Two lossless transmission lines having different characteristic impedances are to be joined end to end. The impedances are \(Z_{01}=100 \Omega\) and \(Z_{03}=25 \Omega\). The operating frequency is \(1 \mathrm{GHz}\). \((a)\) Find the required characteristic impedance, \(Z_{02}\), of a quarter-wave section to be inserted between the two, which will impedance-match the joint, thus allowing total power transmission through the three lines. \((b)\) The capacitance per unit length of the intermediate line is found to be \(100 \mathrm{pF} / \mathrm{m}\). Find the shortest length in meters of this line that is needed to satisfy the impedance-matching condition. ( \(c\) ) With the three-segment setup as found in parts \((a)\) and \((b)\), the frequency is now doubled to \(2 \mathrm{GHz}\). Find the input impedance at the line-1-to-line- 2 junction, seen by waves incident from line \(1 .(d)\) Under the conditions of part \((c)\), and with power incident from line 1 , evaluate the standing wave ratio that will be measured in line 1 , and the fraction of the incident power from line 1 that is reflected and propagates back to the line 1 input.

For a quarter-wave transformer, the impedance transformation formula is given by: $$Z_{02} = \sqrt{Z_{01} \cdot Z_{03}}$$ Substitute the given impedance values \(Z_{01} = 100 \Omega\) and \(Z_{03} = 25 \Omega\) and then calculate \(Z_{02}\): $$Z_{02} = \sqrt{100 \Omega \cdot 25 \Omega} = 50 \Omega$$ So, the required characteristic impedance \(Z_{02}\) of the intermediate line is \(50 \Omega\).

To find the shortest length required for impedance matching, first we need to calculate the wavelength \(\lambda_2\) of the signal in the intermediate line at an operating frequency of \(1 \mathrm{GHz}\) using this formula: $$\lambda_2 = \frac{c}{f_2 \sqrt{\epsilon_{r2}}}$$ Here, \(c \approx 3 \times 10^8 \mathrm{m/s}\) is the speed of light, \(f_2 = 1 \mathrm{GHz}\) is the operating frequency, and \(\epsilon_{r2}\) is the relative permittivity. Now, given the capacitance per unit length \(C_2 = 100 \mathrm{pF/m}\) and characteristic impedance \(Z_{02} = 50 \Omega\), we can find the inductance per unit length \(L_2\) using: $$Z_{02} = \sqrt{\frac{L_2}{C_2}}$$ Solving for \(L_2\): $$L_2 = Z_{02}^2 C_2 = 50^2 \times 100 \times 10^{-12} \mathrm{H/m} = 2.5 \times 10^{-7}\mathrm{H/m}$$ The propagation constant \(\gamma_2\) can be derived from \(L_2\) and \(C_2\) as follows: $$\gamma_2 = \sqrt{\frac{L_2}{C_2}} = \frac{1}{\sqrt{L_2 C_2}} = \frac{1}{\sqrt{2.5 \times 10^{-7}\mathrm{H/m} \times 100 \times 10^{-12}\mathrm{F/m}}} = 2 \times 10^8 \mathrm{rad/s}$$ Now, we can find \(\epsilon_{r2}\) using the formula: $$\gamma_2 = \frac{\omega}{\sqrt{\mu_0 \epsilon_0 \epsilon_{r2}}}$$ We know the angular frequency \(\omega = 2 \pi f_2 = 2 \pi \times 10^9 \mathrm{rad/s}\), and \(\mu_0 \approx 4 \pi \times 10^{-7} \mathrm{H/m}\) and \(\epsilon_0 \approx 8.85 \times 10^{-12} \mathrm{F/m}\) are the permeability and permittivity of free space, respectively. Solve for \(\epsilon_{r2}\): $$\epsilon_{r2} = \frac{\omega^2}{\gamma_2^2 \mu_0 \epsilon_0} = \frac{(2 \pi \times 10^9)^2}{(2 \times 10^8)^2 \times 4 \pi \times 10^{-7} \times 8.85 \times 10^{-12}} \approx 1.158$$ Now, calculate the wavelength \(\lambda_2\) using the formula above: $$\lambda_2 = \frac{3 \times 10^8}{10^9 \sqrt{1.158}} \approx 0.278 \mathrm{m}$$ Since we require a quarter wavelength, the shortest length of the intermediate line for impedance matching is: $$L_2 = \frac{\lambda_2}{4} = \frac{0.278}{4} \approx 0.0695\ \mathrm{m}$$

When the frequency is doubled to \(f_3 = 2 \mathrm{GHz}\), the new angular frequency is calculated as \(\omega_3 = 2 \pi \times 2 \times 10^9 \mathrm{rad/s}\). The new wavelength \(\lambda_3\) can be calculated using the previously found relative permittivity \(\epsilon_{r2}\): $$\lambda_3 = \frac{c}{f_3 \sqrt{\epsilon_{r2}}} = \frac{3 \times 10^8}{2 \times 10^9 \sqrt{1.158}} \approx 0.139\ \mathrm{m}$$ Now calculate the reflection coefficient \(\Gamma_2\) at the junction of line 2 and line 3, using the impedance values \(Z_{02} = 50 \Omega\) and \(Z_{03} = 25 \Omega\): $$\Gamma_2 = \frac{Z_{03}-Z_{02}}{Z_{03}+Z_{02}} = \frac{25 - 50}{25 + 50} = -\frac{1}{3}$$ Notice that the intermediate line is still a quarter-wavelength transformer at the new frequency, as the new length \(L_3 = 0.0695 \mathrm{m}\) is given by: $$L_3 = \frac{\lambda_3}{2} = \frac{0.139}{2} \approx 0.0695\ \mathrm{m}$$ Thus, we can find the input impedance at the junction of line 1 and line 2 when the operating frequency is doubled, using the quarter-wave transformer formula: $$Z_{in} = Z_{01} \frac{1 + \Gamma_2}{1 - \Gamma_2} = 100 \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = 50 \ \Omega$$

To find the standing wave ratio (SWR) in line 1 and the fraction of incident power reflected, first calculate the reflection coefficient \(\Gamma_1\) at the junction of line 1 and line 2, using the impedance values \(Z_{01} = 100 \Omega\) and \(Z_{in} = 50 \Omega\): $$\Gamma_1 = \frac{Z_{in}-Z_{01}}{Z_{in}+Z_{01}} = \frac{50 - 100}{50 + 100} = -\frac{1}{3}$$ Now, calculate the SWR and the fraction of incident power reflected using \(\Gamma_1\): $$\mathrm{SWR} = \frac{1 + |\Gamma_1|}{1-|\Gamma_1|} = \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}} = 2$$ $$\text{Reflected power fraction} = |\Gamma_1|^2 = \left(-\frac{1}{3}\right)^2 = \frac{1}{9}$$ Therefore, the standing wave ratio (SWR) in line 1 is 2, and the fraction of incident power from line 1 that is reflected and propagates back to the line 1 input is \(\frac{1}{9}\).