Laplace transform Modeling systems and processes (11MSP)...Laplace transform Properties { transformation of convolution Theorem (Transformation of convolution) Lff(t) g(t)g= L ˆZ

Fourier transform Laplace transform Examples of using Laplace transform

Laplace transformModeling systems and processes (11MSP)

Bohumil Kovar, Jan Prikryl, Miroslav Vlcek

Department of Applied MahematicsCTU in Prague, Faculty of Transportation Sciences

4th lecture 11MSP2019

verze: 2019-03-25 11:22


Table of content

1 Fourier transform

2 Laplace transform

3 Examples of using Laplace transform


Fourier transformdefinition

Consider a continuous LTI system with a impulse response of h(t)and an input signal of u(t) = est , where s ∈ C:

y(t) =

∫ ∞−∞

h(τ) u(t − τ) dτ =

∫ ∞−∞

h(τ) es(t−τ) dτ.

es(t−τ) = este−sτ and est does not depend on integrand, so

y(t) = est∫ ∞−∞

h(τ) e−sτ dτ.


Table of content

1 Fourier transform

2 Laplace transform

Reasons for use

Definition

Properties

Laplace transform tables



Laplace transformReasons for use

The Laplace transform significantly simplifies some operations inthe analysis of continuous LTI systems, for example

• derivation ⇒ multiplication by variable p

• integration ⇒ division by variable p

• differential equation of n-th order with constant coeficient ⇒algebraic equation of n-th order

• convolution f (t) ∗ g(t) ⇒ multiplication F (p) · G (p)


Laplace transformDefinition

Definition (Laplace transform)

Laplace transform of function f (t), witch is at most polynomialgrowth

f (t) = a0 + a1t + a2t2 + · · ·+ ant

n

is defined by the integral

F (p) =

∫ ∞0

f (t)e−pt dt ≡ L{f (t)} .


Laplace transformDefinition of transform inversion

The inverse Laplace transform has the form of an integral alongthe curve in the complex plane p

f (t) =1

2πi

∫ c+i∞

c−i∞F (p) ept dp ≡ L−1 {F (p)} .


Laplace transformProperties - linearity

Theorem (Linearity)

Laplace transform is a linear:

L

{∑k

ak fk(t)

}=∑k

akL{fk(t)} =∑k

akFk(p)

L−1{∑

m

bmFm(p)

}=∑m

bmL−1 {Fm(p)} =∑m

bmfm(t)


Laplace transformproperties – time scaling

Theorem (Time scaling)

For F (p) = L{f (t)} is

L{f (at)} =1

aF(pa

)Proof.

By substitution at = τ

L{f (at)} =

∫ ∞0

f (at)e−pt dt =

∫ ∞0

f (τ)e−paτ 1

adτ =

1

aF(pa

)


Laplace transformProperties – time shifting

Theorem (Time shifting)

If F (t) = L{f (t)}, then

L{1(t − τ)f (t − τ)} = e−pτL{f (t)} = e−pτF (p).

Proof.

By substitution t − τ = ϑ

L{f (t − τ)} =

∫ ∞0

f (t − τ)e−pt dt =

∫ ∞−τ

f (ϑ)e−p(τ+ϑ) dϑ

= e−pτ∫ −0−τ

f (ϑ)e−pϑ dϑ+ e−pτ∫ ∞0

f (ϑ)e−pϑ dϑ

= e−pτ∫ ∞0

f (ϑ)e−pϑ dϑ ≡ e−pτF (p)


Laplace transformProperties – transformation of convolution

Theorem (Transformation of convolution)

L{f (t) ∗ g(t)} = L{∫ ∞

0f (t − τ) · g(τ) dτ

}= F (p) · G (p)

Proof is easier to do in discrete time.

Consequence

L{y(t) =

∫ ∞0

h(τ) · u(t − τ) dτ

}⇔ Y (p) = H(p) · U(p)


Laplace transformProperties – derivative transformation

Theorem (Derivative transformation)

L{f (t)} = F (p)

L{

d

dtf (t)

}= pF (p)− f (0)

L{

d2

dt2f (t)

}= p2F (p)− pf (0)− d

dtf (0)

...

