A Load Shunted by a Matched Pair of Open and Shorted Stubs: 22-Elec-A7 Electromagnetics, May 2014, Q2

Question 2: A Load Shunted by a Matched Pair of Open and Shorted Stubs (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 data
Quantity	Symbol	Value
Line characteristic impedance	$Z_0$	50 $\Omega$
Propagation velocity (line and stubs)	$v_p$	$2\times10^{8}$ m/s
Load resistor	$R_L$	50 $\Omega$
Stub length (both stubs)	$\ell$	25 cm
Stub terminations	—	one open circuit, one short circuit

Find. (i) the two lowest frequencies at which the 50 Ω resistor receives no power at all, and (ii) the lowest frequency at which the resistor alone terminates the feed line in a match.

Figure 2.1 — The load plane: a 50 Ω resistor in parallel with a 25 cm open-circuited stub and a 25 cm short-circuited stub.

Approach. All three branches sit at the same plane, so work in admittances: add the two stub susceptances, then ask separately when their sum becomes infinite (the load is shorted out) and when it becomes zero (the resistor stands alone).

Input admittance of each stub. A lossless stub of length $\ell$ on a line of admittance $Y_0 = 1/Z_0$ presents$$Y_{oc} = jY_0\tan\beta\ell,\qquad Y_{sc} = -jY_0\cot\beta\ell,$$with $\beta = \omega/v_p = 2\pi f/v_p$. The two are exact complements: whatever one does, the other does the opposite.
Total load admittance. Adding the three parallel branches at the load plane,$$Y_L = \frac{1}{R_L} + jY_0\left(\tan\beta\ell - \cot\beta\ell\right) = Y_0\bigl[1 + jb\bigr],\qquad b = \tan\beta\ell - \cot\beta\ell,$$since $R_L = Z_0$ here. Everything that follows is a statement about the single number $b$.
(i) When does the load get no power? The resistor is starved only when the stub pair becomes a dead short across it, i.e. when $b \to \infty$. That happens whenever either stub is itself a short circuit. The open stub is a short when it is an odd number of quarter-wavelengths long,$$\ell = \frac{\lambda}{4} \;\Rightarrow\; f = \frac{v_p}{4\ell} = \frac{2\times10^{8}}{4(0.25)} = 200\ \text{MHz},$$at which $\lambda = 1$ m and the 25 cm stub is exactly $\lambda/4$.
The shorted stub closes the second window. A shorted stub repeats its termination every half wavelength, so it is a short again when$$\ell = \frac{\lambda}{2} \;\Rightarrow\; f = \frac{v_p}{2\ell} = \frac{2\times10^{8}}{2(0.25)} = 400\ \text{MHz}.$$No lower frequency qualifies, so$$\boxed{f_1 = 200\ \text{MHz}\quad\text{and}\quad f_2 = 400\ \text{MHz}.}$$In general the two stubs alternate in shorting the load out, once every 200 MHz.
(ii) When is the resistor matched? The feed line sees a match when the stub pair contributes nothing, $b = 0$, which requires$$\tan\beta\ell = \cot\beta\ell \;\Rightarrow\; \tan^{2}\beta\ell = 1 \;\Rightarrow\; \beta\ell = \frac{\pi}{4}.$$The capacitive susceptance of the open stub then cancels the inductive susceptance of the shorted one exactly.
Lowest matching frequency. $\beta\ell = \pi/4$ means the stubs are one eighth of a wavelength long, so$$\lambda = 8\ell = 2\ \text{m},\qquad \boxed{f = \frac{v_p}{8\ell} = \frac{2\times10^{8}}{2} = 100\ \text{MHz}.}$$The next matches are at 300 MHz, 500 MHz and so on — the odd multiples — because $\tan\beta\ell = \pm1$ recurs every $\pi/2$ in $\beta\ell$.

The three answers interleave in a way that is worth noticing: the line is matched at 100 MHz, the load is starved at 200 MHz, matched again at 300 MHz, starved again at 400 MHz. The stub pair is therefore a 100 MHz-period comb filter bolted onto an otherwise perfect termination, and the useful bandwidth around each match is set by how fast $b$ climbs away from zero.

Final Results
Quantity	Symbol	Value
Lowest frequency with no power in the load	$f_1$	200 MHz (open stub = $\lambda/4$)
Second such frequency	$f_2$	400 MHz (shorted stub = $\lambda/2$)
General starving condition	—	$f = n\times200$ MHz, $n = 1, 2, 3\ldots$
Lowest matching frequency	$f_m$	100 MHz (stubs = $\lambda/8$)
General matching condition	—	$f = (2n-1)\times100$ MHz

Topic. Reactive stub terminations and the complementary behaviour of an open and a shorted stub of equal length.

Key relations. $Y_{oc} = jY_0\tan\beta\ell$; $Y_{sc} = -jY_0\cot\beta\ell$; both are periodic in $\beta\ell$ with period $\pi$; a short repeats every $\lambda/2$ and inverts every $\lambda/4$.

Why this works. A lossless stub is a pure susceptance whose value is fixed entirely by its electrical length, so sweeping frequency sweeps the susceptance through every value from $-\infty$ to $+\infty$. Placing an open and a shorted stub of the same physical length in parallel makes the two susceptances exact negatives of one another apart from the tangent-versus-cotangent difference, so their sum vanishes at $\beta\ell = \pi/4$ and blows up whenever either one alone becomes a short. No Smith chart is needed: the whole question reduces to the roots and poles of $\tan\beta\ell - \cot\beta\ell$.

Common pitfall. Students often read “no power delivered” as “the load is open-circuited” and hunt for an infinite impedance. It is the opposite: the stub pair short-circuits the resistor, forcing the terminal voltage to zero. The other frequent slip is to give the open stub the half-wavelength condition and the shorted stub the quarter-wavelength one, which swaps 200 MHz and 400 MHz.

Source. Pozar, Microwave Engineering, 4th ed., §2.5 (the quarter-wave transformer and stub tuning); Sadiku, Elements of Electromagnetics, 7th ed., §11.5.

Question 2 of 8: A Load Shunted by a Matched Pair of Open and Shorted Stubs

Question 2: A Load Shunted by a Matched Pair of Open and Shorted Stubs (20 marks)