Question

For a lossless line, the surge is purely resistive and the propagation constant is pure imaginary.

(a) True

(b) False

Short answer

Therefore, the correct option is (a): True.

Step-by-step solution

Determine the expression circuit constant for the transmission line,

The impedance of the transmission line given by,

$z=R+j\omega L\frac{\Omega }{m}$ …… (1)

Here,R is the resistance, L is the inductance, $\omega$is the angular frequency.

The admittance of the transmission line is given by.

$y=G+j\omega C\frac{S}{m}$ …… (2)

Here, G is the conductance and C is the capacitance.

The expression for the surge impedance is given by,

$Z_c=\sqrt{\frac{z}{y}}$ …… (3)

The expression for the propagation constant is given by,

$\gamma =\sqrt{ZY}$ …… (4)

Determine the surge and propagation constant.

For the lossless line, the resistance of the transmission line is zero.

Substitute 0 for R into equation (1).

$\begin{array}{l}z=0+j\omega L\\ z=j\omega L\end{array}$

For the lossless line, the conductance of the transmission line is zero.

Substitute 0 for G into equation (1).

$\begin{array}{l}y=0+j\omega C\\ y=j\omega C\end{array}$

Calculate the surge impedance.

Substitute $j\omega L$for z and $j\omega C$for y into equation (3).

$\begin{array}{l}{Z}_c=\sqrt{\frac{j\omega L}{j\omega C}}\\ Z_c=\sqrt{\frac{L}{C}}\end{array}$

Hence, the surge for the lossless transmission line is$\sqrt{\frac{L}{C}}$ which is purely resistive.

Calculate the propagation constant.

Substitute $j\omega L$for z and $j\omega C$for y into equation (4).

$\begin{array}{l}\gamma =\sqrt{\left(j\omega L\right)\left(j\omega C\right)}\\ \gamma =\sqrt{j^2{\omega }^{2}LC}\\ \gamma =j\omega \sqrt{LC}\end{array}$

Hence the propagation constant is $j\omega \sqrt{LC}$which is purely imaginary.

Therefore, the given statement is true.