## 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.

## 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.

## Impedance transformation

[Click here for a PDF of this post with nicer formatting]

In our final problem set we used the impedance transformation for calculations related to a microslot antenna. This transformation wasn’t familiar to me, and is apparently covered in the third year ECE fields class. I found a derivation of this in [1], but the idea is really simple and follows from the reflection coefficient calculation for a normal reflection configuration.

Consider a normal field reflection between two interfaces, as sketched in fig. 1.

fig. 1. Normal reflection and transmission between two media.

The fields are

\label{eqn:impedanceTransformation:40}
\BE^\textrm{i} = \xcap E_0 e^{-j k_1 z}

\label{eqn:impedanceTransformation:60}
\BH^\textrm{i} = \ycap \frac{E_0}{\eta_1} e^{-j k_1 z}

\label{eqn:impedanceTransformation:80}
\BE^\textrm{r} = \xcap \Gamma E_0 e^{j k_1 z}

\label{eqn:impedanceTransformation:100}
\BH^\textrm{r} = -\ycap \Gamma \frac{E_0}{\eta_1} e^{j k_1 z}

\label{eqn:impedanceTransformation:120}
\BE^\textrm{t} = \xcap E_0 T e^{-j k_2 z}

\label{eqn:impedanceTransformation:140}
\BH^\textrm{t} = \ycap \frac{E_0}{\eta_1} T e^{-j k_2 z}.

The field orientations have been picked so that the tangential component of the electric field is $$\xcap$$ oriented for all of the incident, reflected, and transmitted components. Requiring equality of the tangential field components at the interface gives

\label{eqn:impedanceTransformation:180}
1 + \Gamma = T

\label{eqn:impedanceTransformation:200}
\inv{\eta_1} – \frac{\Gamma}{\eta_1} = \frac{T}{\eta_2}.

Solving for the transmission coefficient gives

\label{eqn:impedanceTransformation:220}
\begin{aligned}
T
&= \frac{2}{ 1 + \frac{\eta_1}{\eta_2} } \\
&= \frac{2 \eta_2}{ \eta_2 + \eta_1 },
\end{aligned}

and for the reflection coefficient

\label{eqn:impedanceTransformation:240}
\begin{aligned}
\Gamma
&= T – 1 \\
&= \frac{2 \eta_2 – \eta_1 – \eta_2}{ \eta_2 + \eta_1 } \\
&= \frac{\eta_2 – \eta_1 }{ \eta_2 + \eta_1 }.
\end{aligned}

The total fields in medium 1 at the point $$z = -l$$ are

\label{eqn:impedanceTransformation:280}
\BE^\textrm{i} + \BE^\textrm{r}
=
\xcap E_0 \lr{ e^{ -j k_1 (-l)} + \Gamma e^{ j k_1 (-l) } }

\label{eqn:impedanceTransformation:300}
\BH^\textrm{i} + \BH^\textrm{r}
=
\ycap \frac{E_0}{\eta_1} \lr{ e^{ -j k_1 (-l)} – \Gamma e^{ j k_1 (-l) }}.

The ratio of the electric field strength to the magnetic field strength is defined as the input impedance

\label{eqn:impedanceTransformation:320}
Z_{\textrm{in}} \equiv \evalbar{\frac{E^\textrm{i} + E^\textrm{r}}{H^\textrm{i} + H^\textrm{r}}}{ z = -l}.

That is

\label{eqn:impedanceTransformation:340}
\begin{aligned}
Z_{\textrm{in}}
&=
\eta_1 \frac{
e^{ j k_1 l} + \Gamma e^{ -j k_1 l }
}{
e^{ j k_1 l} – \Gamma e^{ -j k_1 l }
} \\
&=
\eta_1 \frac{
\lr{ \eta_1 + \eta_2} e^{ j k_1 l} + \lr{ \eta_2 – \eta_1} e^{ -j k_1 l }
}{
\lr{ \eta_1 + \eta_2} e^{ j k_1 l} – \lr{ \eta_2 – \eta_1} e^{ -j k_1 l }
} \\
&=
\eta_1 \frac{
\eta_2 \cos( k_1 l ) + \eta_1 j \sin( k_1 l)
}{
\eta_2 j \sin( k_1 l ) + \eta_1 \cos( k_1 l)
},
\end{aligned}

or
\label{eqn:impedanceTransformation:360}
\boxed{
Z_{\textrm{in}}
=
\eta_1 \frac{
\eta_2 + j \eta_1 \tan( k_1 l)
}{
\eta_1 + j \eta_2 \tan( k_1 l )
}.
}

# References

[1] Constantine A Balanis. Advanced engineering electromagnetics, chapter {Reflection and transmission}. Wiley New York, 1989.

