math and physics play

Electric field due to spherical shell

August 24, 2016 math and physics play , ,

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

Here’s a problem (2.7) from [1], to calculate the field due to a spherical shell. The field is

\begin{equation}\label{eqn:griffithsEM2_7:20}
\BE = \frac{\sigma}{4 \pi \epsilon_0} \int \frac{(\Br – \Br’)}{\Abs{\Br – \Br’}^3} da’,
\end{equation}

where \( \Br’ \) is the position to the area element on the shell. For the test position, let \( \Br = z \Be_3 \). We need to parameterize the area integral. A complex-number like geometric algebra representation works nicely.

\begin{equation}\label{eqn:griffithsEM2_7:40}
\begin{aligned}
\Br’
&= R \lr{ \sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta } \\
&= R \lr{ \Be_1 \sin\theta \lr{ \cos\phi + \Be_1 \Be_2 \sin\phi } + \Be_3 \cos\theta } \\
&= R \lr{ \Be_1 \sin\theta e^{i\phi} + \Be_3 \cos\theta }.
\end{aligned}
\end{equation}

Here \( i = \Be_1 \Be_2 \) has been used to represent to horizontal rotation plane.

The difference in position between the test vector and area-element is

\begin{equation}\label{eqn:griffithsEM2_7:60}
\Br – \Br’
= \Be_3 \lr{ z – R \cos\theta } – R \Be_1 \sin\theta e^{i \phi},
\end{equation}

with an absolute squared length of

\begin{equation}\label{eqn:griffithsEM2_7:80}
\begin{aligned}
\Abs{\Br – \Br’ }^2
&= \lr{ z – R \cos\theta }^2 + R^2 \sin^2\theta \\
&= z^2 + R^2 – 2 z R \cos\theta.
\end{aligned}
\end{equation}

As a side note, this is a kind of fun way to prove the old “cosine-law” identity. With that done, the field integral can now be expressed explicitly

\begin{equation}\label{eqn:griffithsEM2_7:100}
\begin{aligned}
\BE
&= \frac{\sigma}{4 \pi \epsilon_0} \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^\pi R^2 \sin\theta d\theta d\phi
\frac{\Be_3 \lr{ z – R \cos\theta } – R \Be_1 \sin\theta e^{i \phi}}
{
\lr{z^2 + R^2 – 2 z R \cos\theta}^{3/2}
} \\
&= \frac{2 \pi R^2 \sigma \Be_3}{4 \pi \epsilon_0} \int_{\theta = 0}^\pi \sin\theta d\theta
\frac{z – R \cos\theta}
{
\lr{z^2 + R^2 – 2 z R \cos\theta}^{3/2}
} \\
&= \frac{2 \pi R^2 \sigma \Be_3}{4 \pi \epsilon_0} \int_{\theta = 0}^\pi \sin\theta d\theta
\frac{ R( z/R – \cos\theta) }
{
(R^2)^{3/2} \lr{ (z/R)^2 + 1 – 2 (z/R) \cos\theta}^{3/2}
} \\
&= \frac{\sigma \Be_3}{2 \epsilon_0} \int_{u = -1}^{1} du
\frac{ z/R – u}
{
\lr{1 + (z/R)^2 – 2 (z/R) u}^{3/2}
}.
\end{aligned}
\end{equation}

Observe that all the azimuthal contributions get killed. We expect that due to the symmetry of the problem. We are left with an integral that submits to Mathematica, but doesn’t look fun to attempt manually. Specifically

\begin{equation}\label{eqn:griffithsEM2_7:120}
\int_{-1}^1 \frac{a-u}{\lr{1 + a^2 – 2 a u}^{3/2}} du
=
\left\{
\begin{array}{l l}
\frac{2}{a^2} & \quad \mbox{if \( a > 1 \) } \\
0 & \quad \mbox{if \( a < 1 \) } \end{array} \right., \end{equation} so \begin{equation}\label{eqn:griffithsEM2_7:140} \boxed{ \BE = \left\{ \begin{array}{l l} \frac{\sigma (R/z)^2 \Be_3}{\epsilon_0} & \quad \mbox{if \( z > R \) } \\
0 & \quad \mbox{if \( z < R \) } \end{array} \right. } \end{equation} In the problem, it is pointed out to be careful of the sign when evaluating \( \sqrt{ R^2 + z^2 - 2 R z } \), however, I don't see where that is even useful?

