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1/4 lambda stub

  1. Jul 20, 2009 #1

    Here's stupid question....
    Been trying to think about what happens to the current and the voltage when you add a stub, and somehow I expect that one must lag the other, but I have no physics to suggest this.

    Can anyone explain what's going on here?

    Thanks !
  2. jcsd
  3. Jul 20, 2009 #2


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    Staff: Mentor

  4. Jul 20, 2009 #3
    The perfect stub would be a solid short across the line. In practice it will have some loss and appear as a low resistance with V and I in phase.
  5. Jul 21, 2009 #4
    Thanks Berkeman I didn't get chance to read the web page yet but I will do it later today....

    Thanks Pumblechook, that's a start but what about an open ?

    You can in theory neglecting loses etc get the voltage to become infinite due to constructive interference.

    Essentially you make a resonant circuit, similar to an LC circuit. In an LC circuit the Energy is transferred back and forth from the capacitor to the Inductor as Electric Field and Magnetic Fields respectively .

    So in my 1/4 lambda resonant circuit where is the energy (as function of time)?
  6. Jul 22, 2009 #5
    Well I read the microwaves101, and all seems to make sense but couldn't see the answer to my question in there?

    Surely someone knows the answer?
  7. Jul 22, 2009 #6
    A solid short would reflect all the RF back towards the source and you would get standing waves. There will be points on the line where the voltage in high. 1/4 wave either side of those points the voltage will be next to zero. I was wrong, thinking about about it ...V and I are at 90 deg. At the open end of the stub there can't be any current (no where for it to flow to) but there will be a voltage peak. At the junction point where it places at short circuit across the line the V will be low and I high.

    Stubs are rather crude devices in that they might put un desirable impedances across the line at the wanted frequncy(cies) which may have to be compensated for with more stubs or other matching devices.

    Stubs will also put shorts across the line at 3, 5, 7 times the frequency.. The actually frequencies don't quite coincide with exactly 3 times etc.

    You can make the stub slightly short and then fine tune with variable capacitance at the open end.

    You can reduce any mis-match effects at wanted frequencies by capacitively coupling to the line (fraction of 1 pF at VHF). You get a much narrower notch but not as deep. You may need several stubs like this across the line spaced by 1/4 wave (maybe) to get the required unwanted freq rejection.

    I have worked on these things on transmitting stations where you want to pick up another station. You might wnat to receive on 88.3 in the presence of 4 kW at 88.9. The stubs are capacitively coupled and are made of solid copper tube or in a helical filter arrangement.
  8. Aug 17, 2009 #7

    Thanks for the explanation Pumblechook- (I just came back from vacation and read your message). If the open ended stub is capacitively coupled to ground then a displacement current should flow and presumably this current would be 90 off phase with the E field?
  9. Aug 17, 2009 #8
    Hi Ben-
    We also use shorted quarter-wave stubs (called clipping lines) in the time domain. If we have an unsaturated collector (current source) output of a digital (pulse) signal, a Z=50 ohm clipping line in parallel with a 50 ohm signal line represents a 25 ohm output impedance to the collector. Equal signals travel out both lines at about v = 0.67 c (for RG-58). When the signal reaches the short, the voltage inverts, but the current does not invert, so there is an inverted voltage pulse travelling back to the collector. When it reaches the collector, it travels down the signal line and cancels or "clips" the pulse amplitude. So a 20-nsec collector output pulse can be "clipped" to say 2 or 3 nsec.

    [Edit] Do not attempt this on emitter outputs or saturated collector outputs.
    Bob S
    Last edited: Aug 17, 2009
  10. Aug 20, 2009 #9
    Interesting way to get short pulses...

    Can you take that much lower, into the picosec. regime for example?

    I can imagine that 'theoretically' you could end up making aribitarily sharp pulses...
  11. Aug 20, 2009 #10
    The lambda/4 clipping line clips the digital output pulse with a fall time that is nearly as sharp as the risetime. It is just an inverted reflection of the risetime. The source is the collector of an unsaturated npn transistor coupled to a 50 ohm cable and a 50 ohm clipping line, and is switching from 15 milliamps to 45 milliamps, with 15 milliamps going to the lambda/4 shorted stub. So with a 1 nsec riretime, 20 nsec long pulse, we can clip it down to 2 to 3 nsec. I think this technique has been used with high voltage spark-gap initiated pulses on air-filled coax hard-line (50 or 75 ohm), but I don't have details. It only works on high impedance (current) sources.
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