## Updated notes for ece1229 antenna theory

I’ve now posted a first update of my notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

## Notes for Balantis chapter 4: linear wire antennas.

[Click here for a PDF of this post with nicer formatting]

These are notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheriades, covering ch. 4 [1] content.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book.)

## Magnetic Vector Potential.

In class and in the problem set $$\BA$$ was referred to as the Magnetic Vector Potential.  I only recalled this referred to as the Vector Potential.  Prefixing this with magnetic seemed counter intuitive to me since it is generated by electric sources (charges and currents).
This terminology can be justified due to the fact that $$\BA$$ generates the magnetic field by its curl. Some mention of this can be found in [4], which also points out that the Electric Potential refers to the scalar $$\phi$$. Prof. Eleftheriades points out that Electric Vector Potential refers to the vector potential $$\BF$$ generated by magnetic sources (because in that case the electric field is generated by the curl of $$\BF$$.)

## Plots of infinitesimal dipole radial dependence.

In section 4.2 of [1] are some discussions of the $$kr < 1$$, $$kr = 1$$, and $$kr > 1$$ radial dependence of the fields and power of a solution to an infinitesimal dipole system. Here are some plots of those $$k r$$ dependence, along with the $$k r = 1$$ contour as a reference. All the $$\theta$$ dependence and any scaling is left out.

The CDF notebook visualizeDipoleFields.cdf is available to interactively plot these, rotate the plots and change the ranges of what is plotted.

A plot of the real and imaginary parts of $$H_\phi = \frac{j k}{r} e^{-j k r} \lr{ 1-\frac{j}{k r} }$$ can be found in fig. 1 and fig. 2.

fig 1. Radial dependence of Re H_phi

fig 2. Radial dependence of Im H_phi

A plot of the real and imaginary parts of $$E_r = \inv{r^2} \lr{1-\frac{j}{k r}} e^{-j k r}$$ can be found in fig. 3 and fig. 4.

fig 3. Radial dependence of Re E_r

fig 4. Radial dependence of Im E_r

Finally, a plot of the real and imaginary parts of $$E_\theta = \frac{ j k }{r} \lr{1 -\frac{j}{k r} -\frac{1}{k^2 r^2} } e^{-j k r}$$ can be found in fig. 5 and fig. 6.

fig. 5. Radial dependence of Re E_theta

fig. 6. Radial dependence of Im E_theta

## Electric Far field for a spherical potential.

It is interesting to look at the far electric field associated with an arbitrary spherical magnetic vector potential, assuming all of the radial dependence is in the spherical envelope. That is

\label{eqn:chapter4Notes:20}
\BA = \frac{e^{-j k r}}{r} \lr{
\rcap a_r\lr{ \theta, \phi }
+\thetacap a_\theta\lr{ \theta, \phi }
+\phicap a_\phi\lr{ \theta, \phi }
}.

The electric field is

\label{eqn:chapter4Notes:40}
\BE = – j \omega \BA – j \frac{1}{\omega \mu_0 \epsilon_0 } \spacegrad \lr{\spacegrad \cdot \BA }.

The divergence and gradient in spherical coordinates are

\label{eqn:chapter4Notes:80}
\begin{aligned}
&=
\inv{r^2} \PD{r}{} \lr{ r^2 A_r }
+ \inv{r \sin\theta } \PD{\theta}{} \lr{A_\theta \sin\theta}
+ \inv{r \sin\theta } \PD{\phi}{A_\phi}
\end{aligned}

\label{eqn:chapter4Notes:100}
\begin{aligned}
&=
\rcap \PD{r}{\psi}
+\frac{\thetacap}{r} \PD{\theta}{\psi}
+ \frac{\phicap}{r \sin\theta} \PD{\phi}{\psi}.
\end{aligned}

