Instructions: Answer Any Five (5) Questions. 1a. What is time division switching? With the help of block diagram, explain basic time division time switching method.

• Frequency, $f = 10 \text{ MHz} = 10 \times 10^6 \text{ Hz}$
• Inductance, $L = 95.3 \text{ nH/m} = 95.3 \times 10^{-9} \text{ H/m}$
• Capacitance, $C = 68.5 \text{ pF/m} = 68.5 \times 10^{-12} \text{ F/m}$
• Resistance, $R = 0.685 \Omega/\text{m}$
• Conductance, $G = 0$

The angular frequency is:

$$\omega = 2 \pi f = 2 \pi (10 \times 10^6 \text{ Hz}) = 62.83185 \times 10^6 \text{ rad/s}$$

Now, calculate the complex impedance per unit length ($Z$) and admittance per unit length ($Y$):

$$Z = R + j\omega L = 0.685 + j(62.83185 \times 10^6)(95.3 \times 10^{-9}) = 0.685 + j5.9880 \Omega/\text{m}$$ $$Y = G + j\omega C = 0 + j(62.83185 \times 10^6)(68.5 \times 10^{-12}) = j4.3039 \times 10^{-3} \text{ S/m}$$

The characteristic impedance is given by:

$$Z_0 = \sqrt{\frac{Z}{Y}} = \sqrt{\frac{R + j\omega L}{G + j\omega C}}$$ $$Z_0 = \sqrt{\frac{0.685 + j5.9880}{j4.3039 \times 10^{-3}}}$$

To simplify, multiply the numerator and denominator inside the square root by $-j$:

$$Z_0 = \sqrt{\frac{(0.685 + j5.9880)(-j)}{j4.3039 \times 10^{-3}(-j)}} = \sqrt{\frac{5.9880 - j0.685}{4.3039 \times 10^{-3}}}$$ $$Z_0 = \sqrt{1391.26 - j159.15}$$

Convert the complex number to polar form:
Magnitude: $\sqrt{1391.26^2 + (-159.15)^2} = 1400.33$
Angle: $\arctan\left(\frac{-159.15}{1391.26}\right) = -6.51^\circ$
So, $1391.26 - j159.15 = 1400.33 \angle -6.51^\circ$.

$$Z_0 = \sqrt{1400.33 \angle -6.51^\circ} = \sqrt{1400.33} \angle \left(\frac{-6.51^\circ}{2}\right)$$ $$Z_0 = 37.42 \angle -3.26^\circ \Omega$$

In rectangular form:

$$Z_0 = 37.42 \cos(-3.26^\circ) + j37.42 \sin(-3.26^\circ) = 37.36 - j2.12 \Omega$$

The characteristics impedance is: $\boxed{\mathbf{Z_0 = 37.36 - j2.12 \Omega}}$

The propagation constant is given by:

$$\gamma = \sqrt{Z \cdot Y} = \sqrt{(R + j\omega L)(G + j\omega C)}$$ $$\gamma = \sqrt{(0.685 + j5.9880)(j4.3039 \times 10^{-3})}$$ $$\gamma = \sqrt{j(0.685 \times 4.3039 \times 10^{-3}) + j^2(5.9880 \times 4.3039 \times 10^{-3})}$$ $$\gamma = \sqrt{j(2.9481 \times 10^{-3}) - (25.770 \times 10^{-3})}$$ $$\gamma = \sqrt{-0.025770 + j0.0029481}$$

Convert the complex number to polar form:
Magnitude: $\sqrt{(-0.025770)^2 + (0.0029481)^2} = 0.025938$
Angle: $\arctan\left(\frac{0.0029481}{-0.025770}\right) = -6.51^\circ$. Since the real part is negative and the imaginary part is positive, the angle is in the second quadrant: $180^\circ - 6.51^\circ = 173.49^\circ$.
So, $-0.025770 + j0.0029481 = 0.025938 \angle 173.49^\circ$.

$$\gamma = \sqrt{0.025938 \angle 173.49^\circ} = \sqrt{0.025938} \angle \left(\frac{173.49^\circ}{2}\right)$$ $$\gamma = 0.16105 \angle 86.745^\circ$$

In rectangular form:

$$\gamma = 0.16105 \cos(86.745^\circ) + j0.16105 \sin(86.745^\circ)$$ $$\gamma = 0.00913 + j0.1608$$

From $\gamma = \alpha + j\beta$:
The attenuation constant is $\alpha = \text{Re}(\gamma) = 0.00913 \text{ Np/m}$.
The phase constant (space shift constant) is $\beta = \text{Im}(\gamma) = 0.1608 \text{ rad/m}$.

Convert the attenuation constant to dB/m:

$$\alpha_{dB} = 8.686 \times \alpha = 8.686 \times 0.00913 = 0.0793 \text{ dB/m}$$

The attenuation constant is: $\boxed{\mathbf{\alpha = 0.00913 \text{ Np/m} \text{ or } 0.0793 \text{ dB/m}}}$
The space shift constant is: $\boxed{\mathbf{\beta = 0.1608 \text{ rad/m}}}$

The velocity of propagation is given by:

$$v_p = \frac{\omega}{\beta}$$ $$v_p = \frac{62.83185 \times 10^6 \text{ rad/s}}{0.1608 \text{ rad/m}}$$ $$v_p = 390.74 \times 10^6 \text{ m/s}$$ $$v_p = 3.91 \times 10^8 \text{ m/s}$$

The velocity of propagation is: $\boxed{\mathbf{v_p = 3.91 \times 10^8 \text{ m/s}}}$