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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Detect0.mws  20 Aug 95   s
tudent 'starter' version " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 18 "PROBLEM STATEMENT." }}{PARA 0 "" 0 "" {TEXT -1 
1075 "A source of radio waves in the sky is detected by a line of N eq
ually-spaced detectors along the y-axis.  When the source has a positi
ve y-coordinate, the detector with the greatest y-value will be closes
t to the source. Signals from the source will reach this detector (det
ector # 1) first.  Signals reaching the second detector will be out of
 phase with those in detector #1, and so on with the other detectors. \+
 The incoming waves look like y = A cos (kL -wt). The phase of each wa
ve is the argument of cosine, namely phase = kL-wt.  At any time t, ev
ery detector 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 th
e source.  To get a big signal from the array, all detector signals mu
st be in phase, so a phase delay D is introduced between detectors. Wh
en D is set correctly, it compensates for the phase difference due to \+
different L values, and brings the signals back into phase. [D is subt
racted from the phase of dectector #2, 2D is subtracted from the phase
 of detector #3, and so forth.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 247 "In this problem, we calculate the result
ant of all 'phasors' from the detectors, and plot its square vs. the p
hase delay D.  The magnitude of the resultant is the amplitude of the \+
total detector response, something we do not prove in this problem." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 281 "This (d
etect0.ms) is a 'starter' program, intended for students to learn from
 and build on.  The source is initially placed at theta=0 (it is on th
e x-axis). The array output will be a maximum when the phase delay D i
s zero, because waves from the source are all arriving in phase." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 228 "The radi
o source in this problem is located at a distance of 10^9 m from the o
rigin, and its wavelength is 0.21 m. The detector spacing is 0.10 m, a
nd initially there are N=7 detectors.  Later in the problem A is set e
qual 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' s
tatement. Failing to do this often makes the program stop executing, a
nd you have to close 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) Initi
al run-through." }}{PARA 0 "" 0 "" {TEXT -1 220 "Run through the progr
am detect0.ms, with N=7 detectors in the array.  In the plot of detect
or 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 ar
ray size on ability 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.  This value should be less than when you had 7 detector \+
elements.  Your larger array should be able to locate a source more pr
ecisely than before." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 57 "c) Path differences between detectors are quite constan
t." }}{PARA 0 "" 0 "" {TEXT -1 86 "Now move the source in the sky to t
heta=0.600 radians, label its detector response s1." }}{PARA 0 "" 0 "
" {TEXT -1 141 "Substitute numerical values into the distances L[i], a
nd call these LVal[i]. They are distances from the new source to the d
etector elements." }}{PARA 0 "" 0 "" {TEXT -1 106 "Create DeltaL[i], w
hich are differences between LVal[i], and display them. They should be
 rather constant." }}{PARA 0 "" 0 "" {TEXT -1 123 "Create DDeltaL[i], \+
which are differences between DeltaL[i], and display them. They should
 be tiny compared to a wavelength." }}{PARA 0 "" 0 "" {TEXT -1 113 "Re
port the sizes of the path differences.  Compare this to W sin theta, \+
where W is the spacing 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 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 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 dis
tances L[i] from the source 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 46 "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 m
agnitude of resultant; first obtain x and y components of the resultan
t, 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 38 "R_y:=R_y+A*sin(k*L[i]-D*(i-1)):    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 "" {TEXT -1 0 "" }}{PARA 2 "" 0 "" 
{TEXT -1 1 "\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "values:=\{R=1*10^
9,A=1,spacing=0.10,wavelength=0.21\};" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "Substitute problem values
 into the resultant vector squared, except for theta value" }}{PARA 2 
"" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "g:=subs(v
alues,Rsquared): " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "Now create t
he response function from source #1 in the sky" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "s1:=subs(Theta=0.00
0,g):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "plot(s1,D=0..2*Pi,
numpoints=200);" }}}}{MARK "0 0 0" 10 }{VIEWOPTS 1 1 0 1 1 1803 }