ES 442 Homework #8 Solutions (Spring 2017 Due May 1, 2017 ) Print out homework and do work on the printed pages.

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NAME Soutions ES 44 Homework #8 Soutions (Spring 017 Due May 1, 017 ) Print out homework and do work on the printed pages. Textbook: B. P. athi & Zhi Ding, Modern Digita and Anaog Communication Systems, 4 th edition, Oxford University Press, New york, 009.

1 NAME Soutions ES 44 Homework #8 Soutions (Spring 017 Due May 1, 017 ) Print out homework and do work on the printed pages. Textbook: B. P. athi & Zhi Ding, Modern Digita and Anaog Communication Systems, 4 th edition, Oxford University Press, New york, 009. Figure for Probems 1 & : Probem 1 Path engths (30 points) Consider the dua-ray ground refection mode shown in the figure above. A ceuar base station antenna transmits a signa directed to the ce phone of hed by the man in the figure. The ceuar base station transmitter and the ce phone receiving the signa are separated by distance. Two signas are received by the ce phone one is direct ine-of-sight covering a distance os and the other is refected from the ground traveing a sighty greater distance ref. Assume: >> h tran and >> h rcvr. Show that in this dua-ray ground refection mode that the difference in the two path engths, given by = ref os, can be approximatey expressed as htran hrcvr, 1

2 where h tran is the height of the ce base station tower and h rcvr is the distance the ce phone is above ground eve. Assume the ground (i.e., Earth) is fat and eve. Using trigonometry, ( h h ) and ref tran rcvr ( h h ) Therefore, os tran rcvr ( h h h h ) ( h h h h ) ref os tran tran rcvr rcvr tran tran rcvr rcvr For very arge spacing, we use a Tayor's series, 1 x 1 ( h tran hrcvr ( htran hrcvr 1 1 ( htran hrcvr ) ( htran hrcvr ) ( h tran htranhrcvr hrcvr ) ( htran htranhrcvr hrcvr ) htranhrcvr Probem Phase shift between the two signa paths (0 points) For this probem again use the figure on page 1 of this homework set. We know that the height of the receiver is h rcvr = meters and the height of the base station antennas is h tran = 30 meters. We want to restrict the phase shift between the two paths to = 0.5 at most, where is the waveength of the carrier frequency f c. Assume the ceuar system operates in the 900 MHz band (i.e., take f c = 900 MHz). Take the speed of eectromagnetic waves to be 300,000,000 meters per second in air. What distance is needed for the phase shift between the two signa paths to be one-quarter waveength (i.e., /4 equivaent to 90 degrees)? 1 x htranh We use the resut from probem 1, namey, First, we find the distance that wi be one-quarter waveength of phase shift. Given f c, where c is meters/second, and f 900 MHz aows us to cacuate. rcvr

The waveength = 0.3333 meters at 900 MHz, therefore, = /4 = 0.08333 meters. htranhrcvr htranhrcvr (30 m)( m) From ; we find 0.0833 m 10 m 1,440 meters 0.

3 The waveength = meters at 900 MHz, therefore, = /4 = meters. htranhrcvr htranhrcvr (30 m)( m) From ; we find m 10 m 1,440 meters m Probem 3 Dopper Phase/Frequency Shifting (0 points) In eementary physics you ikey covered the Dopper shift in wave phenomena (such as with sound, water, ight and radio waves, etc.). Radio waves in ceuar communications are subject to the Dopper shift. [If you don t remember you can Googe Dopper Shift ).] A typica situation encountered in ce phones is where the mobie unit is moving such as in the iustration beow. For the case shown in the figure the car is moving at veocity v and moving from point A to point B in time t. It is moving toward the ceuar base station which is transmitting to the ce phone in the car. From the theory on the Dopper shift we know that moving toward the base station the frequency wi be appear to be higher because of waveength shortening from the motion. This is sometimes caed bue shift. If the car were moving in the opposite direction, then the frequency woud appear to be ower (this is caed red shift ). As the car moves from A to B it covers a distance d which effectivey shortens the radio signa path by distance. Note aso that the car is not moving directy at the antennas, but at an ange to the ine of signa propagation. Assume that ange is approximatey the same at both points A and B. (a) Show that the Dopper shift in phase is given by cos( ) in radians 3

4 We begin by noting dcos( ) and vt cos( ) because d vt. The change in phase from the Dopper shift is from the difference in path engths between the two vt paths. = cos( ) (b) Show that the Dopper shift change in frequency f is given by f cos( ) in Hz We begin by noting f, then we can write, t v vfc f= cos( ) cos t Probem 4 Dopper Shift Probem (30 points) This probem is a continuation of Probem 3 above. The base station is transmitting a sinusoida carrier signa f c = 1850 MHz. The vehice is moving at 75 mies per hour (mph). [Note: That might be exceeding the speed imit.] (a) Given that the car is far away from the base station, but moving directy at the transmitter (i.e., assumes = 0, so cos() = 1), find the received Dopper shifted carrier frequency as the car moves toward the base station. We begin with f 1,850 MHz. Therefore, the waveength is C 8 c 310 m/sec = = 0.16 meter 6 1 f sec C The vehice speed is 75 mph = m/sec When the car is moving toward the base station, the Dopper shift is positive, hence, m/sec f fc fdopper 1850 MHz 0.16 m f 1,850,000,07 Hz 4

5 (b) Now the car is moving away from the transmitter (again assumes = 0, so cos() = 1), find the received Dopper shifted carrier frequency as the car moves away from the base station. The waveength 0.16 meter but the car is moving in the opposite direction vehice speed is 75 mph = m/sec. When the car is moving away from the base station, the Dopper shift is negative, hence, m/sec f fc fdopper 1850 MHz 0.16 m f 1,849,999,793 Hz (c) The car is at the base station moving under it. What is the Dopper shifted carrier frequency under this condition? In this case 90 degrees; thus, cos( ) 0. There is no Dopper shift in the case where the car is didrecty under the base station antennas. But the ce phone may not receive a signa uness the antennas emit a signa downward. 5

Communication Systems

Communication Systems 1. A basic communication system consists of (1) receiver () information source (3) user of information (4) transmitter (5) channe Choose the correct sequence in which these are arranged