# Quarter Wavelength Transmission Line

by nathangrand
Tags: electrical circuits, transmission line, uni
 P: 40 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
 P: 40 anyone?
 P: 1 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!).

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