Exercise 7.5.5: The best ratio b/a for a fixed upper frequency

About the further exercises of Section 7.5

Five further exercises on the coaxial line, in the spirit of Problems 7.5.1–7.5.3, the worked example and the home exercise, graded from easy to harder. They use only the results listed at the head of this subsection. (Five more, on the circular guide, close Section 7.4.)

Exercise 7.5.5

The best ratio $b / a$ for a fixed upper frequency

Section 7.5.5 optimises $b / a$ at fixed outer radius $b$ . A microwave designer instead fixes the upper frequency, that is $a + b = S$ through $f_{c} \approx v / (π S)$ .

With $x = b / a$ , show that $α_{c} \propto (1 + x)^{2} / (x ln x)$ and $P_{m a x} \propto ln x / (1 + x)^{2}$ .
Find the $x$ of minimum loss and of maximum power, and the corresponding $Z_{0}$ of an air line.
An air line must be single-mode up to $18$ GHz. Find $a$ , $b$ , $α_{c}$ at $10$ GHz and $P_{m a x}$ ( $E_{br} = 3$ MV/m) for $x = 2.09$ , $2.30$ , $3.59$ and $4.68$ .
How does the conclusion “ $50 Ω$ is the compromise” change?

Try it yourself first.

Strategy

Express $a$ and $b$ through $S$ and $x$ : $a = S / (1 + x)$ , $b = S x / (1 + x)$ . Then both figures of merit become functions of $x$ alone, times powers of $S$ . Setting the derivatives of their logarithms to zero gives transcendental equations that are solved numerically.

$\frac{1}{a} + \frac{1}{b} = \frac{1 + x}{S} (1 + \frac{1}{x}) = \frac{( 1 + x ) ^{2}}{S x}$ , so $α_{c} = \frac{R _{s}}{2 η S} \frac{( 1 + x ) ^{2}}{x ln x}$ . With $P_{m a x} = \frac{π a ^{2} E _{br}^{2}}{η} ln x$ and $a^{2} = S^{2} / (1 + x)^{2}$ : $P_{m a x} = \frac{π S ^{2} E _{br}^{2}}{η} \frac{ln x}{( 1 + x ) ^{2}}$ .

Minimum loss: $\frac{d}{d x} [2 ln (1 + x) - ln x - ln ln x] = \frac{2}{1 + x} - \frac{1}{x} - \frac{1}{x ln x} = 0$ , which simplifies to

(x - 1) ln x = x + 1.

At $x = 4.5$ : $5.26 < 5.5$ , at $x = 5$ : $6.44 > 6$ . Bisection gives $x = 4.68$ , $Z_{0} = 60 ln 4.68 = 92.6 Ω$ . Maximum power: $\frac{d}{d x} [ln ln x - 2 ln (1 + x)] = \frac{1}{x ln x} - \frac{2}{1 + x} = 0$ , that is $1 + x = 2 x ln x$ . At $x = 2$ the left side is larger ( $3 > 2.77$ ), at $x = 2.2$ smaller ( $3.2 < 3.47$ ), and the root is $x = 2.09$ , $Z_{0} = 60 ln 2.09 = 44.3 Ω$ .

$S = c / (π \times 18 GHz) = 5.31$ mm. At $10$ GHz $R_{s} = 26.1$ m $Ω$ :

$x = b / a$	$Z_{0}$ [ $Ω$ ]	$a$ , $b$ [mm]
$2.09$	$44.3$	$1.71$ , $3.59$
$2.30$	$50.0$	$1.61$ , $3.70$
$3.59$	$76.7$	$1.16$ , $4.15$
$4.68$	$92.6$	$0.93$ , $4.37$

$x = b / a$	$α_{c}$ [dB/m]	$P_{m a x}$ [kW]
$2.09$	$0.351$	$163$
$2.30$	$0.322$	$161$
$3.59$	$0.260$	$128$
$4.68$	$0.253$	$101$

Example column $x = 2.30$ : $a = 5.305/3.30 = 1.61$ mm, $b = 3.70$ mm, $α_{c} = \frac{0.0261}{754} \times \frac{623 + 270}{0.834} = 0.0370$ Np/m $= 0.322$ dB/m, $V_{m a x} = 3 \times 1 0^{6} \times 1.606 \times 1 0^{- 3} \times 0.834 = 4.02$ kV and $P_{m a x} = 401 8^{2} /100 = 161$ kW.

At fixed $a + b$ the power optimum ( $44 Ω$ ) moves close to $50 Ω$ : a $50 Ω$ line keeps $99 %$ of the best power, but loses $27 %$ more than the $93 Ω$ optimum. The flat loss curve means that $75$ to $95 Ω$ air lines lose almost equally little.

Answer

minimum loss: (x - 1) ln x = x + 1 x = 4.68, Z_{0} = 92.6 Ω maximum power: 1 + x = 2 x ln x x = 2.09, Z_{0} = 44.3 Ω 18 GHz line at 10 GHz : 50 Ω gives 0.322 dB/m, 161 kW

Check

$(4.68 - 1) ln 4.68 = 3.68 \times 1.543 = 5.68 = 4.68 + 1$ and $2 \times 2.09 \times ln 2.09 = 3.08 \approx 1 + 2.09$ . The exact TE $_{11}$ cutoffs of the four lines are $18.3$ to $18.5$ GHz, slightly above $18$ GHz, so the estimate is on the safe side.

Remark

Which ratio is “best” depends on what is held fixed. With the outer size fixed the optima are $30$ and $77 Ω$ , with the frequency limit fixed they are $44$ and $93 Ω$ . In both cases $50 Ω$ lies between them, and in the second case, the one that matters for precision connectors and test cables, it is nearly the high-power design while losing only a quarter more than the best.

cf. Problem 7.5.3