Question 2 of 8: Section length giving a purely resistive input impedance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because the set is a study resource rather than a sat examination. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, from which $\eta_0=\sqrt{\mu_0/\varepsilon_0}\approx 377\ \Omega$ and $c = 3\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, waveguides); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (statics, Faraday's law, plane waves); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short elements, image theory).
Given. A uniform lossless section of line is terminated by a resistor in parallel with a capacitor, and we work at the single frequency at which the capacitive reactance is stated.
Given data
Quantity
Symbol
Value
Characteristic impedance
$Z_0$
$50\ \Omega$
Phase velocity
$v_p$
$2\times10^{8}\ \text{m/s}$
Frequency
$f$
$300\ \text{MHz}$
Shunt resistance
$R$
$50\ \Omega$
Shunt capacitive reactance at $f$
$X_C$
$-100\ \Omega$
Find. The (shortest) length of section for which the input impedance at 300 MHz is purely real, and the value of that resistance.
The terminated section. The 100 $\Omega$ capacitive reactance corresponds to $C=1/(2\pi f\,|X_C|)=5.31\ \text{pF}$ at 300 MHz.
Approach. Combine the shunt pair into one load impedance, then impose $\operatorname{Im}\{Z_{in}\}=0$ on the line-transformation formula. That condition is a quadratic in $\tan\beta l$, whose two roots are the voltage minimum and the voltage maximum of the standing wave; the shorter root is the answer and its impedance is $Z_0/\mathrm{SWR}$.
Combine the load elements. $$Z_L=\frac{R\,(jX_C)}{R+jX_C}=\frac{(50)(-j100)}{50-j100}=40-j20\ \Omega .$$ The parallel combination is not $50\ \Omega$ — the capacitor pulls both the real and the imaginary part down.
Wavelength and phase constant on the section. $$\lambda=\frac{v_p}{f}=\frac{2\times10^{8}}{3\times10^{8}}=0.6667\ \text{m}=66.67\ \text{cm},\qquad \beta=\frac{2\pi}{\lambda}=9.425\ \text{rad/m}.$$
Load reflection coefficient and standing-wave ratio. $$\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}=\frac{-10-j20}{90-j20}=0.2425\,\angle\,{-104.04^{\circ}},\qquad \mathrm{SWR}=\frac{1+|\Gamma_L|}{1-|\Gamma_L|}=1.640 .$$
Impose a real input impedance. With $t=\tan\beta l$ and $Z_L=R_L+jX_L$, $Z_{in}=Z_0\dfrac{Z_L+jZ_0t}{Z_0+jZ_Lt}$ has zero imaginary part when $$Z_0X_L+t\left(Z_0^{2}-R_L^{2}-X_L^{2}\right)-Z_0X_Lt^{2}=0 .$$ Substituting $Z_0=50$, $R_L=40$, $X_L=-20$ gives $-1000+500\,t+1000\,t^{2}=0$, i.e. $2t^{2}+t-2=0$.
Solve for the length. The roots are $t=0.7808$ and $t=-1.2808$. The first gives $\beta l=\arctan(0.7808)=37.98^{\circ}$, so $$l=\frac{37.98^{\circ}}{360^{\circ}}\lambda=0.1055\,\lambda\;\Rightarrow\;\boxed{l = 7.03\ \text{cm}} .$$
Evaluate the input impedance there. Substituting $t=0.7808$ back, $Z_{in}=50\dfrac{(40-j20)+j50(0.7808)}{50+j(40-j20)(0.7808)}=30.48+j0\ \Omega$. This is the voltage-minimum point, where the general result $Z_{in}=Z_0/\mathrm{SWR}$ applies: $$\boxed{Z_{in}=\frac{Z_0}{\mathrm{SWR}}=\frac{50}{1.640}=30.5\ \Omega\ (\text{purely resistive})} .$$
The second solution, and periodicity. The root $t=-1.2808$ corresponds to $\beta l=142.02^{\circ}$, i.e. $l=0.3555\lambda=23.70$ cm, at which $Z_{in}=Z_0\,\mathrm{SWR}=82.0\ \Omega$ (the voltage maximum). Both families repeat every $\lambda/2=33.33$ cm of added line.
Standing-wave envelope on the section. The input impedance is purely real only where the envelope is stationary; the first such point, at $0.1055\lambda$ from the load, is a voltage minimum.
The physical reading is that a mismatched line presents a resistive input only at the two stationary points of its standing wave, and those points alternate every quarter wavelength. Because the load here is capacitive, the first stationary point encountered on moving away from the load is a minimum, which is why the answer is below $Z_0$ rather than above it. A designer who needs the higher resistive value simply adds the extra $\lambda/4 = 16.67$ cm of line.