References

[1] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

Geeking out: Oriented surface of volume element

August 20, 2016 math and physics play

Reading the intro chapters to my Griffiths electrodynamics, I ended up re-deriving the 1,2 and 3 parameter variations of Stokes Theorem (a quick derivation as previously done using Geometric Algebra [PDF], but without looking at my notes).  To understand how to map from the algebraic representation to a geometric one for the 3 parameter volume element, I built the following nice little model:IMG_20160820_231303757 IMG_20160820_231333998 IMG_20160820_231316618_HDR

This is like Fig 1.10 from the pdf notes linked above, but was a lot more fun to construct than the drawing.

Variational principle with two by two symmetric matrix

March 12, 2016 math and physics play , ,

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

I pulled [1], one of too many lonely Dover books, off my shelf and started reading the review chapter. It posed the following question, which I thought had an interesting subquestion.

Variational principle with two by two matrix.

Consider a \( 2 \times 2 \) real symmetric matrix operator \(\BO \), with an arbitrary normalized trial vector

\begin{equation}\label{eqn:variationalMatrix:20}
\Bc =
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix}.
\end{equation}

The variational principle requires that minimum value of \( \omega(\theta) = \Bc^\dagger \BO \Bc \) is greater than or equal to the lowest eigenvalue. If that minimum value occurs at \( \omega(\theta_0) \), show that this is exactly equal to the lowest eigenvalue and explain why this is expected.

Why this is expected is the part of the question that I thought was interesting.

Finding the minimum.

If the operator representation is

\begin{equation}\label{eqn:variationalMatrix:40}
\BO =
\begin{bmatrix}
a & b \\
b & d
\end{bmatrix},
\end{equation}

then the variational product is

\begin{equation}\label{eqn:variationalMatrix:80}
\begin{aligned}
\omega(\theta)
&=
\begin{bmatrix}
\cos\theta & \sin\theta
\end{bmatrix}
\begin{bmatrix}
a & b \\
b & d
\end{bmatrix}
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos\theta & \sin\theta
\end{bmatrix}
\begin{bmatrix}
a \cos\theta + b \sin\theta \\
b \cos\theta + d \sin\theta
\end{bmatrix} \\
&=
a \cos^2\theta + 2 b \sin\theta \cos\theta
+ d \sin^2\theta \\
&=
a \cos^2\theta + b \sin( 2 \theta )
+ d \sin^2\theta.
\end{aligned}
\end{equation}

The minimum is given by

\begin{equation}\label{eqn:variationalMatrix:60}
\begin{aligned}
0
&=
\frac{d\omega}{d\theta} \\
&=
-2 a \sin\theta \cos\theta + 2 b \cos( 2 \theta )
+ 2 d \sin\theta \cos\theta \\
&=
2 b \cos( 2 \theta )
+ (d -a)\sin( 2 \theta )
\end{aligned}
,
\end{equation}

so the extreme values will be found at

\begin{equation}\label{eqn:variationalMatrix:100}
\tan(2\theta_0) = \frac{2 b}{a – d}.
\end{equation}

Solving for \( \cos(2\theta_0) \), with \( \alpha = 2b/(a-d) \), we have

\begin{equation}\label{eqn:variationalMatrix:120}
1 – \cos^2(2\theta) = \alpha^2 \cos^2(2 \theta),
\end{equation}

or

\begin{equation}\label{eqn:variationalMatrix:140}
\begin{aligned}
\cos^2(2\theta_0)
&= \frac{1}{1 + \alpha^2} \\
&= \frac{1}{1 + 4 b^2/(a-d)^2 } \\
&= \frac{(a-d)^2}{(a-d)^2 + 4 b^2 }.
\end{aligned}
\end{equation}

