$\mathrm{A\; complex\; load\; (in}\Omega \text{) is represented as}{\mathrm{ \Gamma }}_L=0.5\angle 30°\text{on the Smith chart.}$

$\text{A co-axial cable with a characteristic impedance of}50\Omega \text{is connected to the load.}$

$\text{The new input impedance of the load now moves to a diametrically opposite point on the same}\Gamma \text{circle.}$

$\text{Which option is the nearest input impedance?}$

The correct option is: 17.7 − j11.8

Here is the step-by-step explanation and derivation to find the new input impedance of the load:

Step 1: Understand the reflection coefficient at the diametrically opposite point
The complex load has a reflection coefficient represented on the Smith chart as:
${\Gamma }_L=0.5\angle 30°$
Moving to a diametrically opposite point on the same constant reflection coefficient circle (constant $\left|\Gamma \right|$ circle) corresponds to a phase shift of 180°. Math mathematically, this is equivalent to multiplying the reflection coefficient by -1. Therefore, the reflection coefficient at the new input location is:
${\Gamma }_{\mathrm{in}}=-{\Gamma }_L=-0.5\angle 30°=0.5\angle (30°-180°)=0.5\angle -150°$

Step 2: Convert the reflection coefficient to rectangular coordinates
Using Euler's relation, we convert the polar representation of the reflection coefficient to rectangular form:
${\Gamma }_{\mathrm{in}}=0.5(cos(-150°)+jsin(-150°\left)\right)$
Since $cos(-150°)=-0.866$ and $sin(-150°)=-0.5$:
${\Gamma }_{\mathrm{in}}=0.5(-0.866-j0.5)=-0.433-j0.25$

Step 3: Calculate the normalized input impedance
The relation between the normalized input impedance $z_{\mathrm{in}}$ and the reflection coefficient ${\Gamma }_{\mathrm{in}}$ is given by:
$z_{\mathrm{in}}=\frac{1+{\Gamma }_{\mathrm{in}}}{1-{\Gamma }_{\mathrm{in}}}$
Substitute ${\Gamma }_{\mathrm{in}}=-0.433-j0.25$ into the equation:
$z_{\mathrm{in}}=\frac{1+(-0.433-j0.25)}{1-(-0.433-j0.25)}=\frac{0.567-j0.25}{1.433+j0.25}$
Multiply the numerator and denominator by the complex conjugate of the denominator ($1.433-j0.25$):
$z_{\mathrm{in}}=\frac{(0.567-j0.25)(1.433-j0.25)}{{1.433}^2+{0.25}^2}$
Calculate the denominator:
${1.433}^2+{0.25}^2=2.0535+0.0625=2.116$
Calculate the numerator:
$(0.567\times 1.433-0.25\times 0.25)-j(0.567\times 0.25+0.25\times 1.433)=(0.8125-0.0625)-j(0.1418+0.3582)=0.75-j0.5$
Therefore, the normalized input impedance is:
$z_{\mathrm{in}}=\frac{0.75-j0.5}{2.116}\approx 0.3544-j0.2363$

Step 4: Find the actual input impedance
Given that the characteristic impedance of the transmission line is $Z_0=50\Omega$, we scale the normalized input impedance back to the actual value:
$Z_{\mathrm{in}}=Z_0\times z_{\mathrm{in}}=50\times (0.3544-j0.2363)\approx 17.72-j11.82\Omega$
Rounding to one decimal place gives approximately 17.7 − j11.8.