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

enter image description here

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.

  • \$\begingroup\$ look up L-section impedence matching. you can easily replace the inductor using stub... \$\endgroup\$ – 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}\$.


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