% 2250-notes-F2008-ch10.txt
% Edit F2010
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10.1-22
Please use Laplace table 10.1.2 to find the transform of
f(t)=sinh^2(3t). The idea is to write
sinh(w) = (exp(w) - exp(-w))/2
and then expand f(t) in terms of exponential functions. Because all
terms of f(t) are in the basic table, this problem can be done without
integrations of any kind.
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10.1-28
This problem is about use of the backward table, that is, use table 10.1.2
backwards or right-to-left. The details would look like
3s + 1
F(s) = --------
s^2 + 4
s 1 2
= 3 ----- + - -----
s^2+4 2 s^2+4
= 3 L(cos 2t) + L(0.5 sin 2t)
= L(3 cos 2t + 0.5 sin 2t)
Lerch's Theorem, or the inverse transform, finishes the problem.
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10.2-6
See the web slides for how to solve this problem and others like it.
You should get
(s^2+4) L(x(t)) = L(cos t)
s
L(x(t)) = -----------------
(s^2 + 1)(s^2 + 4)
Then do partial fractions and apply the backwards table. End with
Lerch's theorem or the inverse transform, finding x(t).
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10.2-14
See 10.2 Example 3 for a similar problem. Here's what should appear
in the details:
1. Transform each differential equation to get a 2x2 linear
system of algebraic equations for the unknowns L(x) and L(y).
2. Solve the system by Cramer's rule for L(x) and L(y).
3. Find x(t) from the formula for L(x), using the backward table,
partial fractions, and the various Laplace rules. The last step
should be Lerch's theorem or take an inverse transform.
4. Find y(t) from the formula for L(y), in the same way.
There are web examples of the method in the Laplace theory manuscript.
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10.2-22
While it is possible to solve for f(t) in this example using Theorem 2
of section 10.2, you are advised to choose a more efficient partial
fraction solution. Because the fraction in this problem has denominator
s(s-3)(s+3), having distinct roots 0, 3, -3, Heavisides coverup method
is efficient to find the partial fraction expansion.
1
L(f(t)) = ---------
s(s-3)(s+3)
A B C
= --- + --- + ---
s s-3 s+3
= A L(1) + B L(exp(3t)) + C L(exp(-3t))
= L(A + B exp(3t) + C exp(-3t))
Lerch's theorem says that the L-symbols cancel leaving
x(t) = A + B exp(3t) + C exp(-3t)
Now evaluate A, B, C by Heaviside's coverup method.
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10.3-6
Details:
s - 1
L(f(t)) = --------
(s+1)^3
u - 2 | [formally subst u=s+1]
= ------ |
u^3 | u=s+1 [prepare for the shifting theorem]
1 2 |
= ---- - --- |
u^2 u^3 | u=s+1
1 2 |
= ---- - --- |
s^2 s^3 | s ==> s+1 [write as a shift]
= L(t - t^2) | s ==> s+1 [backward table used]
= L((t-t^2)exp(-t)) by the first shifting theorem
Finish with Lerch's theorem or inverse transform.
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10.3-16
Details:
1
L(f(t)) = ---------------
(s+3)^2(s-2)^2
A B C D
= --- + ------- + --- + -------
s+3 (s+3)^2 s-2 (s-2)^2
[ A B ]| [ C D ]|
= [ - + ---]| + [ - + ---]|
[ u u^2]| u = s+3 [ u u^2]| u = s-2
[ A B ]| [ C D ]|
= [ - + ---]| + [ - + ---]|
[ s s^2]| s ==> s+3 [ s s^2]| s ==> s-2
= L(A+Bt)| s ==> s+3 + L(C+Dt)| s ==> s-2
= L((A+Bt)exp(-3t)) + L((C+Dt)exp(2t)) Shifting theorem
= L((A+Bt)exp(-3t) + (C+Dt)exp(2t)) Linearity of L
Apply Lerch's theorem or inverse transforms to find f(t). Then
evaluate A,B,C,D by partial fractions.
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10.4-22
The idea:
tf(t) = exp(t)-exp(-t)
L(tf(t)) = L(exp(t)-exp(-t))
(-d/ds)L(f(t)) = L(exp(t)-exp(-t))
(-d/ds)F(s) = 1/(s-1)-1/(s+1)
The latter is a quadrature differential equation in s for F(s)=L(f(t)).
Solve the quadrature equation. Then use the Theorem
lim L(f(t)) = 0 as s approaches infinity
to evaluate the constant of integration. Then L(f(t))=F(s) is found!
