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Antenna Theory Question
"AndyS" wrote in news:1175314795.342922.213610
@o5g2000hsb.googlegroups.com: Andy asks: I have a quick question about the number of microwatts that an antenna will extract from the air and put into a known load. Assume a 70 ohm perfect antenna used for receiving a signal.... Preceived = Pwr density ( watts/sq mtr) X Capture Area (sq mtrs) Further to Roy's notes, keep in mind that for most antennas, you cannot determine the Effective Aperture Area (or Capture Area) using a ruler. This effect is demonstrated if you consider that a short low loss dipole with low loss matching has an Effective Aperture Area nearly as large as the much larger half wave dipole. For many purposes, you would determine the Effective Aperture Area by determining Gain from Directivity and Loss, thence Effective Aperture Area from Gain and Wavelength. Owen |
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Antenna Theory Question
"Owen Duffy" wrote in message ... "AndyS" wrote in news:1175314795.342922.213610 @o5g2000hsb.googlegroups.com: Andy asks: I have a quick question about the number of microwatts that an antenna will extract from the air and put into a known load. Assume a 70 ohm perfect antenna used for receiving a signal.... Preceived = Pwr density ( watts/sq mtr) X Capture Area (sq mtrs) Further to Roy's notes, keep in mind that for most antennas, you cannot determine the Effective Aperture Area (or Capture Area) using a ruler. This effect is demonstrated if you consider that a short low loss dipole with low loss matching has an Effective Aperture Area nearly as large as the much larger half wave dipole. For many purposes, you would determine the Effective Aperture Area by determining Gain from Directivity and Loss, thence Effective Aperture Area from Gain and Wavelength. Owen and hence if you start with the capture area by definition all the power is sent to the matched load. basically you measure the capture area by measuring the power in the load and dividing by the power density. |
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Antenna Theory Question
Owen Duffy wrote: Further to Roy's notes, keep in mind that for most antennas, you cannot determine the Effective Aperture Area (or Capture Area) using a ruler. This effect is demonstrated if you consider that a short low loss dipole with low loss matching has an Effective Aperture Area nearly as large as the much larger half wave dipole. For many purposes, you would determine the Effective Aperture Area by determining Gain from Directivity and Loss, thence Effective Aperture Area from Gain and Wavelength. Owen Andy writes: Thanks to all for the discussion....... I have used the results of these calculations for many years but wanted to see what explanations I would get from others who also have experience in these matters..... For capture area, I use " Gee lambda squared over four pi ", which is the standard definition for a well behaved antenna with a main lobe, and it has always worked well for me. I don't remember doing the derivation for this, but I'm sure I must have done it in the past (olden times)... Yet, I was not comfortable with what seemed like a discrepancy between "all the extracted power goes to the load" and the proposition that the cosmos and antenna acted like a generator and therefore had an internal impedance which must be Thevenin matched for max extracted power.... I resolved it, in my mind, by ignoring the latter explanation (grin)..... and apparently it was the correct thing to do... So, thanks again, guys........ Andy W4OAH |
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Antenna Theory Question
On 31 Mar 2007 04:57:23 -0700, "AndyS" wrote:
Yet, I was not comfortable with what seemed like a discrepancy between "all the extracted power goes to the load" and the proposition that the cosmos and antenna acted like a generator and therefore had an internal impedance which must be Thevenin matched for max extracted power.... I resolved it, in my mind, by ignoring the latter explanation (grin)..... and apparently it was the correct thing to do... Hi Andy, It is a bit of a stumble reading through that, and it gives the impression that you have learned the incorrect thing to do (although I find it hard to imagine how you could be "doing" anything when the mechanics of radiation and reception take care of themselves). 73's Richard Clark, KB7QHC |
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Antenna Theory Question
On 31 mar, 13:57, "AndyS" wrote:
{text deleted} Andy writes: Thanks to all for the discussion....... I have used the results of these calculations for many years but wanted to see what explanations I would get from others who also have experience in these matters..... For capture area, I use " Gee lambda squared over four pi ", which is the standard definition for a well behaved antenna with a main lobe, and it has always worked well for me. I don't remember doing the derivation for this, but I'm sure I must have done it in the past (olden times)... Yet, I was not comfortable with what seemed like a discrepancy between "all the extracted power goes to the load" and the proposition that the cosmos and antenna acted like a generator and therefore had an internal impedance which must be Thevenin matched for max extracted power.... I resolved it, in my mind, by ignoring the latter explanation (grin)..... and apparently it was the correct thing to do... So, thanks again, guys........ Andy W4OAH Hi Andy, As you mentioned, Thevenin does apply, and your effective area formula is correct (gain w.r.t. isotropic radiator). So a thin halve wave dipole antenna has an internal resistance of about 70 Ohms, and half the power is lost, but not as heat in the antenna. The driving (incident) EM-field induces voltage with causes a current to flow in your 75 Ohms load (and in the antenna). 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. As your dipole delivers 1mW to the load, also 1mW is reradiated into space. Behind the antenna, the reradiated field cancels (partially) the incident field (causing a shadow effect). When you remove the load, the output voltage at the antenna doubles and the current in the dipole reduces to a low value, hence less power is reradiated. When you short circuit the dipole, the current will double (w.r.t. to the matched situation). This will also happen to the current in the bars forming the dipole. As the current has been doubled, the reradiated power is 4 times higher (w.r.t. the matched situation). The reradiated power can be detected. Radar and passive UHF RFID use the reradiated power. For example in UHF RFID, the chip on the RF tag changes the load to the dipole (that receives power from the interrogator), and controls the reradiated power. In this way the ID of the tag is transmitted from the tag to the interrogator (the tag has no battery present). The more gain you have in your antenna, the higher the reradiated EIRP in the direction of maximum gain. I hope this clarifies the antenna internal resistance. Best Regards, Wim PA3DJS |
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Antenna Theory Question
As you mentioned, Thevenin does apply, and your effective area formula
is correct (gain w.r.t. isotropic radiator). So a thin halve wave dipole antenna has an internal resistance of about 70 Ohms, and half the power is lost, but not as heat in the antenna. If I may add just a simple observation to the exhaustive Wim's explanation, I would say that we generally tend to consider resistors as things that necessarily dissipate power, turning it into heat. That is not always the case. As a matter of fact, having current in phase with voltage (what we usually call resistance) signify the transformation of electrical power into any form of power, not just heat. For instance: - 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 - an ideal antenna (no ohmic loss) fully transforming the applied electrical power into electromagnetic power. Again it looks like a resistor, but no heat is generated anywhere .. 73 Tony I0JX |
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Antenna Theory Question
Antonio Vernucci wrote:
- 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. -- 73, Cecil http://www.w5dxp.com |
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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 |
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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 |
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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 |
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