Question:
A complex load (in \(\Omega\)) is represented as \(\Gamma_L=0.5\angle30^{\circ}\) on the Smith chart. A co-axial cable with a characteristic impedance of \(50\ \Omega\) is connected to the load. The new input impedance of the load now moves to a diametrically opposite point on the same \(\Gamma\) circle on the Smith chart.
Which option is the nearest input impedance of the cable connected load (in \(\Omega\))?
A complex load (in \(\Omega\)) is represented as \(\Gamma_L=0.5\angle30^{\circ}\) on the Smith chart. A co-axial cable with a characteristic impedance of \(50\ \Omega\) is connected to the load. The new input impedance of the load now moves to a diametrically opposite point on the same \(\Gamma\) circle on the Smith chart.
Which option is the nearest input impedance of the cable connected load (in \(\Omega\))?
Which option is the nearest input impedance of the cable connected load (in \(\Omega\))?
Show Hint
A diametrically opposite point on the same reflection coefficient circle means the sign of Gamma flips.
Updated On: Jul 20, 2026
- \(20.7-j5.1\)
- \(17.7-j11.8\)
- \(97.5-j65.0\)
- \(97.5+j65.0\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Convert the given reflection coefficient to rectangular form.
\[ \Gamma_L=0.5\angle30^{\circ}=0.5\cos(30^{\circ})+j\,0.5\sin(30^{\circ})=0.433+j0.25 \]
Step 2: Interpret "diametrically opposite on the same \(\Gamma\) circle".
On a Smith chart, moving along a lossless transmission line keeps \(|\Gamma|\) fixed and rotates its angle by \(-2\beta l\), tracing a circle centred at the origin. A diametrically opposite point on that same circle is reached when the angle changes by exactly \(180^{\circ}\), which happens for a line length of a quarter wavelength (\(l=\lambda/4\), since \(2\beta l=2\cdot\frac{2\pi}{\lambda}\cdot\frac{\lambda}{4}=\pi\)). So the new reflection coefficient at the cable input is
\[ \Gamma_{in}=\Gamma_L\,e^{-j\pi}=-\Gamma_L \]
Step 3: Compute \(\Gamma_{in}\).
\[ \Gamma_{in}=-(0.433+j0.25)=-0.433-j0.25=0.5\angle210^{\circ} \]
Step 4: Convert \(\Gamma_{in}\) to the input impedance.
\[ Z_{in}=Z_0\cdot\frac{1+\Gamma_{in}}{1-\Gamma_{in}} \]
\[ 1+\Gamma_{in}=0.567-j0.25,\qquad 1-\Gamma_{in}=1.433+j0.25 \]
Multiply numerator and denominator by the conjugate of the denominator:
\[ \frac{1+\Gamma_{in}}{1-\Gamma_{in}}=\frac{(0.567-j0.25)(1.433-j0.25)}{(1.433)^2+(0.25)^2}=\frac{0.75-j0.5}{2.116}=0.354-j0.236 \]
Step 5: Scale by \(Z_0=50\ \Omega\).
\[ Z_{in}=50(0.354-j0.236)=17.7-j11.8\ \Omega \]
Step 6: Analyze the options.
(A) \(20.7-j5.1\): Does not match either the real or the imaginary part found above. Incorrect.
(B) \(17.7-j11.8\): Matches the computed real and imaginary parts. Correct.
(C) \(97.5-j65.0\): This is what results from mistakenly using the conjugate of \(\Gamma_L\) as \(\Gamma_{in}\) instead of applying the diametric flip. Incorrect.
(D) \(97.5+j65.0\): This is what results from forgetting the diametric flip altogether and using \(\Gamma_L\) itself as \(\Gamma_{in}\). Incorrect.
Step 7: Final conclusion.
\[ \boxed{Z_{in}\approx17.7-j11.8\ \Omega} \]
\[ \Gamma_L=0.5\angle30^{\circ}=0.5\cos(30^{\circ})+j\,0.5\sin(30^{\circ})=0.433+j0.25 \]
Step 2: Interpret "diametrically opposite on the same \(\Gamma\) circle".
On a Smith chart, moving along a lossless transmission line keeps \(|\Gamma|\) fixed and rotates its angle by \(-2\beta l\), tracing a circle centred at the origin. A diametrically opposite point on that same circle is reached when the angle changes by exactly \(180^{\circ}\), which happens for a line length of a quarter wavelength (\(l=\lambda/4\), since \(2\beta l=2\cdot\frac{2\pi}{\lambda}\cdot\frac{\lambda}{4}=\pi\)). So the new reflection coefficient at the cable input is
\[ \Gamma_{in}=\Gamma_L\,e^{-j\pi}=-\Gamma_L \]
Step 3: Compute \(\Gamma_{in}\).
\[ \Gamma_{in}=-(0.433+j0.25)=-0.433-j0.25=0.5\angle210^{\circ} \]
Step 4: Convert \(\Gamma_{in}\) to the input impedance.
\[ Z_{in}=Z_0\cdot\frac{1+\Gamma_{in}}{1-\Gamma_{in}} \]
\[ 1+\Gamma_{in}=0.567-j0.25,\qquad 1-\Gamma_{in}=1.433+j0.25 \]
Multiply numerator and denominator by the conjugate of the denominator:
\[ \frac{1+\Gamma_{in}}{1-\Gamma_{in}}=\frac{(0.567-j0.25)(1.433-j0.25)}{(1.433)^2+(0.25)^2}=\frac{0.75-j0.5}{2.116}=0.354-j0.236 \]
Step 5: Scale by \(Z_0=50\ \Omega\).
\[ Z_{in}=50(0.354-j0.236)=17.7-j11.8\ \Omega \]
Step 6: Analyze the options.
(A) \(20.7-j5.1\): Does not match either the real or the imaginary part found above. Incorrect.
(B) \(17.7-j11.8\): Matches the computed real and imaginary parts. Correct.
(C) \(97.5-j65.0\): This is what results from mistakenly using the conjugate of \(\Gamma_L\) as \(\Gamma_{in}\) instead of applying the diametric flip. Incorrect.
(D) \(97.5+j65.0\): This is what results from forgetting the diametric flip altogether and using \(\Gamma_L\) itself as \(\Gamma_{in}\). Incorrect.
Step 7: Final conclusion.
\[ \boxed{Z_{in}\approx17.7-j11.8\ \Omega} \]
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