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1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Unit 1: Ordinary Different ial Equations - Part 1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Chapter 6: The Laplace Transform" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "Section 6.9: convolution \+ products by the convolution theorem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Copyright" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "Copyright * 2001 by Addis on Wesley Longman, Inc." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 302 "All rights reserved. No part of this publication m ay be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording , or otherwise, without the prior written permission of the publisher. Printed in the United States of America." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 15 "Initializations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "res tart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "with(inttrans):\nw ith(plots):\nwith(plottools):\nwith(student):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 49 "Comput ing Convolutions by the Convolution Theorem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 370 "Section 6.8 presented t he definition of the convolution product, and demonstrated how to obta in this product by performing one of two alternate versons of the conv olution integral. In addition, Section 6.8 revealed the convolution t heorem whereby the product of two Laplace transforms is the transform \+ of the convolution. In symbols we wrote both the \"forward\" version " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "L*[f*`*`*g]=F(s)*G(s)" "6#/*&%\"LG\"\"\"7#*(%\"fGF&%\"* GF&%\"gGF&F&*&-%\"FG6#%\"sGF&-%\"GG6#F0F&" }{TEXT -1 1 " " }}{PARA 0 " " 0 "" {TEXT -1 26 "and the \"backward\" version" }}{PARA 262 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L^`-1`*[F(s)*G(s)]=f*`*`*g" "6#/*&)%\"L G%#-1G\"\"\"7#*&-%\"FG6#%\"sGF(-%\"GG6#F.F(F(*(%\"fGF(%\"*GF(%\"gGF(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 34 "Moreover, Section 6.8 showed that " }{XPPEDIT 18 0 "f*`*` *g" "6#*(%\"fG\"\"\"%\"*GF%%\"gGF%" }{TEXT -1 29 ", the convolution pr oduct of " }{XPPEDIT 18 0 "f(t)" "6#-%\"fG6#%\"tG" }{TEXT -1 5 " and \+ " }{XPPEDIT 18 0 "g(t)" "6#-%\"gG6#%\"tG" }{TEXT -1 55 " could be obta ined as the inverse Laplace transform of " }{XPPEDIT 18 0 "F(s)*G(s)" "6#*&-%\"FG6#%\"sG\"\"\"-%\"GG6#F'F(" }{TEXT -1 43 ", the product of t he Laplace transforms of " }{XPPEDIT 18 0 "f(t)" "6#-%\"fG6#%\"tG" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "g(t)" "6#-%\"gG6#%\"tG" }{TEXT -1 133 ". The present section will again demonstrate how to use the conv olution theorem to compute the convolution product of two functions " }{XPPEDIT 18 0 "f(t)" "6#-%\"fG6#%\"tG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g(t)" "6#-%\"gG6#%\"tG" }{TEXT -1 173 ". In particular, we will be interested in cases where at least one of the two functions is a p iecewise-defined function, especially one in which Heaviside functions appear." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "When a factor in a convolution product is piecewise-defined, shift ing the argument from " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 4 " to \+ " }{XPPEDIT 18 0 "t-x" "6#,&%\"tG\"\"\"%\"xG!\"\"" }{TEXT -1 187 " in \+ the convolution integral can be subtle because such functions don't ha ve a single formula for their definition. The inequalities in the pie cewise-defined function and the inequality " }{XPPEDIT 18 0 "0 " 0 "" {MPLTEXT 1 0 32 "f := t^2;\ng := Heaviside(t - 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "The n, obtain " }{XPPEDIT 18 0 "F(s)*G(s)" "6#*&-%\"FG6#%\"sG\"\"\"-%\"GG6 #F'F(" }{TEXT -1 43 ", the product of the Laplace transforms of " } {XPPEDIT 18 0 "f(t)" "6#-%\"fG6#%\"tG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g(t)" "6#-%\"gG6#%\"tG" }{TEXT -1 9 ", getting" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "`L[f*g]` \+ := laplace(f,t,s)*laplace(g,t,s);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "Since this is the " }{TEXT 256 9 "transform" }{TEXT -1 98 " of the convolution, we \+ need only invert this transform to obtain the desired convolution prod uct." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "`f*g` := factor(invlaplace(`L[f*g]`, s, t));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "A graph of the ordinary product " }{XPPEDIT 18 0 "f(t)*g( t)" "6#*&-%\"fG6#%\"tG\"\"\"-%\"gG6#F'F(" }{TEXT -1 29 " and the convo lution product " }{XPPEDIT 18 0 "f*`*`*g" "6#*(%\"fG\"\"\"%\"*GF%%\"gG F%" }{TEXT -1 14 " is revealing." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 220 "p1 := plot(f*g, t=0..3, dis cont=true, color=red, thickness=3, title = `f(t) g(t)`):\np2 := plot(` f*g`, t = 0..3, color=green,thickness=3, title = `f * g`):\ndisplay(ar ray(1..2,[p1,p2]),xtickmarks=2, font=[TIMES,ROMAN,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "The ordinary product, namely, " }{XPPEDIT 18 0 "f(t)*g(t) = t^2 *Heaviside(t-1)" "6#/*&-%\"fG6#%\"tG\"\"\"-%\"gG6#F(F)*&F(\"\"#-%*Heav isideG6#,&F(F)F)!\"\"F)" }{TEXT -1 119 ", appears in red on the left. \+ The convolution, in green, appears on the right. The two products ar e indeed different." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 231 "Doing a convolution integral for piecewise-def ined functions is tricky. Maple allows the convolution integral to be computed directly in terms of Heaviside functions, but it still requi res care to get the answer to come out right." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "q := Int((t -x)^2*Heaviside(x-1),x=0..t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "q1 := value(q);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "The signum function is a mathem atical object that Maple understands and uses. A plot of the signum f unction is the best explanation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot(signum(x),x=-1..1);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "The best way to deal with the signum function in our ans wer is to convert it to Heaviside functions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "q2 := convert(q1, Heaviside);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Now, factoring yields the previous soluti on." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "factor(q2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 35 "Using the Definition o f Convolution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 " " {TEXT -1 46 "Direct evaluation of the convolution integral " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 267 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "L*[t^2*`*`*Heaviside(t-1)]=Int(Heaviside(x-1)*(t-x)^2,x =0..t)" "6#/*&%\"LG\"\"\"7#*(%\"tG\"\"#%\"*GF&-%*HeavisideG6#,&F)F&F&! \"\"F&F&-%$IntG6$*&-F-6#,&%\"xGF&F&F0F&*$,&F)F&F8F0F*F&/F8;\"\"!F)" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "requires a delicacy that starts with an analysis of the \+ integrand. First, it is far easier to shift " }{XPPEDIT 18 0 "t" "6#% \"tG" }{TEXT -1 4 " to " }{XPPEDIT 18 0 "t-x" "6#,&%\"tG\"\"\"%\"xG!\" \"" }{TEXT -1 17 " in the function " }{XPPEDIT 18 0 "t^2" "6#*$%\"tG\" \"#" }{TEXT -1 112 " than it is in the Heaviside function. Second, be cause of the Heaviside function, the integrand is nonzero for " } {XPPEDIT 18 0 "x-1*`>`*0" "6#,&%\"xG\"\"\"*(F%F%%\">GF%\"\"!F%!\"\"" } {TEXT -1 54 ". Third, the very definition of integration requires " } {XPPEDIT 18 0 "0`*0" "6#*(%\"tG\"\"\"%\">GF%\"\"!F%" }{TEXT -1 18 ", the integral is " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 268 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int((t-x)^2,x=1..t)=1/3 " "6#/-% $IntG6$*$,&%\"tG\"\"\"%\"xG!\"\"\"\"#/F+;F*F)*&F*F*\"\"$F," }{TEXT -1 1 " " }{XPPEDIT 18 0 "(t-1)^3" "6#*$,&%\"tG\"\"\"F&!\"\"\"\"$" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "so the convolution is again" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 269 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "1/3" "6#*&\"\"\"F$\" \"$!