lt examples

LAPLACE TRANSFORMS

1. Basic transforms

In this course, Laplace Transforms will be introduced and their properties examined; a table of common trans-forms will be built up; and transforms will be used to solve some di�erential equations by transforming theminto algebraic equations which can be easily solved.Let f(t) be de�ned for t ≥ 0. Then we de�ne the Laplace Transform of f as

F (s) = L(f) =∫ ∞

0

e−stf(t) dt

We also de�ne the inverse transform L−1 byf = L−1F

Remark 1.1. L is an operator on the function f . Whereas f depends on the variable t, F isindependent of t.

Example. f(t) ≡ 1, t ≥ 0

F (s) =∫ ∞

0

e−st dt

= −1s

[e−st

]∞0

=1s

Thus we say that L(1) =1s

1.1. Exponentials. Example f(t) = eat, t ≥ 0, a

L(eat) =∫ ∞

0

e−steat dt

=∫ ∞

0

e(a−s)t dt

=1

a− s

[e(a−s)t

]∞0

=1

s− aif s > a.

Linearity of Laplace TransformLet f(t) and g(t) be de�ned for t ≥ 0 with transforms F (s) = L(f) and G(s) = L(g) respectively, and let a andb be constants. Then

L{af(t) + bg(t)} = aF (s) + bG(s)

1

2 LAPLACE TRANSFORMS

Proof.

L{af(t) + bg(t)} =∫ ∞

0

e−st[af(t) + bg(t)] dt

= a

∫ ∞

0

e−stf(t) dt + b

∫ ∞

0

e−stg(t) dt

= aF (s) + bG(s)

¤

Example. cosh t =eat + e−at

2

L(cosh t) =12{Leat + Le−at}

=12

{1

s− a+

1s + a

}

when s− a > 0 and s + a > 0. We summarise this as

L(cosh at) =s

s2 − a2s > a ≥ 0

1.2. Powers of t.

L(ta) =∫ ∞

0

e−stta dt

=∫ ∞

0

e−x(x

s

)a dx

ssetting x = st

=1

sa+1

∫ ∞

0

e−xxa dx

orL(ta) =

Γ(a + 1)sa+1

where Γ(a) is the Gamma function de�ned by

Γ(a) =∫ ∞

0

e−xxa−1 dx

for a > 0. Integrating by parts:

Γ(a + 1) =∫ ∞

0

e−xxa dx

= − [xae−x

]∞0

+ a

∫ ∞

0

e−xxa−1 dx

= aΓ(a)

Remark 1.2. Γ(1) =∫∞0

e−x dx = − [e−x]∞0 = 1

Let n be a positive integer. Then

Γ(n + 1) = nΓ(n)

= n(n− 1)Γ(n− 1)

= n(n− 1)(n− 2)Γ(n− 2)

...

= n(n− 1)(n− 2) . . . (3)(2)(1)Γ(1)

= n!

1. BASIC TRANSFORMS 3

Thus for a positive integer n we have

L(tn) =n!

sn+1

We also have Γ(1/2) =√

π.Even though the Gamma function is so far only de�ned for positive values of a, we can extend the de�nition to(some) negative values of a using the property Γ(a + 1) = aΓ(a). For example,

Γ(−1/2) =Γ(1/2)−1/2

= −2√

π

and

Γ(−3/2) =Γ(−1/2)−3/2

= 4√

π/3

However, Γ(1) = 0Γ(0) and so Γ(0) = 1/0 which is unbounded. Similarly Γ(n) is unbounded for all negative

integers n.

