Standing-wave ratio and the real-impedance point: 22-Elec-A7 Electromagnetics, May 2016, Q2

Question 2: Standing-wave ratio and the real-impedance point (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 short line section is driven at a single frequency into a parallel resistor-inductor load, so the load impedance is complex and a standing wave is set up on the section.

Given data
Quantity	Symbol	Value
Frequency	$f$	$300\ \text{MHz}$
Characteristic impedance	$Z_0$	$50\ \Omega$
Propagation velocity	$v_p$	$3\times10^{8}\ \text{m/s}$
Section length	$\ell$	$30\ \text{cm}$
Load resistance	$R$	$50\ \Omega$
Load inductance	$L$	$1.59\times10^{-8}\ \text{H}$

Find. The standing-wave ratio on the section, and the value of the purely real input impedance at the point where it occurs, together with that point's position.

Figure 2.1 — the driven section and its parallel $R$–$L$ termination. Distance $l$ is measured from the load plane toward the generator.

Approach. Convert the load to a single complex impedance, form its reflection coefficient, read the standing-wave ratio from the magnitude, then use the angle to locate the points where the rotating reflection coefficient becomes real.

Set the electrical scale. $\lambda = v_p/f = (3\times10^{8})/(3\times10^{8}) = 1.00\ \text{m}$, so the 30 cm section is $0.300\lambda$ long and $\beta = 2\pi/\lambda = 2\pi\ \text{rad/m}$.
Combine the load elements. The inductive reactance is $X_L = \omega L = 2\pi(3\times10^{8})(1.59\times10^{-8}) = 29.97\ \Omega$ (the stated inductance is simply $30/\omega$), and the parallel combination is $$Z_L = \frac{R\,(jX_L)}{R + jX_L} = \frac{50\,(j29.97)}{50 + j29.97} = 13.22 + j22.05\ \Omega$$
Form the reflection coefficient. $$\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{-36.78 + j22.05}{63.22 + j22.05} = 0.6406\,\angle\,129.83^{\circ}$$
Read off the standing-wave ratio. $$S = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|} = \frac{1.6406}{0.3594} = \boxed{4.564}$$
Locate the real-impedance points. Moving a distance $l$ toward the generator rotates the reflection coefficient to $\Gamma(l) = |\Gamma_L|e^{\,j(\theta_L - 2\beta l)}$. The input impedance is real precisely when that phase is $0$ (a voltage maximum) or $\pm180^{\circ}$ (a minimum). The first solution is $$l = \frac{\theta_L}{2\beta} = \frac{2.2660}{4\pi}\ \text{m} = \boxed{18.03\ \text{cm}}$$
Evaluate the impedance there. At a voltage maximum $\Gamma$ is real and positive, so $$Z_{in} = Z_0\,\frac{1 + |\Gamma_L|}{1 - |\Gamma_L|} = Z_0 S = 50 \times 4.564 = \boxed{228.2\ \Omega}$$

Two features of this answer deserve comment. First, the load is inductive ($\operatorname{Im} Z_L \gt 0$, equivalently $0 \lt \theta_L \lt 180^{\circ}$), and an inductive load always puts a voltage maximum nearer to itself than a minimum; a capacitive load reverses that order. Second, the question says “at some point” in the singular, and that is literally correct here: successive real-impedance points are a quarter wavelength (25 cm) apart, so the next one — a voltage minimum with $Z_{in} = Z_0/S = 10.96\ \Omega$ — would sit at 43.03 cm, beyond the far end of a 30 cm section.

Figure 2.2 — standing-wave envelope on the 30 cm section. The single touch of the upper envelope inside the section is the one point at which the impedance is real.

As a check on the whole calculation, the direct line-transformation formula $Z_{in} = Z_0(Z_L + jZ_0\tan\beta l)/(Z_0 + jZ_L\tan\beta l)$ evaluated at $l = 18.03\ \text{cm}$ returns $228.2 - j1\times10^{-13}\ \Omega$: real to machine precision, and equal to $Z_0 S$ as it must be.

Final results
Quantity	Value
Wavelength on the line	$1.00\ \text{m}$
Load impedance $Z_L$	$13.22 + j22.05\ \Omega$
Reflection coefficient	$0.6406\,\angle\,129.83^{\circ}$
Standing-wave ratio	$\mathbf{S = 4.564}$
Position of the real-impedance point	$\mathbf{18.03\ \text{cm}}$ from the load
Impedance there (a voltage maximum)	$\mathbf{Z_{in} = Z_0 S = 228.2\ \Omega}$
Next real point (a minimum), outside the section	$43.03\ \text{cm}$, $Z_0/S = 10.96\ \Omega$

Topic. Standing waves on a mismatched line, and the special role of the voltage maxima and minima as the only points of purely real impedance.

Key relations. $\Gamma_L = (Z_L - Z_0)/(Z_L + Z_0)$; $S = (1+|\Gamma|)/(1-|\Gamma|)$; $\Gamma(l) = \Gamma_L e^{-j2\beta l}$; $Z_{max} = Z_0 S$ and $Z_{min} = Z_0/S$; maxima and minima alternate every $\lambda/4$ and each family repeats every $\lambda/2$.

Why this works. Because the line is lossless, moving along it changes only the phase of the reflection coefficient, never its magnitude. The input impedance therefore traces a circle of constant $|\Gamma|$, and it crosses the real axis exactly twice per half wavelength — once where incident and reflected waves add in phase (the maximum, a resistance larger than $Z_0$) and once where they oppose (the minimum, a resistance smaller than $Z_0$). That is why quoting $S$ is equivalent to quoting the extreme resistances, and it is the whole basis of the slotted-line measurement.

Common pitfall. Reporting $Z_0/S$ when the geometry demands $Z_0 S$. Which one you meet first is decided by the sign of the load reactance, not by convenience: check whether $\theta_L$ is positive (inductive, maximum first) or negative (capacitive, minimum first) before committing. A second trap is forgetting that a short section may simply not contain the second point.

Source. D. M. Pozar, Microwave Engineering, 4th ed., §2.1–2.7 (transmission-line theory, standing waves, impedance transformation). M. N. O. Sadiku, Elements of Electromagnetics, 7th ed., ch. 11 (transmission lines) and ch. 12 (waveguides).

Question 2 of 8: Standing-wave ratio and the real-impedance point

Question 2: Standing-wave ratio and the real-impedance point (20 marks)