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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "Detect1.mws  20 Aug 95   \+
faculty version, with extra stuff done   [student 'starter' version is
 detect0.mws]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 18 "PROBLEM STATEMENT." }}{PARA 0 "" 0 "" {TEXT -1 1075 "A so
urce of radio waves in the sky is detected by a line of N equally-spac
ed detectors along the y-axis.  When the source has a positive x-coord
inate, the detector with the greatest y-value will be closest to the s
ource. Signals from the source will reach this detector (detector # 1)
 first.  Signals reaching the second detector will be out of phase wit
h those in detector #1, and so on with the other detectors.  The incom
ing waves look like y = A cos (kL -wt). The phase of each wave is the \+
argument of cosine, namely phase = kL-wt.  At any time t, every detect
or will have the same k, w, and t value, but the L's are all different
, so there is a phase difference between the signals from the source. \+
 To get a big signal from the array, all detector signals must be in p
hase, so a phase delay D is introduced between detectors. When D is se
t correctly, it compensates for the phase difference due to different \+
L values, and brings the signals back into phase. [D is subtracted fro
m the phase of dectector #2, 2D is subtracted from the phase of detect
or #3, and so forth.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 247 "In this problem, we calculate the resultant of all 'p
hasors' from the detectors, and plot its square vs. the phase delay D.
  The magnitude of the resultant is the amplitude of the total detecto
r response, something we do not prove in this problem." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 274 "detect0.ms is a 'st
arter' program, intended for students to learn from and build on.  The
 source is initially placed at theta=0 (it is on the x-axis). The arra
y output will be a maximum when the phase delay D is zero, because wav
es from the source are all arriving in phase." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 228 "The radio source in this
 problem is located at a distance of 10^9 m from the origin, and its w
avelength is 0.21 m. The detector spacing is 0.10 m, and initially the
re are N=7 detectors.  Later in the problem A is set equal to 1." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "Note: In
 'do' loops, be sure to close the loop with the 'od' statement. Failin
g to do this often makes the program stop executing, and you have to c
lose and then re-enter Maple." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 22 "PARTS OF THE PROBLEM :" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "a) Initial run-through.
" }}{PARA 0 "" 0 "" {TEXT -1 220 "Run through the program detect0.ms, \+
with N=7 detectors in the array.  In the plot of detector response vs \+
phase delay D, notice that the big response comes at D=0.  Record the \+
delay value when Rsquared first equals zero." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "b) Effect of array size on abil
ity to locate a source." }}{PARA 0 "" 0 "" {TEXT -1 295 "Increase the \+
number of detectors N to equal 15.  Rerun and replot s1.  Make a note \+
of the D value at which the detector response first becomes zero.  Thi
s value should be less than when you had 7 detector elements.  Your la
rger array should be able to locate a source more precisely than befor
e." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "c) \+
Path differences between detectors are quite constant." }}{PARA 0 "" 
0 "" {TEXT -1 86 "Now move the source in the sky to theta=0.600 radian
s, label its detector response s1." }}{PARA 0 "" 0 "" {TEXT -1 141 "Su
bstitute numerical values into the distances L[i], and call these LVal
[i]. They are distances from the new source to the detector elements.
" }}{PARA 0 "" 0 "" {TEXT -1 106 "Create DeltaL[i], which are differen
ces between LVal[i], and display them. They should be rather constant.
" }}{PARA 0 "" 0 "" {TEXT -1 123 "Create DDeltaL[i], which are differe
nces between DeltaL[i], and display them. They should be tiny compared
 to a wavelength." }}{PARA 0 "" 0 "" {TEXT -1 113 "Report the sizes of
 the path differences.  Compare this to W sin theta, where W is the sp
acing between detectors." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{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 because of the 10
^9 m distance used later in the problem. It turns out you need at leas
t 4 or 5 more digits available so that the calculated values of kL are
 accurate.  There is a compromise between accuracy and speed.  Using m
ore 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 
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 69 "Determine the distances L[i] from the sou
rce to each detector element" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 20 "for i from 1 to N do" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 45 "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 20 "for
 i from 1 to N do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "R_x:=R_x+A*cos
(k*L[i]-D*(i-1)):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "R_y:=R_y+A*sin
(k*L[i]-D*(i-1)):   od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "
Rsquared:=R_x^2+R_y^2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "v
alues:=\{R=1*10^9,A=1,spacing=0.10,wavelength=0.21\}:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 83 "Substitute problem values into the result
ant vector squared, except for theta value" }}{PARA 2 "" 0 "" {TEXT 
-1 1 "\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "g:=subs(values,Rsquared
): " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "Now create the response fu
nction from source #1 in the sky" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "s1:=subs(Theta=0.600,g):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "plot(s1,D=0..2*Pi,numpoints=
200);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 71 "for i from 1 to N do LVal[i]:=evalf(subs(values,Theta
=0.600,L[i])); od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "for i
 from 1 to N-1 do DeltaL[i]:=LVal[i+1]-LVal[i]; od:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 39 "0.10*sin(0.600):  # spacing  sin(theta)" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "for i from 1 to N-2 do DD
eltaL[i]:=DeltaL[i+1]-DeltaL[i]; od:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 8 "Remarks:" }}{PARA 0 "" 0 "" {TEXT -1 50 "Replace colons by semic
olons to see actual values." }}{PARA 0 "" 0 "" {TEXT -1 79 "The value \+
of W sin theta is extremely close to the path differences calculated.
" }{MPLTEXT 1 0 0 "" }}}}{MARK "20 2 1" 0 }{VIEWOPTS 1 1 0 1 1 1803 }