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "t^2*`*`*Heaviside(t-1)=(t-1) ^3*Heaviside(t-1)" "6#/*(%\"tG\"\"#%\"*G\"\"\"-%*HeavisideG6#,&F%F(F(! \"\"F(*&,&F%F(F(F-\"\"$-F*6#,&F%F(F(F-F(" }{TEXT -1 1 " " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 160 "The hard part of \+ this computation is sketching Figure 6.31 which shows the region of in tegration. The tools Maple provides for this task are now illustrated . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "Th e region of integration is that portion of the " }{XPPEDIT 18 0 "xt" " 6#%#xtG" }{TEXT -1 34 "-plane satisfying the inequalities" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 270 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "0`*0" "6#,&%\"xG\"\"\"*(F%F%%\">GF%\"\"!F%! \"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 45 "The first is from the meaning of the integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 272 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[t^2*`*`*Heaviside(t-1)]=Int(Heaviside(x-1)*(t-x)^2,x=0..t)" " 6#/*&%\"LG\"\"\"7#*(%\"tG\"\"#%\"*GF&-%*HeavisideG6#,&F)F&F&!\"\"F&F&- %$IntG6$*&-F-6#,&%\"xGF&F&F0F&*$,&F)F&F8F0F*F&/F8;\"\"!F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 125 " and the second is from the Heaviside function in the integrand. Where this inequality is valid, the Heaviside function is 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "To create Figure 6.3 1, execute the following Maple code, the essence of which is the " } {TEXT 262 7 "inequal" }{TEXT -1 83 " command which plots a region in t he plane satisfying a set of linear inequalities." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 373 "p1:=inequa l(\{t>x,x-1>0\}, x=0..3,t=0..3,\noptionsfeasible=(color=yellow),\nopti onsexcluded=(color=white),\noptionsopen=(color=black,thickness=3), xti ckmarks=3,ytickmarks=3, labels=[x,t]):\np2:=arrow([1,2.5],[2.47,2.5],. 07,.2,.1,color=black):\np3:=textplot(\{[1.2,.95,`(1,1)`],[2.3,2,`t = x `]\},font=[TIMES,ROMAN,10]):\ndisplay([p||(1..3)], labels=[x,t],labelf ont=[TIMES,ITALIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "The integration requires 0 < \+ " }{TEXT 257 1 "x" }{TEXT -1 3 " < " }{TEXT 258 1 "t" }{TEXT -1 8 ", s o if " }{TEXT 259 1 "t" }{TEXT -1 11 " < 1, then " }{TEXT 260 1 "x" } {TEXT -1 9 " < 1 and " }{XPPEDIT 18 0 "Heaviside(x-1)=0" "6#/-%*Heavis ideG6#,&%\"xG\"\"\"F)!\"\"\"\"!" }{TEXT -1 46 ". Hence, the integral \+ yields the value 0 for " }{TEXT 261 1 "t" }{TEXT -1 15 " < 1. In fact ," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "t^2" "6#*$%\"tG\"\"#" }{TEXT -1 20 " * Heaviside(t-1) \+ = " }{XPPEDIT 18 0 "Int((t-x)^2*Heaviside(x-1),x=0..t)" "6#-%$IntG6$*& ,&%\"tG\"\"\"%\"xG!\"\"\"\"#-%*HeavisideG6#,&F*F)F)F+F)/F*;\"\"!F(" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "PIECEWISE([0, t < 1],[Int((t-x)^2,x = 1 .. t), 1 <= t])" "6#-%*PIECEWISEG6$7$\"\"!2%\"tG\"\"\"7$-%$IntG6$*$ ,&F)F*%\"xG!\"\"\"\"#/F1;F*F)1F*F)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 11 "Since, for " }{XPPEDIT 18 0 "t>=1" "6#1\" \"\"%\"tG" }{TEXT -1 2 ", " }}{PARA 259 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "Int((t-x)^2,x=1..t)=(t-1)^3/3" "6#/-%$IntG6$*$,&%\"tG\" \"\"%\"xG!\"\"\"\"#/F+;F*F)*&,&F)F*F*F,\"\"$F2F," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "we have" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "t^2 " "6#*$%\"tG\"\"#" }{TEXT -1 3 " * " }{XPPEDIT 18 0 "Heaviside(t-1)" " 6#-%*HeavisideG6#,&%\"tG\"\"\"F(!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "PIECEWISE([0, t < 1],[(t-1)^3/3, 1 <= t])" "6#-%*PIECEWISEG6$7$\"\" !2%\"tG\"\"\"7$*&,&F)F*F*!\"\"\"\"$F/F.1F*F)" }{TEXT -1 2 " " }} {PARA 260 "" 0 "" {TEXT -1 50 " \+ = " }{XPPEDIT 18 0 "(t-1)^3/3" "6#*&,&%\"tG\"\"\"F&!\"\"\"\"$ F(F'" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Heaviside(t-1)" "6#-%*HeavisideG 6#,&%\"tG\"\"\"F(!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 179 "Computing a convolution in which one of the factors is piecewise defined requires careful analysis of the inequalities in volved. It is far simpler to use the convolution theorem." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Example 6.35" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Obtain the conv olution product" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 273 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Heaviside(t-1)" "6#-%*HeavisideG6#,&%\" tG\"\"\"F(!\"\"" }{TEXT -1 3 " * " }{XPPEDIT 18 0 "Heaviside(t-2)" "6# -%*HeavisideG6#,&%\"tG\"\"\"\"\"#!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "by using (a) the Con volution Theorem, and (b) the defining integral for convolution." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 26 " By the Convolution Theorem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "To use the convolution theorem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 274 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[f*`*`*g]=F(s)*G(s)" "6#/*&%\"LG\"\"\"7#*(%\"fGF&%\"*GF&%\"gGF &F&*&-%\"FG6#%\"sGF&-%\"GG6#F0F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 13 "or actually, " }}{PARA 275 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "f*`*`*g=L^`-1`*[F(s)*G(s)]" "6#/*(%\"fG\"\"\"%\"*GF&%\" gGF&*&)%\"LG%#-1GF&7#*&-%\"FG6#%\"sGF&-%\"GG6#F2F&F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 6 "obtain" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "F := laplace(Heaviside(t-1) ,t,s);\nG := laplace(Heaviside(t-2),t,s);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "The resulti ng convolution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 276 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Heaviside(t-1)*`*`*Heaviside(t-2)=L^`-1 `*[exp(-3*s)/s^2]" "6#/*(-%*HeavisideG6#,&%\"tG\"\"\"F*!\"\"F*%\"*GF*- F&6#,&F)F*\"\"#F+F**&)%\"LG%#-1GF*7#*&-%$expG6#,$*&\"\"$F*%\"sGF*F+F** $F=F0F+F*" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "(t-3)*Heaviside(t-3)" " 6#*&,&%\"tG\"\"\"\"\"$!\"\"F&-%*HeavisideG6#,&F%F&F'F(F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "is \+ computed by applying the Second Shifting Law in reverse to the transfo rm " }{XPPEDIT 18 0 "L*[t]=1/s^2" "6#/*&%\"LG\"\"\"7#%\"tGF&*&F&F&*$% \"sG\"\"#!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 69 "In Maple, the multiplication and inversio n steps would be as follows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "`f * g` := factor(invlaplace (F*G,s,t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 27 "By the Convolution Integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "The integr and of the convolution integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 277 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(Heaviside(x-1)*He aviside(t-x-2),x=0..t)" "6#-%$IntG6$*&-%*HeavisideG6#,&%\"xG\"\"\"F,! \"\"F,-F(6#,(%\"tGF,F+F-\"\"#F-F,/F+;\"\"!F1" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 34 "is nonzero for that region in the " }{XPPEDIT 18 0 "xt" "6#%#xtG" }{TEXT -1 32 "-pla ne in which the inequalities" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 278 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x-1*`>`*0" "6#,&%\"xG\"\"\" *(F%F%%\">GF%\"\"!F%!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 279 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "t-x-2*`>`*0 " "6#,(%\"tG\"\"\"%\"xG!\"\"*(\"\"#F%%\">GF%\"\"!F%F'" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 146 "are \+ satisfied. This region is shaded in Figure 6.32, created below. A re presentative arrow indicates a path of integration for a fixed value o f " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 25 ". The integrand is 0 i f " }{XPPEDIT 18 0 "t<3" "6#2%\"tG\"\"$" }{TEXT -1 32 ". Hence, the i ntegral is 0 for " }{XPPEDIT 18 0 "t<3" "6#2%\"tG\"\"$" }{TEXT -1 7 ". For " }{XPPEDIT 18 0 "t*`>`*0" "6#*(%\"tG\"\"\"%\">GF%\"\"!F%" } {TEXT -1 23 " the integrand is 1 for" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 280 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "1 " 0 "" {MPLTEXT 1 0 37 "q := Heaviside(x-1)*Heavi side(t-x-2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "presents a significant challenge. Maple \+ finds the integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "`H1 * H2` := Int(q,x=0..t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 "difficult to interpret since it cannot easily unravel the tang led skein of conditionals hiding in this integral. Indeed," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "v alue(`H1 * H2`);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "returns unevaluated." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "In the " }{XPPEDIT 18 0 "xt" "6#%#xtG" }{TEXT -1 103 "-plane, the region over which the i ntegration takes place can be visualized via Figure 6.32, created by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 400 "p1:=inequal(\{t-2>x,x-1>0\}, x=0..5,t=0..7,\noptionsfeasible= (color=yellow),\noptionsexcluded=(color=white),\noptionsopen=(color=bl ack,thickness=3), xtickmarks=5,ytickmarks=7, labels=[x,t]):\np2:=arrow ([1,5],[2.91,5],.07,.2,.1,color=black):\np3:=textplot(\{[1.4,2.8,`(1,3 )`],[4.7,6,`x = t - 2`]\},font=[TIMES,ROMAN,10]):\ndisplay([p||(1..3)] , scaling=constrained, labels=[x,`t `],labelfont=[TIMES,ITALIC,12]); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "The convolution integral is evaluated by first fixin g " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 38 ", and then integrating \+ along the line " }{XPPEDIT 18 0 "t=constant" "6#/%\"tG%)constantG" } {TEXT -1 8 " in the " }{XPPEDIT 18 0 "xt" "6#%#xtG" }{TEXT -1 9 "-plan e. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "T he integrand is 0 if " }{XPPEDIT 18 0 "t<3" "6#2%\"tG\"\"$" }{TEXT -1 32 ". Hence, the integral is 0 for " }{XPPEDIT 18 0 "t<3" "6#2%\"tG\" \"$" }{TEXT -1 37 ". This is expressed by the function " }{XPPEDIT 18 0 "Heaviside(t-3)" "6#-%*HeavisideG6#,&%\"tG\"\"\"\"\"$!\"\"" } {TEXT -1 30 " that appears in the solution " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "`f * g`;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "For " }{XPPEDIT 18 0 "t>3" "6#2\"\"$%\"tG" }{TEXT -1 28 " \+ the integrand is 1 between " }{XPPEDIT 18 0 "x=1" "6#/%\"xG\"\"\"" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "x=t-2" "6#/%\"xG,&%\"tG\"\"\"\"\"#! \"\"" }{TEXT -1 9 ". Hence," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "int(1,x=1..t-2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "That's where the " }{XPPEDIT 18 0 "t-3" "6#,&%\"tG\"\"\"\"\"$! \"\"" }{TEXT -1 28 " comes from in the answer. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Example 6.36" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "Obtain the convolu tion product of two unit pulses of duration " }{XPPEDIT 18 0 "tau=1" " 6#/%$tauG\"\"\"" }{TEXT -1 18 ", one starting at " }{XPPEDIT 18 0 "t=1 " "6#/%\"tG\"\"\"" }{TEXT -1 21 " and one starting at " }{XPPEDIT 18 0 "t=3" "6#/%\"tG\"\"$" }{TEXT -1 98 ". These pulses, graphed in Figu re 6.33, below, are represented in terms of Heaviside functions as" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "f := Heaviside(t-1) - Heaviside(t-2);\ng := Heaviside(t-3) - Hea viside(t-4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "Figure 6.33 is generated by" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "plo t([f,g], t=0..6, discont=true, color=black, scaling=constrained, xtick marks=6, ytickmarks=2, labels=[t,``],labelfont=[TIMES,ITALIC,12]);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 26 "By the Convolution Theorem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "To compute the convolution \+ by Laplace transforms and the convolution theorem, obtain" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 283 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[f(t)]=(exp(-s)-exp(-2*s))/s" "6#/*&%\"LG\"\"\"7#-%\"fG6#%\"tGF&* &,&-%$expG6#,$%\"sG!\"\"F&-F/6#,$*&\"\"#F&F2F&F3F3F&F2F3" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 284 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[g(t)]=(exp(-3*s)-exp(-4*s))/s" "6#/*&%\"LG\"\" \"7#-%\"gG6#%\"tGF&*&,&-%$expG6#,$*&\"\"$F&%\"sGF&!\"\"F&-F/6#,$*&\"\" %F&F4F&F5F5F&F4F5" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 15 "by using Maple:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "F := laplace(f,t, s);\nG := laplace(g,t,s);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Then, invert the product of \+ transforms" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 285 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[f(t)]*L*[g(t)]=(exp(-4*s)-2*exp(-5*s)+exp( -6*s))/s^2" "6#/**%\"LG\"\"\"7#-%\"fG6#%\"tGF&F%F&7#-%\"gG6#F+F&*&,(-% $expG6#,$*&\"\"%F&%\"sGF&!