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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "PHYSICS RESOURCE PACKETS P
ROJECT                               File: ENVELOPE.MWS" }}{PARA 0 "" 
0 "" {TEXT -1 85 "Rose-Hulman Institute of Technology                 \+
                      Author: mjm" }}{PARA 0 "" 0 "" {TEXT -1 97 "http
://www.rose-hulman.edu/~moloney                                   Soft
ware: Maple V Release 4" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 118 "In AM (amplitude-modulated) radio, the amplitude of
 a high-frequency signal is modulated by the lower-frequency audio " }
}{PARA 0 "" 0 "" {TEXT -1 113 "information. AM radios use 'envelope de
tectors' involving RC circuits in order to remove the high-frequency (
the " }}{PARA 0 "" 0 "" {TEXT -1 44 "'carrier wave') and leave the aud
io signal. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 126 "This problem gives you a simple modulated signal. It also give
s you the 'rectified' signal we would see after passing through " }}
{PARA 0 "" 0 "" {TEXT -1 57 "a diode (which lets current flow in one d
irection only.)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 121 "You are to pass the signal through a series RC circuit, \+
first through a resistor R and then a capacitor C to ground. The " }}
{PARA 0 "" 0 "" {TEXT -1 60 "'envelope' signal is to be the voltage ac
ross the capacitor." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 79 "Letting v be the incoming signal, the equation to be so
lved is v -IR -q/C = 0 ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 128 "You must replace the current I by a function of q
, the charge on the capacitor. This (call it eq1) will be the generic \+
equation " }}{PARA 0 "" 0 "" {TEXT -1 122 "to be solved for q the char
ge on the capacitor as a function of time. By substituting values of R
 and C into eq1, you get " }}{PARA 0 "" 0 "" {TEXT -1 109 "eq2. This n
ew equation is to be solved numerically for q (the charge on the capac
itor) as a function of time." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 88 "a) Get the rectified signal first, by using the
 Heaviside function. The input signal is " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "     v := sin(96 Pi t) *sin(12000 \+
Pi t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
120 "b) Set up the differential equation, using q(t) for the charge on
 the capacitor C . C is connected to ground and to the " }}{PARA 0 "" 
0 "" {TEXT -1 96 "resistor R . R has the input voltage on one side and
 C on the other. [The current will be dq/dt]" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "c) Substitute values into the \+
DE to make a new DE which will be solved numerically. Start off by set
ting RC equal to 1/10 " }}{PARA 0 "" 0 "" {TEXT -1 39 "the period of t
he 'carrier frequency'. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 121 "d) Solve the new DE numerically. Plot (use odeplo
t) and see what this gives you for an envelope function Compare this t
o " }}{PARA 0 "" 0 "" {TEXT -1 36 "the plot of the modulated function.
 " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart; with(plots):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "th:=6000*2*Pi*t; # carrier f
requency is 6000 hz" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "z:=s
in(th*8/1000)*sin(th); # amplitude-modulated (AM) signal (modulated at
 48 hz)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "v:=z*Heaviside(z
); # rectified signal after going through a diode" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 67 "plot(v,t=0..0.015,numpoints=400, title=`plot
 of modulated signal`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "From h
ere on down, the students should be doing the work." }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 31 "eq1:=v-R*diff(q(t),t)-q(t)/C=0;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 31 "eq2:=subs(R=45,C=1/100000,eq1);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "RC is 0.45 ms. carrier pd= 0.17 ms
. modulation pd .= 20 ms." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "sol:=d
solve(\{eq2,q(0)=0\}, q(t),numeric); " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 88 "odeplot(sol,[t,q(t)*100000/1],0..15/1000,numpoints=40
0,title=`signal across capacitor`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 94 "Note that we don't get all of the signal back, nor is the shape
 a perfect fit to the envelope." }}}}{MARK "9 0 0" 2 }{VIEWOPTS 1 1 0 
1 1 1803 }