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Quarter Wavelength Transmission Line 
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#1
Dec2811, 01:55 PM

P: 40

An openended quarterwavelength, airspaced, parallelwire 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(kzwt)) 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
Dec2911, 07:36 AM

P: 40

anyone?



#3
Jan712, 03:26 PM

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 quarterwavelength, 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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