1.3. Trigonometric functions. We already have L(eat) =1

s− aand so

L(eiωt) =1

s− iω=

s + iω

s2 + ω2

=s

s2 + ω2+ i

ω

s2 + ω2

and since eiωt = cos ωt + i sin ωt we have

L(cos ωt + i sin ωt) =s

s2 + ω2+ i

ω

s2 + ω2

and so, equating real and imaginary parts, we have

L(cos ωt) =s

s2 + ω2

and

L(sin ωt) =ω

s2 + ω2

1.4. Step function.

f(t) =

k, 0 ≤ t < c;

0, c ≤ t

L{f(t)} =∫ ∞

0

e−stf(t) dt

=∫ c

0

ke−st dt = −k

s

[e−st

]c

0

=k

s[1− e−sc]


1.5. Laplace transforms of derivatives. Let f(t) be di�erentiable for t ≥ 0 with transform L{f(t)}.Then L{f ′(t)} =

∫∞0

e−stf ′(t) dt Integrating by parts, with

u = e−st dv = f ′(t) dt

du = −s e−st dt v = f(t)

⇒ Lf ′(t) =[f(t)e−st

]∞0

+ s

∫ ∞

0

e−stf(t) dt

= sLf(t)− f(0)

Similarly, if f is twice di�erentiable, then

L{f ′′(t)} = sLf ′(t)− f ′(0)

= s [sLf(t)− f(0)]− f ′(0)

= s2Lf(t)− sf(0)− f ′(0)

and in generalL{f (n)(t)} = snL{f(t)} − sn−1f(0)− sn−2f ′(0)− . . .− f (n−1)(0)

To verify this, we can see for example that if f(t) = t3, then

f(0) = f ′(0) = f ′′(0) = 0 f ′′′(t) = 6

Then

L{f ′′′(t)} = L(6) = 6/s

= s3L{f(t)} − s2f(0)− sf ′(0)− f ′′(0) = s3L(t3)

and soL(t3) = 6/s4 = 3!/s4

as already seen.Example (Initial Value Problem)

y′′ + 5y′ + 6y = 0 y(0) = 2, y′(0) = 3

Let Y (s) = L{y(t)} Then

L{y′(t)} = sY − y(0) = sY − 2

L{y′′(t)} = s2Y − sy(0)− y′(0) = s2Y − 2s− 3

and transforming the equation gives

s2Y − 2s− 3 + 5(sY − 2) + 6Y = 0

or(s2 + 5s + 6)Y = 2s + 13

which is called the subsidiary equation. This has solution

Y =2s + 13

(s + 3)(s + 2)

which now must be inverted to obtain the solution y(t).


Using partial fractions

Y =9

s + 2− 7

s + 3and thus

y = L−1(Y ) = 9L−1{ 1s + 2

} − 7L−1{ 1s + 3

} = 9e−2t − 7e−3t

1.6. The shifting theorems. To solve more complicated problems we need the following.

Theorem 1.3. (First Shifting Theorem.) If L{f(t)} = F (s) for s > γ; then L{eatf(t)} = F (s − a) fors > γ + a

Proof. F (s) =∫∞0

e−stf(t) dt

F (s− a) =∫ ∞

0

e−(s−a)tf(t) dt

=∫ ∞

0

e−steatf(t) dt

= L{eatf(t)}

¤

Example. L(1) = 1/s = F (s) for s > 0. Thus L(eat) = F (s− a) = 1/(s− a) for s > a as seen already.

Example.y′′ + 2y′ + 5y = 0, y(0) = 2, y′(0) = −4

Transforming ⇒

s2Y − 2s + 4 + 2(sY − 2) + 5Y = 0

⇒ (s2 + 2s + 5)Y = 2s

⇒ Y =2s

(s + 1)2 + 22

= 2s + 1

(s + 1)2 + 22− 2

(s + 1)2 + 22

From before, we know that

L−1{ s

s2 + 22} = cos 2t

and

L−1{ 2s2 + 22

} = sin 2t

and so, using the �rst shifting theorem this implies

L−1{ s + 1(s + 1)2 + 22

} = e−t cos 2t

and

L−1{ 2(s + 1)2 + 22

} = e−t sin 2t

and so y = L−1(Y ) = e−t[2 cos 2t− sin 2t]


The analogue of the �rst shifting theorem for inverse transforms is the second shifting theorem. This makes useof the unit step function:

ua(t) =

0, t < a

1, t ≥ a

This is also sometimes written as u(t− a) or as the Heaviside function Ha(t) or H(t− a).

Theorem 1.4. (Second shifting theorem) If L−1(F (s)) = f(t), then

L−1(e−asF (s)) = f̃(t) =

0, t < a

f(t− a), t ≥ a

= f(t− a)ua(t)

Proof.

e−asF (s) = e−as

∫ ∞

0

e−szf(z) dz

=∫ ∞

0

e−s(a+z)f(z) dz

t = a + z =∫ ∞

a

e−stf(t− a) dt

=∫ ∞

0

e−stf(t− a)ua(t) dt

= L[f(t− a)ua(t)]

¤

Example. L−1

[e−3s

s2 + 1

]Since L−1

[1

s2 + 1

]= sin t, the second shifting theorem implies that

L−1

[e−3s

s2 + 1

]= sin(t− 3)u3(t)

=

0, t < 3

sin(t− 3), t ≥ 3

Remark 1.5.

L{ua(t)} =∫ ∞

0

e−stua(t) dt

=∫ ∞

a

e−st dt

=[−1

se−st

]∞

a

=e−as

s

Remark 1.6. The �delta function� δa(t) (also written δ(t − a)) which is in�nite at the point t = a and zeroelsewhere is the �rst derivative of the unit step function ua(t). Thus

L{δa(t)} = L{u′a(t)} = sL{ua(t)} − ua(0) = e−as


1.7. Transforms of integrals.

Theorem 1.7. If f has transform L(f) = F (s) then

L[∫ t

0

f

]=

F (s)s

Proof. Let g(t) =∫ t

0

f . Then g′(t) = f(t) and g(0) = 0 and so

L{f(t)} = L{g′(t)} = sL{g(t)} − g(0)

= sL{g(t)}

and so

L{g(t)} = L[∫ t

0

f

]=

1sL{f(t)}

¤

Example. Invert s− 1s2(s + 1)

.

L−1

[s− 1

s2(s + 1)

]= L−1

[1

s(s + 1)

]− L−1

[1

s2(s + 1)

]

and

L−1

[1

s + 1

]= e−t

⇒ L−1

[1

s(s + 1)

]=

∫ t

0

e−z dz = 1− e−t

⇒ L−1

[1

s2(s + 1)

]=

∫ t

0

(1− e−z) dz

=[z + e−z

]t

0= t + e−t − 1

⇒ L−1

[s− 1

s2(s + 1)

]= 2(1− e−t)− t

We can use these results to solve more complicated initial value problems, which could not be solved without

the use of transforms:

Example. Response of an undamped system to a single square wave.

y′′ + 2y = r(t)

y(0) = y′(0) = 0

r(t) =

1, t < 1

0, t > 1

Thus r(t) = 1− u1(t) and L{r(t)} =1− e−s

s. Setting Y = L(y) and transforming the equation we obtain

s2Y + 2Y =1s(1− e−s)


and so

Y =1

s(s2 + 2)[1− e−s]

To �nd y we �rst invert

L−1

[1

s2 + 2

]=

1√2

sin√

2t

and so

L−1

[1

s(s2 + 2)

]=

1√2

∫ t

0

sin√

2z dz

= −12

[cos

√2z

]t

0

=12

(1− cos

√2t

)

We set f(t) = 12

(1− cos

√2t

)and therefore, using the second shifting theorem,

L−1

[e−s

s(s2 + 2)

]= u1(t)f(t− 1)

=12

[1− cos

√2(t− 1)

]u1(t)

Thus

y =12

[1− cos

√2t

]− 1

2

[1− cos

√2(t− 1)

]u1(t)

For t < 1, u1(t) = 0 and so

y =12

[1− cos

√2t

]

whereas, for t ≥ 1, u1(t) = 1 and so

y =12

[cos

√2(t− 1)− cos

√2t

]

Remark 1.8. Instead of using integration, we could have used partial fractions:

1s(s2 + 2)

=A

s+

Bs + C

s2 + 2

or

1 = A(s2 + 2) + (Bs + C)s

For s = 0, 1 = 2a and so A = 1/2. Equating coe�cients of powers of s gives

s2 : 0 = A + B ⇒ B = −1/2

s : 0 = C

Thus1

s(s2 + 2)=

12

[1s− s

s2 + 2

]

and so

L−1

[1

s(s2 + 2)

]=

12

[1− cos

√2t

]

as before.


Example. RC circuit

Ri(t) +1C

∫ t

0

i = v(t)

where v(t) =

0, t < a

V0, a < t < b

0, t > b

That is, v(t) = V0[ua(t)− ub(t)] If we set I(s) = L[i(t)] then the transformed equation is

RI +I

sC=

V0

s

[e−as − e−bs

]

and so

I = F (s)(e−as − e−bs)

where F (s) =V0

Rs + 1/C

L−1

[V0

Rs + 1/C

]= L−1

[V0/R

s + 1/RC

]

=V0

RL−1

[1

s + 1/RC

]

=V0

Re−t/RC

Thus

i(t) = L−1[I(s)] = L−1[e−asF (s)]− L−1[e−bsF (s)]

=V0

R

{e−(t−a)/RCua(t)− e−(t−b)/RCub(t)

}

So, for t < a, ua = ub = 0 and so i(t) = 0. For a < t < b ua = 1 but ub = 0 still and so

i(t) =[V0

Rea/RC

]e−t/RC

and �nally, for t > b, ua = ub = 1 and thus

V0

R

[ea/RC − eb/RC

]e−t/RC

1.8. Periodic functions. A function f(t) has period T if f(t + T ) = f(t) for all t. For example, sin t

and cos t have period 2π. We already know the transforms of these functions; but to transform more general(including discontinuous) periodic functions we do the following:

L[f(t)] =∫ ∞

0

e−stf(t) dt

=∫ T

0

e−stf(t) dt +∫ 2T

T

e−stf(t) dt +∫ 3T

2T

e−stf(t) dt + . . .

. . . +∫ nT

(n−1)T

e−stf(t) dt + . . .

Setting t = u + T in the �rst integral, t = u + 2T in the second, and so on, with t = u + (n − 1)T in the nth,we obtain


L[f(t)] =∫ T

0

e−suf(u) du +∫ T

0

e−s(u+T )f(u) du +∫ T

0

e−s(u+2T )f(u) du+

. . . +∫ T

0

e−s(u+(n−1)T )f(u) du + . . .

=[1 + e−sT + e−2sT + . . .

] ∫ T

0

e−suf(u) du

(using the periodicity of f). Noting that the term in square brackets is a geometric series, we can sum it toobtain �nally

L[f(t)] =1

1− e−sT

∫ T

0

e−stf(t) dt

Example. Periodic square wave

f(t) =

k, 2na < t < (2n + 1)a

−k, (2n + 1)a < t < 2(n + 1)a

for n = 0, 1, 2, . . ., which has period 2a.

k

0

-k

Figure 1. f(t)

L[f(t)] =1

1− e−2as

[∫ a

0

ke−st dt−∫ 2a

a

ke−st dt

]

=k[1− e−as − e−as + e−2as]

s(1− e−2as)

=k

s

[1− e−as]2

[1− e−as][1 + e−as]

=k

s

1− e−as

1 + e−as(=

k

stanh

as

2)

1.9. Di�erentiation of transforms. If F (s) =∫∞0

e−stf(t) dt then

F ′(s) = −∫ ∞

0

e−sttf(t) dt = −L[tf(t)]

Hence, L[tf(t)] = −F ′(s)

Example. Find the inverse transform of s(s2+a2)2


Noting that L{sin at} = as2+a2 , we have

L[t sin at] =2as

(s2 + a2)2

and soL−1

[s

(s2 + a2)2

]=

t

2asin at

Example.

F (s) = lns + a

s− a= ln(s + a)− ln(s− a)

Di�erentiating:

− d

ds

[ln

s + a

s− a

]=

1s− a

− 1s + a

=2a

s2 − a2

= L[eat − e−at] = 2L[sinh at]

⇒ L−1[F (s)] =eat − e−at

t= 2

sinh at

t

1.10. Integration of transforms. From this rule for di�erentiation of transforms, we can obtain thefollowing rule for integration:

Theorem 1.9. If F (s) = L[f(t)] and limt→0+

f(t)t

exists, then

L[f(t)

t

]=

∫ ∞

s

F

Proof.∫ ∞

s

F (u) du =∫ ∞

s

[∫ ∞

0

e−utf(t) dt

]du

=∫ ∞

0

f(t)[∫ ∞

s

e−ut du

]dt

=∫ ∞

0

e−st f(t)t

dt = L[f(t)

t

]

¤

Example. We know that L{sin at} =a

s2 + a2and lim

t→0

sin at

t= a. Thus

L{sinat

t} =

∫ ∞

s

a

u2 + a2du

= −[cot−1 u

a

]∞s

= cot−1 s

a

At this stage we can summarise all the transforms which have been established in the following table.


Table 1. Table of Laplace Transforms

f(t) F (s)

0 0

11s

kk

s

t1s2

tnn!

sn+1

tαΓ(α + 1)

sα+1

eat 1s− a

sin ata

s2 + a2

cos ats

s2 + a2

sinh ata

s2 − a2

cosh ats

s2 − a2

eat sin btb

(s− a)2 + b2

eat cos bts− a

(s− a)2 + b2

eattnn!

(s− a)n+1

eatf(t) F (s− a)

ua(t)f(t− a) e−asF (s)

t sin at2as

(s2 + a2)2

t cos ats2 − a2

(s2 + a2)2

tf(t) −F ′(s)

tnf(t) (−1)nF (n)(s)

f ′(t) sF (s)− f(0)

f ′′(t) s2F (s)− sf(0)− f ′(0)

f (n)(t) snF (s)− sn−1f(0)− . . .− f (n−1)(0)∫ t

0f(u) du

F (s)s

f(t)t

∫ ∞

s

F (u) du

1.11. Systems of linear di�erential equations. These present no special di�culties for Laplace trans-forms. The transformed system of linear algebraic equations is solved and then inverted.

Example. A circuit is described by the system of two equations:

L1i′1 + i1R1 = Mi′2 + v(t)

L2i′2 + i2R2 = Mi′1


If L1 = L2 = 2, M = 1 and R1 = R2 = 3, then this becomes

2i′1 + 3i1 = i2 + v(t)

2i′2 + 3i2 = i′1

Assuming i1(0) = i2(0) = 0, the transformed equation is then

2sI1 + 3I1 = sI2 + V (s)

2sI2 + 3I2 = sI1

or

(2s + 3)I1 − sI2 = V (s) (1)

−sI1 + (2s + 3I2) = 0 (2)

Equation 2 ⇒ [(2s + 3)2

s− s

]I2 = V (s)

and soI2 =

sV (s)(2s + 3)2 − s2

=sV (s)

3(s + 1)(s + 3)Consider now the case where v(t) = E sin t for some constant E.Then V (s) =

E

s2 + 1and so

I2 =E

3

[s

(s + 1)(s + 3)(s2 + 1)

]

Using partial fractions:s

(s + 1)(s + 3)(s2 + 1)=

A

s + 1+

B

s + 3+

Cs + D

s2 + 1or

s = A(s + 3)(s2 + 1) + B(s + 1)(s2 + 1) + (Cs + D)(s + 1)(s + 3)

Then s = −1 ⇒ −1 = 4A and so A = −1/4

s = −3 ⇒ −3 = −20B ⇒ B = 3/20

Equating the coe�cients of s3 we have: 0 = A + B + C and so C = −A−B = (5− 3)/20 = 1/10

and �nally setting s = 0 we obtain 0 = 3A + B + 3D or D = −1/3(3A + B) = −13

(−15 + 320

)= 1/5

ThusI2 =

E

60

[− 5

s + 1+

3s + 3

+2s

s2 + 1+

4s2 + 1

]

and inverting we obtaini2 =

E

60[−5e−t + 3e−3t + 2 cos t + 4 sin t

]

Similarly,

I1 =2s + 3

sI2 =

E

3

[2s + 3

(s + 1)(s + 3)(s2 + 1)

]

=E

60

[5

s + 1+

3s + 3

− 8s

s2 + 1+

14s2 + 1

]

andi1 =

E

60[5e−t + 3e−3t − 8 cos t + 14 sin t

]


