To start the discussion, I refer you to Jeff, K6JCA's blog:
https://k6jca.blogspot.com/search/label/Networks%3A%20L
Look at the 7th Smith Chart on this page. One part of the chart is colored yellow and the other half not colored. The basic idea behind matching the load to the source is that if the load is in the yellow region, we add capacitance in parallel with the load to reach the r = 1 circle and then add inductance in series with the load to get to the Smith Chart origin. If the load is in the region not colored, we add inductance in series with the load until we intersect the g = 1 circle and then add capacitance in parallel with the source until we reach the Smith Chart origin.
Now that the basic idea is clear, let's delve into the equations. We need to calculate the magnitude of the reflection coefficient for SWR values of 1.5, 2, 3, 4, 5, 6, 7, 8, 9, and 10. We also need to assign values to the phase of the reflection coefficient from 0 to 359 in one degree increments. So, let's state the value of the SWR in terms of the reflection coefficient:
\begin{equation}S=\frac{1+\left | \Gamma \right |}{1-\left | \Gamma \right |}\end{equation}
And solve it for the magnitude of the reflection coefficient:
\begin{equation}\left | \Gamma \right |=\frac{S-1}{S+1}\end{equation} This describes a family of circles around the origin of the Smith Chart. \begin{equation} \Gamma =\left | \Gamma \right |e^{j\theta } \end{equation}
\begin{equation} \Gamma _{r}=\left | \Gamma \right |cos\theta \end{equation}
\begin{equation} \Gamma _{i}=\left | \Gamma \right |sin\theta\end{equation}
The magnitude and phase of the reflection coefficient are directly measured by the directional coupler and from that the real and imaginary values of the reflection coefficient are calculated. Then we need to find the resistance and the reactance of the load:
\begin{equation}\Gamma =\frac{Z_{L}-Z_{0}}{Z_{L}+Z_{0}}\end{equation}
We can normalize this by the system (real) impedance by setting:
\begin{equation}z=\frac{Z_{L}}{Z_{0}}\end{equation}Which will result in
\begin{equation}\Gamma =\frac{z-1}{z+1}\end{equation}Now we will solve this for z
\begin{equation}z=\frac{1+\Gamma }{1-\Gamma }\end{equation}
\begin{equation}\Gamma =\Gamma _{r}+j\Gamma _{i}\end{equation}
\begin{equation}z=\frac{1+\Gamma _{r}+j\Gamma _{i}}{1-\Gamma _{r}-j\Gamma _{i}}\end{equation}
Which after some algebra becomes:
\begin{equation}z=\frac{1-\left | \Gamma \right |^{2}}{(1-\Gamma _{r})^{2}+\Gamma _{i}^{2}}+j\frac{2\Gamma _{i}}{(1-\Gamma _{r})^{2}+\Gamma _{i}^{2}}\end{equation}
\begin{equation}r_{0}=\frac{1-\left | \Gamma \right |^{2}}{(1-\Gamma _{r})^{2}+\Gamma _{i}^{2}}\end{equation}
\begin{equation}x_{0}=\frac{2\Gamma _{i}}{(1-\Gamma _{r})^{2}+\Gamma _{i}^{2}}\end{equation}
Since the magnitude of the reflection coefficient is less than or equal to one, the real part of the impedance will always be non-negative. There is a second way of writing these expressions that sometimes is easier to manipulate.
\begin{equation}r_{0}=\frac{1-\left | \Gamma \right |^{2}}{1+\left | \Gamma \right |^{2}-2\Gamma _{r}}\end{equation}
\begin{equation} x_{0}=\frac{2\Gamma _{i}}{1+\left | \Gamma \right |^{2}-2\Gamma _{r}}\end{equation}
Where the subscript 0 indicates the measured location of the load on the Smith Chart.
Knowing the resistance and the reactance of the load, we now want to determine if the load is in the part of the Smith Chart colored in yellow (see K6JCA blogpost referenced above), which we will call Region 1 or in the part of the Smith Chart that is not colored, which we will call Region 2.
