22-Elec-A7 Electromagnetics · May 2013

Question 2 of 8: Stub and Quarter-Wave Matching Network at 300 MHz

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2013 — 07-Elec-A7, Electromagnetics. Three hours, closed book; one of two approved calculators (Casio or Sharp) permitted. Eight questions, all of equal value; any five constitute a complete paper and only the first five presented are marked. All eight are solved here, since the set is a study resource. Aids printed on the cover page: ε₀ = 8.85 × 10⁻¹² F/m and μ₀ = 4π × 10⁻⁷ H/m; Question 2 additionally supplies the quarter-wave relation.

Reference texts (22-Elec-A7 Electromagnetics).

D. M. Pozar, Microwave Engineering, 4th ed. — transmission lines, matching, waveguides.
M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. — plane waves, guided waves, radiation.
W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. — field theory and transients on lines.
C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. — short elements, image theory.
F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. — oblique incidence, surface currents.
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. — rotating-loop induction and torque (Question 3).

Conventions used throughout. The paper itself quotes propagation velocities as $3\times10^{8}\ \text{m/s}$, so free space is taken as $c=3.00\times10^{8}\ \text{m/s}$ and the intrinsic impedance as $\eta_0=\mu_0 c=120\pi=376.99\ \Omega$. Lines are treated as lossless, and "rms" is stated explicitly wherever the exam asks for it.

Question 2: Stub and Quarter-Wave Matching Network at 300 MHz (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 at 300 MHz must deliver maximum power to a parallel RC load, using only an adjustable-length short-circuited stub of the same 50 Ω line and a fixed 25 cm section whose characteristic impedance may be chosen freely.

Given data
Quantity	Symbol	Value
Main line characteristic impedance	$Z_0$	50 Ω
Propagation velocity (both sections)	$v_p$	$3\times10^{8}$ m/s
Operating frequency	$f$	300 MHz
Load resistor	$R_L$	100 Ω
Load capacitive reactance (in parallel)	$X_C$	$-j50$ Ω
Adjustable section length (fixed)	$\ell_T$	25 cm
Tuning element	—	short-circuited 50 Ω stub, adjustable length

Find. The stub length and the characteristic impedance of the 25 cm section that together present 50 Ω to the feeder, so that the load absorbs the maximum available power.

Figure 2.1 — The completed matching network: a short-circuited shunt stub tunes out the load susceptance, and the 25 cm quarter-wave transformer converts the residual 100 Ω to 50 Ω.

Approach. Work in admittance because the load and the stub are both in parallel: cancel the load susceptance with the short-circuited stub so that a pure conductance remains, then use the supplied quarter-wave relation to transform that resistance to 50 Ω.