L{

dn

dtnf (t)

}= pnF (p)− pn−1f (0)− pn−2

d

dtf (0) . . .− f (n−1)(0)


Laplace transform – properties

Proof.

Integrating par-partes,∫ ba u′v = [uv ]ba −

∫ ba uv ′:

L{

d

dtf (t)

}=

∫ ∞0

d

dtf (t)e−pt dt

=

[f (t)e−pt

]∞0

− (−p)

∫ ∞0

f (t)e−pt dt

= −f (0) + pF (p).

By repeating this process, we obtain

L{

d2

dt2f (t)

}= p2F (p)− pf (0+)− d

dtf (0+)


Laplace transform – properties

Theorem (Time domain inegration)

L{∫ t

0f (τ) dτ

}=

1

pF (p)

Proof.

Integrating per partes,∫ ba uv ′ = [uv ]ba −

∫ ba u′v :

L{∫ t

0f (τ) dτ

}=

∫ ∞0

(∫ t

0f (τ) dτ

)e−pt dt

=1

−p

[∫ t

0f (τ) dτe−pt

]∞0

− 1

−p

∫ ∞0

f (t)e−pt dt

=1

pF (p).


Laplace transform tables (1/4)

f (t) = L−1 {F (p)} F (p) = L{f (t)}

f (t) =1

2πi

c+i∞∫c−i∞

F (p) ept dp F (p) =

∞∫0

f (t) e−pt dt

δ(t) 1

1(t)1

p



f (t) = L−1 {F (p)} F (p) = L{f (t)}

e−αt1

p + α

sinωtω

p2 + ω2

cosωtp

p2 + ω2



f (t) = L−1 {F (p)} F (p) = L{f (t)}

e−αt sinωtω

(p + α)2 + ω2

e−αt cosωtp + α

(p + α)2 + ω2

tnn!

pn+1



f (t) = L−1 {F (p)} F (p) = L{f (t)}

tne−αtn!

(p + α)n+1

t cosωtp2 − ω2

(p2 + ω2)2

t sinωt2ωp

(p2 + ω2)2


Table of content

1 Fourier transform

2 Laplace transform


RC integrator

Impulsnı odezva LTI systemu


Laplace transform – example 1

The response of RC integrator to input signal.

R

Cu1(t) uC(t)

Differential equation is

RCd

dtuC(t) + uC(t) = u1(t).



For α =1

RCand input u1(t) = U0 · 1(t) is

d

dty(t) + αy(t) = αU0 · 1(t).

Because it is the differential equations with constant coefficients,we can use the Laplace transform and its properties

L{

d

dty(t)

}+ L{αy(t)} = L{αU0 · 1(t)} ,

and we get an algebraic equation for the unknown function Y (p)

pY (p)− y(0) + αY (p) = αU0 ·1

p.



We rewrite the equation so that the unknown Y (p) will be on theleft and all known constants on the right

(p + α)Y (p) =αU0

p+ y(0).

and find a solution in the plane p

Y (p) =αU0

p(p + α)+

y(0)

p + α=

U0

p− U0

p + α+

y(0)

p + α

With the help of tables we can find for t > 0 solution

y(t) = U0 (1− e−αt) + y(0)e−αt



0 5 10 150

1

2

3

4

5

6

7

8

9

10

t

u C(t

)



Consider a LTI system that is described by the input values fort > 0

u(t) = e−t + e−3t

and outputy(t) = te−3t .

How to find impulse response?



Because

U(p) =1

p + 1+

1

p + 3= 2

p + 2

(p + 1)(p + 3)

Y (p) =1

(p + 3)2

andY (p) = H(p) · U(p),

is

H(p) =Y (p)

U(p)=

1

2

p + 1

(p + 2)(p + 3)=

1

2

[2

p + 3− 1

p + 2

].



With the help of tables we can find solutions for t > 0

h(t) = e−3t − 1

2e−2t .

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

t

h(t)


Have a nice spring day

Documents

Laplace transform Modeling systems and processes (11MSP)...Laplace transform Properties { transformation of convolution Theorem (Transformation of convolution) Lff(t) g(t)g= L ˆZ