V_0^{+} – V_0^{-} } \\
&= Z_0 \frac{ 1 + \Gamma_{\textrm{L}} }{1 – \Gamma_{\textrm{L}} },
\end{aligned}

where

\label{eqn:uwaves4TransmissionLines:900}
\Gamma_{\textrm{L}} \equiv \frac{V_0^{-} }{V_0^{+}},

is the reflection coefficient at the load.

The phasors for the signals take the form

\label{eqn:uwaves4TransmissionLines:920}
\begin{aligned}
V(z) &= V_0^{+} \lr{ e^{-j \beta z} + \Gamma_{\textrm{L}} e^{+j \beta z} } \\
I(z) &= \frac{V_0^{+}}{Z_0} \lr{ e^{-j \beta z} – \Gamma_{\textrm{L}} e^{+j \beta z} }.
\end{aligned}

Observe that we can rearranging for $$\Gamma_{\textrm{L}}$$ in terms of the impedances

\label{eqn:uwaves4TransmissionLines:940}
\lr{ 1 – \Gamma_{\textrm{L}} } Z_{\textrm{L}} = Z_0 \frac{ 1 + \Gamma_{\textrm{L}} },

or
\label{eqn:uwaves4TransmissionLines:960}
\Gamma_{\textrm{L}} \lr{ Z_0 + Z_{\textrm{L}} } = Z_{\textrm{L}} – Z_0,

or
\label{eqn:uwaves4TransmissionLines:980}
\Gamma_{\textrm{L}}
= \frac{Z_{\textrm{L}} – Z_0}
{ Z_0 + Z_{\textrm{L}} } .

### Power

The average (time) power on the line is

\label{eqn:uwaves4TransmissionLines:1000}
\begin{aligned}
P_{ \textrm{av}}
&= \inv{2} \textrm{Re}\lr{ V(Z) I^\conj(z) } \\
&=
\inv{2} \textrm{Re}
\lr{
V_0^{+} \lr{ e^{-j \beta z} + \Gamma_{\textrm{L}} e^{+j \beta z} }
\lr{\frac{V_0^{+}}{Z_0}}^\conj \lr{ e^{j \beta z} – \Gamma_{\textrm{L}}^\conj e^{-j \beta z} }
} \\
&= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \textrm{Re}\lr{
1 + \Gamma_{\textrm{L}} e^{2 j \beta z} – \Gamma_{\textrm{L}}^\conj e^{-2 j \beta z} – \Abs{\Gamma_{\textrm{L}}}^2
} \\
&= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \lr{
1 – \Abs{\Gamma_{\textrm{L}}}^2
}.
\end{aligned}

where we’ve made use of the fact that $$Z_0 = \sqrt{L/C}$$ is real for the lossless line, and the fact that a conjugate difference $$A – A^\conj = 2 j \textrm{Im}(A)$$ is purely imaginary.

This can be written as

\label{eqn:uwaves4TransmissionLines:1020}
P_{ \textrm{av}} = P^{+} – P^{-},

where

\label{eqn:uwaves4TransmissionLines:1040}
\begin{aligned}
P^{+} &= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \\
P^{+} &= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \Abs{\Gamma_{\textrm{L}}}^2.
\end{aligned}

This difference is the power delivered to the load. This is not z-dependent because we are considering the lossless case. Maximum power is delivered to the load when $$\Gamma_{\textrm{L}} = 0$$, which occurs when the impedances are matched.

## Return loss and insertion loss. Defined.

Return loss (dB) is defined as

\label{eqn:uwaves4TransmissionLines:1060}
\begin{aligned}
\textrm{RL}
&= 10 \log_{10} \frac{P_{\textrm{inc}}}{P_{\textrm{refl}}} \\
&= 10 \log_{10} \inv{\Abs{\Gamma}^2} \\
&= -20 \log_{10} \Abs{\Gamma}.
\end{aligned}

Insertion loss (dB) is defined as

\label{eqn:uwaves4TransmissionLines:1080}
\begin{aligned}
\textrm{IL}
&= 10 \log_{10} \frac{P_{\textrm{inc}}}{P_{\textrm{trans}}} \\
&= 10 \log_{10} \frac{P^{+}}{P^{+} – P^{-}} \\
&= 10 \log_{10} \inv{1 – \Abs{\Gamma}^2} \\
&= -10 \log_{10} \lr{ 1 – \Abs{\Gamma}^2 }.
\end{aligned}

