Unformatted text preview: ME375 Handouts Laplace Transform
• Motivation
• Laplace Transform
–
–
–
– Review of Complex Numbers
Definition
Time Domain vs sDomain
Important Properties • Inverse Laplace Transform
• Solving ODEs with Laplace Transform School of Mechanical Engineering
Purdue University ME375 Laplace  1 Motivation – Solving Differential Eq.
Differential Equations (ODEs)
+
Initial Conditions (ICs) y(t): Solution in
Time Domain (Time Domain) L −1 [ • ] L[ •] Algebraic Equations
( sdomain ) Y(s): Solution in
Laplace Domain School of Mechanical Engineering
Purdue University ME375 Laplace  2 1 ME375 Handouts Review of Complex Numbers
• The Many Faces of a Complex Number:
– Coordinate Form:
Form: Img. z = x + jy z – Phasor (Euler) Form:
Form:
z = A e j φ = A (c o s φ + j sin φ )
• Moving Between Representations
– Phasor (Euler) Form → Coordinate Form
x = A cos φ
e jφ = cos φ + j sin φ
y = A sin φ Real R
S
T – Coordinate Form → Phasor (Euler) Form
⎧ A = x2 + y2
⎪
⎨
⎪φ = atan2( y , x )
⎩ ⎧ tan −1 ( y x )
when z is in the 1st or 4th quadrant
⎪ −1 y
atan2( y , x ) ≡ ⎨ tan ( x ) + π when z is in the 2nd quadrant
⎪ −1 y
⎩ tan ( x ) − π when z is in the 3rd quadrant
School of Mechanical Engineering
Purdue University ME375 Laplace  3 Definition
• Laplace Transform
– One Sided Laplace Transform F ( s ) = L[ f (t ) ] = ∫ ∞ f (t ) e − st dt 0 where s is a complex variable that can be represented by s = σ + j ω
and f (t) is a continuous function of time that equals 0 when t < 0.
– Laplace Transform converts a function in time t into a function of a
complex variable s. • Inverse Laplace Transform
f (t ) = L−1 [ F ( s ) ] = 1
2 πj ∫ c + j∞ c − j∞ F ( s ) e st ds School of Mechanical Engineering
Purdue University ME375 Laplace  4 2 ME375 Handouts s  Domain vs Time Domain
• Two Representations of a Signal (System Response)
The response or input of a system can have two representations:
– Time Domain f (t)
Represents the value of the response at time t, which is a function of time.
– s  Domain F(s)
A function of a complex variable s.
f (t) , t > 0 F(s) Time Domain s  Domain Signals
e.g. sin(ωt), …
sin(ω
Systems
e.g. ODEs, ...
School of Mechanical Engineering
Purdue University ME375 Laplace  5 Important Properties
• Linearity
Given • Differentiation
Given F1 ( s ) = L[ f1 (t ) ] F ( s ) = L[ f ( t ) ]
The Laplace transform of the derivative
of f (t) is: F2 ( s ) = L[ f 2 (t ) ]
a and b are arbitrary constants,
then
L[ a f1 (t ) + b f 2 (t ) ]
= d
L ⎡ dt f ( t ) ⎤ =
⎣
⎦ ∫ ∞ 0 ( df ) e − st dt
dt =
L ⎡ f (t ) ⎤ =
⎣
⎦
⎤
L ⎡ f (t ) ⎦ =
⎣ – For zero initial condition:
condition:
Q: If u(t) = u1(t) + 4 u2(t) what is the Laplace
u(
transform of u(t) ?
u( d
•
dt
Differentiation School of Mechanical Engineering
Purdue University L⎡ ⎤ ⎣ ⎦
⎯⎯⎯
→
←⎯[⎯
⎯
L −1 ] s L [• ]
Multiply by s ME375 Laplace  6 3 ME375 Handouts Important Properties
• Integration
Given F ( s) = L[ f (t )] The Laplace transform of the definite
integral of f (t) is:
t
∞
t
L ⎡ ∫ f (λ )d λ ⎤ = ∫ ⎡ ∫ f (λ )d λ ⎤ e− st dt
⎢0
⎥ 0 ⎢0
⎥
⎣
⎦
⎣
⎦ Q : Given that the Laplace transforms of a unit
step function u(t) = 1 and f(t) = sin(2t) are
u(
f(
sin(2t
1
U ( s ) = L[1] =
s
2
F ( s ) = L[sin(2t ) ] = 2
s +4
What is the Laplace transform of = e ( t ) = 2 + 3t + 5t 2 + 4 cos( 2 t ) – Conclusion:
L⎡ ⎤ ∫ t 0 • dλ Integration ⎣ ⎦
⎯⎯⎯
→ ←⎯⎯
⎯
L−1 [ ] L [• ]
s
Divide by s School of Mechanical Engineering
Purdue University ME375 Laplace  7 Important Properties
• Some Laplace Transform Pairs: • Initial Value Theorem
f (0 + ) = lim sF ( s ) U nit Im pulse δ ( t ) s→∞ U nit S tep u ( t ) = 1 • Final Value Theorem 1 ⇔ 1
s
1
s2
n!
s n +1
1
s+a
1
(s + a)2 Q: Given that F(s) = L [f (t)], how would you
F(
find the initial slope of the response f (t)
without knowing f (t)? ⇔ tn ⇔ e − at ⇔
⇔ sin(ω t ) s→ 0 t te − at f ( ∞ ) = lim f ( t ) = lim sF ( s )
t→∞ ⇔ ⇔ School of Mechanical Engineering
Purdue University ω
s2 + ω 2 ME375 Laplace  8 4 ME375 Handouts Signals and Systems
An s  domain representation F(s) can represent a signal or an operation. Most of
the time its representation can be inferred from the context. However, when doing
However,
algebraic manipulations, one should be very careful in distinguishing the signals
distinguishing
from the operators (systems).
