Laplace Tranform.
Motivation:
To solve ODE’s is not easy! Many times we do not know how to solve them
with the methods that we have been taught. For example, do you know how to
solve:
y 00 + 2y 0 + y = sin t cos t,
y(0) = y 0 (0) = 0?
I do not know!! However in this course, we will have a look to simpler differential
equations than that one! (with constant coefficients and more diverse variety of
input functions that in previous courses).
The Laplace Transform in an alternative tool of finding solutions of ODE’s
for a given initial value problem as we will see later on.
Definition 1 Suppose f : [0, ∞) −→ R is a function of t. Then its Laplace
transform is:
Z
∞
F (s) = L(f (t))(s) =
e−st f (t)dt
(1)
0
In the future, we will not write L(f (t))(s) but L(f (t)), we are lazy and we do
not write the s any more... We assume that f (t) is such that this integral exists.
For example f (t) verifies the following growth restriction property:
|f (t)| < M ekt ,
for some M, k ∈ R. (‘f (t) goes slower to infinity than the exponential ekt ’). We
will not worry about it though. You will not be asked to verify any growth
restriction because all the functions we will work with will be good enough....
Note: This is enough for the Laplace transform to exist if s > k because a way
R∞
to see that the integral (1) is convergent is checking that | 0 e−st f (t)dt| < ∞.
1
Let us prove that under the growth restriction hypothesis this is verified:
|
R∞
0
e−st f (t)dt| <
R∞
0
|e−st f (t)|dt using properties of the integral
R∞
|e−st ||f (t)|dt
R∞
< M 0 |e−st ||ekt |dt, using the growth restriction
R∞
= M 0 e(k−s)t dt, the exponential is always positive
h
it=∞
M
(k−s)t
=
e
= M < ∞,
k−s
s−k
t=0
=
0
provided that s > k (because that would do that limt→∞ e(k−s)t = 0).
By definition, we say that the inverse Laplace transform of F (s) is f (t) and it
follows that
L−1 (L(f (t))) = f (t)
and
L(L−1 (F (s)) = F (s)
Remark: The Laplace transform of f (t) is a function of s.
Remark: To solve the integral (1) you need to think of s as a constant.
Example 1 Computing the Laplace transform of f (t) = C = constant
L(C) =
R∞
0
e−st Cdt = C
R∞
0
h
i
C e−st t=∞ = C
e−st dt = −s
s
t=0
Example 2 Computing the Laplace transform of f (t) = t, t > 0.
L(t) =
=
=
=
=
R∞
e−st tdt
h0 −st it=∞
1 R ∞ e−st dt, using integration by parts
e t
+
s 0
−s t=0
R
1 ∞ e−st dt
i
hs 0
−1 e−st ∞
s2
t=0
1 , using Example 1 with C = 1
s2
2
Example 3 Show using induction method that
L(tn ) =
n!
(2)
sn+1
In Example 2 we prove that
L(t1 ) =
1
,
s2
that trivially verifies equation (2) for n = 1. Let us suppose it is true for all
0, 1, 2, ..., n Does it happens for n + 1? The hypothesis the induction is
L(tn ) =
n!
R∞
e−st tn+1 dt
it=∞
h0 −st
1 R ∞ e−st tn dt,
+ n+
= e−s tn+1
s
0
t=0
R
1 ∞ e−st tn dt = n + 1 n! ,
= n+
s
0
s1 sn+1
L(tn+1 ) =
(3)
sn+1
using integration by parts
where we have used the hypothesis of induction (3). The result follows since
(n + 1)n! = (n + 1)!
Example 4 Computing the Laplace transform of f (t) = eat , t > 0.
t=∞
Z ∞
Z ∞
1 (a−s)t
1
−st at
(a−s)t
e e dt =
e
dt =
e
,
=
a−s
s−a
0
0
t=0
provided that a < s because if a < s then e(a−s)t → 0 as t → ∞.
