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Antenna Theory Question
- an ideal electrical engine (no ohmic loss, no friction) fully
transforming the applied electrical power into mechanical power. It looks like a resistor, but no heat is generated anywhere No, it looks like a resistance, not a resistor. There is one IEEE definition for resistor. There are two IEEE definitions for resistance. A resistor with current dissipates power. A resistance may or may not dissipate power. One definition of resistance in the IEEE describes a dissipationless resistance. There is no such thing as a dissipationless resistor. It IS a resistance, but it LOOKS like a resistor because, from the exterior, one has no means to determine whether energy gets dissipated or transformed in some other form. 73 Tony I0JX |
Antenna Theory Question
Antonio Vernucci wrote:
It IS a resistance, but it LOOKS like a resistor because, from the exterior, one has no means to determine whether energy gets dissipated or transformed in some other form. Appearances can be deceiving. It doesn't look like a resistor to me until I hold it in my hand. -- 73, Cecil http://www.w5dxp.com |
Antenna Theory Question
Cecil Moore wrote: There is no such thing as a dissipationless resistor. -- 73, Cecil http://www.w5dxp.com Andy comments: Not to make too fine a point on this, Cec, but I , personally, would consider a perfect 52 ohm antenna to be a dissipationless resistor......... although, in the cosmic sense, the power that is fed into it will eventually reside as "heat" in the bowels of the cosmos.....somewhere...... Actually, .... "heat" is simply EM at a different frequency.... ..... Hmmmmm... I see a deep, philosophical discussion brewing here..... .. ... Me brain hoits......!!!! Andy W4OAH |
Antenna Theory Question
On 1 Apr 2007 09:16:46 -0700, "AndyS" wrote:
I see a deep, philosophical discussion brewing here.... Only if it starts out with: "Who's on first?" |
Antenna Theory Question
AndyS wrote:
Not to make too fine a point on this, Cec, but I , personally, would consider a perfect 52 ohm antenna to be a dissipationless resistor......... although, in the cosmic sense, the power that is fed into it will eventually reside as "heat" in the bowels of the cosmos.....somewhere...... Yep, a common misconception. The feedpoint impedance of a standing-wave antenna, like a 1/2WL dipole, is a *virtual* impedance equal to (Vfor+Vref)/(Ifor+Iref). It may be a resistance, but it is not a resistor. It is essentially the same entity as the impedance looking into a stub. For a resonant 1/2WL dipole the equation becomes: Feedpoint resistance = (|Vfor|-|Vref|)/(|Ifor|+|Iref|) With Vfor at a reference angle of zero, Vref is at 180 degrees. With Ifor at zero degrees, Iref is also at zero degrees. Thus the ability to add and subtract magnitudes directly. Some guru once challenged me to make a measurement at the shack and tell the difference between a 50 ohm dipole and a 50 ohm resistor. Told him all I needed was a field strength meter. :-) The words "impedor" and "resistor" are reserved for real physical devices. The words "impedance" and "resistance" have two meanings and can mean either real devices or virtual V/I's. So says the IEEE Dictionary. -- 73, Cecil http://www.w5dxp.com |
Antenna Theory Question
On Sun, 01 Apr 2007 18:24:33 GMT, Cecil Moore wrote:
AndyS wrote: Not to make too fine a point on this, Cec, but I , personally, would consider a perfect 52 ohm antenna to be a dissipationless resistor......... although, in the cosmic sense, the power that is fed into it will eventually reside as "heat" in the bowels of the cosmos.....somewhere...... Yep, a common misconception. The feedpoint impedance of a standing-wave antenna, like a 1/2WL dipole, is a *virtual* impedance equal to (Vfor+Vref)/(Ifor+Iref). It may be a resistance, but it is not a resistor. It is essentially the same entity as the impedance looking into a stub. For a resonant 1/2WL dipole the equation becomes: Feedpoint resistance = (|Vfor|-|Vref|)/(|Ifor|+|Iref|) With Vfor at a reference angle of zero, Vref is at 180 degrees. With Ifor at zero degrees, Iref is also at zero degrees. Thus the ability to add and subtract magnitudes directly. Some guru once challenged me to make a measurement at the shack and tell the difference between a 50 ohm dipole and a 50 ohm resistor. Told him all I needed was a field strength meter. :-) The words "impedor" and "resistor" are reserved for real physical devices. The words "impedance" and "resistance" have two meanings and can mean either real devices or virtual V/I's. So says the IEEE Dictionary. Amen to that, Cecil Walt, W2DU |
Antenna Theory Question
The current that flows into the antenna structure (two bars) radiate.
So part of the incident field is reradiated by the dipole. It doesn't matter whether the antenna current is caused by a transmitter or incident field. Win, your explanation is very good and convincing. I believe that most of us would read with interest some further remarks on what happens in the case of a shorted element (e.g. a director or a reflector), instead of the driven element. In those cases all received energy must be re-radiated, but more details on that mechanism would be welcome. 73 Tony I0JX |
Antenna Theory Question
Antonio Vernucci wrote:
Win, your explanation is very good and convincing. I believe that most of us would read with interest some further remarks on what happens in the case of a shorted element (e.g. a director or a reflector), instead of the driven element. In those cases all received energy must be re-radiated, but more details on that mechanism would be welcome. The impinging field induces a current on the element. The current produces a field which radiates. An AC current flowing on a conductor creates a field regardless of how that current is produced. It doesn't matter whether the current is due to conduction from a source, by induction from a field, or any combination of the two -- the field resulting from the current is exactly the same. Yagis and other "parasitic" arrays depend entirely on this phenomenon for their operation. Roy Lewallen, W7EL |
Antenna Theory Question
In article , Roy Lewallen
wrote: Antonio Vernucci wrote: Win, your explanation is very good and convincing. I believe that most of us would read with interest some further remarks on what happens in the case of a shorted element (e.g. a director or a reflector), instead of the driven element. In those cases all received energy must be re-radiated, but more details on that mechanism would be welcome. The impinging field induces a current on the element. The current produces a field which radiates. An AC current flowing on a conductor creates a field regardless of how that current is produced. It doesn't matter whether the current is due to conduction from a source, by induction from a field, or any combination of the two -- the field resulting from the current is exactly the same. Yagis and other "parasitic" arrays depend entirely on this phenomenon for their operation. Roy Lewallen, W7EL Yes, and with the corollary that you can never extract more than half of the power/energy transported by an EM wave. That maximum (available power) is given by Voc^2/(4 * Ra) where Voc and Ra are the open circuit voltage magnitude and real part of the extractor (receive antenna) impedance, respectively, measured at the antenna terminals at the frequency of interest. The remaining energy is scattered (reradiated) from the antenna. Sincerely, and 73s from N4GGO, John Wood (Code 5550) e-mail: Naval Research Laboratory 4555 Overlook Avenue, SW Washington, DC 20375-5337 |
Antenna Theory Question
Roy Lewallen wrote: The impinging field induces a current on the element. The current produces a field which radiates. An AC current flowing on a conductor creates a field regardless of how that current is produced. It doesn't matter whether the current is due to conduction from a source, by induction from a field, or any combination of the two -- the field resulting from the current is exactly the same. Yagis and other "parasitic" arrays depend entirely on this phenomenon for their operation. Roy Lewallen, W7EL Andy asks: Roy, is the field radiated by the parasitic element shifted in phase relative to the incident field. If not, what determines the phase shift ? Andy W4OAH |
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