For the assumed potential, the divergence is

\label{eqn:chapter4Notes:120}
\begin{aligned}
&=
\frac{a_r}{r^2} \PD{r}{} \lr{ r^2 \frac{e^{-j k r}}{r} }
+ \inv{r \sin\theta } \frac{e^{-j k r}}{r} \PD{\theta}{} \lr{\sin\theta a_\theta}
+ \inv{r \sin\theta } \frac{e^{-j k r}}{r} \PD{\phi}{a_\phi} \\
&=
a_r
e^{-j k r}
\lr{
\inv{r^2}
-j k \inv{r}
}
+ \inv{r^2 \sin\theta } e^{-j k r} \PD{\theta}{} \lr{\sin\theta a_\theta}
+ \inv{r^2 \sin\theta } e^{-j k r} \PD{\phi}{a_\phi} \\
&\approx
-j k \frac{a_r}{r}
e^{-j k r}.
\end{aligned}

The last approximation dropped all the $$1/r^2$$ terms that will be small compared to $$1/r$$ contribution that dominates when $$r \rightarrow \infty$$, the far field.

The gradient can now be computed

\label{eqn:chapter4Notes:140}
\begin{aligned}
&\approx
-j k
\lr{
\frac{a_r}{r}
e^{-j k r}
} \\
&=
-j k \lr{
\rcap \PD{r}{}
+\frac{\thetacap}{r} \PD{\theta}{}
+ \frac{\phicap}{r \sin\theta} \PD{\phi}{}
}
\frac{a_r}{r}
e^{-j k r} \\
&=
-j k \lr{
\rcap a_r \PD{r}{} \lr{
\frac{1}{r}
e^{-j k r}
}
+\frac{\thetacap}{r^2}
e^{-j k r}
\PD{\theta}{a_r}
+
e^{-j k r}
\frac{\phicap}{r^2 \sin\theta}
\PD{\phi}{a_r}
} \\
&=
-j k \lr{
-\rcap \frac{a_r}{r^2} \lr{
1
+ j k r
}
+\frac{\thetacap}{r^2}
\PD{\theta}{a_r}
+
\frac{\phicap}{r^2 \sin\theta}
\PD{\phi}{a_r}
}
e^{-j k r} \\
&\approx
– k^2 \rcap \frac{a_r}{r}
e^{-j k r}.
\end{aligned}

Again, a far field approximation has been used to kill all the $$1/r^2$$ terms.

The far field approximation of the electric field is now possible

\label{eqn:chapter4Notes:160}
\begin{aligned}
\BE
&= – j \omega \BA – j \frac{1}{\omega \mu_0 \epsilon_0 } \spacegrad \lr{\spacegrad \cdot \BA } \\
&=
– j \omega
\frac{e^{-j k r}}{r} \lr{
\rcap a_r\lr{ \theta, \phi }
+\thetacap a_\theta\lr{ \theta, \phi }
+\phicap a_\phi\lr{ \theta, \phi }
}
+ j \frac{1}{\omega \mu_0 \epsilon_0 }
k^2 \rcap \frac{a_r}{r}
e^{-j k r} \\
&=
– j \omega
\frac{e^{-j k r}}{r} \lr{
\rcap a_r\lr{ \theta, \phi }
+\thetacap a_\theta\lr{ \theta, \phi }
+\phicap a_\phi\lr{ \theta, \phi }
}
+ j \frac{c^2}{\omega }
\lr{\frac{\omega}{c}}^2 \rcap \frac{a_r}{r}
e^{-j k r}
\\
&=
– j \omega
\frac{e^{-j k r}}{r} \lr{
\thetacap a_\theta\lr{ \theta, \phi }
+\phicap a_\phi\lr{ \theta, \phi }
}.
\end{aligned}

Observe the perfect, somewhat miraculous seeming, cancellation of all the radial components of the field. If $$\BA_{\textrm{T}}$$ is the non-radial projection of $$\BA$$, the electric far field is just

\label{eqn:chapter4Notes:180}
\boxed{
\BE_{\textrm{ff}} = -j \omega \BA_{\textrm{T}}.
}

