**Let’s consider the most general case of a circuit, where we have the arbitrary loads Z _{1, }Z_{2 }and an arbitrary generator.**

The input impedance of the terminated transmission line is ${Z}_{i}n={Z}_{0}\frac{{e}^{j}\beta l+{\Gamma}_{l}{e}^{(\u2013j\beta l})}{{e}^{j}\beta l\u2013{\Gamma}_{l}{e}^{(\u2013j\beta l}},{\Gamma}_{l}=\frac{{Z}_{1}\u2013{Z}_{0}}{{Z}_{1}+{Z}_{0}}$. The voltage on the line $V\left(z\right)={V}_{0}({e}^{\u2013j\beta z}+{\Gamma}_{l}{e}^{\u2013j\beta z})$ . To receive the voltage of the generator we must take into consideration $z=\u2013l,so{V}_{0}=\frac{{Z}_{0}{V}_{g}}{{Z}_{0}+{Z}_{2}}\frac{{e}^{\u2013j\beta l}}{1\u2013{\Gamma}_{l}{\Gamma}_{g}{e}^{\u20132j\beta l}}.{\Gamma}_{g}=\frac{{Z}_{2}\u2013{Z}_{0}}{{Z}_{2}+{Z}_{0}}$ is the reflection coefficient of the generator. The standing wave ratio here is $SWR=\frac{1+\left|{\Gamma}_{L}\right|}{1\u2013\left|{\Gamma}_{L}\right|}$ . So the power delivered to the load is: $P=\frac{1}{2}\left|{V}_{i}n{I}_{in}\right|=\frac{1}{2}|{V}_{g}{|}^{2}\frac{{R}_{in}}{({R}_{in}+{R}_{2}{)}^{2}+({Z}_{in}+2{)}^{2}}$ where ${Z}_{in}={R}_{in}+j{Z}_{in},{Z}_{2}={R}_{2}+j{Z}_{2}$.

1. When the load is matched to the line, ${Z}_{in}={Z}_{0},{\u0413}_{l}=0$ and $P=\frac{1}{2}|{V}_{g}{|}^{2}\frac{{Z}_{0}}{({Z}_{0}+{R}_{2}{)}^{2}+{X}_{2}^{2}}$ .

2. When the load is matched to the line, ${Z}_{in}={Z}_{2}$ the load power is $P=\frac{1}{2}|{V}_{g}{|}^{2}\frac{{R}_{2}}{4}\frac{1}{({R}_{2}^{2}+{X}_{2}^{2})}.$

3. Assuming that the generator impedance is fixed, the condition $\frac{dP}{d{Z}_{in}}\to 0$ leads to the maximum power delivered to the load.

The conjugate matching condition is ${R}_{(in}={R}_{g},{X}_{in}=\u2013{X}_{g},or{Z}_{in}={Z}_{g}^{*}$ . This condition is conjugate matching, that leads to the maximum power delivered to the load for a fixed generator impedance $P=\frac{1}{2}|{V}_{g}{|}^{2}\frac{1}{4{R}_{g}}$. Figure 1 depicts the general case of transmission line circuit for mismatched load and the generator.