Question 2 of 10: Steady-State Standing Wave on a Complex-Loaded Line

Question 2: Steady-State Standing Wave on a Complex-Loaded Line (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.

Find. The forward-wave amplitude $|V_0^+|$ launched at $z=-l$; the standing-wave ratio $S$; the locations of all voltage maxima on $-l\le z\le0$.

Approach. Because $R_g=Z_0$ (a matched source), the forward wave launched onto the line has a fixed amplitude independent of the load or line length — show this from the general terminal relations. Then get $\Gamma_L$, $S$, and use the standard maxima condition (round-trip phase a multiple of $2\pi$) to locate the maxima.

1. Part (a) — Forward-wave amplitude at the input. Write $V(z)=V_0^+e^{-j\beta z}+V_0^-e^{j\beta z}$, $I(z)=(V_0^+e^{-j\beta z}-V_0^-e^{j\beta z})/Z_0$. At any point, $V_0^+e^{-j\beta z}=(V(z)+Z_0I(z))/2$. Evaluated at $z=-l$ using the source-side circuit ($I_{in}=V_g/(R_g+Z_{in})$, $V_{in}=I_{in}Z_{in}$): $$V_0^+e^{j\beta l}=\frac{V_{in}+Z_0I_{in}}{2}=\frac{I_{in}(Z_{in}+Z_0)}{2}=\frac{V_g}{2}\cdot\frac{Z_{in}+Z_0}{R_g+Z_{in}}.$$ Since $R_g=Z_0$, the fraction is exactly 1 for ANY $Z_{in}$ (i.e. any load or line length): $$\boxed{|V_0^+|=\frac{|V_g|}{2}=\frac{5}{2}=2.5\ \text{V}.}$$ (As a check: here $\beta l=(2\pi/\lambda)(4)=2\pi(4/2)=4\pi$, an exact multiple of $2\pi$ since $\lambda=v/f=2$ m and $l=2\lambda$, so $Z_{in}=Z_L=30+j50\ \Omega$ independently confirms $V_{in}=I_{in}Z_L$ is consistent with the boxed result above.)
2. Part (b) — Reflection coefficient and SWR. $$\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}=\frac{-20+j50}{80+j50}=0.101+j0.562=0.571\angle79.8^\circ.$$ $$S=\frac{1+|\Gamma_L|}{1-|\Gamma_L|}=\frac{1.571}{0.429}=\boxed{3.66}.$$
3. Locations of the voltage maxima. $|V(z)|=|V_0^+|\,|1+\Gamma_Le^{2j\beta z}|$ is maximum where the round-trip phase $\theta_L+2\beta z=0\ (\text{mod}\ 2\pi)$, i.e. at $z=-d$ with $d=\theta_L/(2\beta)-n\lambda/2$ chosen inside $0\le d\le l$. With $\theta_L=79.8^\circ=1.393$ rad and $\beta=\pi$ rad/m ($\lambda=2$ m so maxima repeat every $\lambda/2=1$ m): $$d_0=\frac{\theta_L}{2\beta}=\frac{1.393}{2\pi}=0.222\ \text{m}.$$ $$\boxed{d=0.222,\ 1.222,\ 2.222,\ 3.222\ \text{m (from the load)}}$$ (equivalently $z=-0.222,-1.222,-2.222,-3.222$ m), giving $|V|_{max}=|V_0^+|(1+|\Gamma_L|)=2.5(1.571)=3.93$ V at each.