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#31
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On Oct 2, 10:33 am, Jim Kelley wrote:
Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: A foot of wire with reflections at one GHz has the same current throughout the circuit? A simple series circuit can be expected to behave as a simple series circuit. Other circuits can be expected to behave differently. Which do you think applies? An *ordinary prudent man* would think that one foot of wire is a "simple series circuit" and it is in a DC circuit. Most ordinary prudent men that I know wouldn't characterize a one foot length of wire as a series circuit. ac6xg The fact that it's described as having a physical dimension suggests that the physical extent matters. I don't know what an "ordinary prudent man" is, but I suppose most people I stopped on the street wouldn't have a clue what I was talking about and their eyes would kind of glaze over with the mention of "wire" and "circuit." Cheers, Tom (presently playing with circuits below 100MHz where 1mm can make a huge difference in performance...who finds folk who say we don't understand such things as distributed circuits deeply offensive, especially when they make it abundantly clear they don't understand such circuits themselves...) |
#32
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Jim Kelley wrote:
Cecil Moore wrote: An *ordinary prudent man* would think that one foot of wire is a "simple series circuit" and it is in a DC circuit. Most ordinary prudent men that I know wouldn't characterize a one foot length of wire as a series circuit. More silly word games - when it is in a DC circuit the current is constant. When it exists in a simple series GHz circuit with reflections, the current is not constant. -- 73, Cecil http://www.w5dxp.com |
#33
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![]() "Cecil Moore" wrote in message news ![]() Jim Kelley wrote: Cecil Moore wrote: An *ordinary prudent man* would think that one foot of wire is a "simple series circuit" and it is in a DC circuit. Most ordinary prudent men that I know wouldn't characterize a one foot length of wire as a series circuit. More silly word games - when it is in a DC circuit the current is constant. When it exists in a simple series GHz circuit with reflections, the current is not constant. -- 73, Cecil http://www.w5dxp.com you don't even need reflections. a properly terminated piece of wire has a different current at every point along the wire. |
#34
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![]() Dave wrote: "Cecil Moore" wrote in message news ![]() Jim Kelley wrote: Cecil Moore wrote: An *ordinary prudent man* would think that one foot of wire is a "simple series circuit" and it is in a DC circuit. Most ordinary prudent men that I know wouldn't characterize a one foot length of wire as a series circuit. More silly word games - when it is in a DC circuit the current is constant. When it exists in a simple series GHz circuit with reflections, the current is not constant. -- 73, Cecil http://www.w5dxp.com you don't even need reflections. a properly terminated piece of wire has a different current at every point along the wire. At any given instant. :-) ac6xg |
#35
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Dave wrote:
you don't even need reflections. a properly terminated piece of wire has a different current at every point along the wire. That's true but the attenuation factor in one foot of wire might be hard to measure. Introduce reflections and even a lossless wire will vary the current from max to min every few inches at GHz frequencies. -- 73, Cecil http://www.w5dxp.com |
#36
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![]() "Cecil Moore" wrote in message ... Dave wrote: you don't even need reflections. a properly terminated piece of wire has a different current at every point along the wire. That's true but the attenuation factor in one foot of wire might be hard to measure. Introduce reflections and even a lossless wire will vary the current from max to min every few inches at GHz frequencies. -- 73, Cecil http://www.w5dxp.com you don't need any loss either. a properly matched lossless line will have different current at each point along the line. |
#37
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Dave wrote:
you don't need any loss either. a properly matched lossless line will have different current at each point along the line. Different RMS current? -- 73, Cecil http://www.w5dxp.com |
#38
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![]() "Jim Kelley" wrote in message ... Dave wrote: "Cecil Moore" wrote in message news ![]() Jim Kelley wrote: Cecil Moore wrote: An *ordinary prudent man* would think that one foot of wire is a "simple series circuit" and it is in a DC circuit. Most ordinary prudent men that I know wouldn't characterize a one foot length of wire as a series circuit. More silly word games - when it is in a DC circuit the current is constant. When it exists in a simple series GHz circuit with reflections, the current is not constant. -- 73, Cecil http://www.w5dxp.com you don't even need reflections. a properly terminated piece of wire has a different current at every point along the wire. At any given instant. :-) ac6xg TADA! And Jim gets the cigar! Cecil, you take too narrow a view. at any given instant in time if you measure the current along a properly matched wire you will measure a different current and voltage all along the wire (repeating every wavelength minus losses of course). nothing was stated that required rms, other average, peak, or phasor representation. |
#39
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Dave wrote:
nothing was stated that required rms, other average, peak, or phasor representation. Nothing was stated that required the signal source to be turned on either. -- 73, Cecil http://www.w5dxp.com |
#40
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Dave wrote:
"You don`t even need reflections. A properly terminated piece of wire has a different current at every point along the wire." Instantaneous values, of course. But, proper termination means no reflection. The only source of variation along a line other than attenuation is reflection. In a uniform line, attenuation causes a steady decline of energy as energy travels. With a lossless line, properly terminated, variation of rnergy along a line is only the phase produced instantaneois values along the line. These are resolved during the period of a cycle by the root mean square calculation. Best regards, Richard Harrison, KB5WZI |
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