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Unformatted text preview: Darrell Cain, Zack Anderson Below is an analysis of the parameter to be maximized along with the constraints on the system. Based on these equations and the fixed parameters it is possible to calculate the range of efficiencies of the overall rectenna panel based upon the efficiency of the climber. This gives the mathematical foundation to build a Matlab simulation which would calculate the range of efficiencies the rectenna would need to fall in based upon the speed of the climber, the size of the dish, the area of antennas and various other parameters. Important parameter: ε = P out P in Estimate Efficiency: 70 percent Constraints: Input: Microwave Field Size of Dish: 2 meter diameter Base generator: 800 W 5.8 GHz generator Beam: Circularly polarized and directional The following observations can be made about Rectennas  Two things cause microwave signals to weaken o Spreading over a larger sphere of influence o Attenuation due to the atmosphere  Necessary to determine how the field will change in saturation based upon height o Since beam is directional, the beam does not spread and remains constant (same amount of flux) o Attenuation is minimal 1 Area of dish = A = pi * 1^2 = 3.14 m^2;  ρ µ = 800/3.14 = 254.7 W/m^3  At first glance, P in = ρ µ * A = 800 Watts  However o Specs of current design base unit = 46.915 mm X 47.26 mm o Area of current design base unit = .00221 m^2 o Area of antenna’s in a base unit = .000581 m^2 o It’s possible to calculate the area of antenna which is exposed to the radiation ¡ Ratio of Area of Antenna Design Base to Area of Current Design = .2628 ¡ Useable area = A useable = Ratio * A = .825 m^2 ¡ Therefore P in = ρ µ * A useable = 210.25 Watts 1 http://www.profc.udec.cl/~gabriel/tutoriales/rsnote/cp3/cp32.htm Figure 3.2.1 ε = .7 ¡ With a P out = 147.18 o Moving at 2 m/s with a 25 kg structure total work done by the climber = .5 * 25 * 2^2 = 50 Watts o Therefore if everything works correctly the multiple of the other efficiencies must be greater than .3397  Brainstormed Solutions o Multiple layers of rectenna, aligned so as to allow the climber to receive microwave on each level  The following observation can be made about the constraints imposed by the capacitors on board o P c ≤ .25* m climber gh total o m climber gh total = .25*25*9.8*100 = 6125 J Power of capacitors can also be represented by t o P c = ∫ P out − W climber / ε climber dt o From t = 0 to t = 50 o Therefore t ¡ 0 ≤ ∫ P out − W climber / ε climber dt ≤ 6125 J o If we use ¡ v=constant = 2 m/s W climber = 50 J ¡ P out = 147.18 J ¡ In simplified version, neither is time dependent, therefore max is at when t = 50 0 ≤ 7359 J − 2500 J / ε climber ≤ 6125 J − 7359 ≤ − 2500/ ε climber ≤ − 1234 .4936 ≤ 1/ ε climber ≤ 2.9436 ¡ .3397 ≤ ε climber ≤ 2.0259 with physical constraints .3397 ≤ ε climber ≤ 1 ¡ This was found earlier by examining a very simple equation o This is a simplified version. This is a simplified version....
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This note was uploaded on 11/08/2011 for the course AERO 16.810 taught by Professor Olivierdeweck during the Winter '07 term at MIT.
 Winter '07
 OlivierdeWeck

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