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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "traksrc.mws  20 Aug 95   \+
 An array of detectors tracks a rotating source of radio waves. [No de
tector phase delay.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 377 "A source of radio waves in the sky is rotating through
 the sky at an angular rate of 7.27*10^(-5) rad/sec at a distance of 4
.2 *10^9 m from a linear array of detectors.  The detector array is al
ong the y-axis, and the source passes through the x-axis at t=0. The s
pacing between detectors is 2.75 m,  the source wavelength is 0.21 m, \+
and the initial number of detectors is N=7." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 98 "a) Determine the distance L[i] \+
from the moving source to each detector, as a function of the time." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 291 "b) Sinc
e the radio waves oscillate around 10^9 times/sec, the distances from \+
source to detector elements change little in the time of one radio-fre
quency oscillation.  An easy way to sum up the detector responses is t
o select a time t when wt = 2*Pi*(integer).  Then from detector 1 we h
ave " }}{PARA 0 "" 0 "" {TEXT -1 344 "y = A cos(kL[1]-wt) =  A cos(kL[
1]-2*Pi*integer) = A cos(kL[1]).  At this instant of time it will be f
airly simple to add up the detector phasors into a resultant phasor. T
his resultant phasor rotates at around 10^9 times/sec, but its magnitu
de will change only slowly as all the L[i] gradually change, due to th
e source moving through the sky." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 152 "Find the resultant phasor amplitude by s
umming x and y components of the N detectors. This should be straightf
orward because you already have the L[i]. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 316 "c) Plot the phasor amplitude s
quared vs the time to observe the detector response as the source move
s across the sky. Let the time run from -1/2 hour to + 1/2 hour.  You \+
should see an expected peak at t=0. At what other times do you see a v
ery large peak, as large as the one at t=0?  Why do these other peaks \+
occur?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 29 "assume(k>0,wavelength>0,t>0);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 342 "The number of digits is set to 14 becaus
e of the 10^9 m distance used later in the problem. It turns out you n
eed at least 4 or 5 more digits available so that the calculated value
s of kL are accurate.  There is a compromise between accuracy and spee
d.  Using more digits improves accuracy, but calculations (and plots) \+
can take a lot longer." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 11 "Digits:=14:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "k:=2*Pi/wavelength; " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "Theta:=(727/100)*10^(-5)*t;" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 16 "X:=R*cos(Theta);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "Y:=R*sin(Theta);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "N:=7;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "m:
=trunc(N/2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 69 "Determine the distances L[i] from the source to each de
tector element" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "for i from 1 to N  do" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 47 "L[i]:=sqrt( (Y-spacing*(N-i-m))^2 + X^2);   od:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "R_x:=0:  R_y:=0:" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 102 "Let t=0 and get magnitude of resultant; first \+
obtain x and y components of the resultant, then square." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i from 1
 to N  do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "R_x:=R_x+A*cos(k*L[i])
;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "R_y:=R_y+A*sin(k*L[i]);       \+
od:" }}}{EXCHG {PARA 2 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "Rsquared:=R_x^2+R_y^2:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 52 "values:=\{R=1*10^9,A=1,spacing=11/4,wavelength=0.21\}
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
104 "Substitute problem values into the resultant vector squared. This
 will depend on time since theta does. " }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "g:=subs(values,Rsquared): " }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "plot(g,t=-1800..1800,numpoi
nts=200);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "Remarks." }}{PARA 0 "
" 0 "" {TEXT -1 652 "The max's are spaced about 1000 sec apart, or aro
und 15 min. [Actually, the angular velocity of the source is that of t
he rotating Earth. The apparent motion of the source is due to the Ear
th's rotation.  The source is placed beyond the Moon, but closer than \+
the Sun.  This is too close to be realistic, but it keeps the computat
ion time down for doing the plot.] The additional max's occur because \+
the detector spacing is much larger than a wavelength.  (We are gettin
g 'aliasing' of the detector response.) If the detector array were a t
ransmitter array, we would be seeing different 'orders' of interferenc
e from the 'grating' at the various angles." }}}}{MARK "16 1 0" 601 }
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