## Standing wave ratio

Consider again the lossless loaded configuration of fig. 9. Now let $$z = – l$$, where $$l$$ is the distance from the load. The phasors at this point on the line are

\label{eqn:uwaves4TransmissionLines:1100}
\begin{aligned}
V(-l) &= V_0^{+} \lr{ e^{j \beta l} + \Gamma_{\textrm{L}} e^{-j \beta l} } \\
I(-l) &= \frac{V_0^{+}}{Z_0} \lr{ e^{j \beta l} – \Gamma_{\textrm{L}} e^{-j \beta l} } \\
\end{aligned}

The absolute voltage at this point is
\label{eqn:uwaves4TransmissionLines:1120}
\begin{aligned}
\Abs{V(-l)}
&= \Abs{V_0^{+}} \Abs{ e^{j \beta l} + \Gamma_{\textrm{L}} e^{-j \beta l} } \\
&= \Abs{V_0^{+}} \Abs{ 1 + \Gamma_{\textrm{L}} e^{-2 j \beta l} } \\
&= \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}}} e^{-2 j \beta l} },
\end{aligned}

where the complex valued $$\Gamma_{\textrm{L}}$$ is given by $$\Gamma_{\textrm{L}} = \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}}}$$.

This gives
\label{eqn:uwaves4TransmissionLines:1140}
\Abs{V(-l)}
= \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} e^{j(\Theta_{\textrm{L}} -2 \beta l)} }.

The voltage magnitude oscillates as one moves along the line. The maximum occurs when $$e^{j (\Theta_{\textrm{L}} -2 \beta l)} = 1$$

\label{eqn:uwaves4TransmissionLines:1160}
V_{\mathrm{max}} = \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} }.

This occurs when $$\Theta_{\textrm{L}} – 2 \beta l = 2 k \pi$$ for $$k = 0, 1, 2, \cdots$$. The minimum occurs when $$e^{j (\Theta_{\textrm{L}} -2 \beta l)} = -1$$

\label{eqn:uwaves4TransmissionLines:1180}
V_{\mathrm{min}} = \Abs{V_0^{+}} \Abs{ 1 – \Abs{\Gamma_{\textrm{L}}} },

which occurs when $$\Theta_{\textrm{L}} – 2 \beta l = (2 k – 1 )\pi$$ for $$k = 1, 2, \cdots$$. The standing wave ratio is defined as

\label{eqn:uwaves4TransmissionLines:1200}
\textrm{SWR} = \frac{V_{\mathrm{max}}}{V_{\mathrm{min}}} = \frac{ 1 + \Abs{\Gamma_{\textrm{L}}} }{ 1 – \Abs{\Gamma_{\textrm{L}}} }.

This is a measure of the mismatch of a line. This is sketched in fig. 10.

../../figures/ece1236/deck4TxlineFig10: fig. 10. SWR extremes.

Notes:

• Since $$0 \le \Abs{\Gamma_{\textrm{L}}} \le 1$$, we have $$1 \le \textrm{SWR} \le \infty$$. The lower bound is for a matched line, and open, short, or purely reactive termination leads to the infinities.
• The distance between two successive maxima (or minima) can be determined by setting $$\Theta_{\textrm{L}} – 2 \beta l = 2 k \pi$$ for two consecutive values of $$k$$. For $$k = 0$$, suppose that $$V_{\mathrm{max}}$$ occurs at $$d_1$$

\label{eqn:uwaves4TransmissionLines:1220}
\Theta_{\textrm{L}} – 2 \beta d_1 = 2 (0) \pi,

or
\label{eqn:uwaves4TransmissionLines:1240}
d_1 = \frac{\Theta_{\textrm{L}}}{2 \beta}.

For $$k = 1$$, let the max occur at $$d_2$$

\label{eqn:uwaves4TransmissionLines:1260}
\Theta_{\textrm{L}} – 2 \beta d_2 = 2 (1) \pi,

or
\label{eqn:uwaves4TransmissionLines:1280}
d_2 = \frac{\Theta_{\textrm{L}} – 2 \pi}{2 \beta}.

