Question 2 of 10: Standing-wave pattern on a mismatched 50 Ω line

Question 2: Standing-wave pattern on a mismatched 50 Ω line

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 4 m, 50 Ω line carrying a 100 MHz signal at $2\times10^8$ m/s, fed from a 5 V source through a 50 Ω internal resistance into a complex load.

Find. The amplitude of the forward wave launched at $z = -\ell$; the standing-wave ratio; the positions of every voltage maximum on the line; and a plot of $|V(z)|$.

[Figure not reproduced: Figure 2 (redrawn). The 4 m line is exactly two wavelengths long at 100 MHz, and R_g = Z_0 so the generator absorbs every returning wave. See the official exam paper.]

Approach. Reduce the line to its wavelength first — it turns out to be exactly two wavelengths long — then use the fact that $R_g = Z_0$ makes the generator reflection coefficient zero, so a single forward wave and its single echo describe the whole steady state.

1. Convert the electrical length. The wavelength on the line is $$\lambda = \frac{v_p}{f} = \frac{2\times10^8}{100\times10^6} = 2.00\ \text{m}, \qquad \beta = \frac{2\pi}{\lambda} = \pi\ \text{rad/m}$$ so $\ell/\lambda = 4/2 = 2$ exactly and $\beta\ell = 4\pi$ rad. A line an integer number of half-wavelengths long repeats its load, so $Z_{in} = Z_L = 30 + j50\ \Omega$ without any transformation — a deliberate simplification built into the numbers.
2. Find the load reflection coefficient. $$\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{-20 + j50}{80 + j50} = \frac{900 + j5000}{8900} = 0.1011 + j0.5618$$ In polar form $\Gamma_L = 0.5708\angle 79.80^\circ$, so $|\Gamma_L| = 0.5708$ and $\theta_\Gamma = 1.3929$ rad.
3. (a) Launch the forward wave. Because $R_g = Z_0 = 50\ \Omega$ the generator reflection coefficient $\Gamma_g = (R_g - Z_0)/(R_g + Z_0) = 0$: whatever returns from the load is absorbed and never re-launched. The forward wave therefore reduces to the simple matched divider $$V_0^+ = \tilde{V}_g\,\frac{Z_0}{R_g + Z_0}\, \frac{e^{-j\beta\ell}}{1 - \Gamma_g\Gamma_L e^{-j2\beta\ell}} = 5\left(\frac{50}{100}\right)e^{-j4\pi}$$ and since $e^{-j4\pi} = 1$, $$\boxed{|V_0^+| = 2.50\ \text{V} \text{ at } \angle 0^\circ}$$ Cross-check without that shortcut: $V_{in} = \tilde{V}_g Z_{in}/(R_g + Z_{in}) = 3.090\angle 27.03^\circ$ V, and $V_0^+ = V_{in}/(1 + \Gamma_L) = 2.50\angle 0^\circ$ V. The two routes agree, which confirms both $\Gamma_L$ and the electrical length.
4. (b) Standing-wave ratio. $$S = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|} = \frac{1.5708}{0.4292} \Rightarrow \boxed{S = 3.66}$$ The envelope extremes follow directly: $|V|_{max} = |V_0^+|(1 + |\Gamma_L|) = 3.927$ V and $|V|_{min} = |V_0^+|(1 - |\Gamma_L|) = 1.073$ V, whose ratio is $S$ as it must be.
5. Locate the maxima. Writing $d = -z$ for distance back from the load, $|V(d)| = |V_0^+|\,|1 + |\Gamma_L|e^{j(\theta_\Gamma - 2\beta d)}|$, which peaks whenever the phase term vanishes modulo $2\pi$: $$d_{max} = \frac{\theta_\Gamma + 2n\pi}{2\beta} = \frac{1.3929}{2\pi} + \frac{n\lambda}{2} = 0.2217 + n(1.000)\ \text{m}$$ Keeping only the values that land on the 4 m line, $$\boxed{d_{max} = 0.222,\ 1.222,\ 2.222,\ 3.222\ \text{m from the load}}$$ equivalently $z = -0.222, -1.222, -2.222, -3.222$ m. Successive maxima are $\lambda/2 = 1.000$ m apart, and minima sit a further quarter wave along at $d = 0.722,\ 1.722,\ 2.722,\ 3.722$ m.
6. Confirm with the impedance at those planes. At a voltage maximum the line impedance is purely real and equal to $Z_0 S = 50(3.660) = 183.0\ \Omega$; at a minimum it is $Z_0/S = 13.66\ \Omega$. Evaluating the full transmission-line impedance $Z(d) = Z_0(Z_L + jZ_0\tan\beta d)/(Z_0 + jZ_L\tan\beta d)$ at $d = 0.222$ m returns $183.0 + j0.0\ \Omega$, so the located maxima are right.