Name:GTID:ECE 6390: Satellite Communications and Navigation SystemsTEST 1 (Fall 2005)• Please read all instructions before continuing with the test.• This is a closed notes, closed book, closed friend, open mind test. On your desk you shouldonly have writing instruments and a calculator.• Show all work. (It helps me to give partial credit.) Work all problems in the spaces below theproblem statement. If you need more room, use the back of the page. DO NOT use or attachextra sheets of paper for work.• Work intelligently – read through the exam and do the easiest problems first. Save the hardones for last.• All necessary mathematical formulas are included either in the problem statements or the lastpage of this test.• You have 50 minutes to complete this examination. When the proctor announces a “last call”for examination papers, he will leave the room in 5 minutes. The fact that the proctor doesnot have your examination in hand will not stop him.• I will not grade your examination if you fail to 1) put your name and GTID number in theupper left-hand blanks on this page or 2) sign the blank below acknowledging the terms of thistest and the honor co de policy.• Have a nice day!Pledge Signature:I acknowledge the above terms for taking this examination. I have neither given nor received unau-thorized help on this test. I have followed the Georgia Tech honor code in preparing and submittingthe test.11. Short Answer Section (35 points)(a)(1) (2) (3) (4)The Helmholtz (scalar) wave equation holds for a medium that has the following fourproperties:Answer 1 , Answer 2 , Answer 3 ,and Answer 4 .(b)Earth Answer cause gravitational irregularities in an otherwise ideal orbit.(c)(1) (2)A satellite in elliptical orbit is furthest away from a planet atAnswer 1 and closest atAnswer 2 .(d)A transmitter sends 100 mW into an antenna with peak linear gain of 4 with respect toan isotropic radiator; the EIRP of this system isAnswer dBm.(e)The cheapest location on earth from which to launch a satellite is the Answer .(f)(1) (2) (3) (4)List 4 types of power supplies for a satellite.(g) Famous Dates: Match the dates below to the events.1630a) First satellite Sputnik launched by USSR1945b) I started taking ECE 63901957c) Arthur C. Clarke publishes “Extra-Terrestrial Relays”1958d) Telestar I and I I launched by Bell Labs1962e) Explorer I, first US satellite is launched1969f) First mobile satellite telephone networks launched1980sg) Johannes Kepler born2000h) Moon landing2005i) Global Positioning System launched22. Conspiracy Theory: An engineer files a complaint with the FCC claiming that it maybe possible to interrogate RFID tags from an LEO satellite (500 km overhead), violating theprivacy of anyone who happens to be wearing a Wal-Mart shirt with an inventory tag still at-tached or driving their car with an automatic toll-paying tag on the windshield. There are twotypes of proposed RFID tags for inventory control, tollway payments, and other identificationpurposes. The first type is a battery-assisted tag that uses a small, internal battery to drive itsmodulating circuitry; all power backscattered by this type of tag is due to the partial reflectionof an incident wave. Highway toll tags are a common example of this type of tag. The secondtype is a purely passive tag that actually powers its modulating circuitry from the incidentradio wave. Inventory tags are a common example of purely passive RFID tags. Both typesof RFID tags operate in the 915 MHz ISM band and have antenna gains of approximately 0dBi. Assume that a 915 MHz satellite interrogator can transmit and receive with 10 dBi ofantenna gain. Based on this scenario, answer the following questions. (30 points)SatelliteRFID TagGtagGsatr=500 kmDownlinkUplinkBackscatterGsatGtag(a) For a passive tag, at least 1µW must be received by the tag from the satellite interrogatorin order for the tag to power-up. How much transmit power must be sent by the interro-gating satellite to make this happen? Does this seem reasonable? (15 points)(b) At least -110 dBm of back-scattered power must be received by the satellite in order toread any RFID tag’s data. How much transmit power must be sent by the interrogatingsatellite to make this happen? Does this seem reasonable? Hint: In a backscatter radiolink, power is transferred from satellite to RFID tag for the first leg of the link; the re-ceived power then becomes the transmit power for the tag (assuming no internal losses)and is send back to the satellite on a second link. (15 points)33. Launch a GPS Satellite!: The US Air Force is planning to launch another replacementGPS satellite into space. The first stage of the launch rocket places the s atellite into a low-earthorbit (LEO) exactly 1000 km above the surface of the earth with an inclination that matchesthe final orbit (see diagram below). To make the jump to final orbit, a smaller second-stagerocket fires for a very brief time at Point A to increase the velocity and place the satellite intoan elliptical transit orbit. At Point B the thrusters of the second stage fire briefly for the lasttime, transferring the satellite to its final circular orbit of 20,200 km above the Earth’s surface.Answer the questions below based on this scenario. (35 points)Point ATarget OrbitTransit OrbitSatellitePoint Br1r2RE= 6380 kmr1= 1000 kmr2= 20,200 kmEarth(a) Draw an arrow at point A that points in the direction of thrust for the second-stagerocket (keeping in mind that thrust is a force opposite the direction of rocket discharge).(5points)(b) Draw an arrow at point B that points in the direction of thrust for the second-stagerocket. (5points)(c) What is the eccentricity of the transit ellipse? (10 points)(d) Estimate the transit time for the satellite between points A and B. (10 points)(e) How fast is the satellite travelling in its final orbit? (5points)4Cheat Sheetλf = cc=3× 108m/sPR= PT+ GT+ GR− 20 log104πλ− 20 log10(r) − Additional Loss in dB¨r = r˙θ2−GMPr2¨θ = −2˙r˙θrT2=4π2a3µµ = GMpG =6.672 × 10−11Nm2/kg2ME=5.974 × 1024kgV = µRb = a1 − e2perigee = (1 − e)a apogee = (1 + e)aG = ηA4πλ2AeG ≈30, 000θHPBWφHPBW(angles in degrees)PN= kT B k =1.3807 ×