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Frank Gilliland wrote:
Wes Stewart wrote: In a word, baloney. The error is independent of length. A zero length bridge calibrated at 75 ohm is in error when measuring in a 50 ohm system. Period. Prove it. A 75 ohm bridge is expecting the ratio of voltage to current to be 75 for a matched system. In a 50 ohm matched system, the ratio of voltage to current will be 50. Therefore, the 75 ohm bridge won't be balanced. A 50 ohm bridge would be balanced. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#2
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On Wed, 29 Jun 2005 23:10:25 -0500, Cecil Moore
wrote in : Frank Gilliland wrote: Wes Stewart wrote: In a word, baloney. The error is independent of length. A zero length bridge calibrated at 75 ohm is in error when measuring in a 50 ohm system. Period. Prove it. A 75 ohm bridge is expecting the ratio of voltage to current to be 75 for a matched system. In a 50 ohm matched system, the ratio of voltage to current will be 50. Therefore, the 75 ohm bridge won't be balanced. A 50 ohm bridge would be balanced. The bridge is calibrated to the impedance of the directional coupler (which is usually built to match the expected line impedance, but cannot be "zero length" in the present state of reality). If the impedance of the signal is different than what is expected by the bridge then your power measurements will probably be wrong (to what extent they are wrong may or may not be important). But if that's the case then any error will be the same by percentage and sign for both forward =AND= reflected power because the impedance of the signal is the same for both forward and reflected power. IOW, the ratio is the same -despite- the impedance. If you don't believe me, try it yourself. ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#3
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On Wed, 29 Jun 2005 22:53:24 -0700, Frank Gilliland
wrote: cannot be "zero length" in the present state of reality). If the impedance of the signal is different than what is expected by the bridge then your power measurements will probably be wrong (to what extent they are wrong may or may not be important). But if that's the case then any error will be the same by percentage and sign for both forward =AND= reflected power because the impedance of the signal is the same for both forward and reflected power. IOW, the ratio is the same -despite- the impedance. Lets make an assumption that we are talking about lossless lines. (If you are not, then the reflectometer does not provide an accurate indication of forward and reverse power.) If you use an ideal 50 ohm reflectometer (that means it is a negligibly short 50 ohm through line and it is nulled to show zero reflected power when connected to a 50+j0 load) to measure conditions in a line, the power flow at that point is the indicated Pf-Pr. If you had placed an ideal 75 ohm instrument in that spot, the readings are not necessarily in the same ratio (they are unlikely to be so), but the difference between Pf and Pr will be the same. The only other inference you can make from one instrument with regard to the other will be if one of the instruments shows zero reflected power, then you know the VSWR that the other instrument will indicate. Real instruments aren't of zero length, but some types of design are so close to it at low HF frequencies, you will not detect the error that is introduced. Owen -- |
#4
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Owen wrote:
Real instruments aren't of zero length, but some types of design are so close to it at low HF frequencies, you will not detect the error that is introduced. The error that we are talking about has nothing to do with the length of the directional coupler. The error that we are talking about has everything to do with an infinite error in the measurement of reflected power. Infinite errors are hard to sweep under the rug. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#5
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Updated: On Wed, 29 Jun 2005 22:53:24 -0700, Frank Gilliland wrote: cannot be "zero length" in the present state of reality). If the impedance of the signal is different than what is expected by the bridge then your power measurements will probably be wrong (to what extent they are wrong may or may not be important). But if that's the case then any error will be the same by percentage and sign for both forward =AND= reflected power because the impedance of the signal is the same for both forward and reflected power. IOW, the ratio is the same -despite- the impedance. Lets make an assumption that we are talking about distortionless lines. (If you are not, then the reflectometer does not provide an accurate indication of forward and reverse power.) If you use an ideal 50 ohm reflectometer (that means it is a negligibly short 50 ohm through line and it is nulled to show zero reflected power when connected to a 50+j0 load) to measure conditions in a line, the power flow at that point is the indicated Pf-Pr. If you had placed an ideal 75 ohm instrument in that spot, the readings are not necessarily in the same ratio (they are unlikely to be so), but the difference between Pf and Pr will be the same. The only other inference you can make from one instrument with regard to the other will be if one of the instruments shows zero reflected power, then you know the VSWR that the other instrument will indicate. Real instruments aren't of zero length, but some types of design are so close to it at low HF frequencies, you will not detect the error that is introduced. Owen -- |
