<Text-field style="Heading 1" size="14" layout="Heading 1"><Font size="18"> Inverse Laplace transform, pag. 392</Font></Text-field>
<Text-field style="_cstyle1" layout="_pstyle2">#3. shifting.</Text-field>DO #5 Replacing s+1 by s=(s+1)-1 multiplies f(t) by 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 .with(inttrans):
with(student):
L:=(s+1)/(s^2+2*s+10);
numer(L)/completesquare(denom(L));subs(s+1=s,%);At this point, we could do a table look up instead of invlaplace.exp(-t)*invlaplace(%,s,t);Comparing:invlaplace(L,s,t);
<Text-field style="_cstyle1" layout="_pstyle2">#11. partial fractions.</Text-field>DO # 12 First, we do the partial fraction decomposition using Maple's identity command. We write out the partial fraction expansion (denominators are linear, so the numerators are constants), then make the equation into an identity in s. Finally, we solve for a, b, and c.restart:
a:='a':
b:='b':
c:='c':
L:=(s^2-26*s-47)/((s-1)*(s+2)*(s+5));
eq:=L=a/(s-1)+b/(s+2)+c/(s+5);
identity(%,s);
csol:=solve(%,{a,b,c});
subs(csol,eq);Alternatively, we can use the convert( ,parfrac,s) command.convert(L,parfrac,s);
<Text-field style="_cstyle1" layout="_pstyle2">#19. partial fractions.</Text-field>DO # 16restart:
L:=1/((s-3)*(s^2+2*s+2));
Check irreducibility: 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 does not factor. factor(op(2,denom(L)));Since the second factor in the denominator ,op(2,denom),is a quadratic, irreducible over the rationals, i.e., and will not decompose into a product of linear factors , the numerator 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 is of degree one.eq:=L=a/(s-3)+(b*s+c)/(s^2+2*s+2);
identity(%,s);
csol:=solve(%,{a,b,c});
subs(csol,eq);Comparing with the output of the Maple command:convert(L,parfrac,s);
<Text-field style="_cstyle1" layout="_pstyle2">#24. inverse Laplace transform from partial fractions.</Text-field>DO #25restart:
with(inttrans):
L:=(7*s^2-41*s+84)/((s-1)*(s^2-4*s+13));
PF:=convert(L,parfrac,s);
We separate the two terms and, by the linearity of the Laplace transform, take their inverse transforms separately. Note that op(1, ) selects the first operand. In this case, the expression is a sum, so the first operand is the first term.first:=op(1,PF);
second:=op(2,PF);We complete the square on the term with the quadratic, then express the numerator in terms of (s-2) as well.with(student):
completesquare(denom(second));taylor(numer(second),s=2);Replacing s-2 by s=(s-2)+2 multiplies f(t) by 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 . (Note that in the following we need to copy the numerator in by hand: although it appears to be a polynomial, it is really a Taylor series for which the "big O" term is zero. Hence,the expand command will not split the fraction into two terms, each over its own denominator, as it does with polynomials.)(-15+2*s)/(s^2+9);
At this stage, we could use a table look up, rather than invlaplace.exp(2*t)*invlaplace(%,s,t)+invlaplace(first,s,t);
Comparing with the output from Maple invlaplace command:invlaplace(L,s,t);
<Text-field style="_cstyle1" layout="_pstyle2">#33. inverse Laplace, derivative by s.</Text-field>DO 34 Taking a derivative with respect to s divides f(t) by -t.restart:
F:=ln((s+20)/(s-5));
diff(%,s):
expand(%);
convert(%,parfrac,s);At this stage, we could do a table look up, rather than invlaplace.(-t)^(-1)*invlaplace(%,s,t);Comparing with the output from Maple's invlaplace command:invlaplace(F,s,t);