antenna area

Effective Area of an Antenna PhL 2.24.03 There are many ideas here that are new to me, so I have tried to number each of the steps to keep them from blurring into each other. I had guidance from: http://farside.ph.utexas.edu/~rfitzp/teaching/jk1/lectures/node83.html 1. Radiation resistance Rrad of a Hertzian Dipole antenna. From Jackson, we know the power and pattern of a "Hertzian Dipole" antenna, a center-fed antenna which is d/2 in length on each side, and where d << . That power is, from 9.29, P = Io 2 (kd)2 / 12c = Io 2/2 [ 2*4 2 / 12 c * (d/)2] = Io 2 /2 * Rrad Here, Io is a peak current, so Io 2 /2 = I2rms and Rrad must then be the radiation resistance of this antenna, of course here in cgs units. Secondly, we know the power dipole radiation pattern is sin2. 2. Converting Rrad to mks units. 1 Coulomb = c(cgs)/10 esu so 1 amp = c(cgs)/10 statamp 1 volt = 1/300 statvolt 1 ohm = 1 volt / 1 amp = 1/300 statvolt / [c(cgs)/10 statamp ] = 1/[30c(cgs) ] statohms => R(ohms) = 30 c(cgs) R(statohms) Thus, for the Hertzian antenna we have Rrad (statohms) = [ 2 2 / 3 c * (d/)2] Rrad (ohms) = [ 20 2 * (d/)2] = 197 (d/)2 3. Model for a receiving antenna. When any antenna is configured for reception, it is modeled as an induced voltage in series with Rrad, and your best bet for power into a load is to have Rload = Rrad . If the antenna emf source generates Vo peak voltage, half of that will then be on the load, so the load receives power of Pload = (1/2) (Vo/2)2 /Rrad = (1/8) Vo2 / Rrad, where the first factor of 1/2 comes because we want to talk rms power, not peak power. In any matched antenna, therefore, fully half the received power is re-radiated into Rrad ! 4. Pload For the Hertzian Dipole antenna. We know that Vo= Eo (d/2) [ or twice this for peak to peak] , just from definition of the field, where Eo is the peak electric field amplitude. Thus, from the previous paragraph, Pload = (1/8) Vo2 / Rrad = (1/32) Eo2 d2 / Rrad On the other hand, from 2 paragraphs ago we know that for this antenna, in mks units, Rrad (ohms) = 20 2 * (d/)2 Therefore, the power delivered to the matched load of a short dipole antenna is Pload = 1/[ 32*20*2] * Eo2 * 2 which is a fundamental result, saying power received is proportional to 2 and to Eo2 . We suspect that the Eo2 proportionality will hold for any antenna, not just a short dipole. 5. Effective area concept. For any antenna, we can imagine an "effective antenna of area A" that scoops up 100% of the Poynting flux it receives and converts that 100% to Pload . We would then say Pload = A * S 6. Poynting flux in mks units and Z0. In cgs units, for a plane wave in free space we know that the time averaged S is given by S = (c/8) E x H* => S = (c/8) Eo2 Converting this to mks as on page 619 Jackson requires c 1/ Eo Eo so we would expect to get S = (1/8) 4o Eo2 = (1/) Eo2/2 = [ Eo2/2 ] / Zo => S = [ Eo2/2 ] / Zo where Zo = 376.7 = 120 ohms = 1/(oc). 7. Area of a Hertzian Dipole antenna. Combining the previous two items, we find that Pload = A * S = A * [ Eo2/2 ] / Zo being the definition of this effective area A. But for the matched Hertzian Dipole we know that Pload = 1/[ 32*20*2] * Eo2 * 2 from which we may find the effective area A * [ Eo2/2 ] / Zo = 1/[ 32*20*2] * Eo2 * 2 A = 1/[ 32*20*2] * 2 * 2 Zo = 1/[ 32*20*2] * 2 * 240 = 12/[ 32*] * 2 = (3/8) 2 => A(short dipole) = (3/8) 2 8. Gain of the Hertzian dipole antenna. Now the Hertzian dipole is not an isotropic radiator, but has a sin2 dependence for dP/d . Let's compare a dipole radiator to an isotropic one: dP/d = D sin2 => P = D*8/3 dP/d = I => P = I * 4 If we want both these radiators to radiate the same total power, we must have D*8/3 = I * 4 => D * 2/3 = I Now if we consider the dipole in its peak radiating direction, we have dP/d = D, so the dipole antenna therefore has a GAIN over the same-total-power isotropic antenna of G = 3/2. 9. The hypothetical isotropic antenna. Imagine a transmitter sending a signal to our Hertzian antenna from a point in the peak of the beam of our receiving antenna. We have Pload = A * S in our receiver load. Now, suppose we replace our receiving antenna with a hypothetical isotropic one. From reciprocity, we know that this antenna for the same incoming S is going to have less Pload by exactly the power gain factor of the Hertz antenna. So we would write Pload = A * S // Hertz antenna P'load = A' * S // isotropic antenna Pload / P'load = G = gain Therefore, A/A' = Pload / P'load = G So you imagine that you start with an isotropic receiving antenna of area A0 and the antenna's gain makes the effective area larger by A = G * A0. 10. Effective area A0 of the hypothetical isotropic antenna. So, at this point we computed the A for a Hertzian short dipole to be A =(3/8)2, and we know that the gain for this antenna is G = 3/2. If we replaced this antenna with a hypothetical isotropic receiving antenna, that hypothetical antenna would have a smaller effective area Ao and would have a smaller power delivered to the receiving antenna load. We would have Ao = (3/8)2/G = (3/8)2 *2/3 = 2 /4 = 2 = /2 11. Interaction between two antennas. Imagine now a transmitting antenna driven by Pxmit all of which is delivered into the transmitting radiation resistance. The flux from this antenna at distance r is going to be G x the flux from an isotropic radiator of the same power, so S = Gt * Pxmit / (4r2) The power delivered to the load in a receiving antenna will be Pload = Ar * S = Gr Ao * { Gt * Pxmit / (4r2) } = Gr Gt Pxmit (2 /4 ) / (4r2) = Gr Gt [ /4r ]2 Pxmit = [Gr 2/4] [Gt 2/4 ] Pxmit/r22 = [ Ar At /2] Pxmit / r2 This is the Friis Transmission Formula, symmetric in both directions of course. 12. Conclusions about Area Therefore we arrive at these conclusions: (1) The effective area A0 of a hypothetical (and non-existent) isotropic antenna is a 2 which is the area of a circle of radius . The meaning of this area is that, if you actually had such a receiving antenna, and put it into the flux S of some transmitted plane wave, and if you matched this antenna to its load, the power delivered to the load would be S* A0. (2) Any real antenna has an effective area A = G A0 = G 2, where G is the directivity or gain factor in presumably the best direction, which is assumed to be the way it would be used. Again, the power delivered to a matched load would be S* A.