#6
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On Thu, 30 Jun 2005 06:47:07 GMT, Owen wrote:
The only other inference you can make from one instrument with regard to the other will be if one of the instruments shows zero reflected power, then you know the VSWR that the other instrument will indicate. As Cecil has hinted ), you can also infer that the impedance at the point where you have two sets of Pf,Pr readings is that the impedance (R, X) is one of two values (which can be visualised as the intersection of two circles on a Smith Chart), one only in the cases VSWR1*VSWR2=Zo1/Zo2 (or the inverse) (which can be visualised as the kissing of two circles on a Smith Chart). The latter includes the case above where one or other instruments indicates VSWR=1, but is more exact because one of the circles is infinitely small. These cases are not reliably of practical value, the conversion of error in the measurements made with each instrument, into error in the estimated R and X at the point could be huge. So, I "correct" my statement to "The only other inference that you can reasonably reliably make from one instrument with regard to the other will be if one of the instruments shows zero reflected power, then you know R and X, and the VSWR that the other instrument should indicate. Thanks Cecil. Owen -- |
#7
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I'm not quite sure what you are trying to say Frank.
Frank Gilliland wrote: On Wed, 29 Jun 2005 23:10:25 -0500, Cecil Moore wrote in : The bridge is calibrated to the impedance of the directional coupler (which is usually built to match the expected line impedance, but cannot be "zero length" in the present state of reality). The direction coupler samples voltage across and current through a given point. There is always a current transformer of some type and a voltage sample through some type of divider. The "voltages" representing E and I are summed before detection (conversion to dc). The "directivity" comes because the current phase sample is reversed 180 degrees from the summing phase, causing voltages to subtract. This means the directional coupler is calibrated for a certain ratio of voltage and current, so when they exist you have twice the voltage in the direction where E and I add, and zero voltage where they subtract. If the impedance of the signal is different than what is expected by the bridge then your power measurements will probably be wrong (to what extent they are wrong may or may not be important). But if that's the case then any error will be the same by percentage and sign for both forward =AND= reflected power because the impedance of the signal is the same for both forward and reflected power. IOW, the ratio is the same -despite- the impedance. ?What does that mean? If the directional coupler is calibrated at 50 ohms and you use it in a 75 ohm system you won't get a total reflected null even if the 75 ohm line has a 1:1 SWR. But if you subtract reflected power from forward power readings you will get the correct power, within linearity and calibration limits of the "meter system". This has nothing to do with standing waves. It has only to do with the relationship between current and voltage at the point where the directional coupler is inserted. I'm not sure if you are saying that or not. 73 Tom |
#8
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Frank Gilliland wrote:
Cecil Moore wrote: A 75 ohm bridge is expecting the ratio of voltage to current to be 75 for a matched system. In a 50 ohm matched system, the ratio of voltage to current will be 50. Therefore, the 75 ohm bridge won't be balanced. A 50 ohm bridge would be balanced. The bridge is calibrated to the impedance of the directional coupler (which is usually built to match the expected line impedance, but cannot be "zero length" in the present state of reality). If the impedance of the signal is different than what is expected by the bridge then your power measurements will probably be wrong (to what extent they are wrong may or may not be important). But if that's the case then any error will be the same by percentage and sign for both forward =AND= reflected power because the impedance of the signal is the same for both forward and reflected power. IOW, the ratio is the same -despite- the impedance. The error is NOT the same percentage. In a matched 50 ohm system, the 75 ohm bridge reflected power reading will be off by an infinite percentage, i.e. division by zero. If you don't believe me, try it yourself. I have tried it and you are wrong. Maybe you should try it. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
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