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10.4-28
F(s) = L(f(t)) as given is already a partial fraction, therefore
the theory of partial fractions is of no immediate help.
The simplistic plan is to use
L((-t)g(t)) = (d/ds)L(g(t))
to work out formulas for the Laplace transforms of the functions
t sin t, t cos t, t^2 cos t.
You will find out that F(s) is a linear combination of these answers.
Another way is to use the convolution theorem, although that involves
integration which is not so easy.
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10.5-4
The equation L(f(t))=(exp(-s)-exp(2-2s))/(s-1) is solved for f(t)
by using the second shifting theorem in one of the forms
L(h(t-a)step(t-a))=exp(-as)L(h(t)), valid for a >= 0.
L(g(t)step(t-a))=exp(-as)L(g(t+a)), valid for a >= 0.
DIVIDE AND CONQUER. Let L(f1(t)) = exp(-s)/(s-1) and
L(f2(t)) = -exp(2-2s)/(s-1).Then f =f1 + f2 is the answer.
SOLVE for f1. Because exp(-s)/(s-1) must match exp(-as)L(h(t)),
then exp(-s)=exp(-as) and a=1. Because 1/(s-1) must match L(h(t)),
then L(exp(t))=L(h(t))) and h(t)=exp(t). Replace t by t-a to obtain
h(t-a) = exp(t-a) = exp(t-1) [because a=1]
Then
L(f1)=exp(-s)/(s-1)=exp(-as)L(h(t)))=L(h(t-a)step(t-a))
implies
L(f1)=L(exp(t-1)step(t-1))
and finally
f1(t)=exp(t-1)step(t-1) [step(t)=u(t)=unit step=Heaviside]
The calculation for f2(t) is similar. A shortcut is to use the first
shifting theorem, then the second shifting theorem:
L(f2)=-exp(-2s)/s shifted s --> s-1
=-exp(-2s)L(1) shifted s --> s-1
=L((-1)step(t-2))
MAPLE ANSWER CHECK. The code:
with(inttrans):
F:=exp(-s)/(s-1)-exp(2-2*s)/(s-1);
invlaplace(F,s,t);
# ans: exp(t)*(Heaviside(2-t)-1)+exp(t-1)*(-Heaviside(1-t)+1)
# Mystery: why does this equal the answer reported above?
# Most hand computation gets exp(t-1)u(t-1)-exp(t)u(t-2)
laplace(exp(t-1)*Heaviside(t-1),t,s)-laplace(exp(t)*Heaviside(t-2),t,s);
Answer to mystery: Heaviside(2-t)-1 == Heaviside(t-2)
Heaviside(1-t)-1 == Heaviside(t-1)
================================================================
10.5-22
First, write the function f(t) in terms of pulses and then steps:
f(t) = t^3 pulse(t,1,2)
= t^3(step(t,1)-step(t,2))
L(f) = L(t^3 step(t-1))-L(t^3 step(t-2))
L(t^3 step(t-1)) = L(g(t)step(t-1))
= exp(-s) L(g(t+1)) second shifting theorem
= exp(-s) L((t+1)^3) [g(t)=t^3, g(t+1)=(t+1)^3]
= exp(-s) L(t^3+3t^2+3t+1)
L(t^3 step(t-2)) = L(g(t)step(t-2))
= exp(-2s) L(g(t+2)) second shifting theorem
= exp(-2s) L((t+2)^3) [g(t)=t^3, g(t+2)=(t+2)^3]
= exp(-2s) L(t^3+6t^2+12t+8)
MAPLE ANS CHECK:
#ans: 1/s^4*(-2*(4*s^3+6*s^2+6*s+3)*exp(-2*s)
+(s^3+3*s^2+6*s+6)*exp(-s))
with(inttrans):
laplace((t+1)^3,t,s);
laplace((t+2)^3,t,s);
laplace(t^3*Heaviside(t-1)-t^3*Heaviside(t-2),t,s);
================================================================
10.5-28
Use the periodic function theorem L(f)=P/(1-exp(-Ts)) where T=period
and P=int(f(t)exp(-st),t=0..T).
The period is T=2a. And f(t)=t for 0 <= t <= a, zero elsewhere.
Then P is the integral over [0,a] instead of [0,T].
The bottom of the fraction is 1-exp(-Ts) or 1-exp(-2as).
MAPLE COMPUTATION:
P:=int(t*exp(-s*t),t=0..a);
#ans: P := -(-1+exp(-s*a)+exp(-s*a)*s*a)/s^2
#the laplace package has no periodic function theorem support.
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