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#1
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We're talking about the instantaneous common mode current of the real
system. That's the current actually flowing in the same direction in both legs of the stub. |
#2
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#4
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wrote:
We're talking about the instantaneous common mode current of the real system. That's the current actually flowing in the same direction in both legs of the stub. But standing waves of common mode current can and probably do exist. The amplitude of the common mode current varies depending upon its location in the distributed network. At a common mode standing wave node, the common mode current may measure zero. 1/4WL away, it may measure an appreciable magnitude. I have measured such on my own transmission lines. I suspect a dead short forces a common mode standing wave node. What else could possibly happen? -- 73, Cecil http://www.qsl.net/w5dxp |
#5
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In the case where the J-pole is isolated (crossbanding HT at the
feedpoint?) there's absolutely a node for the common mode current at the short, just like there's a node for the current on a wire at the end where the current can't flow. If you put a feedline on it, though, the standing wave pattern of common mode current changes. The feedline + jpole develops some pattern of standing waves on it. The common mode current is induced on the entire length of coax and stub... The sum of the common mode currents at point B is always as stated before... but the magnitude of ALL THREE of them depends on the details of the coax, whether or not there is a choke balun on it, the length and boundary condition on the other end of the coax. That is to say the common mode impedance at the top of the stub where the halfwave is connected depends on the details of the feedline. Stub vs. Stub + Coax: different common mode standing wave pattern, different common mode impedance at the top of the stub, different magnitude of current induced on the stub+coax than on the plain stub... Dan |
#6
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wrote:
If you put a feedline on it, though, the standing wave pattern of common mode current changes. +-----A------------X----------------------------- | +-----B------------Y On my diagram, the coax is always connected at A and B so *nothing has changed*. I just don't see any reason for the common mode current at 'X' to be exactly the same as the common mode current at 'B'. Do you? -- 73, Cecil http://www.qsl.net/w5dxp |
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