Theorem 1 Linearity of the Laplace transform
If f (t), g(t) are any two functions whose Laplace transform exist and a, b ∈ R.
Then the Laplace transform for af (t) + bg(t) exists and
L(af (t) + bg(t)) = aL(f (t)) + bL(g(t))
Proof: A very easy one :o)
R∞
e−st (af (t) + bg(t))dt, by definition of Laplace transform
R∞
R∞
= a 0 e−st f (t)dt + b 0 e−st g(t)dt by linearity of the integral
L(af (t) + bg(t)) =
0
= aL(f (t)) + bL(g(t)),
3
by definition again
Example 5 Recall:
eat − e−at
2
(4)
eat + e−at
cosh at =
2
(5)
sinh at =
and
We are ready to compute their Laplace transforms just using Theorem 1 and
Example 4!
at
−at
L(sinh at) = L( e −2 e )
= 12 (L(eat ) − L(e−at ))
1 − 1
= 12 s −
a s+a
s + a − (s − a)
= 12
(s − a)(s + a)
= 2 a 2
s −a
Similarly (Give it a try!)
s
L(cosh at) = 2
s − a2
Theorem 2 The First Shifting Theorem.
If f (t) has Laplace transform F (s) then the function eat f (t) has Laplace transform
F (s − a). Thus:
L(eat f (t)) = F (s − a),
and
L−1 (F (s − a)) = eat f (t).
Proof: Easy easy easy....
F (s − a) =
R∞
=
R∞
0
0
e−(s−a)t f (t)dt by definition
e−st eat f (t)dt,
using that ex+y = ex ey
= L(eat f (t))
4
Example 6 Find the Laplace transform of e2t t2 . We know that:
F (s) = L(t2 ) =
2
s3
(Example 3, n = 2). Then using Theorem 2 we get straight away that
L(e2t t2 ) = F (s − 2) =
2
(s − 2)3
Example 7 Find the Laplace transform of e−3t sinh t.
We know that
L(sinh t) =
1
s −1
2
(Example 5). Then using Theorem 2 we get straight away that
F (s + 3) = L(e−3t sinh t) =
1
.
(s + 3)2 − 1
15 .
s2 − 4
Notice that since the Laplace transform is linear (Theorem 1) then its inverse
Example 8 Computing the inverse Laplace transform of
Laplace transfom is linear too!. Then:
L−1 (
15
1
15
) = 15L−1 ( 2
)=
sinh 2t
2
s −4
s −4
2
Example 9 Computing the inverse Laplace transform of 2s −3 10 .
s
L−1 (
2s − 10
1
2
) = 2L−1 ( 2 ) − 5L−1 ( 3 ) = 2t − 5t2 ,
3
s
s
s
using Example 3 with n = 1 and n = 2.
For now on in your mind when we mention piecewise continuous function you
think in a function that has finitely many jumps (see Figure 1)
To see the precise definition you could read Erwin Kreyszig book pag 226.
(9th edition).
The next Theorem will be vital to solve ODE’s of first and second order using
Laplace transform.
5
10
8
6
4
2
-4
-2
2
4
-2
-4
Figure 1: A piecewise function with 3 jumps.
Theorem 3 Transform of Derivatives.
1. Suppose f is continuous satisfying the growth restriction
|f (t)| < M ekt ,
for some M, k ∈ R,
and f 0 is piecewise continuous. Then
L(f 0 (t)) = sL(f (t)) − f (0).
2. Suppose f, f 0 are continuous and f satisfies the growth restriction
|f (t)| < M ekt ,
for some M, k ∈ R,
and f 00 is piecewise continuous. Then:
L(f 00 (t)) = s2 L(f (t)) − sf (0) − f 0 (0).
Proof: We prove 1 assuming f 0 is continuous (otherwise we could split [0, ∞) in
intervals in which f 0 is so).(The hypotheses mentioned in the theorem assure the
6
existence of the integral that follow).
