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#1
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#2
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#3
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#5
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Tom Bruhns wrote:
"If you want dissipated power in a TEM line, then P = Irms^2*R + Erms^2*G (at a particular frequency where R and G have fixed values)." Sure. Total loss is the sum of the series and shunt losses. Tom also wrote: "Since Irms and Erms are functions of position along the line, P is a per-unit-length quantity like R and G and total power is found by integrating the incremental P over the length of the length of the line you are interested in." Yes. Loss is a dB per 100 ft. quantity and loss is cumulative over its length. But, I wouldn`t worry about the volts and amps produced by SWR. The forward and reflected waves don`t oscillate in value. Only their interference pattern does that and it is of no consequence. The forward and reflected waves are smoothly attenuated by "alpha", the attenuation constant, which is a function of frequency. Alpha has the same value for the incident and reflected waves, but though the dB per 100 ft is the same for a wave traveling in either direction on the line, the reflected wave is likely much smaller than the incident wave, and the loss produced by the reflected wave will be much smaller too. The total loss is the sum of the losses produced by the incident and reflected waves. As was shown yesterday, the ARRL Antenna Book has charts to determine the added loss caused by standing waves (actually caused by the reflected power that produces SWR). These convenient charts almost eliminate arithmetic in determining additional loss to be expected given the SWR. Best regards, Richard Harrison, KB5WZI |
#6
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(Garvin) wrote in message om...
(Tom Bruhns) wrote in message om... .... If you want dissipated power in a TEM line, then P=Irms^2*R+Erms^2*G (at a particular frequency where R and G have fixed values). Since Irms and Erms are functions of position along the line, I would dissagree with this last statement. The root mean square is a type of averaging (not really just the average though), so how could it be a function of position?? Well, not surprising. You seem to dissssagree with just about everything... The RMS is of course an average, "the square root of the mean [average] squared value...," but it's a time average, not a position average. There is an RMS current associated with every point along the line, and because of standing waves, it's not the same everywhere. Similarly with RMS voltage. If you wish, you can use instantaneous current and voltage and integrate over time as well, but that's just performing the RMS function. Putting it another way that's even easier to see, would you expect the RMS current in my refrigerator power cord to be the same as in my blender power cord? Clearly, RMS current CAN be a function of location. And with standing waves, or with attenuation along the line, or both, it SHOULD be pretty clear that it can be a function of position along a TEM line. |
#7
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Keith wrote:
"Consider a slotted line used to measure voltages for the computation of VSWR." O.K. I`m looking at my PRD 250-A. Its slot is about 10 inches long or about 25.4 cm. The distance between a maximum and a minimum voltage point in an standing wave pattern is 1/4-wavelength. The velocity factor is not quite as high as that of free-space. This tends to shorten the distance between maxima and minima in the slotted line. If you could get a good voltage sample anywhere within a 25 cm slot, you could get one maximum and one minimum in a standing wave pattern at a frequency where the slot was at least 1/4-wavelength, but you might have to adjust feedline length to place the pattern in a favorable slotted line location. What frequency has a 1/4-wavelength of 25 cm? My calculation says: 300 MHz. Slotted lines are called trough lines in the U.K. I believe. SWR is more easily determined with an SWR meter or a wattmeter. These don`t require a slotted line`s 1/4-wave or more minimum of space for operation. The maximum and minimum voltages on a transmission line are not as significant as the forward and reflected powers because the difference between these powers is the power delivered to the load and is also the power supplied by the transmitter as the transmission line has no storage capacity beyond that required to completely energize the line in both directions by the traveling wave. Best regards, Richard Hsarrison, KB5WZI |
#8
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Tom Bruhns wrote:
"Would you then say that R3 is dissipating---0.50 watts?" Certainly not. The two 1.5 V cells don`t put any current through each other because they have no potential difference. The two resistors, one in series with each cell, each drop 0.5 wolts from 1/2 amp through 1 ohm in each case. So, there`s one volt across the common 1-ohm resistor, R3. 1 V x 1 A = 1 watt dissipated in R3. Best regards, Richard Harrison, KB5WZI |
#9
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#10
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