\"\"F&*&\"\"#F&-F36#,$*&\"\"&F&F8F&F9F&F9-F3 6#,$*&\"\"'F&F8F&F9F&F&*$F8F;F9" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "The Second Shifting Law a pplied in reverse to the transform " }{XPPEDIT 18 0 "L*[t]=1/s^2" "6#/ *&%\"LG\"\"\"7#%\"tGF&*&F&F&*$%\"sG\"\"#!\"\"" }{TEXT -1 36 " then giv es the desired convolution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "In Maple, this is done via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "`f * g` := \+ collect(invlaplace(F*G,s,t), Heaviside);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "Moreover, by converting to a piecewise format we can see the convolution product d efined by subintervals. Thus," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "convert(`f * g`, piecewise); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "A graph of the convolution " }{XPPEDIT 18 0 "f*`*`*g " "6#*(%\"fG\"\"\"%\"*GF%%\"gGF%" }{TEXT -1 46 " is found in Figure 6. 34, produced in Maple by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 142 "plot(`f * g`, t = 0..7, thickness= 3, color = red, scaling=constrained, xtickmarks=7, ytickmarks=2, label s=[t,``],labelfont=[TIMES,ITALIC,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 27 "By the Conv olution Integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Evaluating the convolution integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 286 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "f* `*`*g=Int([Heaviside(x-1)-Heaviside(x-2)]*[Heaviside(t-x-3)-Heaviside( t-x-4)],x=0..t)" "6#/*(%\"fG\"\"\"%\"*GF&%\"gGF&-%$IntG6$*&7#,&-%*Heav isideG6#,&%\"xGF&F&!\"\"F&-F06#,&F3F&\"\"#F4F4F&7#,&-F06#,(%\"tGF&F3F4 \"\"$F4F&-F06#,(F>F&F3F4\"\"%F4F4F&/F3;\"\"!F>" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "requires \+ knowing where, in the " }{XPPEDIT 18 0 "xt" "6#%#xtG" }{TEXT -1 13 "-p lane, both " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g(t-x)" "6#-%\"gG6#,&%\"tG\"\"\"%\"xG!\"\"" }{TEXT -1 53 " are simultaneously nonzero. Hence, the inequalities" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 287 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x*`>`*1" "6#*(%\"xG\"\"\"%\">GF%F%F%" }{TEXT -1 1 " " }}{PARA 288 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x<2" "6#2%\"xG\"\"#" } {TEXT -1 1 " " }}{PARA 289 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "t-x* `>`*3" "6#,&%\"tG\"\"\"*(%\"xGF%%\">GF%\"\"$F%!\"\"" }{TEXT -1 1 " " } }{PARA 290 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "t-x<4" "6#2,&%\"tG\" \"\"%\"xG!\"\"\"\"%" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 148 "must all be satisfied. This region is \+ shaded in Figure 6.35, generated below. Horizontal lines represent ty pical paths of integration along lines " }{XPPEDIT 18 0 "t=constant" " 6#/%\"tG%)constantG" }{TEXT -1 27 ". Any such line for which " } {XPPEDIT 18 0 "t<4" "6#2%\"tG\"\"%" }{TEXT -1 4 " or " }{XPPEDIT 18 0 "t*`>`*6" "6#*(%\"tG\"\"\"%\">GF%\"\"'F%" }{TEXT -1 90 " yields a zero integrand, and a value of zero for the convolution. Any such line be tween " }{XPPEDIT 18 0 "t=4" "6#/%\"tG\"\"%" }{TEXT -1 5 " and " } {XPPEDIT 18 0 "t=5" "6#/%\"tG\"\"&" }{TEXT -1 42 " yields an integrand of 1 and a value for " }{XPPEDIT 18 0 "f*`*`*g" "6#*(%\"fG\"\"\"%\"*G F%%\"gGF%" }{TEXT -1 41 " which will be determined by the integral" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 291 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "Int(1,x=1..t-3)=t-4" "6#/-%$IntG6$\"\"\"/%\"xG;F',&%\"t GF'\"\"$!\"\",&F,F'\"\"%F." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "Any such line between " } {XPPEDIT 18 0 "t=5" "6#/%\"tG\"\"&" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "t=6" "6#/%\"tG\"\"'" }{TEXT -1 47 " also yields an integrand of 1 a nd a value for " }{XPPEDIT 18 0 "f*`*`*g" "6#*(%\"fG\"\"\"%\"*GF%%\"gG F%" }{TEXT -1 41 " which will be determined by the integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 292 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(1,x=t-4..2)=6-t" "6#/-%$IntG6$\"\"\"/%\"xG;,&%\"tGF'\"\"%!