Question:medium
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
The Correct Option is B
Solution and Explanation
Step 1: Recognize the physical meaning of the flip.
A point diametrically across a $|\Gamma|$ circle is what you reach after moving a quarter wavelength down a lossless line, since a quarter-wave section always maps $Z_{in}=Z_0^2/Z_L$. So instead of rotating $\Gamma$ by hand, this problem is really asking for a quarter-wave transformed impedance.
Step 2: Turn $\Gamma_L$ into $Z_L$ first.
\[ \Gamma_L=0.5\angle30^\circ=0.433+j0.25 \]
\[ Z_L=Z_0\frac{1+\Gamma_L}{1-\Gamma_L}=50\cdot\frac{1.433+j0.25}{0.567-j0.25} \]
Multiplying top and bottom by the conjugate of the denominator, $0.567+j0.25$, the denominator becomes $0.567^2+0.25^2=0.384$ and the numerator becomes $0.75+j0.5$. So
\[ Z_L=50\cdot\frac{0.75+j0.5}{0.384}=97.7+j65.1\ \Omega \]
Step 3: Apply the quarter-wave rule.
\[ Z_{in}=\frac{Z_0^2}{Z_L}=\frac{2500}{97.7+j65.1} \]
Multiply by the conjugate over itself:
\[ Z_{in}=\frac{2500(97.7-j65.1)}{97.7^2+65.1^2}=\frac{2500(97.7-j65.1)}{13783} \]
\[ Z_{in}\approx0.1814(97.7-j65.1)\approx17.7-j11.8\ \Omega \]
Step 4: Conclude.
The cable-connected input impedance comes out the same way as a direct $\Gamma$ rotation would give.
\[ \boxed{Z_{in}\approx17.7-j11.8\ \Omega} \]
A point diametrically across a $|\Gamma|$ circle is what you reach after moving a quarter wavelength down a lossless line, since a quarter-wave section always maps $Z_{in}=Z_0^2/Z_L$. So instead of rotating $\Gamma$ by hand, this problem is really asking for a quarter-wave transformed impedance.
Step 2: Turn $\Gamma_L$ into $Z_L$ first.
\[ \Gamma_L=0.5\angle30^\circ=0.433+j0.25 \]
\[ Z_L=Z_0\frac{1+\Gamma_L}{1-\Gamma_L}=50\cdot\frac{1.433+j0.25}{0.567-j0.25} \]
Multiplying top and bottom by the conjugate of the denominator, $0.567+j0.25$, the denominator becomes $0.567^2+0.25^2=0.384$ and the numerator becomes $0.75+j0.5$. So
\[ Z_L=50\cdot\frac{0.75+j0.5}{0.384}=97.7+j65.1\ \Omega \]
Step 3: Apply the quarter-wave rule.
\[ Z_{in}=\frac{Z_0^2}{Z_L}=\frac{2500}{97.7+j65.1} \]
Multiply by the conjugate over itself:
\[ Z_{in}=\frac{2500(97.7-j65.1)}{97.7^2+65.1^2}=\frac{2500(97.7-j65.1)}{13783} \]
\[ Z_{in}\approx0.1814(97.7-j65.1)\approx17.7-j11.8\ \Omega \]
Step 4: Conclude.
The cable-connected input impedance comes out the same way as a direct $\Gamma$ rotation would give.
\[ \boxed{Z_{in}\approx17.7-j11.8\ \Omega} \]
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