## Magnetic Far field for a spherical potential.

Application of the same assumed representation for the magnetic field gives
\label{eqn:chapter4Notes:220}
\begin{aligned}
\BB
&=
\spacegrad \cross \BA \\
&=
\frac{\rcap}{r \sin\theta} \partial_\theta \lr{A_\phi \sin\theta}
+ \frac{\thetacap}{r} \lr{ \inv{\sin\theta} \partial_\phi A_r – \partial_r \lr{r A_\phi}}
+ \frac{\phicap}{r} \lr{ \partial_r\lr{r A_\theta} – \partial_\theta A_r} \\
&=
\frac{\rcap}{r \sin\theta} \partial_\theta \lr{
\frac{e^{-j k r}}{r} a_\phi
\sin\theta}
+ \frac{\thetacap}{r} \lr{ \inv{\sin\theta} \partial_\phi \lr{
\frac{e^{-j k r}}{r} a_r
} – \partial_r \lr{r
\frac{e^{-j k r}}{r} a_\phi
}
}
+ \frac{\phicap}{r} \lr{ \partial_r\lr{r
\frac{e^{-j k r}}{r} a_\theta
} – \partial_\theta
\lr{
\frac{e^{-j k r}}{r} a_r
}
} \\
&=
\frac{\rcap}{r \sin\theta}
\frac{e^{-j k r}}{r}
\partial_\theta \lr{
a_\phi
\sin\theta}
+ \frac{\thetacap}{r} \lr{ \inv{\sin\theta}
\frac{e^{-j k r}}{r}
\partial_\phi
a_r
– \partial_r \lr{
e^{-j k r}
}
a_\phi
}
+ \frac{\phicap}{r} \lr{
\partial_r
\lr{
e^{-j k r}
}
a_\theta

\frac{e^{-j k r}}{r}
\partial_\theta
a_r
}
\approx
j k \lr{ \thetacap a_\phi

\phicap a_\theta
}
\frac{e^{-j k r}}{r} \\
&=
-j k \rcap \cross \lr{
\thetacap a_\theta
+\phicap a_\phi
}
\frac{e^{-j k r}}{r} \\
&=
\inv{c} \BE_{\textrm{ff}}.
\end{aligned}

The approximation above drops the $$1/r^2$$ terms. Since

\label{eqn:chapter4Notes:240}
\inv{\mu_0 c} = \inv{\mu_0} \sqrt{\mu_0\epsilon_0} = \sqrt{\frac{\epsilon_0}{\mu_0}} = \inv{\eta},

the magnetic far field can be expressed in terms of the electric far field as
\label{eqn:chapter4Notes:260}
\boxed{
\BH = \inv{\eta} \rcap \cross \BE.
}

## Plane wave relations between electric and magnetic fields

I recalled an identity of the form \ref{eqn:chapter4Notes:260} in [3], but didn’t think that it required a far field approximation.
The reason for this was because the Jackson identity assumed a plane wave representation of the field, something that the far field assumptions also locally require.

Assuming a plane wave representation for both fields

\label{eqn:chapter4Notes:300}
\boldsymbol{\mathcal{E}}(\Bx, t) = \BE e^{j \lr{\omega t – \Bk \cdot \Bx}}

\label{eqn:chapter4Notes:320}
\boldsymbol{\mathcal{B}}(\Bx, t) = \BB e^{j \lr{\omega t – \Bk \cdot \Bx}}

The cross product relation between the fields follows from the Maxwell-Faraday law of induction

\label{eqn:chapter4Notes:340}
0 = \spacegrad \cross \boldsymbol{\mathcal{E}} + \PD{t}{\boldsymbol{\mathcal{B}}},

or

\label{eqn:chapter4Notes:360}
\begin{aligned}
0
&=
\Be_r \cross \BE \partial_r e^{j\lr{ \omega t – \Bk \cdot \Bx}}
+
j \omega \BB e^{j \lr{\omega t – \Bk \cdot \Bx}} \\
&=
-j \Be_r k_r \cross \BE e^{j \lr{\omega t – \Bk \cdot \Bx}}
+
j \omega \BB e^{j \lr{\omega t – \Bk \cdot \Bx}} \\
&=
\lr{ – \Bk \cross \BE + \omega \BB } j
e^{j \lr{\omega t – \Bk \cdot \Bx}},
\end{aligned}

or

\label{eqn:chapter4Notes:380}
\begin{aligned}
\BH
&= \frac{ k}{k c \mu_0 } \kcap \cross \BE \\
&= \inv{ \eta } \kcap \cross \BE,
\end{aligned}

which also finds \ref{eqn:chapter4Notes:260}, but with much less work and less mess.