Establish the wavelength and identify the fixed section. $$\lambda=\frac{v_p}{f}=\frac{3\times10^{8}}{300\times10^{6}}=1.00\ \text{m}$$ so the 25 cm section is exactly $\lambda/4$ — which is precisely why the exam supplies the relation $Z(s)Z(s\pm\lambda/4)=Z_0^{2}$. It is a quarter-wave transformer, not an arbitrary length of line.
Write the load as an admittance. For a resistor in parallel with a capacitive reactance the admittances simply add: $$Y_L=\frac{1}{R_L}+\frac{1}{-jX}=\frac{1}{100}+\frac{1}{-j50}=0.0100+j0.0200\ \text{S}$$ equivalently $Z_L=1/Y_L=20-j40\ \Omega$. The conductance is the resistor and the positive (capacitive) susceptance is the reactive branch.
Cancel the susceptance with the stub. A short-circuited line of length $\ell$ presents $Z_{stub}=jZ_0\tan\beta\ell$, so its admittance is $Y_{stub}=-j\cot(\beta\ell)/Z_0$. Setting $Y_{stub}=-j0.0200$ S requires $$\cot\beta\ell=Z_0\times0.0200=1\quad\Rightarrow\quad\beta\ell=45^\circ\quad\Rightarrow\quad \ell=\frac{\lambda}{8}=\boxed{12.5\ \text{cm}}$$ placed in shunt directly across the load. Any $\ell=12.5\ \text{cm}+n\times50\ \text{cm}$ works equally well, since the stub admittance repeats every half wavelength.
Read off the resistance that remains. With the susceptance gone, $$Y=Y_L+Y_{stub}=0.0100+j0\ \text{S}\quad\Rightarrow\quad R=\frac{1}{0.0100}=100\ \Omega$$ which is real, as required before a quarter-wave transformer can be used.
Size the transformer with the supplied aid. The relation $Z(s)Z(s\pm\lambda/4)=Z_{0T}^{2}$ says that a quarter-wave section of impedance $Z_{0T}$ maps $R$ into $Z_{0T}^{2}/R$. Demanding that this equal the feeder impedance, $$Z_{0T}=\sqrt{R\,Z_0}=\sqrt{100\times50}=\sqrt{5000}=\boxed{70.7\ \Omega}$$
Verify the match. Looking into the 25 cm section, $$Z_{in}=\frac{Z_{0T}^{2}}{R}=\frac{5000}{100}=50\ \Omega\quad\Rightarrow\quad \Gamma=\frac{50-50}{50+50}=0,\qquad \text{SWR}=1$$ No power returns to the generator, so the load receives the maximum available power — the design goal.

The order matters: the stub must sit at the load so that the transformer only ever sees a real impedance. Reversing the two elements would leave the transformer working into $20-j40\ \Omega$, which no real characteristic impedance can match on its own.

Final results — Question 2
Quantity	Value
Wavelength on both sections at 300 MHz	1.00 m (so 25 cm = $\lambda/4$)
Load admittance / impedance	$0.0100+j0.0200$ S; $20-j40\ \Omega$
Short-circuited shunt stub, at the load	12.5 cm = $\lambda/8$ (+ any multiple of $\lambda/2$)
Impedance seen after the stub	100 Ω, purely real
Characteristic impedance of the 25 cm section	70.7 Ω
Input impedance presented to the feeder	50 Ω; $\Gamma=0$, SWR = 1

Topic. Single-stub tuning combined with a quarter-wave transformer — a two-element conjugate match.

Key relations. Short-circuited stub $Z_{stub}=jZ_0\tan\beta\ell$; open/short stub admittance $Y_{stub}=-j\cot(\beta\ell)/Z_0$; quarter-wave transformer $Z_{in}=Z_{0T}^{2}/Z_L$, hence $Z_{0T}=\sqrt{Z_{in}Z_L}$ for real loads; $\lambda=v_p/f$; SWR $=(1+|\Gamma|)/(1-|\Gamma|)$.

Why this works. Maximum power transfer on a line means eliminating the reflected wave, which requires the input impedance to equal $Z_0$. Any load can be brought to that condition in two moves: first make it real, then scale it. A shunt element can only change the susceptance seen at a node, so working in admittance makes the first move a single subtraction — and a short-circuited stub is the natural adjustable susceptance because its length maps continuously over all inductive and capacitive values. Once the node is purely conductive, a quarter-wave section acts as an impedance inverter, and the geometric mean of the two resistances is the only characteristic impedance that lands exactly on the target. The exam's aid is that inverter property written as a product.

Common pitfall. Adding the stub reactance to the load reactance: elements in parallel add in admittance, and $-j40\ \Omega$ of load reactance is not cancelled by $+j40\ \Omega$ of shunt stub. A second slip is forgetting that a short-circuited stub shorter than $\lambda/4$ is inductive, so a capacitive load correctly calls for a stub below 25 cm here.

Source. Pozar, Microwave Engineering, 4th ed., §2.5 (quarter-wave transformer) and §5.1 (single-stub tuning); Sadiku, Elements of Electromagnetics, 7th ed., Ch. 11.

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