Certainly! Here is the rewritten version of the solution given in the picture: --- ### Solution: #### The following are given: - \( l = 2\, \text{cm} = 0.02\, \text{m} \) - \( Z_0 = 50\, \Omega \) - \( f = 2\, \text{GHz} = 2 \times 10^9\, \text{Hz} \) - \( R = 5\, k\Omega = 5 \times 10^3\, \Omega \) (in parallel with the inductor) --- #### (a) At resonance, the TL (transmission line) inductance is maximum, when the reactance of the inductor cancels the TL reactance of the line. For a very short line, \[ Z_{in} = jZ_0 \tan(\beta l) \] where \(\beta = \frac{2\pi}{\lambda} = \frac{2\pi f}{v}\) For an insulated coaxial line: \(v_p = c = 3 \times 10^8\, \text{m/s}\) \[ \beta = \frac{2\pi \times 2 \times 10^9}{3 \times 10^8} = \frac{4\pi}{0.3} \approx 41.89\, \text{rad/m} \] \[ \beta l = 41.89 \times 0.02 = 0.8378\, \text{rad} \] Now, the impedance of the transmission line as seen at the input: \[ Z_{TL} = jZ_0 \tan(\beta l) = j50 \tan(0.8378) \] \[ = j \times 50 \times 1.118 \] \[ = j 55.9\, \Omega \quad [\because \tan(0.8378) = 1.118] \] This reactance is cancelled by the inductor at resonance, \[ X_L = 55.9\, \Omega \implies \omega L = X_L \] \[ \Rightarrow L = \frac{X_L}{\omega} = \frac{55.9}{2\pi \times 2 \times 10^9} \] \[ = \frac{55.9}{4 \times 3.14 \times 10^9} \] \[ = \frac{55.9}{12.56 \times 10^9} \] \[ \approx 4.45 \times 10^{-9} \] \[ = 4.45\, \text{nH} \quad \boxed{A_1} \] --- #### (b) For a QLC circuit, we know: \[ Q = \frac{R}{\omega L} = \frac{5 \times 10^3}{2 \pi \times 2 \times 10^9 \times 4.45 \times 10^{-9}} \] \[ = \frac{5 \times 10^3}{4 \times 3.14 \times 10^9 \times 4.45 \times 10^{-9}} \] \[ = \frac{5 \times 10^3}{55.88} \] \[ \approx 89.5 \quad \boxed{A_2} \] ---

Certainly! Here is the rewritten version of the solution based on the given image: --- ### **Solution:** **Given Data:** - Length of the transmission line, \( l = 2\, \text{cm} = 0.02\, \text{m} \) - Characteristic impedance, \( Z_0 = 50\, \Omega \) - Frequency, \( f = 2\, \text{GHz} = 2 \times 10^9\, \text{Hz} \) - Load resistance, \( R = 5\, \text{k}\Omega = 5000\, \Omega \) --- ### **(a) Determining the Inductance \( L \) at Resonance** At resonance, the inductive reactance cancels the reactive part of the transmission line impedance. **Step 1:** Calculate the phase constant \(\beta\): \[ \beta = \frac{2\pi f}{v} \] Since it’s a coaxial cable with velocity \( v \approx c = 3 \times 10^8\, \text{m/s} \): \[ \beta = \frac{2\pi \times 2 \times 10^9}{3 \times 10^8} = \frac{4\pi \times 10^9}{3 \times 10^8} = \frac{4\pi}{0.3} \approx 41.89\, \text{rad/m} \] **Step 2:** Calculate \(\beta l\): \[ \beta l = 41.89 \times 0.02 = 0.8378\, \text{rad} \] **Step 3:** Find the input reactance \(X_{TL}\): \[ X_{TL} = Z_0 \tan(\beta l) = 50 \times \tan(0.8378) \] \[ \tan(0.8378) \approx 1.118 \] \[ X_{TL} \approx 50 \times 1.118 = 55.9\, \Omega \] **Step 4:** Calculate the inductance \(L\) that cancels this reactance at the operating angular frequency \(\omega = 2\pi f\): \[ L = \frac{X_{TL}}{\omega} = \frac{55.9}{2\pi \times 2 \times 10^9} \] \[ L = \frac{55.9}{12.566 \times 10^9} \approx 4.45 \times 10^{-9}\, \text{H} = \boxed{4.45\, \text{nH}} \] --- ### **(b) Calculating the Quality Factor \(Q\)** \[ Q = \frac{R}{\omega L} \] \[ \omega = 2\pi \times 2 \times 10^9 = 12.566 \times 10^9\, \text{rad/sec} \] \[ Q = \frac{5000}{12.566 \times 10^9 \times 4.45 \times 10^{-9}} \] \[ Q = \frac{5000}{55.88} \approx \boxed{89.5} \] --- ### **Final Answers:** - **(a)** Inductance \( L \approx \boxed{4.45\, \text{nH}} \) - **(b)** Quality factor \( Q \approx \boxed{89.5} \) ---