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[ •] L −1 [ • ] 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:
Img. z = x + jy z – Phasor (Euler) 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 ) ] = ∫ ∞ 0 f (t ) e − st dt where s is a complex variable that can be represented by s = σ + j ω is 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 Time Domain
Signals e.g. sin(ωt), … Systems e.g. ODEs, ...
School of Mechanical Engineering Purdue University ME375 Laplace  5 F(s) s  Domain Important Properties
• Linearity
Given F1 ( s ) = L[ f1 (t ) ] • Differentiation
Given
F ( s ) = L[ f ( t ) ] The Laplace transform of the derivative of f (t) is:
d L ⎡ dt f ( t ) ⎤ = ⎣ ⎦ F2 ( s ) = L[ f 2 (t ) ] a and b are arbitrary constants, are then L[ a f1 (t ) + b f 2 (t ) ]
= ∫ ∞ 0 ( df ) e − st dt dt = L ⎡ f (t ) ⎤ = ⎣ ⎦ ⎤ L ⎡ f (t ) ⎦ = ⎣ – For zero initial condition: zero
Q: If u(t) = u1(t) + 4 u2(t) what is the Laplace transform of u(t) ? d • dt
Differentiation ⎣⎦ ⎯⎯⎯ → ←⎯[⎯ ⎯ L −1 ] L⎡ ⎤ s L [• ]
Multiply by s School of Mechanical Engineering Purdue University 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 are 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: ∫ t 0 • dλ ⎣⎦ ⎯⎯⎯ → L⎡ ⎤ Integration ←⎯⎯ ⎯ L−1 [ ] L [• ] s
Divide by s School of Mechanical Engineering Purdue University ME375 Laplace  7 Important Properties
• Initial Value Theorem f (0 + ) = lim sF ( s )
s→∞ • Some Laplace Transform Pairs:
U nit Im pulse δ ( t ) U nit S tep u ( t ) = 1 t tn e − at te − at sin(ω t ) ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ 1 1 s 1 s2 n! s n +1 1 s+a 1 (s + a)2 • Final Value Theorem
f ( ∞ ) = lim f ( t ) = lim sF ( s )
t→∞ s→ 0 Q: Given that F(s) = L [f (t)], how would you find the initial slope of the response f (t) without knowing f (t)? ω
s2 + ω 2 School of Mechanical Engineering Purdue University 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’
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. 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[ •] L −1 [ • ] 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 B θ τ (t ) 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 Obtaining
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.
z2 M2 K2 M1 K1 B1 B2 z1 M1z1 + ( B1 + B2 ) z1 + ( K1 + K2 ) z1 − B2z2 − K2z2 = f (t) M2z2 + B2z2 + K2 z2 − B2 z1 − K2z1 = 0 f(t) Input: 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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This note was uploaded on 12/17/2009 for the course IE 383 taught by Professor Leyla,o during the Spring '08 term at Purdue.
 Spring '08
 Leyla,O

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