Stub-Loaded Termination and Standing-Wave Ratio: 22-Elec-A7 Electromagnetics, December 2013, Q2

Question 2: Stub-Loaded Termination and Standing-Wave Ratio (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 50 Ω feeder is terminated by three elements in parallel: a 50 cm open-circuited stub, a 50 cm short-circuited stub, and a 50 Ω resistor.

Given data
Quantity	Symbol	Value
Characteristic impedance	$Z_0$	$50\ \Omega$
Propagation velocity	$v_p$	$3\times10^{8}$ m/s
Stub length (each)	$\ell$	0.50 m
Terminating resistor	$R$	$50\ \Omega$
Second frequency of interest	$f_2$	50 MHz

Find. The lowest frequency at which the feeder is perfectly matched (SWR = 1), and the SWR at 50 MHz.

Figure 2.1 — The terminating network: an open stub, a shorted stub and a 50 Ω resistor, all in parallel across the feeder.

Approach. Because the three branches are in parallel, work in admittance. Add the two stub susceptances to the resistor conductance, impose $Y = Y_0$ for a match, and evaluate the reflection coefficient at 50 MHz.

Input admittances of the two stubs. For a lossless stub of electrical length $\beta\ell$,$$Y_{oc} = jY_0\tan\beta\ell,\qquad Y_{sc} = -jY_0\cot\beta\ell,$$with $Y_0 = 1/Z_0 = 20$ mS. The open stub is capacitive for $\beta\ell \lt 90^{\circ}$ and the shorted stub inductive, so the two susceptances oppose each other.
Total terminating admittance. Adding the resistor's conductance $G = 1/50 = Y_0$ to the two susceptances,$$Y_L = Y_0\Big[\,1 + j\big(\tan\beta\ell - \cot\beta\ell\big)\Big] \equiv Y_0\,(1 + jb).$$The normalised susceptance $b = \tan\beta\ell - \cot\beta\ell$ carries the whole frequency dependence.
Condition for a perfect match. The line is matched only when $Y_L = Y_0$, i.e. $b = 0$:$$\tan\beta\ell = \cot\beta\ell \;\Longrightarrow\;\tan^{2}\beta\ell = 1 \;\Longrightarrow\;\beta\ell = \frac{\pi}{4} + \frac{n\pi}{2}.$$The lowest positive root is $\beta\ell = \pi/4$, i.e. each stub is one eighth of a wavelength long. Physically the two stubs then present equal and opposite susceptances ($+jY_0$ and $-jY_0$) which cancel, leaving only the matched resistor.
Lowest matched frequency. Setting $\ell = \lambda/8$ gives $\lambda = 8\ell = 4.00$ m, so$$f_1 = \frac{v_p}{\lambda} = \frac{3\times10^{8}}{4.00},$$$$\boxed{f_1 = 75\ \text{MHz}\quad(\text{SWR} = 1)}$$
Electrical length at 50 MHz. At the second frequency $\lambda = (3\times10^{8})/(50\times10^{6}) = 6.00$ m, so$$\beta\ell = \frac{2\pi\ell}{\lambda} = \frac{2\pi(0.50)}{6.00} = \frac{\pi}{6} = 30^{\circ}.$$
Normalised load at 50 MHz. Substituting that electrical length,$$b = \tan 30^{\circ} - \cot 30^{\circ} = 0.5774 - 1.7321 = -1.1547,$$$$y_L = 1 - j1.1547 \;\Longrightarrow\; z_L = \frac{1}{y_L} = 0.4286 + j0.4949,$$i.e. $Z_L = 21.43 + j24.74\ \Omega$. The net susceptance is inductive because the shorted stub dominates below the matching frequency.
Reflection coefficient and SWR. For a load written as $y_L = 1+jb$ the reflection coefficient simplifies neatly:$$\Gamma = \frac{1-y_L}{1+y_L} = \frac{-jb}{2+jb}\;\Longrightarrow\;|\Gamma| = \frac{|b|}{\sqrt{4+b^{2}}} = \frac{1.1547}{\sqrt{4+1.3333}} = 0.500.$$Hence$$\text{SWR} = \frac{1+|\Gamma|}{1-|\Gamma|} = \frac{1.5}{0.5},$$$$\boxed{\text{SWR}(50\ \text{MHz}) = 3.00}$$

Final Results
Quantity	Symbol	Value
Stub electrical length for a match	$\beta\ell$	$45^{\circ}$ ($\ell = \lambda/8$)
Lowest frequency with SWR = 1	$f_1$	75 MHz
Electrical length at 50 MHz	$\beta\ell$	$30^{\circ}$
Terminating impedance at 50 MHz	$Z_L$	$21.43 + j24.74\ \Omega$
Reflection coefficient magnitude	$\|\Gamma\|$	0.500
Standing-wave ratio at 50 MHz	SWR	3.00

Topic. Stub susceptance, matching, and the standing-wave ratio.

Key relations. $Y_{oc} = jY_0\tan\beta\ell$; $Y_{sc} = -jY_0\cot\beta\ell$; $\Gamma = (Z_L-Z_0)/(Z_L+Z_0)$; $\text{SWR} = (1+|\Gamma|)/(1-|\Gamma|)$; $\beta = 2\pi f/v_p$.

Why this works. An open stub and a shorted stub of the same length are exact complements: their susceptances are the tangent and the negative cotangent of the same angle, so they cancel wherever the tangent equals the cotangent, which first happens at $\ell = \lambda/8$. At that one frequency the reactive pair vanishes and the feeder sees only the matched resistor. Away from it the residual susceptance detunes the termination, and because the conductance stays at exactly $Y_0$ the reflection coefficient depends on that residual susceptance alone.

Common pitfall. Adding the three branches as impedances rather than admittances is the usual error; the parallel combination only adds cleanly in $Y$. It is also easy to forget that $\cot\beta\ell$ blows up as $\beta\ell \to 0$, so at very low frequencies the shorted stub short circuits the line and the SWR is enormous rather than small.

Source. Pozar, Microwave Engineering, 4th ed., §2.3–2.5 (the lossless terminated line, stubs and the Smith chart); Ulaby, Fundamentals of Applied Electromagnetics, 8th ed., §2.8–2.10.

Question 2 of 8: Stub-Loaded Termination and Standing-Wave Ratio

Question 2: Stub-Loaded Termination and Standing-Wave Ratio (20 marks)