# How can I replace this inductor with a transmission line?

I'm new in transmission line theory, hope you can help me. I have this circuit:

And I have the admittance; Yin=(1-j) x10^-3 [S] and its frequency f=100 [MHz].

What I need its first to find the values of L and C, so that the admittance of entry Yin be the mentioned before. Then I need to replace the inductor L by transmission line in open circuit. The characteristic impedance is 50[ohm] and the phase velocity is 80% the speed of light.

• look up L-section impedence matching. you can easily replace the inductor using stub... – hassan789 Mar 29 '14 at 2:36

The admittance of this circuit can be written as :

Y = $\dfrac{1}{sL} + \dfrac{1}{R + \dfrac{1}{sC}}\ = \dfrac{CLs^2 + CRs + 1}{Ls(CRs + 1)}$.

Substituting $s = j\omega$, multiplying the denominator by its complex cojugated and simplifying into real and imaginary parts gives us:

$\dfrac{R}{\frac{1}{C^2 \omega^2} + R^2} + j\left(\dfrac{1}{C\omega\left(\frac{1}{C^2\omega^2} + R^2 \right)} - \dfrac{1}{L\omega}\right)$.

A complex admittance consists of a conductance (real part) and a susceptance (imaginary part).

Substituting the value of the resistance and frequency, we want

$\dfrac{50}{\frac{1}{C^2 (2*\pi*10^9)^2} + 50^2} = 10^{-3}S.$

Solving for C gives C $\approx$ 0.73 pF.

Plugging this value of C and R into

$j\left(\dfrac{1}{C\omega\left(\frac{1}{C^2\omega^2} + R^2 \right)} - \dfrac{1}{L\omega}\right)\ = -j10^{-3}$

and solving for L gives L $\approx$ 30 nH.

The admittance of this inductance is $\approx 5.3*10^{-9}$ S.

Looking here as a reference, the equation for the length of an open-circuited transmission line to act as an inductor is:

l = ${\frac {1}{\beta }}\left[\pi(n+1) -\operatorname{arccot} \left({\frac {\omega L}{Z_{0}}}\right)\right]$, where $L = 30*10^{-9}$$, \beta = \dfrac{2\pi f }{c_l}$, $f = 10^9$, and $c_{l} \approx 0.8*3.0*10^9 \frac{m}{s}$.