So,

\begin{equation}\label{eqn:variationalMatrix:200}
\begin{aligned}
\cos(2 \theta_0) &= \frac{ \pm (a-d) }{\sqrt{ (a-d)^2 + 4 b^2 }} \\
\sin(2 \theta_0) &= \frac{ \pm 2 b }{\sqrt{ (a-d)^2 + 4 b^2 }},
\end{aligned}
\end{equation}

Substituting this back into \( \omega(\theta_0) \) is a bit tedious.
I did it once on paper, then confirmed with Mathematica (quantumchemistry/twoByTwoSymmetricVariation.nb). The end result is

\begin{equation}\label{eqn:variationalMatrix:160}
\omega(\theta_0)
=
\inv{2} \lr{ a + d \pm \sqrt{ (a-d)^2 + 4 b^2 } }.
\end{equation}

The eigenvalues of the operator are given by

\begin{equation}\label{eqn:variationalMatrix:220}
\begin{aligned}
0
&= (a-\lambda)(d-\lambda) – b^2 \\
&= \lambda^2 – (a+d) \lambda + a d – b^2 \\
&= \lr{\lambda – \frac{a+d}{2}}^2 -\lr{ \frac{a+d}{2}}^2 + a d – b^2 \\
&= \lr{\lambda – \frac{a+d}{2}}^2 – \inv{4} \lr{ (a-d)^2 + 4 b^2 },
\end{aligned}
\end{equation}

so the eigenvalues are exactly the values \ref{eqn:variationalMatrix:160} as stated by the problem statement.

Why should this have been anticipated?

If the eigenvectors are \( \Be_1, \Be_2 \), the operator can be diagonalized as

\begin{equation}\label{eqn:variationalMatrix:240}
\BO = U D U^\T,
\end{equation}

where \( U = \begin{bmatrix} \Be_1 & \Be_2 \end{bmatrix} \), and \( D \) has the eigenvalues along the diagonal. The energy function \( \omega \) can now be written

\begin{equation}\label{eqn:variationalMatrix:260}
\begin{aligned}
\omega
&= \Bc^\T U D U^\T \Bc \\
&= (U^\T \Bc)^\T D U^\T \Bc.
\end{aligned}
\end{equation}

We can show that the transformed vector \( U^\T \Bc \) is still a unit vector

\begin{equation}\label{eqn:variationalMatrix:280}
\begin{aligned}
U^\T \Bc
&=
\begin{bmatrix}
\Be_1^\T \\
\Be_2^\T \\
\end{bmatrix}
\Bc \\
&=
\begin{bmatrix}
\Be_1^\T \Bc \\
\Be_2^\T \Bc \\
\end{bmatrix},
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:variationalMatrix:300}
\begin{aligned}
\Abs{
U^\T \Bc
}^2
&=
\Bc^\T \Be_1
\Be_1^\T \Bc
+
\Bc^\T \Be_2
\Be_2^\T \Bc \\
&=
\Bc^\T \lr{ \Be_1 \Be_1^\T
+
\Be_2
\Be_2^\T } \Bc \\
&=
\Bc^\T \Bc \\
&= 1,
\end{aligned}
\end{equation}

so the transformed vector can be written as

\begin{equation}\label{eqn:variationalMatrix:320}
U^\T \Bc =
\begin{bmatrix}
\cos\phi \\
\sin\phi
\end{bmatrix},
\end{equation}

for some \( \phi \). With such a representation we have
\begin{equation}\label{eqn:variationalMatrix:340}
\begin{aligned}
\omega
&=
\begin{bmatrix}
\cos\phi & \sin\phi
\end{bmatrix}
\begin{bmatrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{bmatrix}
\begin{bmatrix}
\cos\phi \\
\sin\phi
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos\phi & \sin\phi
\end{bmatrix}
\begin{bmatrix}
\lambda_1 \cos\phi \\
\lambda_2 \sin\phi
\end{bmatrix} \\
&=
\lambda_1 \cos^2\phi + \lambda_2 \sin^2\phi.
\end{aligned}
\end{equation}

