22-Elec-A7 Electromagnetics · May 2013

Question 1 of 8: Step Response of an Infinite Line with a Shunt Resistor

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Step Response of an Infinite Line with a Shunt Resistor (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 27 V step source of 100 Ω internal impedance drives a lossless 50 Ω line; 10 km out, a 50 Ω resistor is bridged across the line and the line continues to infinity beyond it.

Find. (i) the duration and the energy of the first segment of the wave that travels away on the infinite section, and (ii) the power that section receives once all transients have died away.

Approach. Launch the initial step through the generator/line divider, treat the shunt resistor in parallel with the infinite continuation as a single 25 Ω junction load to obtain the reflection and transmission there, take the generator round trip as the length of the first plateau, and finally solve the DC steady state directly.

1. Launch the first step onto the line. At $t=0^{+}$ the generator sees only the line's characteristic impedance, so the source resistance and $Z_0$ form a simple divider: $$V_1^{+}=E\,\frac{Z_0}{R_g+Z_0}=27\times\frac{50}{100+50}=9.00\ \text{V}$$ carrying $I_1^{+}=V_1^{+}/Z_0=0.180$ A.
2. Find the one-way transit time. The step needs $$t_d=\frac{d}{v_p}=\frac{10\times10^{3}}{2\times10^{8}}=50\ \mu\text{s}$$ to reach the shunt resistor.
3. Reduce the junction to one load. Looking to the right at the tap point, the 50 Ω resistor is in parallel with the infinite line, which itself always presents $Z_0$: $$Z_J=\frac{R_{sh}Z_0}{R_{sh}+Z_0}=\frac{50\times50}{100}=25\ \Omega ,\qquad \Gamma_J=\frac{Z_J-Z_0}{Z_J+Z_0}=\frac{25-50}{25+50}=-\tfrac{1}{3}$$ This is the step most often missed: the shunt resistor does not match the line, because the line beyond it is still there.
4. Transmit into the infinite section. The voltage that continues past the junction is the incident wave times $1+\Gamma_J$: $$V_{\infty}=V_1^{+}\left(1+\Gamma_J\right)=9.00\times\tfrac{2}{3}=\boxed{6.00\ \text{V}}$$ while $-3.00$ V heads back toward the generator.
5. Time the end of the first segment. That $-3.00$ V wave reaches the generator at $t=2t_d=100\ \mu\text{s}$, where the source mismatch $\Gamma_g=(100-50)/(100+50)=+1/3$ returns $-1.00$ V toward the load; this new step arrives at the junction at $t=150\ \mu\text{s}$ and changes the amplitude on the infinite section. The 6.00 V plateau therefore lives from 50 μs to 150 μs: $$T_{seg}=\frac{2d}{v_p}=\boxed{100\ \mu\text{s}}$$ Travelling at $2\times10^{8}$ m/s, that segment is $v_p T_{seg}=20$ km long in space.
6. Convert the segment to energy. While the plateau lasts, the infinite section carries a pure travelling wave, so its power is $$P_{seg}=\frac{V_{\infty}^{2}}{Z_0}=\frac{6.00^{2}}{50}=0.720\ \text{W}$$ and the energy carried away in the first segment is $$W_{seg}=P_{seg}\,T_{seg}=0.720\times100\times10^{-6}=\boxed{72.0\ \mu\text{J}}$$
7. Solve the steady state. After many round trips the excitation is simply DC. A lossless line of finite length is then just a pair of wires, so the generator sees $R_g$ in series with $Z_J$: $$V_{J}=E\,\frac{Z_J}{R_g+Z_J}=27\times\frac{25}{125}=5.40\ \text{V}$$ The infinite section still presents 50 Ω at that node, so $$P_{\infty}=\frac{V_J^{2}}{Z_0}=\frac{5.40^{2}}{50}=\boxed{0.583\ \text{W}}$$
8. Confirm with the bounce series. Successive forward waves at the junction are $9.00,\ -1.00,\ +0.111,\dots$, a geometric series of ratio $\Gamma_J\Gamma_g=-1/9$. Summing gives $9.00/(1+1/9)=8.10$ V incident in total, of which $8.10\times\tfrac{2}{3}=5.40$ V is transmitted — the same steady state, which confirms both the reflection coefficients and the DC reduction.

Note that the shunt resistor absorbs the difference: at steady state it draws $5.40^{2}/50=0.583$ W as well, and the generator delivers $27\times(5.40/25)=5.83$ W, most of which is dissipated in its own 100 Ω internal resistance.