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The real part of the load impedance is rarely equal to the transmission-line impedance. In most cases, we have to transform the real part of the impedance as well.

For example, load impedance $Z_L=25+j 50 \Omega$ represents a series connection of a $25\Omega$ resistor and a $7.96$ nH inductor at 1 GHz. To match $Z_L=25+j 50 \Omega$ impedance to the transmission-line impedance $Z_0=50 \Omega$, we first normalize the load impedance to transmission-line impedance.

This impedance is shown in Figure fig:PointSC. Figure 1: Load impedance $Z_L=0.5+j1$ on Smith Chart.

#### SWR circle

Then, we identify an SWR circle that this impedance is on, as shown in Figure fig:SWRfor25Ohm. The point where the SWR circle intersects the green circle is of interest because the real part of the input admittance is equal to one $Y_1=1$. This second point, where $Y=1+j1.6$, will give us the length of the line that we have to add to the load impedance.

#### Length of the line that will transform the real part of load impedance

To find the length of the line that will transform the real part of $z_L$ to $y_{1}=1$, we identify the position of the load impedance $Z_L=0.5+j1$, and the input impedance $Z_1=0.3-j0.45$ at the Wavelengths Towards Generator (WTG) scale. The reason we picked impedance $Z_1=0.3-j0.45$ is because the real part of the admittance $y_1=1/z_1$ is equal to one $\Re (y_{1})=1$. Load impedance $Z_L$ is at $0.135 \lambda$, and the input impedance $Z_1$ is at $0.425 \lambda$. The difference between these two positions gives us the length of the line $0.29 \lambda$. In electrical degrees, this length is approximately $105^0$. The input admittance to the line is now $Y_L=1+1.6$. Figure 2: SWR circle for impedance $Z_L=0.5+j1$.

#### Adding a lumped-element to remove the susceptance

The final step is to add a susceptance that will remove the imaginary part of the input admittance $Y_1=1+j 1.6$. We see that to get the final admittance of $Y_M=1$, numerically, we have to add an admittance of $Y_{add}=-j1.6$. This represents an inductance. Since we are adding the two admittances $Y_1+Y_{add}$, they have to be in parallel (as we know that when elements are in parallel, we add their admittances). Figure 3: The result of impedance matching.

#### Other possible solutions

Graphically, there are several different mixed or transmission-line impedance matching circuits that we can make for a specific impedance. For example, for impedance $Z_L=25+j50 \Omega$, $z_L=0.5+j1$, there are four different circuits that we can make, as shown in Figure fig:MixedVariety. In this paragraph, we used the green path on the Smith Chart, with intermediate admittance $Y_2$. Figure 4: A variety of possible impedance matching circuits for impedance $Z_L=25+j50 \Omega$.