As an example, recall that the Laplace transform of a unit step function (signal) is:
u (t ) = 1, t > 0 ⇒ L[u (t )] = 1
s However, note that the integration of a function in time is equivalent to multiplying
equivalent
the function’s Laplace transform by 1/s:
function’
1/s:
u(t ) = 1, t > 0 ⇒ U(s) =
v(t ) = z t 0 1
u(τ ) dτ = t ⇒ V (s) = ⋅ U(s) =
s School of Mechanical Engineering
Purdue University ME375 Laplace  9 Inverse Laplace Transform
Given an sdomain function F(s), the inverse Laplace transform is
used to obtain the corresponding time domain function f (t).
Procedure:
– Write F(s) as a rational function of s.
– Use long division to write F(s) as the sum of a strictly proper
rational function and a quotient part.
– Use PartialFraction Expansion (PFE) to break up the strictly
Partialproper rational function as a series of components, whose
inverse Laplace transforms are known.
– Apply inverse Laplace transform to individual components.
School of Mechanical Engineering
Purdue University ME375 Laplace  10 5 ME375 Handouts Use Laplace Transform to Solve ODEs
Differential Equations (ODEs)
+
Initial Conditions (ICs) y(t): Solution in
Time Domain (Time Domain) L −1 [ • ] L[ •] Algebraic Equations
( sdomain ) Y(s): Solution in
Laplace Domain School of Mechanical Engineering
Purdue University ME375 Laplace  11 Examples
Q: Use LT to solve the free response of a
1st Order System.
τ y+ y=0
y (0) = y 0 Q: Use LT to find the step response of a
1st Order System. τ y + y = K ⋅1 y (0) = 0 Q: What is the step response when the initial
condition is not zero, say
y(0) = 5.
School of Mechanical Engineering
Purdue University ME375 Laplace  12 6 ME375 Handouts Use LT and ILT to Solve for System Responses
Find the free response of a 2nd order system with two distinct real characteristic
real
roots (e.g., a motor rotor turning inside its bearings):
y + 9 y + 18 y = 0 where y (0) = 0 and y (0) = 3 J
J θ τ (t) B School of Mechanical Engineering
Purdue University ME375 Laplace  13 Use LT and ILT to Solve for System Responses
Find the free response of a 2nd order system with two identical real characteristic
roots (e.g., boxcar arrester):
y + 10 y + 25 y = 0 w here y (0) = 1 and y (0) = 5 School of Mechanical Engineering
Purdue University ME375 Laplace  14 7 ME375 Handouts Using MATLAB to Compute Residues
>> [R,P,K]=residue(3,[1 9 18])
R =
1
1
P =
6
3
K = >> [R,P,K]=residue([1 15],[1 10 25])
R =
1
10
P =
5
5
K = >>
School of Mechanical Engineering
Purdue University ME375 Laplace  15 Use LT and ILT to Solve for System Responses
Find the free response of a 2nd order system with complex characteristic roots
characteristic
(e.g., vehicle suspension system):
y + 6 y + 25 y = 0 where y (0) = 1 and y (0) = − 3 School of Mechanical Engineering
Purdue University ME375 Laplace  16 8 ME375 Handouts Use LT and ILT to Solve for System Responses
Find the unit step response of a 2nd order system:
y + 6 y + 25 y = u + 25 u w here y (0) = 0 and y (0) = 0 School of Mechanical Engineering
Purdue University ME375 Laplace  17 Use LT and ILT to Solve for System Responses
Find the unit step response of a 2nd order system:
y + 6 y + 8 y = u + 3u w here y (0) = 0 and y (0) = 0 School of Mechanical Engineering
Purdue University ME375 Laplace  18 9 ME375 Handouts Obtaining I/O Model Using LT
Obtain the I/O model for the vibration absorber shown below. The input is the
The
force f(t) and the output is the displacement of mass m2. M1z1 + ( B1 + B2 ) z1 + ( K1 + K2 ) z1 − B2z2 − K2z2 = f (t) z2
M2
K2 M2z2 + B2z2 + K2 z2 − B2 z1 − K2z1 = 0
B2
z1 M1
K1 f(t) B1 Input:
Input:
Output:
Output:
School of Mechanical Engineering
Purdue University ME375 Laplace  19 Obtaining I/O Model Using LT School of Mechanical Engineering
Purdue University ME375 Laplace  20 10 ...
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 Spring '10
 Meckle
 Mechanical Engineering, Laplace, Purdue University

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