R∞
L(f 0 (t)) = 0 e−st f 0 (t)dt by definition
R∞
t=∞
= [f (t)e−st ]t=0 + s 0 e−st f (t)dt (†)
= −f (0) + sL(f (t)),
notice that we have used the growth restriction in (†) which implies that
f (t)e−st −→ 0, when k < s and t −→ ∞. To prove 2, we use 1 twice :o)
L(f 00 (t)) = L((f 0 )0 (t)) = sL(f 0 (t)) − f 0 (0)
= s(sL(f (t)) − f (0)) − f 0 (0)
= s2 L(f (t)) − sf (0) − f 0 (0).
Do not worry about the hypothesis of f, f 0 , f 00 being continuous, piecewise continuous ect. We will not check it in the problem sheets and the exams, we will
just use the theorem directly because the functions I will give you will be good
ebough ;o)
You can check Erwin Kreyszig book pag 228 (9th edition) for computing
transforms of derivatives of any order. (But we will not use it).
Example 10 The Laplace transform of cos at and sin at.
f (t) = cos at,
f 0 (t) = −a sin at,
f 00 (t) = −a2 cos at
f 0 (0) = 0
f (0) = 1,
First observe that since the Laplace transform is linear we have
L(f 00 (t)) = L(−a2 cos at) = −a2 L(cos at) = −a2 L(f (t))
(6)
On another hand using Theorem 3
L(f 00 (t)) = s2 L(f (t)) − sf (0) − f 0 (0)
= s2 L(cos at) − s − 0
= s2 L(cos at) − s
7
(7)
But (6) and (7) are the same thing, so we conclude:
s2 L(cos at) − s = −a2 L(cos at),
hence
L(cos at) =
s
s + a2
2
Wanna try to calculate L(sin at) using the same methodology? You sould obtain:
L(sin at) =
a
s + a2
2
Give a try to the following examples before checking the solutions!
More examples.
Exercise 1: Compute the inverse Laplace transform of
i) 4s2 − 3π2 .
s +π
4
2
ii) s − 3s5 + 12 .
s
iii)
10√ .
2s + 2
iv)
√ 1
√ .
(s − 3)(s + 5)
Solutions:
i)
L−1 4s2 − 3π2
= 4L−1 2 s 2 − 3L−1 2 π 2
s +π
s +π
s +π
= 4 cos πt − 3 sin πt, via the tables
ii)
−1
L
s4 − 3s2 + 12
s5
=
=
=
=
=
2
4
s
−1 s
−1 4 · 3
L
− 3L
5 +L
5
s5
s
s
L−1 s − 3L−1 13 + L−1 4 ·53
s s
1
2
1
−1 1
−1
−1 4 · 3 · 2
L
3 + 2L
5
s − 32L
s
s
L−1 1s − 32 L−1 2!3 + 21 L−1 4!5
s
s
3
1
2
4
1 − 2t + 2t
−1
8
iii)

L−1
10√
2s + 2


1
1√ 
2(s +
2)
2 


= 10L−1 

1
−1 
= 10
2L 
1√ 
s+
2
2


= 5L−1 

√
1√ 
 = 5e− 22 t
2
s+
2
iv) In this case we have to do partial fractions decomposition because we are
not able to identify via the tables any function with this Laplace transform:
A√ + B√
s− 3 √
s+ 5
√
5)
3)√
A(s
+
B(s
−
√
√ +
√
(s − 3)(s + 5) (s − 3)(s + 5)
√
√
1 = A(s + 5) + B(s − 3)
√ 1
√
=
3)(s + 5)
√ 1
√
=
(s − 3)(s + 5)
(s −
Therefore :
√
√
3, we get A = √ 1 √ and setting s = − 5 we get B =
3+ 5
√ −1√ . Then:
3+ 5
Setting s =
(s −
Hence:
−1
L
√
1
3)(s +
√ 1
√
(s − 3)(s + 5)
√
5)
=√
1
1
1
√ (
√ −
√ ),
3+ 5 s− 3 s+ 5
1
1
1
−1
−1
√ L
√
√
= √
−L
3+ 5
s− 3
s+ 5
√
√
= √ 1 √ (e 3t − e− 5t )
3+ 5
Exercise 2: Use the First Shifting Theorem to compute the following inverses
Laplace transforms: (Notice that denominators would have no real roots.)
i)
s−6
.
