Quarter Wavelength Transmission Line

[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

 

[nathangrand]

anyone?

 

[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!).