I recently had an assignment where I had to use open stub series tuning to find a way to maximize power delivered to the load from the generator. I understand that power delivered to the load is maximized when the input resistance is equal to the generator resistance and that reactance is zero. I know that when you are doing open stub series tuning you choose a length l for the stub such that it cancels any reactance at the position of the series open stub. What I don't understand is how I would calculate this length analytically (in this assignment we had to do it analytically without a smith chart). Below I have included a picture of the solution our teacher provided to us. I don't understand how he took the input impedance equation and arrived to an equation that was dependent on cotangent.
Electronic – Understanding series open stub tuning
electromagneticimpedance-matchingtransmission line
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Best Answer
We know that the impedance presented by a (lossless) transmission line with characteristic impedance equal to \$Z_0\$, phase constant \$\beta\$, length \$l\$ and load impedance \$Z_l\$ equals
$$ Z_{in}(l) = Z_0 \frac{1+\Gamma \exp(-j 2 \beta l)}{1 - \Gamma \exp(-j 2 \beta l)} $$
Where \$\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}\$
Now, for a short circuit, we know that \$Z_L=+\infty\$ and thus \$\Gamma = +1\$. This gives us that:
$$ Z_{open}(l) = Z_0 \frac{1+ \exp(-j 2 \beta l)}{1 - \exp(-j 2 \beta l)} $$
Now we will multiply the entire thing by \$ \frac{\exp(j \beta l)}{\exp(j \beta l)} =1 \$ which gives us:
$$ Z_{open}(l) = Z_0 \frac{\exp(j \beta l)+ \exp(-j 2 \beta l)\cdot \exp(j \beta l)}{\exp(j \beta l) - \exp(-j 2 \beta l) \cdot \exp(j \beta l)} $$
$$ =Z_0 \frac{\exp(j \beta l)+ \exp(-j 2 \beta l + j \beta l)}{\exp(j \beta l)- \exp(-j 2 \beta l + j \beta l)} $$
$$= Z_0 \frac{\exp(j \beta l)+ \exp(-j \beta l)}{\exp(j \beta l)- \exp(-j \beta l)} $$
We can use the trigonometric identities
$$ \sin(x) = \frac{\exp(jx)-\exp(-jx)}{2j} \hspace{3mm} \text{and}\hspace{3mm} \cos(x) = \frac{\exp(jx) +\exp(-jx)}{2} $$
to simplify the earlier experssion for \$Z_{open}(l)\$ by multiplying it by \$\frac{2j}{2j}=1\$:
$$ Z_{open}(l) = Z_0 \frac{2j(\exp(j \beta l)+ \exp(-j \beta l))}{2j(\exp(j \beta l)- \exp(-j \beta l))} $$
And we can now find our \$\color{red}{\sin(x)}\$ and \$\color{blue}{\cos(x)}\$ identities:
$$ Z_{open}(l) = Z_0 \frac{\color{red}{2j}\color{blue}{(\exp(j \beta l)+ \exp(-j \beta l))}}{\color{blue}{2}j\color{red}{(\exp(j \beta l)- \exp(-j \beta l))}} = Z_0 \frac{\color{blue}{\cos(\beta l)}}{j\color{red}{\sin(\beta l)} }$$
Now use \$\cot(x) = \frac{\cos(x)}{\sin(x)}\$ and \$\frac{1}{j} = -j\$ to get
$$ Z_{open}(l) = Z_0 \cdot -j \cot(\beta l) = -j Z_0 \cot(\beta l)$$
the original equation given by your sollution.