Quarter Wavelength Transmission Line

1. Dec 28, 2011

nathangrand

An open-ended quarter-wavelength, air-spaced, parallel-wire transmission line is found to
be in resonance with an oscillator when its length is 0.25 m. When a capacitance of 1 pF
is connected across the open end, it is found that the length of the line must be reduced
to 0.125 m to obtain resonance. Show that the characteristic impedance of the line is
approximately 530 *Ohms
[Remember that, using V proportional to exp(i(kz-wt)) the impedance of a capacitor is i/wc]

The main problem here is I don't really understand the question!!! Can someone explain what it actually means/wants me to do? It's the found to be in resonance part that I don't get - does it mean is impedance matched to?

I think that the two circumstances will give me two equations which should be easier enough to manipulate to find the characteristic impedance.

For a quarter wavelength line Zin/Z = Z/Zt where Zin is the input impedance, Z the characteristic impedance and Zt the terminating impedance.

For an open circuited line Zin/Z = -jcot(ka) where j is the imaginary unit, and a the length of the line

2. Dec 29, 2011

anyone?

3. Jan 7, 2012

WaldoXerxes

Dunno if you're still looking for an answer, but hopefully this will be useful to someone - I had exactly the same question to solve...
I think the resonance bit just means they are impedance matched.

From Zin/Z = -jcot(ka) you can work out that Zin must be zero (since cot ka=0). We also know that since a quarter wavelength is 0.25m, the wavelength must be 1m.

Remember that when the capacitor is added the waveguide is shortened so is no longer a quarter-wavelength, and therefore the quarter wavelength equations don't apply any more. The general equation Zin/Z = (Zt cos(ka) + jZ sin(ka)) / (Z cos (ka) + jZt sin(ka)) is useful (remembering that a is different now!).