How would I use a Smith Chart and L-network to set input admittance of a circuit network?

Question

I'm doing the practice problem pictured below:

Find d1 and d2 with:

$$Z_o = 50 \Omega \hspace{5mm} Z_L = 100 \Omega \hspace{5mm} f=10GHz$$

So that: $$Y_{in}=20\angle30^o\hspace{1mm}mS$$

I summed up the load impedance, capacitor, and inductor in series:

$$Z_c = Z_L + j\omega fL + \frac{-1}{j2\omega fC} = 100 +j105.8 \hspace{1mm} \Omega$$

Then I attempted to used a smith chart to determine the length of d2 and d1 so that the input admittance would be: $$20 \angle 30^{o} \hspace{1mm} mS$$

The actions done on the Smith Chart are as follows:

(1) I plotted Zc on the chart and drew the circle.

(2) I chose d2 to be: $$0.177 \lambda$$ which would make the normalized impedance from node B to Ground: $$1.7-j0.5 \hspace{1mm} \Omega$$

However, this is where I became stuck. I do not know how to proceed from here, or if my previous steps are conducive towards solving the problem.

Thank you

\$\begingroup\$ Maybe with a differential equation? \$\endgroup\$

user317139
– user317139

2024-09-23 05:25:17 +00:00
Commented Sep 23, 2024 at 5:25 — user317139
– user317139, Commented Sep 23, 2024 at 5:25

a360pilot · Accepted Answer · 2024-09-28 20:41:18Z

Your initial effort in calculating \$ Z_c \$ is correct. Next, you must recognize that parallel stubs contribute to the imaginary part of the admittance or impedance, through: $$ Z_{stub} = Z_0 \frac{Z_L + jZ_0\tan(\beta d)}{Z_0 + jZ_L\tan(\beta d)}, $$ reducing to \$ Z_{stub} = Z_0/j\tan(\beta d) \$ for an open stub (\$ Z_L = \infty\$). Thus, it will change the impedance on the constant-conductance circles. The next steps to solve the problem would be:

Calculate \$ d_2 \$ to get the correct real part for \$ Y_{in} \$. I.e. \$ G_{in} = 20 \, \text{mS} \times \cos(30^\circ) = 17.32 \, \text{mS}\$. Since \$ d_2 \$ corresponds to a rotation on the Smith chart centered around the origin, you can readily calculate \$ d_2 = 0.37 \lambda \$.
Next, use the open stub to set the correct imaginary part for your impedance. Changing the length of \$ d_1 \$ will rotate you on a constant-\$ G \$ circle on the admittance Smith chart. Then, you can calculate \$ d_1 = 0.18 \lambda \$.

Here is the final matching result using the Smith Chart Utility in Keysight's ADS:

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How would I use a Smith Chart and L-network to set input admittance of a circuit network?

I'm doing the practice problem pictured below:

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How would I use a Smith Chart and L-network to set input admittance of a circuit network?

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