Laplace Transform

# Laplace Transform - ME375 Handouts Laplace Transform •...

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Unformatted text preview: ME375 Handouts Laplace Transform • Motivation • Laplace Transform Transform – – – – Review of Complex Numbers Definition Time Domain vs s-Domain Important Properties • Inverse Laplace Transform 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 (Time Domain) L −1 [ • ] L[ •] Algebraic Equations ( s-domain ) 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 s in φ ) • Moving Between Representations – Phasor (Euler) Form → Coordinate Form Coordinate 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 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) Domain 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 Domain s - Domain Signals e.g. sin(ωt), … 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 ) ] Th The Laplace transform of the derivative of f (t) is: F2 ( s ) = L[ f 2 (t ) ] a and b are arbitrary constants, are then L[ a f1 (t ) + b f 2 (t ) ] = d L ⎡ dt f ( t ) ⎤ = ⎣ ⎦ ∫ ∞ 0 ( df ) e − st dt dt = L ⎡ && ( t ) ⎤ = ⎣f ⎦ ⎡f ⎤ L ⎣ &&&( 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 School of Mechanical Engineering Purdue University L ⎡o ⎤ ⎣⎦ ⎯⎯⎯ → ←⎯[⎯ ⎯ L −1 o ] 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 The Laplace transform of the definite integral 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 • d λ 123 Integration L ⎡o ⎤ ⎣⎦ ⎯⎯⎯ → ←⎯⎯ ⎯ L −1 [o] 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 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 operation. the time its representation can be inferred from the context. However, when doing context. algebraic manipulations, one should be very careful in distinguishing the signals from the operators (systems). (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 the function’s Laplace transform by 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 s-domain 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 Partial-Fraction Expansion (PFE) to break up the strictly Partialproper rational function as a series of components, whose ti 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 (Time Domain) L −1 [ • ] L[ •] Algebraic Equations ( s-domain ) Y(s): Solution in Laplace Domain School of Mechanical Engineering Purdue University ME375 Laplace - 11 Examples (first order system) 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 Example (2nd order system with distinct roots) Find the free response of a 2nd order system with two distinct real characteristic roots (e.g., a motor rotor turning inside its bearings): && + 9 y + 18 y = 0 & y where y (0) = 0 a nd y ( 0) = 3 and & (0) J J θ τ (t) B School of Mechanical Engineering Purdue University ME375 Laplace - 13 Example (Cont) Another way to find a and b (by Partial Fraction Expansion): School of Mechanical Engineering Purdue University ME375 Laplace - 14 7 ME375 Handouts Partial Fraction Expansion Formula I For For real distinct roots School of Mechanical Engineering Purdue University ME375 Laplace - 15 Example (2nd order system with repeated roots) Find the free response of a 2nd order system with two identical real characteristic roots (e.g., boxcar arrester): && + 10 y + 25 y = 0 & y where y (0) = 1 and y ( 0) = 5 a nd & (0) School of Mechanical Engineering Purdue University ME375 Laplace - 16 8 ME375 Handouts Partial Fraction Expansion Formula II For For repeated roots School of Mechanical Engineering Purdue University ME375 Laplace - 17 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 - 18 9 ME375 Handouts Example (2nd order system with complex roots) Find the free response of a 2nd order system with complex characteristic roots (e.g., vehicle suspension system): && + 6 y + 25 y = 0 & y where & y (0) = 1 and y (0) = − 3 School of Mechanical Engineering Purdue University ME375 Laplace - 19 Partial Fraction Expansion Formula III For complex roots School of Mechanical Engineering Purdue University ME375 Laplace - 20 10 ME375 Handouts Which case to apply in PFE School of Mechanical Engineering Purdue University ME375 Laplace - 21 Examples Find the unit step response of a 2nd order system: && + 6 y + 25 y = u + 25 u & & y w here & y (0) = 0 and y (0) = 0 School of Mechanical Engineering Purdue University ME375 Laplace - 22 11 ME375 Handouts Examples Find the unit step response of a 2nd order system: && + 6 y + 8 y = u + 3u & & y w here & y (0) = 0 and y (0) = 0 School of Mechanical Engineering Purdue University ME375 Laplace - 23 12 ...
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## This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue.

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