## Transverse only nature of the far-field fields

Also observe that its possible to tell that the far field fields have only transverse components using the same argument that they are locally plane waves at that distance. The plane waves must satisfy the zero divergence Maxwell’s equations

\label{eqn:chapter4Notes:420}
\spacegrad \cdot \boldsymbol{\mathcal{E}} = 0

\label{eqn:chapter4Notes:440}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = 0,

so by the same logic

\label{eqn:chapter4Notes:480}
\Bk \cdot \BE = 0

\label{eqn:chapter4Notes:500}
\Bk \cdot \BB = 0.

In the far field the electric field must equal its transverse projection

\label{eqn:chapter4Notes:520}
\BE = \textrm{Proj}_\T \lr{-j \omega \BA
– j \frac{1}{\omega \mu_0 \epsilon_0 } \spacegrad \lr{\spacegrad \cdot \BA } }.

Since by \ref{eqn:chapter4Notes:140} the scalar potential term has only a radial component, that leaves

\label{eqn:chapter4Notes:540}
\BE = -j \omega \textrm{Proj}_\T \BA,

which provides \ref{eqn:chapter4Notes:180} with slightly less work.

## Vertical dipole reflection coefficient

In class a ground reflection scenario was covered for a horizontal dipole. Reading the text I was surprised to see what looked like the same sort of treatment section 4.7.2, but ending up with a quite different result. It turns out the difference is because the text was treating the vertical dipole configuration, whereas Prof. Eleftheriades was treating a horizontal dipole configuration, which have different reflection coefficients. These differing reflection coefficients are due to differences in the polarization of the field.

To understand these differences in reflection coefficients, consider first the field due to a vertical dipole as sketched in fig. 7, with a wave vector directed from the transmission point downwards in the z-y plane.

fig. 7. vertical dipole configuration.

The wave vector has direction

\label{eqn:chapter4Notes:560}
\kcap = \zcap e^{\zcap \xcap \theta} = \zcap \cos\theta + \ycap \sin\theta.

Suppose that the (magnetic) vector potential is that of an infinitesimal dipole

\label{eqn:chapter4Notes:580}
\BA = \zcap \frac{\mu_0 I_0 l}{4 \pi r} e^{-j k r} %= \frac{A_r}{4 \pi r} e^{-j k r}

The electric field, in the far field, can be computed by computing the normal projection to the wave vector direction

\label{eqn:chapter4Notes:600}
\begin{aligned}
\BE
&= -j \omega \lr{\BA \wedge \kcap} \cdot \kcap \\
&= -j \omega \frac{\mu_0 I_0 l}{4 \pi r} \lr{\zcap \wedge \lr{\zcap \cos\theta
+ \ycap \sin\theta} } \lr{\zcap \cos\theta + \ycap \sin\theta} \\
&= -j \omega \frac{\mu_0 I_0 l}{4 \pi r} \lr{ \zcap \ycap \sin\theta }
\lr{\zcap \cos\theta + \ycap \sin\theta} \\
&= -j \omega \frac{\mu_0 I_0 l}{4 \pi r} \sin\theta \lr{-\ycap \cos\theta +
\zcap \sin\theta} \\
&= j \omega \frac{\mu_0 I_0 l}{4 \pi r} \sin\theta \ycap e^{\zcap \ycap \theta}.
\end{aligned}

This is directed in the z-y plane rotated an additional $$\pi/2$$ past $$\kcap$$. The magnetic field must then be directed into the page, along the x axis. This is sketched in fig. 8.

fig. 8. Electric and magnetic field directions

Referring to [2] (\eqntext 4.40) for the coefficient of reflection component

\label{eqn:chapter4Notes:620}
R
=
\frac{
n_t \cos\theta_i – n_i \cos\theta_t
}
{
n_i \cos\theta_i + n_t \cos\theta_t
}

This is the Fresnel equation for the case when
that corresponds to

$$\BE$$ lies in the plane of incidence, and the magnetic field is completely parallel to the plane of reflection). For the no transmission case, allowing $$v_t \rightarrow 0$$, the index of refraction is $$n_t = c/v_t \rightarrow \infty$$, and the reflection coefficient is $$1$$ as claimed in section 4.7.2 of [1]. Because of the symmetry of this dipole configuration, the azimuthal angle that the wave vector is directed along does not matter.

