Understanding series open stub tuning

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.

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.

• Thank you Joren this helped a lot Apr 18 '20 at 18:48