Question:
The standing wave ratio on a $50\ \Omega$ lossless transmission line terminated in an unknown load impedance is found to be $2.0$. The distance between successive voltage minima is $30\ \text{cm}$ and the first minimum is located at $10\ \text{cm}$ from the load. $Z_L$ can be replaced by an equivalent length $l_m$ and terminating resistance $R_m$ of the same line. The value of $R_m$ and $l_m$, respectively, are _____________
The standing wave ratio on a $50\ \Omega$ lossless transmission line terminated in an unknown load impedance is found to be $2.0$. The distance between successive voltage minima is $30\ \text{cm}$ and the first minimum is located at $10\ \text{cm}$ from the load. $Z_L$ can be replaced by an equivalent length $l_m$ and terminating resistance $R_m$ of the same line. The value of $R_m$ and $l_m$, respectively, are _____________
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For SWR problems: compute $|\Gamma|$ from SWR, find possible resistive loads using $\Gamma=\tfrac{R_L-Z_0}{R_L+Z_0}$, and use the given minima position to determine equivalent line lengths modulo $\lambda/2$.
Updated On: Feb 12, 2026
- $R_m=100\ \Omega,\ l_m=20\ \text{cm}$
- $R_m=25\ \Omega,\ l_m=20\ \text{cm}$
- $R_m=100\ \Omega,\ l_m=5\ \text{cm}$
- $R_m=25\ \Omega,\ l_m=5\ \text{cm}$
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The Correct Option is A
Solution and Explanation
Step 1: Reflection coefficient from SWR
The standing wave ratio is given by \[ S = \frac{1+|\Gamma|}{1-|\Gamma|}. \] With \(S=2.0\): \[ |\Gamma| = \frac{S-1}{S+1} = \frac{1}{3}. \]
Step 2: Relation between reflection coefficient and load impedance
\[ \Gamma = \frac{Z_L - Z_0}{Z_L+Z_0}, \quad Z_0=50\ \Omega. \] For purely resistive loads, this yields two solutions depending on whether \(Z_L > Z_0\) or \(Z_L < Z_0\).
Step 3: Solve for possible load resistances
Case 1 (if \(Z_L > Z_0\)): \[ \frac{R_m - 50}{R_m+50} = \frac{1}{3} \;\;\Rightarrow\;\; R_m=100\ \Omega. \] Case 2 (if \(Z_L < Z_0\)): \[ \frac{50 - R_m}{R_m+50} = \frac{1}{3} \;\;\Rightarrow\;\; R_m=25\ \Omega. \] Thus, possible values: \(R_m = 100\ \Omega\) or \(25\ \Omega\).
Step 4: Use position of minima
The spacing between voltage minima is \(\lambda/2 = 30\ \text{cm} \Rightarrow \lambda = 60\ \text{cm}\). The phase constant is \(\beta = \tfrac{2\pi}{\lambda} = \tfrac{\pi}{30}\ \text{rad/cm}\).
The first minimum occurs at 10 cm from the load. The equivalent normalized length for stub matching can be taken as 20 cm, or by periodicity, 5 cm.
Step 5: Final conclusion
Thus, the valid combinations are: \[ R_m = 100\ \Omega,\;\; l_m = 20\ \text{cm}\quad (\text{Option A}) \] \[ R_m = 100\ \Omega,\;\; l_m = 5\ \text{cm}\quad (\text{Option C}) \]
Final Answer:
\[ \boxed{\text{Correct choices: (A) and (C)}} \]
The standing wave ratio is given by \[ S = \frac{1+|\Gamma|}{1-|\Gamma|}. \] With \(S=2.0\): \[ |\Gamma| = \frac{S-1}{S+1} = \frac{1}{3}. \]
Step 2: Relation between reflection coefficient and load impedance
\[ \Gamma = \frac{Z_L - Z_0}{Z_L+Z_0}, \quad Z_0=50\ \Omega. \] For purely resistive loads, this yields two solutions depending on whether \(Z_L > Z_0\) or \(Z_L < Z_0\).
Step 3: Solve for possible load resistances
Case 1 (if \(Z_L > Z_0\)): \[ \frac{R_m - 50}{R_m+50} = \frac{1}{3} \;\;\Rightarrow\;\; R_m=100\ \Omega. \] Case 2 (if \(Z_L < Z_0\)): \[ \frac{50 - R_m}{R_m+50} = \frac{1}{3} \;\;\Rightarrow\;\; R_m=25\ \Omega. \] Thus, possible values: \(R_m = 100\ \Omega\) or \(25\ \Omega\).
Step 4: Use position of minima
The spacing between voltage minima is \(\lambda/2 = 30\ \text{cm} \Rightarrow \lambda = 60\ \text{cm}\). The phase constant is \(\beta = \tfrac{2\pi}{\lambda} = \tfrac{\pi}{30}\ \text{rad/cm}\).
The first minimum occurs at 10 cm from the load. The equivalent normalized length for stub matching can be taken as 20 cm, or by periodicity, 5 cm.
Step 5: Final conclusion
Thus, the valid combinations are: \[ R_m = 100\ \Omega,\;\; l_m = 20\ \text{cm}\quad (\text{Option A}) \] \[ R_m = 100\ \Omega,\;\; l_m = 5\ \text{cm}\quad (\text{Option C}) \]
Final Answer:
\[ \boxed{\text{Correct choices: (A) and (C)}} \]
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