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#31
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Harry wrote:
Would you please explain the difference between these two formulas (4-37 and 4-93)? Equation 4-37 is from section "4.3 SMALL DIPOLE", i.e. shorter than 1/2WL. The dipole is so short that its current distribution is triangular, not sinusoidal. Quoting section 4.3: "The radiation resistance of the antenna is strongly dependent upon the current distribution." The "1/2" on the diagram does NOT mean 1/2WL. Equation 4-93 is from section "4.6 HALF-WAVELENGTH DIPOLE". The current distribution on the thin-wire 1/2WL dipole is considered to be sinusoidal. See Figure 4.8. Note the triangular current distribution for L = 1/4WL and the sinusoidal current distribution for L = 1/2WL. -- 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 =---- |
#32
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See Figure 4.8. Note the triangular current distribution for
L = 1/4WL and the sinusoidal current distribution for L = 1/2WL. Very good,..... thanks a lot! Your figure 4.8 (second edition) is figure 4.7 in the first edition. These current distribution curves (1/4, 1/2, 1, 3/2, and 2 Lamda length) are interesting to study. -- Harry |
#33
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| "Roy Lewallen"
| wrote in message ... | | Just about any antenna textbook will show you the calculation | [...] | ln(gamma * x) - (x^2)/(2!2) + (x^4)/(4!4) - (x^6)/(6!6). . . | | Roy Lewallen, W7EL [... ] and after all that, we finally draw a figure such the one at: http://antennas.ee.duth.gr/Antennas_...II_2005/23.htm pezSV7BAXdag |
#34
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On Thu, 15 Sep 2005 21:26:04 +0300, news-z-non-exist-z wrote:
| "Roy Lewallen" | wrote in message ... | | Just about any antenna textbook will show you the calculation | [...] | ln(gamma * x) - (x^2)/(2!2) + (x^4)/(4!4) - (x^6)/(6!6). . . | | Roy Lewallen, W7EL [... ] and after all that, we finally draw a figure such the one at: http://antennas.ee.duth.gr/Antennas_...II_2005/23.htm pezSV7BAXdag Yes, but: http://antennas.ee.duth.gr/Antennas_...II_2005/22.htm explains it best of all! HI!HI! 73 Jonesy -- Marvin L Jones | jonz | W3DHJ | linux Pueblo, Colorado | @ | Jonesy | OS/2 __ 38.24N 104.55W | config.com | DM78rf | SK |
#35
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"Allodoxaphobia" wrote in message ...
| On Thu, 15 Sep 2005 21:26:04 +0300, news-z-non-exist-z wrote: | [...] | Yes, but: | http://antennas.ee.duth.gr/Antennas_...II_2005/22.htm | | explains it best of all! HI!HI! | | 73 | Jonesy d;^) |
#36
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In message , Jim Kelley
writes Cecil Moore wrote: Ian Jackson wrote: Just a quick question. What is the impedance at the centre of an infinitely long dipole (in free space)? Same as a terminated dipole in an anechoic chamber? 600-800 ohms? Not at zero Hz. ac6xg What is a 'terminated dipole'? And why 600-800 ohms? No, Mine will be just a normal dipole (but long). I'll be in a spaceship, miles from anywhere, and I intend to put out a really long antenna so I can work the universe on all amateur bands. (I don't care about the polar diagram - there's bound to be someone out there in one of the major lobes). I intend to throw out an infinitely long wire either side of the ship, and use a balanced tuner connected directly to the antenna (no feeder required). Because of weight restrictions, I can only take one tuner, and I want to make sure that the one I do take will cope with the feed impedance of the antenna. I think the impedance will be the same at all frequencies (maybe even at 'zero Hz'). But what will it be? Ian. -- |
#37
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Ian Jackson wrote:
What is a 'terminated dipole'? In an anechoic chamber, the dipole has a resistor to chassis at both ends. When the value of the resistor equals the characteristic impedance of the dipole, the dipole becomes a traveling wave antenna instead of a standing wave antenna. Your infinite dipole in free space will be a traveling wave antenna, i.e. no reflections. And why 600-800 ohms? That's a ballpark Z0 for traveling wave antennas not located near ground. It's the Z0 of a 1/2WL dipole during the transient state before the arrival of the first reflections from the ends. The resonant feedpoint impedance of a center-fed 1/2WL dipole is (the difference between the values of the forward voltage and the reflected voltage) divided by (the sum of the values of the forward current and reflected current). Note: the two voltages are out of phase and the two currents are in phase. This is the (B) definition of "impedance" in the "IEEE Dictionary", i.e. impedance as a *result*, not a cause. Destructive wave interference at the feedpoint of a center-fed 1/2WL dipole is what causes a Z0-match to a low resistive value, e.g. 50 ohms or 70 ohms. -- 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! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
#38
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What is the impedance at the centre of an infinitely long dipole
(in free space)? =============================== Its not very different from - Zin = 120 * Ln( Wavelength / d ) ohms. where d = conductor diameter, both measured in metres. Thus, at wavelength = 80 metres with 14 gauge copper wire, input impedance = 1300 ohms approx. If you don't believe me, just measure it. ---- Reg. |
#39
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Ian Jackson wrote:
In message , Jim Kelley writes Cecil Moore wrote: Ian Jackson wrote: Just a quick question. What is the impedance at the centre of an infinitely long dipole (in free space)? Same as a terminated dipole in an anechoic chamber? 600-800 ohms? Not at zero Hz. Now#1, an infinitely long wire has an infinite self inductance. Now#2, an infinitely long wire has an infinite self capacitance. Now#3, frequency does not enter into the equation Zo = SQRT[L(infinite)/C(infinite)] Now, from my old college days, infinity divided by infinity is undefined. The SQRT of an undefined value is by definition undefined. Conclusion: the impedance of an infinitely long wire in free space is not the SAME as a terminated dipole in an anechoic chamber. NOTE: an infinitely long wire in free space cannot be terminated. A terminated wire in an anechoic chamber is not the same as a free space model. A terminated wire in in an anechoic chamber is a standard math model with Zo varying as the ratio of the wire diameter and the distance to the 'space cloth/cones' [which are not perfect rf absorbers ... hence some reflection]. Also, the termination destroys the symmetry. ac6xg What is a 'terminated dipole'? And why 600-800 ohms? No, Mine will be just a normal dipole (but long). I'll be in a spaceship, miles from anywhere, and I intend to put out a really long antenna so I can work the universe on all amateur bands. (I don't care about the polar diagram - there's bound to be someone out there in one of the major lobes). I intend to throw out an infinitely long wire either side of the ship, and use a balanced tuner connected directly to the antenna (no feeder required). Because of weight restrictions, I can only take one tuner, and I want to make sure that the one I do take will cope with the feed impedance of the antenna. I think the impedance will be the same at all frequencies (maybe even at 'zero Hz'). But what will it be? Ian. |
#40
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You had better go back to your old college days and start again with
new teachers. The input impedance of an infinite dipole is as I have already simply mathematically described. It changes a little with frequency and wire diameter. Anything less than infinite length will be altogether different. Now you can stop trying to pull our legs. ---- Reg. |
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