VSWR on a 50 ohm line terminated in 100 ohms
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A lossless transmission line of characteristic impedance $Z_0 = 50\ \Omega$ is terminated in a purely resistive load $Z_L = 100\ \Omega$.
The voltage standing wave ratio on the line is ________.
Show the step-by-step derivation
Answer
2 unitless unitless
Step-by-step derivation
- Compute the voltage reflection coefficient at the load: $$\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{100 - 50}{100 + 50} = \frac{50}{150} = \frac{1}{3}.$$ It is real and positive, as expected for a purely resistive load larger than $Z_0$.
- VSWR is the ratio of maximum to minimum voltage magnitude along the line. The forward and reflected waves add in phase at the maxima and subtract at the minima: $$VSWR = \frac{|V|_{max}}{|V|_{min}} = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|}.$$
- Substitute $|\Gamma_L| = 1/3$: $$VSWR = \frac{1 + 1/3}{1 - 1/3} = \frac{4/3}{2/3} = \frac{4}{3}\cdot\frac{3}{2} = \mathbf{2}.$$
- Shortcut for resistive loads. When $Z_L$ is purely real, VSWR is just the ratio of the larger to the smaller impedance: $$VSWR = \frac{Z_L}{Z_0} = \frac{100}{50} = 2 \quad(\text{or } Z_0/Z_L \text{ if } Z_L < Z_0).$$ Same answer, no reflection coefficient needed. ✓
- Power check. The fraction of incident power reflected is $|\Gamma_L|^2 = 1/9 = 11.1\%$, so $88.9\%$ is delivered - a return loss of $-20\log_{10}(1/3) = 9.5$ dB.
The trap this question is built around
Forgetting the magnitude bars. For a complex or capacitive load, $\Gamma_L$ is complex, and VSWR uses $|\Gamma_L|$ only - the phase sets where the maxima sit along the line, not how large they are.