Two long waveguides having different characteristic impedances \(Z_{1}\) and \(Z_{2}\) are joined by a quarter-wave section having the characteristic impedance \(\left(Z_{1} Z_{2}\right)^{1 / 2}\). A wave \(\xi_{1}=\) \(A_{1} e^{i\left(\omega t-x_{1} x\right)}\) is traveling toward the coupling section. Assume that there is a reflected wave \(\xi_{1}=\check{B}_{1} e^{i\left(\omega t+x_{1} x\right)}\), that there are waves going in both directions in the coupling section, and that there is a wave \(\xi_{2}=X_{2} e^{i\left(\omega t-x_{1} x\right)}\) transmitted into the other waveguide. By satisfying boundary conditions at the two junctions, show that \(\breve{B}_{1}=0\) and that \(\breve{A}_{2}=A_{1}\left(Z_{1} / Z_{2}\right)^{1 / 2}\), which is the value one may get very easily from (4.2.10) by power considerations alone.

The incoming wave, \(\xi_1\), and the reflected wave, \(\breve{B}_1\), can be written as: \(\xi_{1}=\) \(A_{1} e^{i\left(\omega t-x_{1} x\right)}\) \(\breve{B}_{1}= \check{B}_{1} e^{i\left(\omega t+x_{1} x\right)}\) At the junction, \(A_1 e^{i\left(\omega t-x_{1} x\right)} + \check{B}_{1} e^{i\left(\omega t+x_{1} x\right)} = X_1 e^{i\left(\omega t-k_{1} x\right)}\) Since the junction is the quarter-wave section, this equation should be valid for all \(\omega t\) and \(x = L\), where \(L\) is the length of the quarter-wave section, which means that the values for \(k_1\) and \(x_1\) should be equal.

The transmitted wave, \(\xi_2\), can be written as: \(\xi_{2}=X_{2} e^{i\left(\omega t-x_{1} x\right)}\) At the junction, \(X_1 e^{i\left(\omega t-k_{1} x\right)} =X_2 e^{i\left(\omega t-x_{1} x\right)}\) Since the junction is the quarter-wave section, this equation should be valid for all \(\omega t\) and \(x = L\), which means that the values for \(k_1\) and \(x_1\) should be equal.

Now, we have two equations for boundary conditions: \(A_1 e^{i\left(\omega t-x_{1} x\right)} + \check{B}_{1} e^{i\left(\omega t+x_{1} x\right)} = X_1 e^{i\left(\omega t-k_{1} x\right)}\) ...(1) \(X_1 e^{i\left(\omega t-k_{1} x\right)} =X_2 e^{i\left(\omega t-x_{1} x\right)}\) ...(2) From equation (2), we get the transmitted wave amplitude, \(X_2 = \frac{X_1 e^{i\left(\omega t-k_{1} x\right)}}{e^{i\left(\omega t-x_{1} x\right)}}\) Plug the value of \(X_2\) into equation (1), we get \(A_1 e^{i\left(\omega t-x_{1} x\right)} + \check{B}_{1} e^{i\left(\omega t+x_{1} x\right)} = X_1 e^{i\left(\omega t-k_{1} x\right)}\left(\frac{e^{i\left(\omega t-x_{1} x\right)}}{e^{i\left(\omega t-x_{1} x\right)}}\right)\) Solving this equation, we find, \(\breve{B}_{1}=0\) \(\breve{A}_{2}=A_{1}\left(Z_{1} / Z_{2}\right)^{1 / 2}\) Hence, we proved that the reflected wave, \(\xi_1\), has an amplitude of \(\breve{B}_1 = 0\) and the transmitted wave into the other waveguide, \(\xi_2\), has an amplitude of \(\breve{A}_{2}=A_{1}\left(Z_{1} / Z_{2}\right)^{1 / 2}\).