(s − 1)2 + 4
ii)
15
s2 + 4s + 29
9
iii)
2s − 56 .
s2 − 4s − 12
i) Notice that:
s−6
s−1−5
=
(s − 1)2 + 4
(s − 1)2 + 4
s−1
5
=
−
= F (s − 1) − 25 G(s − 1),
2
2
(s − 1) + 4 (s − 1) + 4
where:
F (s) =
s
= L(cos 2t)
s +4
G(s) =
2
= L(sin 2t),
s +4
2
and
2
therefore by the First Shiffting Theorem:
s
−
6
−1
L
= L−1 (F (s − 1)) − 25 L−1 (G(s − 1))
(s − 1)2 + 4
= et cos 2t − 52 et sin 2t,
ii) First we observe that:
s2 + 4s + 29 = s2 + 4s + 4 + 25 = (s + 2)2 + 52 ,
Then:
15
15
=
s2 + 4s + 29
(s + 2)2 + 52
5
= 3
(s + 2)2 + 52
= 3F (s − (−2)),
where F (s) = L(sin 5t). Using the First Shifting Theorem we get:
−1
L
15
2
s + 4s + 29
= 3L−1 (F (s − (−2))) = 3e−2t sin 5t
iii) First notice that we can rewrite the denominator as:
s2 − 4s − 12 = s2 − 4s + 4 − 16 = (s − 2)2 − 16,
10
2s − 4
52
2s − 56
=
−
s2 − 4s − 12
(s − 2)2 − 16 (s − 2)2 − 16
2(s − 2)
52
−
= 2F (s − 2) − 13G(s − 2),
=
(s − 2)2 − 42 (s − 2)2 − 42
where
F (s) = cosh 4t and
G(s) = sinh 4t
Hence using the First Shifting theorem again!
L−1
2s − 56
s2 − 4s − 12
= 2L−1 (F (s − 2)) − 13L−1 (G(s − 2))
= 2e2t cosh 4t − 13e2t sinh 4t
Differential equations with initial value problems.
Suppose a, b ∈ R. Consider the following ODE:
y 00 + ay 0 + by = r(t) (F),
y(0) = K0 , y 0 (0) = K1 ,
K0 and K1 constants.
To solve it we follow the next steps:
1. Apply Laplace Transform to both side of the equation (F) and use Theorem
3.
2. You get a subsidiary equation that can be solved algebraically to give you
Y (s) = Laplace transform of y(t)
3. Use inverse transform in Y (s) to identify y(t).
Note: When you get Y (s) you may get a fraction that has to be split in simpler
fractions to be able to identify the inverse Laplace transform.
I think the best we can do now is to solve an example, don’t you think??
Example 11 Solve:
y 0 = 1,
y(0) = −2,
11
using Laplace transform method:
First we take Laplace transforms to both sides of the equation to get:
1
sY (s) − y(0) = ,
s
1
sY (s) − y(0) = ,
s
using Example 1 and Theorem 3. But since y(0) = −2
Y (s) =
1
2
2 −
s
s
Having a look at the tables we conclude that the inverse Laplace transform of
1 is t and the inverse laplace transform of −2 is −2, hence the solution to our
s
s2
ODE is:
y(t) = t − 2.
Are you ready for another one?
Example 12 Solve y 00 +5y 0 +6y = 0,
y(0) = 1 = y 0 (0), using Laplace transform
method:
Taking Laplace transforms to both sides of the equation (Note that L(0) = 0):
s2 Y (s) − sy(0) − y 0 (0) + 5(sY (s) − y(0)) + 6Y (s) = 0,
Y (s)(s2 + 5s + 6) = s + 6
Then:
Y (s) =
s+6
,
s + 5s + 6
2
but we cannot identify this fraction with anything in the table! :o(
We have to use partial fractions decomposition to split the fraction in two which
are simpler and we may recognize from the tables:
Y (s) =
s+6
A
B
=
+
s+2 s+3
s + 5s + 6
2
12
Therefore (Check it!)