## Horizontal dipole reflection coefficient

In the class notes, a horizontal dipole coming out of the page is indicated. With the page representing the z-y plane, this is a magnetic vector potential directed along the x-axis direction

\label{eqn:chapter4Notes:640}
\BA = \xcap \frac{\mu_0 I_0 l}{4 \pi r} e^{-j k r}.

For a wave vector directed in the z-y plane as in \ref{eqn:chapter4Notes:560}, the electric far field is directed along

\label{eqn:chapter4Notes:660}
\begin{aligned}
\lr{ \xcap \wedge \kcap } \cdot \kcap
&=
\xcap – \lr{ \xcap \cdot \kcap } \kcap \\
&=
\xcap – \lr{ \xcap \cdot \lr{
\zcap \cos\theta + \ycap \sin\theta
} } \kcap \\
&= \xcap.
\end{aligned}

The electric far field lies completely in the plane of reflection. From [2] (\eqntext 4.34), the Fresnel reflection coefficients is

\label{eqn:chapter4Notes:680}
R =
\frac{
n_i \cos\theta_i – n_t \cos\theta_t
}
{
n_i \cos\theta_i + n_t \cos\theta_t
},

which approaches $$-1$$ when $$n_t \rightarrow \infty$$. This is consistent with the image theorem summation that Prof. Eleftheriades used in class.

### Azimuthal angle dependency of the reflection coefficient

Now consider a horizontal dipole directed along the y-axis. For the same wave vector direction as avove, the electric far field is now directed along

\label{eqn:chapter4Notes:700}
\begin{aligned}
\lr{ \ycap \wedge \kcap } \cdot \kcap
&=
\ycap – \lr{ \ycap \cdot \kcap } \kcap \\
&=
\ycap – \lr{ \ycap \cdot \lr{
\zcap \cos\theta + \ycap \sin\theta
} } \kcap \\
&=
\ycap – \kcap \sin\theta \\
&=
\ycap – \sin\theta \lr{
\zcap \cos\theta + \ycap \sin\theta
} \\
&=
\ycap \cos^2 \theta – \sin\theta \cos\theta \zcap \\
&= \cos\theta \lr{ \ycap \cos\theta – \sin\theta \zcap } \\
&= \cos\theta \ycap e^{ \zcap \ycap \theta }.
\end{aligned}

That is

\label{eqn:chapter4Notes:720}
\BE =
-j \omega \frac{\mu_0 I_0 l}{4 \pi r} e^{-j k r}
\cos\theta \ycap e^{ \zcap \ycap \theta }.

This far field electric field lies in the plane of incidence (a direction of $$\thetacap$$ rotated by $$\pi/2$$), not in the plane of reflection. The corresponding magnetic field should be directed along the plane of reflection, which is easily confirmed by calculation

\label{eqn:chapter4Notes:740}
\begin{aligned}
\kcap \cross
\lr{ \ycap \cos\theta – \sin\theta \zcap }
&=
\lr{ \zcap \cos\theta + \ycap \sin\theta } \cross
\lr{ \ycap \cos\theta – \sin\theta \zcap } \\
&=
-\xcap \cos^2 \theta – \xcap \sin^2\theta \\
&= -\xcap.
\end{aligned}

The far field magnetic field is seen to be

\label{eqn:chapter4Notes:721}
\BH =
j \omega \frac{I_0 l}{4 \pi r} e^{-j k r}
\cos\theta \xcap,

so a reflection coefficient of $$1$$ is required to calculate the power loss after a single ground reflection signal bounce for this relative orientation of antenna to the target.

I fail to see how the horizontal dipole treatment in section 4.7.5 can use a single reflection coefficient without taking into account the azimuthal dependency of that reflection coefficient.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[2] E. Hecht. Optics. 1998.

[3] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

[4] Wikipedia. Magnetic potential — Wikipedia, The Free Encyclopedia, 2015. URL https://en.wikipedia.org/w/index.php?title=Magnetic_potential&oldid=642387563. [Online; accessed 5-February-2015].