</table>
== A Summary of Series Stub Matching Theory == The design of the series tuning stub utilized in the previous part is now explained. Let the load admittance be Y<sub>L</sub> = 1 / Z<sub>L</sub> = G<sub>L</sub> + j B<sub>L</sub>. The admittance looking into a T-Line segment of length d terminated by the load Y<sub>L</sub> is given by: <math>Y_{in} = Y_0 \frac{ \left(G_L + j B_L \right) + jY_0 t}{Y_0 + j \left( G_L + j B_L \right) t}</math> where t = tan (β d). The impedance at this point can then be written as: <math>Z_{in} = \frac{1}{Y_{in}} = R_{in} + j X_{in} = \frac {G_L ( 1+ t^2)}{G_L^2+(B_L + Y_0 t)^2} + j \frac {G_L^2 t - (Y_0 - B_L t)(B_L + Y_0 t)}{Y_0 \left[ G_L^2+(B_L + Y_0 t)^2 \right]} </math> Now we choose the T-Line segment length d such that R<sub>in</sub> = Z<sub>0</sub>. This results in a quadratic equation for t: <math> Y_0(G_L - Y_0)t^2 -2B_L Y_0 t + (G_L Y_0 t -G_L^2 -B_L^2) = 0 </math> which can be solved for t. In the special case G<sub>L</sub> = Y<sub>0</sub>, there is one solution t = -B<sub>L</sub> / 2Y<sub>0</sub>. Otherwise, two distinct roots for t are found. Once t is found, the segment length can be calculated by solving t = tan (2π d / λ<sub>g</sub>). The stub reactance X is chosen such that X = -X<sub>in</sub>. From this condition, you can find the length of the stub depending on whether you use a short or open stub. Recall that the impedance of an open-ended or shorted transmission line segment of length L is given by: <math> Z_{open} = -jZ_0 cot(\beta L) </math> and <math> Z_{short} = jZ_0 tan(\beta L) </math> Therefore, the equations for the length L<sub>s</sub> of the open stub or short stub are given by: <math> X_{open} = -Z_0 cot (2\pi L_s / \lambda_g) = -X_{in} </math> <math> X_{short} = Z_0 tan (2\pi L_s / \lambda_g) = -X_{in} </math> == Single-Stub Impedance Matching Using a Series Open Stub==
As a first example, you will design a series open stub to match the inductive load you already encountered at the end of Tutorial Lesson 3. The following is a list of parts needed for this part of the tutorial lesson:
</table>
To verify this, run a network analysis of the circuit with the tuning stub matching network as a one-port. Use Node 2 (at the input of T-Line segment XTL1) as Port 1. Use the [[parameters]] given in the table below:
{| border="0"
|}
Plot First, plot the s11 S11 data on the Smith chart. Use the same start and stop frequencies as in the previous tutorial lesson, i.e. 1GHz and 3GHz, respectively, with a step size of 100MHz. You can see from the figure below that at f = 2GHz, the plot passes through the center of the Smith chart. It would be informative to plot the return lossThen, i.e. |s11| as a function of frequency on a Cartesian graph. In in the Output tab of the Network Analysis Test Panel, choose "Cartesian (Amplitude)" as the graph typewith the "Decibels" box unchecked. To obtain This will produce a smooth curvegraph of the return loss, set the i.e. |s11|, as a function of frequency step size to 10MHzon a Cartesian graph. The results are depicted in the figure belowAgain, showing you can see a vanishing return loss quite good impedance match at 2GHz.
<table>
</tr>
</table>
== A Summary of Series Stub Matching Theory ==
The design of the series tuning stub utilized in the previous part is now explained. Let the load admittance be Y<sub>L</sub> = 1 / Z<sub>L</sub> = G<sub>L</sub> + j B<sub>L</sub>. The admittance looking into a T-Line segment of length d terminated by the load Y<sub>L</sub> is given by:
<math>Y_{in} = Y_0 \frac{ \left(G_L + j B_L \right) + jY_0 t}{Y_0 + j \left( G_L + j B_L \right) t}</math>
where t = tan (β d). The impedance at this point can then be written as:
<math>Z_{in} = \frac{1}{Y_{in}} = R_{in} + j X_{in} = \frac {G_L ( 1+ t^2)}{G_L^2+(B_L + Y_0 t)^2} + j \frac {G_L^2 t - (Y_0 - B_L t)(B_L + Y_0 t)}{Y_0 \left[ G_L^2+(B_L + Y_0 t)^2 \right]} </math>
Now we choose the T-Line segment length d such that R<sub>in</sub> = Z<sub>0</sub>. This results in a quadratic equation for t:
<math> Y_0(G_L - Y_0)t^2 -2B_L Y_0 t + (G_L Y_0 t -G_L^2 -B_L^2) = 0 </math>
which can be solved for t. In the special case G<sub>L</sub> = Y<sub>0</sub>, there is one solution t = -B<sub>L</sub> / 2Y<sub>0</sub>. Otherwise, two distinct roots for t are found. Once t is found, the segment length can be calculated by solving t = tan (2π d / λ<sub>g</sub>).
The stub reactance X is chosen such that X = -X<sub>in</sub>. From this condition, you can find the length of the stub depending on whether you use a short or open stub. Recall that the impedance of an open-ended or shorted transmission line segment of length L is given by:
<math> Z_{open} = -jZ_0 cot(\beta L) </math>
and
<math> Z_{short} = jZ_0 tan(\beta L) </math>
Therefore, the equations for the length L<sub>s</sub> of the open stub or short stub are given by:
<math> X_{open} = -Z_0 cot (2\pi L_s / \lambda_g) = -X_{in} </math>
<math> X_{short} = Z_0 tan (2\pi L_s / \lambda_g) = -X_{in} </math>
== Single-Stub Impedance Matching Using a Shunt Short Stub==