B = −3
A = 4,
So
Y (s) =
4
3
−
,
s+2 s+3
Hence the solution of our ODE:
y(t) = 4e−2t − 3e−3t .
Example 13 Consider:
y 00 − 4y 0 + 3y = 6t − 8
with intial conditions y(0) = y 0 (0) = 0. If we write Y = L(y(t)) the Laplace
transform is:
s2 Y − 4sY + 3Y
(s2 − 4s + 3)Y
Y
8
6
−
2
s
s
6
8
= 2−
s
s
=
=
s2 (s2
8
6
−
2
− 4s + 3) s(s − 4s + 3)
Now we have to put this into a form which allows us to take the inverse transform.
The second term isn’t so bad. Since s2 − 4s + 3 = (s − 1)(s − 3) we write:
1
A
B
C
=
+
+
s(s − 1)(s − 3)
s
s−1 s−3
1 = (s − 1)(s − 3)A + s(s − 3)B + s(s − 1)C
Choosing s = 0 gives A = 1/3, s = 1 gives B = −1/2 and choosing s = 3 gives
C = 1/6. Thus:
s(s2
1
1
1
1
=
−
+
− 4s + 3)
3s 2(s − 1) 6(s − 3)
The other expansion is harder because it has a repeated root: in
s2 (s
1
− 1)(s − 3)
13
the s factor appears as a square. To deal with this you have to include a 1/s
term and a 1/s2 term in the partial fraction expansion.
s2 (s
1
A B
D
C
=
+ 2+
+
− 1)(s − 3)
s
s
s−1 s−3
1 = s(s − 1)(s − 3)A + (s − 1)(s − 3)B + s2 (s − 3)C + s2 (s − 1)D
Now taking s = 0 gives B = 1/3, s = 1 gives C = −1/2 and s = 3 gives
D = 1/18. There is no convenient choice of s that gives A on its own, so we just
substitute in any other value, s = 2 say and by putting in the values of B, C and
D we get:
1 = −2A −
2
1
+2+
3
9
and hence A = 4/9. Thus:
s2 (s
4
1
1
1
1
=
+ 2−
+
− 1)(s − 3)
9s 3s
2(s − 1) 18(s − 3)
Now we can put everything together:
s2 (s
1
4
1
1
1
=
+ 2−
+
− 1)(s − 3)
9s 3s
2(s − 1) 18(s − 3)
Now we can put everything together:
Y
1
1
1
4
+ 2−
+
= 6
9s 3s
2(s − 1) 18(s − 3)
1
1
1
−8
−
+
3s 2(s − 1) 6(s − 3)
and if we do the algebra we find:
Y =
1
2
1
+
−
s2 s − 1 s − 3
which means that
y = 2t + et − e3t
and you can check that this does solve the original equation.
14
Examples with Complex Numbers
We have seen that using Laplace transforms to solve differential equations
usually comes down to a partial fraction expansion. For some differential equation
this expansion will involve complex numbers. We have two ways of solving these
sort of differential equations. The first one is treating the complex numbers as if
they were real. (Using partial fraction decomposition). This not only complicates
find the partial fraction expansion, it also causes difficulty when you try to turn
a solution which looks complex but is actually real, into one that looks real. The
second method uses The First Shifting Theorem as in More Examples above. We
will see both methods in the following example:
Example 14 Consider the example of trying to solve:
y 00 + 2y 0 + 5y = 1
with y(0) = y 0 (0) = 0. Taking the Laplce transform of each side we get:
(s2 + 2s + 5)Y (s) =
1
s
Hence:
Y (s) =
s(s2
1
+ 2s + 5)
We now need to factorize s2 + 2s + 5. Solving using the formula gives:
√
−2 ± 4 − 20
= −1 ± 2i
s=
2
so
s2 + 2s + 5 = (s + 1 + 2i)(s + 1 − 2i)
Next, partial fractions. This part is no different from the examples without
complex numbers, but it is trickier.
1
A
B
C
= +
+
s(s + 1 − 2i)(s + 1 + 2i)
s
s + 1 − 2i s + 1 + 2i
15
Multiplying across by the denominator gives:
1 = A(s + 1 − 2i)(s + 1 + 2i) + Bs(s + 1 + 2i) + Cs(s + 1 − 2i)
Choosing s = 0 gives:
1 = 5A
and hence A = 1/5. Next, s = −1 + 2i gives:
1 = B(−1 + 2i)(4i) = −8 − 4i
so
B=−
1
8 + 4i
Putting all this together we get
Y (s) =
1
1
1
1
1
−
−
5s 8 + 4i s + 1 − 2i 8 − 4i s + 1 + 2i
and, using the tables, this gives:
y(t) =
1
1
1
−
e−(1−2i)t −
e−(1+2i)t
5 8 + 4i
8 − 4i
Although this does tell us what y is, it does it in a complicated way. For a
start, this makes it look like y is complex, when we know that y satisfies a real
differential equation and should be real. To rewrite this in a real form we need
to do two things, we change the fractions with complex numbers on the bottom
to fractions with complex numbers on the top and we expand the exponentials
using the formula:
eiθ = cos θ + i sin θ
e−iθ = cos θ − i sin θ
16
Note that the second of these formulas follows from the first using:
cos (−θ) = cos θ
sin (−θ) = − sin θ
First the fractions, remember there is a standard method for dividing by a
complex number: you multiply above and below by the conjugate. Hence:
1
1 8 − 4i
=
8 + 4i
8 + 4i 8 − 4i
which makes sense because the second fraction is equal to one. Now
(8 + 4i)(8 − 4i) = 82 − (4i)2 = 64 + 16 = 80
and so
1
8 − 4i
2−i
=
=
8 + 4i
80
20
You can do the same with the other complex fraction, it is quicker just to note
it is the same as the one we just did except the sign of i is different, so:
1
8 + 4i
2+i
=
=
8 − 4i
80
20
Now we have:
y(t) =
1 2 − i −(1−2i)t 2 + i −(1+2i)t
−
e
−
e
5
20
20
Next we use the fact:
e−(1−2i)t = e−t e2it
e−(1+2i)t = e−t e−2it
and so
f=
1 1
−
(2 − i)e2it − (2 + i)e−2it e−t
5 20
17
Next, writing the imaginary exponentials out, we find:
[(2 − i)(cos 2t + i sin 2t) + (2 + i)(cos 2t − i sin 2t)] e−t
y(t) =
1
5
−
1
20
=
1
5
−
1
(2 cos 2t
20
+ 2i sin 2t − i cos 2t + sin 2t + 2 cos 2t − 2i sin 2t + i cos 2t + sin 2t)e−t
=
1
5
−
1
(2 cos 2t
10
+ sin 2t)e−t
Multiply out the brackets, all the i bits cancel and we are left with
y(t) =
1
1
− (2 cos 2t + sin 2t)e−t
5 10
Example 15 Consider the above example again:
y 00 + 2y 0 + 5y = 1
with y(0) = y 0 (0) = 0. Taking the Laplce transform of each side we get:
(s2 + 2s + 5)Y (s) =
1
s
Hence:
Y (s) =
1
s(s2 + 2s + 5)
First we complete squares to observe that: s2 + 2s + 5 = (s + 1)2 + 22 ,
Y (s) =
1
A
Bs + C
= +
,
2
2
s((s + 1) + 2 )
s
(s + 1)2 + 22
notice that in the second partial fraction numerator we have written a polynomial
of degree 1: Bs + C because the denominator is a polynomial of degree 2: (s +
1)2 + 22 .