The difference is

\label{eqn:uwaves4TransmissionLines:1300}
\begin{aligned}
d_1 – d_2
&= \frac{\Theta_{\textrm{L}}}{2 \beta} – \frac{\Theta_{\textrm{L}} – 2 \pi}{2 \beta} \\
&= \frac{\pi}{\beta} \\
&= \frac{\pi}{2 \pi/\lambda} \\
&= \frac{\lambda}{2}.
\end{aligned}

The distance between two consecutive maxima (or minima) of the SWR is $$\lambda/2$$.

## Impedance Transformation.

Referring to fig. 11, let’s solve for the impedance at the load where $$z = 0$$ and at $$z = -l$$.

../../figures/ece1236/deck4TxlineFig11: fig. 11. Configuration for impedance transformation.

At any point on the line we have

\label{eqn:uwaves4TransmissionLinesCore:1320}
V(z) = V_0^{+} e^{-j \beta z} \lr{ 1 + \Gamma_{\textrm{L}} e^{2 j \beta z} },

so at the load and input we have

\label{eqn:uwaves4TransmissionLinesCore:1340}
\begin{aligned}
V_{\textrm{L}} &= V_0^{+} \lr{ 1 + \Gamma_{\textrm{L}} } \\
V(-l) &= V^{+} \lr{ 1 + \Gamma_{\textrm{L}}(-1) },
\end{aligned}

where

\label{eqn:uwaves4TransmissionLinesCore:1360}
\begin{aligned}
V^{+} &= V_0^{+} e^{ j \beta l } \\
\Gamma_{\textrm{L}}(-1) &= \Gamma_{\textrm{L}} e^{-2 j \beta l}
\end{aligned}

Similarly

\label{eqn:uwaves4TransmissionLinesCore:1380}
I(-l) = \frac{V^{+}}{Z_0} \lr{ 1 – \Gamma_{\textrm{L}}(-1) }.

Define an input impedance as
\label{eqn:uwaves4TransmissionLinesCore:1400}
\begin{aligned}
Z_{\textrm{in}}
&= \frac{V(-l)}{I(-l)} \\
&= Z_0 \frac{1 + \Gamma_{\textrm{L}}(-1)}{1 – \Gamma_{\textrm{L}}(-1)}
\end{aligned}

This is analogous to

\label{eqn:uwaves4TransmissionLinesCore:1420}
Z_{\textrm{L}}
= Z_0 \frac{1 + \Gamma_{\textrm{L}}}{1 – \Gamma_{\textrm{L}}}

From \ref{eqn:uwaves4TransmissionLines:980}, we have

\label{eqn:uwaves4TransmissionLinesCore:1440}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{Z_0 + Z_{\textrm{L}} + \lr{Z_{\textrm{L}} – Z_0} e^{-2 j \beta l}}{Z_0 + Z_{\textrm{L}} – \lr{Z_{\textrm{L}} – Z_0} e^{-2 j \beta l}} \\
&= Z_0 \frac{\lr{Z_0 + Z_{\textrm{L}}} e^{j\beta l} + \lr{Z_{\textrm{L}} –
Z_0} e^{- j \beta l}}{\lr{Z_0 + Z_{\textrm{L}}} e^{j\beta l} – \lr{Z_{\textrm{L}} – Z_0} e^{- j \beta l}} \\
&= Z_0
\frac
{Z_{\textrm{L}} \cos( \beta l ) + j Z_0 \sin(\beta l ) }
{Z_0 \cos( \beta l ) + j Z_{\textrm{L}} \sin(\beta l ) },
\end{aligned}

or
\label{eqn:uwaves4TransmissionLinesCore:1460}
\boxed{
Z_{\textrm{in}} =
\frac
{Z_{\textrm{L}} + j Z_0 \tan(\beta l ) }
{Z_0 + j Z_{\textrm{L}} \tan(\beta l ) }.
}

This can be thought of as providing a reflection coefficient function along the line to the load at any point as sketched in fig. 12.

../../figures/ece1236/deck4TxlineFig12: fig. 12. Impedance transformation reflection on the line.

# References

[1] David M Pozar. Microwave engineering. John Wiley \& Sons, 2009.