All of the points inside the r = 1 circle (normalized resistance larger than one) are in region 1. All of the points in the upper half of the Smith Chart, except the points inside the g = 1 circle are also included in Region 1. Analytically, this is expressed as:
\begin{equation}r> 1\ \ or \ \ x>0\ \ and\ \ r< \left | z \right |^{2}\end{equation}
The inequality on the left side of the "or" clause is obvious (inside of r = 1 circle). The conditions on the right side of the or clause need an explanation. The x > 0 indicates in the top half of the Smith Chart. The second term to the right of the and clause indicates outside of the g = 1 according to the following logic:
\begin{equation}z = r + jy\end{equation}
\begin{equation}y=\frac{1}{z}=\frac{1}{r+jx}\frac{r-jx}{r-jx}=\frac{r-jx}{r^{2}+x^{2}}=\frac{r-jx}{\left | z \right |^{2}}\end{equation}
\begin{equation}y=g+jb=\frac{r}{\left | z \right |^{2}}-j\frac{x}{\left | z \right |^{2}}\end{equation}
\begin{equation}g=\frac{r}{\left | z \right |^{2}} \ \ and \ \ b = -\frac{x}{\left | z \right |^{2}}\end{equation}
\begin{equation}g< 1 \ \ implies \ \ r< \left | z \right | ^{2}\end{equation}
We can similarly derive the criteria for a load being in Region 2 or assume that if it is not in region 1, it will be in region 2.
There is a technicality that a point might be exactly on the r = 1 in the lower half of the Smith Chart or g = 1 circle in the upper half of the Smith Chart. In the former case, only a series inductance with the load will be needed. In the latter case, only a parallel capacitance with the source is required. We can call these two possibilities Region 3 and Region 4 and in the simulator (and in real life operation) handle them accordingly. Special numerical treatment will be needed since the simulator will be using 64 bit floating point numbers.
Now that we have calculated the load impedance from the measurement of the reflection coefficient and located it in a region of the Smith Chart, we will find the equation of the appropriate conductance or resistance circle going through the load point and find its intersection with the r = 1 or g = 1 circle. These equations are analytically derived on a Maxim (now a part of Analog Devices) web page. I will not re-derive them here, but just restate them. You can find the derivation here:
The equations for the resistance and reactance circles in the Smith Chart frame of reference are:
\begin{equation}(\Gamma _{r}-\frac{r}{1+r})^{2}+\Gamma _{i}^{2}=(\frac{1}{1+r})^{2}\end{equation}
\begin{equation}(\Gamma _{r}-1)^2+(\Gamma _{i}-\frac{1}{x})^{2}=\frac{1}{x^{2}}\end{equation}
The equations for the conductance and and susceptance in the same reference frame are:
\begin{equation}(\Gamma _{r}+\frac {g}{g+1}) ^{2}+\Gamma _{i} ^{2}= (\frac {1}{1+g}) ^{2}\end{equation}
\begin{equation}(\Gamma _{r}+1) ^{2} + (\Gamma _{i}+\frac {1}{b}) ^{2} = \frac {1}{b ^{2}}\end{equation}
First, we will deal with the case when the load is in Region 1. that requires us to solve the two simultaneous equations of the conductance circle of the load with the r = 1 circle.
\begin{equation}(\Gamma _{r}+\frac {g _{0}}{g _{0}+1}) ^{2}+\Gamma _{i} ^{2}= (\frac {1}{1+g _{0}}) ^{2}\end{equation}
\begin{equation}(\Gamma _{r}-\frac{1}{2})^{2}+\Gamma _{i}^{2}=\frac{1}{4}\end{equation}With a little simplification:
\begin{equation}\Gamma _{r} ^{2} + (\frac {g _{0}}{1+g _{0}}) ^2 + 2(\frac {g _{0}}{1+g _{0}})\Gamma _{r} + \Gamma _{i} ^{2} = (\frac {1}{1+g _{0}} ) ^{2}\end{equation}\begin{equation}\Gamma _{r} ^{2} + \frac {1}{4} - \Gamma _{r} + \Gamma _{i} ^{2} = \frac {1}{4}\end{equation}
Cancelling the 1/4 from both sides of the second equation and subtracting it from the first we get:
\begin{equation}(\frac {g _{0}}{1+g _{0}}) ^{2} + 2(\frac {g _{0}}{1+g _{0}})\Gamma _{r} + \Gamma _{r} = (\frac {1}{1+g _{0}}) ^{2}\end{equation}
Rearranging terms:
\begin{equation}\left [ 2(\frac {g _{0}}{1+g _{0}})+1 \right ]\Gamma _{r} = (\frac {1}{1+g _{0}}) ^{2} - (\frac {g _{0}}{1+g _{0}}) ^{2} = \frac {1-g _{0} ^{2}}{(1+g _{0}) ^{2}}\end{equation}
\begin{equation}(3g _{0}+1)\Gamma _{r} = \frac {(1+g _{0})(1-g _{0})}{1+g _{0}}=1-g _{0}\end{equation}
\begin{equation}\Gamma _{r1} = \frac {1-g _{0}}{1+3g _{0}}\end{equation}
Since this point is on the r = 1 circle, we can plug this value into the second equation and get:
\begin{equation}\Gamma _{i1}=\pm \sqrt{\Gamma _{r1}-\Gamma _{r1} ^{2}}\end{equation}
Since the intersection of the conductance circle with the r = 1 circle will always be in the lower half of the Smith Chart (see the figure referred to above), we will take the negative square root as the answer. We must also show that the value under the square root is positive:
\begin{equation}\Gamma = \frac {z-1}{z+1}=\frac {r+jx-1}{r+jx+1}=\frac {(r-1)+jx}{(r+1)+jx}\times \frac {(r+1)-jx}{(r+1)-jx}\end{equation}
\begin{equation}\Gamma = \frac {r ^{2}-1+x ^{2}+jx(r+1-r+1)}{(r+1) ^{2} +x ^{2}}= \frac {r ^{2}+x ^{2}-1}{(r+1) ^{2} + x ^{2}}+j\frac {2x}{(r+1) ^{2} + x ^{2}}\end{equation}
\begin{equation}\Gamma _{r1} = \frac {r ^{2} + x ^{2}-1}{(r+1) ^{2} + x ^{2}}\end{equation}
Since this point is on the r = 1 circle, we can plug this value in:
\begin{equation}\Gamma _{r1} = \frac {x _{1} ^{2}}{4+x _{1} ^{2}}\end{equation}
The real part of gamma is positive and less than one, its square will be less than itself. Hence, the value under the square root will always be positive.
Now armed with the real and imaginary values of the reflection coefficient for the point of intersection between the conductance circle going through the load point and the r = 1 circle, as before, like we did for the location of the load impedance, we can calculate the resistance, reactance, conductance, and susceptance of the intersection point.
Now, knowing r (resistance), x (reactance), g (conductance) and b (susceptance) for both the Smith Chart coordinate of the load (subscript zero) and the intersection of the load conductance circle with the r = 1 circle (subscript 1) we can calculate the following values:
\begin{equation}Bcomp = \frac{b _{1} - b _{0}}{Z _{0}}=\omega C\end{equation}
\begin{equation}Xcomp=Z _{0}x _{1}=\omega L\end{equation}
We add a capacitor with the susceptance of Bcomp in parallel to the load and an inductor with the reactance of Xcomp in series with the load and we are done. Note that we are rotating clockwise on the conductance circle by adding capacitance and also rotating clockwise on the resistance circle by adding inductance. They are both consistent with the Smith Chart rules.
If the load resistance and reactance put the initial point in Region 2, we need to solve for the intersection of the resistance circle going through this point and the g = 1 circle:
\begin{equation}(\Gamma _{r} - \frac {r _{0}} {1+r _{0}}) ^{2} + \Gamma _{i} ^{2} = (\frac {1}{1+r _{0}}) ^2\end{equation}
\begin{equation}(\Gamma _{r}+\frac {1}{2}) ^{2}+ \Gamma _{i} ^{2}=\frac {1}{4}\end{equation}
After some algebra, just like we did in region 1, we will get:
\begin{equation}\Gamma _{r1} = \frac {r _{0}-1}{3r _{0}+1}\end{equation}
\begin{equation}\Gamma _{i1} = \pm \sqrt{-\Gamma _{r1} - \Gamma _{r1} ^{2}}\end{equation}
Since the intersection point is in the upper half of the Smith Chart, this time, we will take the positive root. In the same way that we showed the value under the square root was positive in Region 1, it can be shown that the value under the square root is positive in this case, except this time, one needs to start with the reactance rather than impedance formula for the reflection coefficient (1-y)/(1+y).
Now armed with the real and imaginary values of the reflection coefficient for the point of intersection between the resistance circle going through the load point and the g = 1 circle, as before, like we did for the location of the load impedance, we can calculate the resistance, reactance, conductance, and susceptance of the intersection point.
Knowing r (resistance), x (reactance), g (conductance) and b (susceptance) for both the Smith Chart coordinate of the load (subscript zero) and the intersection of the load susceptance circle with the g = 1 circle (subscript 1) we can calculate the following values:
\begin{equation} Xcomp = Z _{0}(x _{1}- x _{0})=\omega L\end{equation}
\begin{equation} Bcomp = \frac {b _{1}}{Z _{0}}=\omega C\end{equation}
We add an inductor with the reactance value of Xcomp in series with the load and a capacitor with the susceptance value of Bcomp in parallel with the source. Again note that we are rotating clockwise on the resistance circle to add inductance and rotating clockwise on the conductance circle to add capacitance, both consistent with the Smith Chart rules.
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