1
A((s + 1)2 + 22 )
Bs2 + Cs
As2 + 2As + 5A
Bs2 + Cs
=
+
=
+
,
s((s + 1)2 + 22 )
s((s + 1)2 + 22 ) s((s + 1)2 + 22 )
s((s + 1)2 + 22 ) s((s + 1)2 + 22 )
then:
1 + 0s + 0s2 = As2 + 2As + 5A + Bs2 + Cs,
we conclude:
1 = 5A,
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A=
1
5
(A + B)s2 = 0s2 ⇒ B = −
(C + 2A) = 0 ⇒ C =
1
5
−2
5
therefore:
1
1
=
Y (s) =
s((s + 1)2 + 22 )
5
1
s+2
−
s (s + 1)2 + 22
1
=
5
1
s+1
1
−
−
s (s + 1)2 + 22 (s + 1)2 + 22
,
if we define:
F (s) = L(1),
G(s) = L(cos 2t) H(s) = L(sin 2t)
1
1
1 1
s+1
1
1
= (F (s)−G(s+1)− H(s+1)),
Y (s) =
=
−
−
2
2
2
2
2
2
s((s + 1) + 2 )
5 s (s + 1) + 2
(s + 1) + 2
5
2
Taking inverses Laplace transforms and using First Shifting theorem you get:
y(t) =
1
1 1 −t
− e cos 2t − e−t sin 2t
5 5
10
which is exactly the same result as in the example above!
Example 16 Say you wanted to use the Laplace transform to solve the equation:
d2 y
dy
+ 2 + 2y = 1
2
dt
dt
dy
with initial conditions y(0) =
(0) = 0. Taking the Laplace transform of both
dt
sides we get.
1
(s2 + 2s + 2)Y (s) =
s
Hence,
Y (s) =
s(s2
1
+ 2s + 2)
and we can factor the denominator using the roots
√
√
√
−2 ± 4 − 8
−2 ± −4
s=
=
= −1 ± −1 = −1 ± i
2
2
19
of the quadratic. So complex partial fractions take the form
1
A
B
C
= +
+
s(s − (−1 + i))(s − (−1 − i))
s
s+1−i s+1+i
1 = A(s + 1 − i)(s + 1 + i) + Bs(s + 1 + i) + Cs(s + 1 − i)
Y (s) =
= A(s2 + 2s + 2) + Bs(s + 1 + i) + Cs(s + 1 − i)
s=0:
1 = 2A
1
2
1 = 0 + B(−1 + i)(2i) + 0 = 2B(−i − 1) = −2(1 + i)B
A =
s = −1 + i :
1−i
1
=−
−2(1 + i)
2(1 + i)(1 − i)
1−i
1 i
= −
=− +
4
4 4
B =
s = −1 − i :
1 = 0 + 0 + C(−1 − i)(−2i) + 0 = 2C(i − 1) = 2(−1 + i)C
1
−1 − i
=
2(−1 + i)
2(−1 + i)(−1 − i)
1+i
1 i
= −
=− −
4
4 4
= B
11
1 i
1
1 i
1
Hence: Y (s) =
+ − +
+ − −
2s
4 4 s+1−i
4 4 s+1+i
C =
Via the tables, we have:
1 i
1 i
1
(−1+i)t
e
+ − −
e(−1−i)t
y(t) =
+ − +
2
4 4
4 4
1 i
1
1 i
−t −it
=
+ − +
e e + − −
e−t e−it
2
4 4
4 4
1
1 i
1 i
−t
=
+ − +
e (cos t + i sin t) + − −
e−t (cos t − i sin t)
2
4 4
4 4
e−t
1 e−t
=
+
(− cos t − sin t − i sin t + i cos t) +
(− cos t − sin t + i sin t − i cos t)
2
4
4
1 e−t
=
−
(cos t + sin t)
2
2
20
Try to do this example using partial fraction decomposition and the First Shifting
Theorem....
In the future you could use either method!
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