Question 2 of 8: Standing Wave Ratio and Real-Impedance Point on a Line Feeding a Complex Load

Question 2: Standing Wave Ratio and Real-Impedance Point on a Line Feeding a Complex Load (equal value)

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. $f=300$ MHz, $Z_0=50\ \Omega$, $v=3\times10^8$ m/s, line length $l=30$ cm; load = $R=50\ \Omega$ in parallel with $L=1.59\times10^{-8}$ H.

Find. The standing wave ratio (SWR) on the 30 cm section, and the real input impedance at the point on that section where it occurs.

Approach. Compute the load impedance $Z_L=R\parallel j\omega L$, its reflection coefficient $\Gamma_L$, hence SWR; the input impedance becomes real wherever the round-trip phase $\theta_\Gamma-2\beta d$ is a multiple of $\pi$ (a voltage maximum or minimum), so locate the nearest such $d$ within the 30 cm section and read off $Z_0\cdot\text{SWR}$ or $Z_0/\text{SWR}$ accordingly.

1. Wavelength and electrical length. $$\lambda=\frac{v}{f}=\frac{3\times10^8}{300\times10^6}=1\ \text{m},\qquad \beta=\frac{2\pi}{\lambda}=2\pi\ \text{rad/m}.$$ The 30 cm section is therefore $l=0.3\lambda$ long.
2. Load impedance. $\omega=2\pi f=1.885\times10^{9}$ rad/s, so $X_L=\omega L=29.97\ \Omega\approx30\ \Omega$. Combining the parallel resistor and inductor, $$Z_L=\frac{1}{\frac{1}{R}+\frac{1}{jX_L}}=\frac{R\,(jX_L)}{R+jX_L}=13.22+j22.05\ \Omega .$$
3. Reflection coefficient and SWR. $$\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}=-0.410+j0.492=0.641\angle129.8^\circ .$$ $$\text{SWR}=\frac{1+|\Gamma_L|}{1-|\Gamma_L|}=\frac{1+0.641}{1-0.641}=\boxed{4.56} .$$
4. Locate the nearest real-impedance point. The input impedance looking toward the load from a distance $d$ is real where the reflection coefficient seen there, $\Gamma(d)=\Gamma_L e^{-j2\beta d}$, is real — i.e. where $\theta_\Gamma-2\beta d=0$ (a voltage maximum, $\Gamma(d)>0$) or $=\pi$ (a voltage minimum, $\Gamma(d)<0$). With $\theta_\Gamma=129.8^\circ=2.266$ rad, $$d=\frac{\theta_\Gamma}{2\beta}=\frac{2.266}{2(2\pi)}=0.1803\ \text{m}=18.03\ \text{cm},$$ which is $\Gamma(d)=+|\Gamma_L|$ (a voltage maximum) and lies inside the 30 cm driving section ($18.03\ \text{cm}<30\ \text{cm}$), so this is the point asked for. (The next candidate, a voltage minimum a further $\lambda/4=25$ cm toward the generator, falls outside the 30 cm section.)
5. Real impedance at that point. At a voltage maximum the input impedance is the largest real value on the line, $$Z_{in}(d=18.03\ \text{cm})=Z_0\cdot\text{SWR}=50\times4.56=\boxed{228.2\ \Omega}.$$ Direct substitution into $Z_{in}=Z_0\dfrac{Z_L+jZ_0\tan\beta d}{Z_0+jZ_L\tan\beta d}$ at this $d$ reproduces $228.2+j0\ \Omega$, confirming the location and value.