\" \"\"\"#,&\"\"'F'F,F." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "These values are consistent with the re sults from the convolution theorem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 82 "To determine the region of integration s hown in Figure 6.35, we again use Maple's " }{TEXT 263 7 "inequal" } {TEXT -1 18 " command to obtain" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 192 "inequal(\{x>1,x<2,t-x-3>0,t -x-4<0\}, x=0..5, t=0..7, optionsfeasible=(color=red), optionsopen=(co lor=blue,thickness=3), optionsexcluded=(color=yellow), labels=[x,t], l abelfont=[TIMES,ROMAN,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "in which the support of th e integrand is the red parallelogram whose bounding edges are the line s " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 293 "" 0 "" {XPPEDIT 18 0 "x=1,x=2,x=t-3,x=t-4" "6&/%\"xG\"\"\"/F$\"\"#/F$,&%\"tGF%\"\"$!\"\"/F$ ,&F*F%\"\"%F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 192 "and whose vertices are the points (1,4), (2,5), (2,6), and (1,5). The red parallelogram is the region where t he integrand has the value 1. Outside this region, the integrand has \+ the value 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "We then embellish the figure to create Figure 6.35, as follows. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 634 "p5 := polygon([[1,4],[2,5],[2,6],[1,5],[1,4]], color =yellow, thickness=3):\np7 := plot(\{[[0,4.5],[5,4.5]],[[0,5.5],[5,5.5 ]]\},color=black):\np8 := plot([x,x+4,x=0..6.5],color=black, thickness =3):\np9 := plot([x,x+3,x=0..3.5],color=black, thickness=3):\np10:= pl ot([1,x,x=0..7],color=black, thickness=3):\np11:= plot([2,x,x=0..7],co lor=black, thickness=3):\np6 := textplot(\{[1.65,3.75,`(1,4)`], [2.7,4 .9,`(2,5)`], [1.9,6.1,`(2,6)`], [.85,5.1,`(1,5)`], [4.5,5.7,`t = 5.5`] , [4.5,4.2,`t = 4.5`]\}, align=LEFT):\ndisplay([p||(5..11)], view=[0.. 5,0..7], labels=[x,t], labelfont = [TIMES,ITALIC,12], scaling=constrai ned, xtickmarks=5, ytickmarks=7);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Integration t akes place along a line " }{XPPEDIT 18 0 "t=constant" "6#/%\"tG%)const antG" }{TEXT -1 27 ". Any such line for which " }{XPPEDIT 18 0 "t<4" "6#2%\"tG\"\"%" }{TEXT -1 4 " or " }{XPPEDIT 18 0 "t>6" "6#2\"\"'%\"tG " }{TEXT -1 90 " yields a zero integrand, and a value of zero for the \+ convolution. Any such line between " }{XPPEDIT 18 0 "t=4 " "6#/%\"tG \"\"%" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "t=5" "6#/%\"tG\"\"&" } {TEXT -1 42 " yields an integrand of 1 and a value for " }{XPPEDIT 18 0 "f*`*`*g" "6#*(%\"fG\"\"\"%\"*GF%%\"gGF%" }{TEXT -1 41 " which will \+ be determined by the integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "q1 := Int(1,x=1..t-3);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "which evaluates to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "q2 := value(q1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Any such line between " }{XPPEDIT 18 0 "t=5" "6#/%\"tG\"\"&" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "t=6" "6#/%\"tG\"\"'" }{TEXT -1 63 " also yields an integrand of 1 and a value for the convolution " } {XPPEDIT 18 0 "f*`*`*g" "6#*(%\"fG\"\"\"%\"*GF%%\"gGF%" }{TEXT -1 41 " which will be determined by the integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "q3 := Int(1,x=t-4..2 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 18 "which evaluates to" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "q4 := value(q3);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Hence, a piecewise definition of the convolution would be " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "q3 := piecewise(t<=4,0, t<=5,q2, t<=6,q4, t>6,0);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "which is precisely what we got by use of the convolution \+ theorem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}}{MARK "1" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }