EGGN307: Laplace Transform Review∗

Tyrone Vincent

Lecture 4

Contents1 Motivation 1

2 Laplace Transform Definition 22.1 Euler’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Quiz Yourself 53.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1 MotivationMotivation

• Differential equations model dynamic systems

• Control system design requires simple methods for solving these equations!

• Laplace Transforms allow us to

– systematically solve linear time invariant (LTI) equations for arbitrary inputs– easily combine coupled differential equations– use block diagrams to represent systems that are made up of smaller subsystems

Transforms

• A transform is a different but equivalent representation for a mathematical object

• Recall that complex numbers can be represented in two ways:

Rectangular Form Polar Forms = a+ jb s = r∠θ

• Given a complex number represented in rectangular form, we can transform it to polar form (and vice-versa)

(a, b)→ (r, θ)

• Some operations are easier in polar form: multiplication and division∗This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/ or send a letter to Creative

Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA.

1

2 Laplace Transform DefinitionLaplace Transforms

• We will introduce the Laplace Transform for functions defined for t > 0.

L[f(t)]→ F (s)

• f(t) is transformed to the function F (s), where s is a complex number

• Operations such as derivatives and convolution are easier in this domain.

Laplace Transform Definition

Let f(t) be a function of time defined for t > 0. The one-sided Laplace Transform of f(t) is defined to be

F (s) := L[f(t)] :=∫ ∞0−

f(t)e−stdt

• 0− is shorthand for the lower limit approaching 0 from the left (always negative)

• The Laplace Transform exists if the integral converges for any value of s

– Region of convergence is not as important for “one-sided” Laplace transforms

Important facts you should already know

• Complex Exponentials (Euler’s Formula): ejθ = cos(θ) + j sin(θ).

• Differentiation and integration with s a complex number:

d

dtest = sest

∫ b

a

estdt =1

sest∣∣∣∣ba

For derivations, see the appendix.

2.1 Finding Laplace Transforms using the defintionExample 1. Find the one-sided Laplace Transform of the function

f(t) = 3

Solution: Plug the function into the Laplace Transform definition

F (s) =

∫ ∞0−

3e−stdt

=1

−s3e−stdt

∣∣∣∣∞0−

=3

−s

[limt→∞

e−st − limt→0−

e−st]

• Laplace Transform defined if it converges for any value of s.

2

– Second limit converges to 1 for all values of s– First limit converges to zero if the real part of s is less than zero

F (s) =3

−s[0− 1]

=3

s

• Note: the value of the function for negative time does not matter (other than what affects the limit as t→ 0.)

• In order to identify a Laplace Transform with a single function, we will pick the function that is equal to zerofor negative time.

Our First Laplace Transform Pair

3u(t)L←→ 3

swhere u(t) is the unit step function

u(t) =

{1 t ≥ 0

0 otherwise

−2 −1 0 1 2−0.5

0

0.5

1

1.5

Time (s)

Mag

nitu

de

Example 2. Find the Laplace Transform of the function

f(t) = Ae−atu(t)

where A and a are fixed constants.

Solution: Plug the function into the Laplace Transform definition

F (s) =

∫ ∞0−

Ae−ate−stdt

=

∫ ∞0−

Ae−(a+s)tdt

=A

−(a+ s)e−(a+s)t

∣∣∣∣∞0−

=A

−(a+ s)

[e−(a+s)∞ − e−(1+s)0

]=

A

−(a+ s)[0− 1]

=A

(a+ s)

Second Laplace Transform Pair

Ae−atu(t)L←→ A

a+ s

3

3 Laplace Transform PropertiesThe Laplace Transform has useful properties that can help us find the Laplace Transform of more complicated func-tions.

3.1 Scaling and LinearityThe first two of these follow directly from the linearity of integral defining the Laplace Transform. Remember, weusing the notation L{f(t)} to indicate the Laplace Transform of f(t).

Scaling L{Af(t)} = AL{f(t)}

Linearity L{f(t) + g(t)} = L{f(t)}+ L{g(t)}

A useful application of the linearity property is to find the Laplace Transform of cos(ωt)u(t) from the sum ofcomplex exponentials

Example 3. cos(ωt)u(t) can be represented as the sum of complex exponentials

cos(ωt)u(t) =1

2

(ejωt + e−jωt

)u(t)

Using the linearity and scaling properties,

L{cos(ωt)u(t)} = 1

2L{ejωtu(t)}+ 1

2L{e−jωtu(t)}.

We have already derived the Laplace Transform pair

e−atu(t)L←→ 1

a+ s

Using this, we have

L{cos(ωt)u(t)} =1

2

1

s− jωt+

1

2

1

s+ jωt

=s

s2 + ω2

By subtracting two complex exponentials, the same method can used to find the Laplace Transform of sine

Sine and Cosine Laplace Transform Pair

cos(ωt)u(t)L←→ s

s2 + ω2

sin(ωt)u(t)L←→ ω

s2 + ω2

3.2 Time and frequency shiftA common occurrence in control systems is for a signal to be delayed due to transmission delays or other effects.When a signal is delayed, this has the effect of shifting the signal in time. To model this effect, we can use the timeshift theorem given below. Many Laplace Transform rules have a “dual” form with symmetry between the time domainand transform domain, and we list the dual to the time shift property here, which is called the frequency shift property.In these definitions F (s) = L{f(t)}.

4

Time shift L{f(t− t0)u(t− t0)} = e−st0F (s)

Frequency shift L{e−s0tf(t)} = F (s+ s0)

Example 4. Use the frequency shift property to find the Laplace Transform of a decaying exponential function

g(t) = e−at cos(ωt)u(t)

Solution: Sincecos(ωt)u(t)

L←→ s

s2 + ω2

The frequency shift property implies

L{e−at cos(ωt)u(t)} = s

s2 + ω2

∣∣∣∣s=s+a

=s+ a

(s+ a)2 + ω2

The same method can be used to show

L{e−at sin(ωt)u(t)} = ω

(s+ a)2 + ω2

3.3 Differentiation and IntegrationTaking the derivative of a function is equivalent in the Laplace domain to multiplying by s. This property will beimportant when solving differential equations.

Differentiation L{ddtf(t)

}= sL{f(t)} − f(0−)

Integration L{∫ t

0f(τ)dτ

}= 1

sL{f(t)}

Remember, the notation f(0−) indicates the limit as we approach 0 from the left, or

f(0−) = limt→0, t<0

f(t)

The proof of these theorems is actually pretty easy - see the appendix for the differentiation theorem. The differentia-tion theorem will be used in the next lecture to solve differential equations.

Example 5. (using Integration theorem.) Find the Laplace Transform of the function f(t) = tu(t).

Solution: We can use the integration formula because a ramp is the integral of a step. That is,∫ t

0−u(τ)dτ = tu(t).

Thus,

L{tu(t)} = 1

sL{u(t)} ,

=1

s

1

s=

1

s2.

5

4 The Impulse FunctionA useful yet tricky signal that finds a lot of use in control systems is the impulse function. It is an approximation to asignal which lasts for a short period of time - much shorter that the natural time scales of the system. It turns out thatthe Laplace Transform of an impulse function is very simple. However, before looking at an impulse function, firstlets consider a pulse:

Example 6. Find the Laplace Transform of the pulse function

Definition of pulse function

pt0(t) =

{1t0

0 < t < t00 otherwise

where t0 is a constant.

1

t0

Time (s)

Magnitude

0 t0

Note that the area of the pulse function is 1 for any value of t0.Solution: The pulse function can be represented as a superposition of steps

pt0(t) =1

t0u(t)− 1

t0u(t− t0)

so the Laplace Transform of pt0(t) is

L{pt0(t)} = L{

1

t0u(t)

}− L

{1

t0u(t− t0)

}=

1

t0s− 1

t0se−st0

=1

t0s

(1− e−st0

)

The impulse function is a pulse that gets shorter and shorter, but whose magnitude gets larger and larger, whilemaintaining the same area under the curve.

Definition of impulse function

δ(t) := limt0→0

pt0(t)

6

1

t0

Time (s)

Magnitude

0 t0

Example 7. Find the Laplace Transform of the impulse function δ(t).

Solution: We can take the limit of the Laplace Transform of the pulse function using L’Hopital’s rule:

L{δ(t)} = limt0→0

1

t0s

(1− e−st0

)= lim

t0→0

ddt0

(1− e−st0)ddt0

(t0s)=s

s= 1

The simplest Laplace Transform pair

δ(t)L←→ 1

The impulse function can also be defined via these two properties:

δ(t) = 0 t 6= 0∫ 0+

0−δ(t)dt = 1

Any function which has these properties (including the limit of the pulse function) has Laplace Transform equal to one

L{δ(t)} =∫ ∞0−

δ(t)e−stdt

=

∫ 0+

0−δ(t)e−stdt

= e−s0∫ 0+

0−δ(t)dt

= 1

5 Table of Laplace Transform PairsHere is a table of the Laplace Transform Pairs derived in this lecture that you are expected to know.

7

f(t) F (s)

δ(t) 1

u(t) 1s

tu(t) 1s2

eatu(t) 1s+a

cos(ωt)u(t) ss2+ω2

sin(ωt)u(t) ωs2+ω2

e−at cos(ωt)u(t) s+a(s+a)2+ω2

e−at sin(ωt)u(t) ω(s+a)2+ω2

6 Appendix

6.1 Euler’s FormulaThe definition of the Laplace Transform includes the function e−st, where s is a complex number. Let’s rememberhow to integrate this function. The function ex is very important, as it appears in many places in engineering subjects.The reason that it is so ubiquitous is that is satisfies the relationship

d

dxex = ex

and thus appears in situations when quantities increase in proportion to how much is already there. We want to extendthis function to complex numbers. In this case, we will need to calculate ez when z is complex.

• The key formula extending exponentials to complex numbers is Euler’s Formula:

ejθ = cos(θ) + j sin(θ)

Let’s check that with this definition, ejθ satisfies ddθ e

jθ = jejθ

d

dθejθ =

d

dθ(cos(θ) + j sin(θ))

= − sin(θ) + j cos(θ)

= j (cos(θ) + j sin(θ)) [switch terms, pull out j and use j2 = −1]

= jejθ

Now, considering dest

dt we can write s = a+ jb, so that

dest

dt=de(a+jb)t

dt

=d(eatejbt)

dt

= aeatejbt + jbeatejbt

= (a+ jb)eatejbt

= sest

Thus, taking the derivative of est with respect to time simply brings down the factor s, just as would occur if s werepurely real. This also means that ∫ b

a

estdt =1

−sest∣∣∣∣ba

8

6.2 Proof of Differentiation Theorem

L{d

dtf(t)

}=

∫ ∞0−

df(t)

dte−stdt

Choose u = e−st and dv = df(t)dt dt. Then du = −se−stdt and v = f(t) so that

L{d

dtf(t)

}= f(t)e−st

∣∣∞0−

+

∫ ∞0−

f(t)se−stdt

= limt→∞�

����:0

f(t)e−st − f(0−) + s

∫ ∞0−

f(t)e−stdt

= sL{f(t)} − f(0−)

7 Quiz Yourself

7.1 Questions1. Find the one-sided Laplace Transform for the following signals

(a) x(t) =

{1 0 ≤ t ≤ 1

0 otherwise

(b) x(t) = (1 + e−t)

(c) x(t) = (1 + e−t)u(t)

(d) x(t) = t

2. Using the step function and the integration formula, find the Laplace Transform of the following function

x(t) = t2u(t)

3. Using the frequency shift property, find the Laplace Transform of the following function

x(t) = e−tt2u(t)

7.2 Solutions

1.

(a)

9

(b)

(c)

10

(d)

2.

11

3.

12