This has it’s minimums where \( 0 = \sin(2 \phi)( \lambda_2 – \lambda_1 ) \). For the non-degenerate case, two zeros at \( \phi = n \pi/2 \) for integral \( n \). For \( \phi = 0, \pi/2 \), we have

\begin{equation}\label{eqn:variationalMatrix:360}
\Bc =
\begin{bmatrix}
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
1
\end{bmatrix}.
\end{equation}

We see that the extreme values of \( \omega \) occur when the trial vectors \( \Bc \) are eigenvectors of the operator.

References

[1] Attila Szabo and Neil S Ostlund. Modern quantum chemistry: introduction to advanced electronic structure theory. Dover publications, 1989.

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

February 12, 2016 ece1236

[Click here for a PDF of this post with nicer formatting and figures] or [Click here for my notes compilation for this class]

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.

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

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

  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:

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

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

where \( \omega_c \) is the cutoff frequency. This gives

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

Triangular Taper

\begin{equation}\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.
\end{equation}

\begin{equation}\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.
\end{equation}

In this case

\begin{equation}\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 ).
\end{equation}

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:

\begin{equation}\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,
\end{equation}

where

\begin{equation}\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.
\end{equation}

Here \( I_1(x) \) is the modified Bessel function. Note that
\begin{equation}\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}
\end{equation}

The resulting reflection coefficient is

\begin{equation}\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., \end{equation} where as usual \begin{equation}\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). \end{equation} The passband is defined by \( \beta L \ge A \) and the maximum ripple in the passband is \begin{equation}\label{eqn:continuumAndOtherTransformersCore:1120} \Gamma_m = \frac{\Gamma_0}{\cosh A}. \end{equation}

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:

    \begin{equation}\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.
    \end{equation}

    The resulting \( \Gamma \) is

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

  • Exponential taper:

    \begin{equation}\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}
    \end{equation}

  • Klopfenstein taper:
    \begin{equation}\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}
    \end{equation}

    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

February 11, 2016 ece1236 , , , , , , ,

[Click here for a PDF of this post with nicer formatting] or [Click here for my notes compilation for this class]

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

\begin{equation}\label{eqn:uwavesDeck7MultisectionTransformersCore:20}
Z_{\textrm{in}} = R + j X
\end{equation}
\begin{equation}\label{eqn:uwavesDeck7MultisectionTransformersCore:40}
Y_{\textrm{in}} = G + j B
\end{equation}

  • \( Z_{\textrm{in}} \) : 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

\begin{equation}\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},
\end{equation}

where

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

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

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

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

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

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

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

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

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

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

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

The reflection coefficient can now be expanded using the binomial theorem

\begin{equation}\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}
\end{equation}

Recall that

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

providing a symmetric set of values

\begin{equation}\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}
\end{equation}

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

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

Approximation for \( Z_k \)

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

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

so for \( x \) near unity a first order approximation of a logarithm is

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

Assuming that \( Z_{k+1}/Z_k \) is near unity we have

\begin{equation}\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}
\end{equation}

Using this approximation, we get

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

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

\begin{equation}\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}
\end{equation}

for \( \Theta_m < \pi/2 \). Then \begin{equation}\label{eqn:uwavesDeck7MultisectionTransformersCore:420} \Theta_m = \arccos\lr{ \inv{2} \lr{ \frac{\Gamma_{\mathrm{m}}}{\Abs{A}}}^{1/N} } \end{equation} The relative width of the interval is \begin{equation}\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} \end{equation}

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
\begin{equation}\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}
\end{equation}

Now use

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

starting from

  • \( n = 0 \).

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

    or
    \begin{equation}\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}
    \end{equation}

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

  • \( n = 1 \)

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

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

  • \( n = 2 \)

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

    so

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

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

\begin{equation}\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}
\end{equation}

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.