2. Convolution

The convolution of two functions f(t) and g(t), denoted f ∗ g is de�ned by

(f ∗ g)(t) =∫ t

0

f(u)g(t− u) du

Convolution has many of the properties of a product, but not all. For example, it is commutative: f ∗ g = g ∗ f

since

(f ∗ g)(t) =∫ t

0

f(u)g(t− u) du and with v = t− u

=∫ 0

t

f(t− v)g(v) (−dv) =∫ t

0

g(v)f(t− v) dv = (g ∗ f)(t)

However, note that 1 ∗ 1 6= 1, since

1 ∗ 1 =∫ t

0

du = t

Also, f ∗ f can be negative, e.g., if f(t) = cos t

cos t ∗ cos t =∫ t

0

cos u cos(t− u) du

=12

∫ t

0

[cos t + cos(2u− t)] du

=12

[u cos t +

12

sin(2u− t)]t

0

=12

{t cos t +

12

sin t− 12

sin(−t)}

=12

[t cos t + sin t]

This is negative at, for example, t = 3π/2.The value of convolutions in the context of Laplace transforms arises from the following:

Theorem 2.1. If f(t) and g(t) have Laplace transforms F (s) and G(s) respectively, then

L(f ∗ g)(t) = F (s)G(s)

Proof.

F (s)G(s) =[∫ ∞

0

e−suf(u) du

] [∫ ∞

0

e−svf(v) dv

]

=∫ ∞

0

[∫ ∞

0

e−s(u+v)f(u)g(v) dv

]du

Now, let t = u + v and so dt = dv (in the inner integral). Then

F (s)G(s) =∫ ∞

0

[∫ ∞

u

e−stf(u)g(t− u) dt

]du

=∫ ∞

0

[∫ t

0

e−stf(u)g(t− u) du

]dt

(switching the order of integration by taking note of the area of integration)Thus

2. CONVOLUTION 15

F (s)G(s)) =∫ ∞

0

e−st

[∫ t

0

f(u)g(t− u) du

]dt

=∫ ∞

0

e−st(f ∗ g)(t) dt

= L(f ∗ g)(t)

Example. f(t) = g(t) = cos t ⇒F (s) = G(s) =

s

s2 + 1and so

F (s)G(s) =s2

(s2 + 1)2

ThusL−1

[s2

(s2 + 1)2

]= cos t ∗ cos t =

12

[t cos t + sin t]

Note that we can arrive at this result in an alternative fashion by using

L[t cos t] = − d

ds

[s

s2 + 1

]

=−(s2 + 1) + s(2s)

(s2 + 1)2=

s2 − 1(s2 + 1)2

Thus

L−1

[s2

(s2 + 1)2

]= L−1

[ 12 (s2 − 1) + 1

2 (s2 + 1)(s2 + 1)2

]

=12L−1

[s2 − 1

(s2 + 1)2+

1s2 + 1

]

=12

[t cos t + sin t]

(as before)

2.1. Initial value problems. Convolution can obviously be used as an alternative way of solving di�er-ential equations. One advantage can be seen in the following:

Example.

y′′ + ω2y = h(t) y(0) = A, y′(0) = B

Transforming ⇒s2Y −As−B + ω2Y = H(s)

and thus(s2 + ω2)Y = H(s) + As + B

orY =

H(s)s2 + ω2

+As

s2 + ω2+

B

s2 + ω2

and thusy =

1ω

sin ωt ∗ h(t) + A cos ωt +B

ωsin ωt

Thus we can obtain a solution to a problem in this general form using convolutions without knowing h(t) andif we do know h(t) we can use this formula to calculate y without having to transform r.


2.2. Integral equations. Convolutions can also be used together with Laplace transforms to solve someintegral equations very simply.

Example.y(t) = sin 2t +

∫ t

0

y(u) sin 2(t− u) du

Transforming ⇒

Y =2

s2 + 22+

2Y

s2 + 22

⇒[1− 2

s2 + 4

]Y =

2s2 + 4

⇒ Y =s2 + 4s2 + 2

(2

s2 + 4

)

=2

s2 + 2

and thus y(t) =√

2 sin√

2t.

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