5 - TransferFunctions5 filled

5 - TransferFunctions5 filled - ME 575 Handouts Transfer...

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Unformatted text preview: ME 575 Handouts Transfer Functions • Laplace Transform and Inverse Laplace Transform • Free and Forced Response • Transfer Functions • Stability ME575 Session 5 – Transfer Functions School of Mechanical Engineering Purdue University Slide 1 Motivation – Solving Differential Eq. Differential Equations (ODEs) + Initial Conditions (ICs) y(t): Solution in Time Domain (Time Domain) L −1 [ • ] L[ •] Algebraic Equations ( s-domain ) ME575 Session 5 – Transfer Functions Y(s): Solution in Laplace Domain School of Mechanical Engineering Purdue University Slide 2 1 ME 575 Handouts Review of Complex Numbers • The Many Faces of a Complex Number: – Coordinate Form: z = x + jy Img. z – Phasor (Euler) Form: z = Ae jφ = A(cos 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 – Coordinate Form → Phasor (Euler) Form ⎧ A = x2 + y 2 ⎪ ⎨ ⎪φ = atan2( y , x ) ⎩ ⎧ tan −1 ( y x ) when z is in the 1st or 4th quadrant ⎪ atan2( y , x ) ≡ ⎨ tan −1 ( y x ) + π when z is in the 2nd quadrant ⎪ tan −1 ( y ) − π when z is in the 3rd quadrant x ⎩ School of Mechanical Engineering Purdue University ME575 Session 5 – Transfer Functions Slide 3 Definition of Laplace Transform • 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 ) ] = ME575 Session 5 – Transfer Functions 1 2 πj ∫ c + j∞ c − j∞ F ( s ) e st ds School of Mechanical Engineering Purdue University Slide 4 2 ME 575 Handouts Important Properties • Linearity Given • Differentiation F1 ( s ) = L[ f1 (t ) ] F2 ( s ) = L[ f 2 (t ) ] a and b are arbitrary constants, are then L [ af1 (t ) + bf 2 (t ) ] = Given F ( s ) = L[ f ( t ) ] The Laplace transform of the derivative of f (t) is: ∞ d L ⎡ dt f ( t ) ⎤ = ∫ − ( df ) e − st dt dt ⎣ ⎦ 0 • Integration Given = F ( s ) = L[ f (t ) ] L ⎡ && ( t ) ⎤ = ⎣f ⎦ ⎤ L ⎡ &&&( t ) ⎦ = ⎣f t L ⎡ ∫ − f (λ ) d λ ⎤ = ⎢0 ⎥ ⎣ ⎦ – For zero initial condition: zero t ∫0 • d λ 123 Integration L ⎡o ⎤ ⎣⎦ ⎯⎯⎯ → ←⎯⎯ ⎯ L −1 [ o] 1 L [• ] s2 13 Division by s L ⎡o ⎤ ⎣⎦ ⎯⎯⎯ → d • dt { Differentiation ←⎯[⎯ ⎯ L −1 o ] s L [• ] { Multiplication by s School of Mechanical Engineering Purdue University ME575 Session 5 – Transfer Functions Slide 5 Important Properties • Some Laplace Transform Pairs: U nit Im pulse δ ( t ) ⇔ 1 • Initial Value Theorem f (0 + ) = lim sF ( s ) s→∞ U nit S tep u ( t ) = 1 • Final Value Theorem ⇔ ⇔ ⇔ te − at ⇔ sin(ω t ) ⇔ cos(ω t ) ME575 Session 5 – Transfer Functions ⇔ e − at t→∞ t tn f ( ∞ ) = lim f ( t ) = lim sF ( s ) ⇔ s→ 0 School of Mechanical Engineering Purdue University 1 s 1 s2 n! s n +1 1 s+a 1 (s + a)2 ω s2 + ω 2 s 2 s +ω2 Slide 6 3 ME 575 Handouts Free & Forced Responses • Free Response (u(t) = 0 & nonzero ICs) – The response of a system to zero input and nonzero initial zero nonzero conditions. – Can be obtained by • Let u(t) = 0 and use LT and ILT to solve for the free response. • Forced Response (zero ICs & nonzero u(t)) – The response of a system to nonzero input and zero initial nonzero zero conditions. – Can be obtained by • Assume zero ICs and use LT and ILT to solve for the forced response (replace differentiation with s in the I/O ODE). School of Mechanical Engineering Purdue University ME575 Session 5 – Transfer Functions Slide 7 Transfer Function Given a general nth order system model: & & y ( n ) + a n −1 y ( n −1) + L + a1 y + a 0 y = bm u ( m ) + bm −1 u ( m −1) + L + b1 u + b0 u The forced response (zero ICs) of the system due to input u(t) is: – Taking the LT of the ODE: L ⎡ y ( n ) ⎤ = s nY ( s) ⎣ ⎦ ∴ ( WHY? ) n −1 s Y ( s) + an −1s Y ( s) + L + a1sY ( s) + a0Y ( s) = bm s mU ( s) + bm−1s m−1U ( s) + L + b1sU ( s) + b0U ( s) n ⇒ ⎡ s n + an−1s n −1 + L + a1s + a0 ⎤ ⋅ Y ( s) = ⎡bm s m + bm−1s m−1 + L + b1s + b0 ⎤ ⋅ U ( s) ⎣ ⎦ ⎣ 14444 244444 4 3 144444 244444 ⎦ 3 D( s ) ⇒ Y ( s) = N (s) m −1 bm s + bm−1s + L + b1s + b0 N ( s) ⋅ U ( s) = ⋅U ( s) = G ( s) ⋅U ( s) n n −1 s + an −1s + L + a1s + a0 D( s) 14444 244444 4 3 m G(s) Transfer Function ME575 Session 5 – Transfer Functions School of Mechanical Engineering Purdue University Slide 8 4 ME 575 Handouts Poles and Zeros Given a transfer function (TF) of a system: G (s) ≡ • Poles bm s m + bm −1 s m −1 + L + b1s + b0 N ( s ) = s n + an −1s n −1 + L + a1s + a0 D(s) The roots of the denominator of the TF, i.e. the roots of the characteristic equation. • Zeros The roots of the numerator of the TF. N ( s ) = bm s m + bm−1s m−1 + L + b1s + b0 D ( s) = s n + an−1s n−1 + L + a1s + a0 = bm ( s − z1 )( s − z2 )L ( s − zm ) = 0 ⇒ z1 , z2 , L , zm : m zeros of the TF = ( s − p1 )( s − p2 )L ( s − pn ) = 0 ⇒ p1 , p2 , L , pn : n poles of the TF G ( s) ≡ bm s m + bm−1s m−1 + L + b1s + b0 N ( s) = = s n + an−1s n−1 + L + a1s + a0 D( s ) ME575 Session 5 – Transfer Functions School of Mechanical Engineering Purdue University Slide 9 General Form of Free Response Given a general nth order system model: & & y( n) + an−1 y( n−1) + L+ a1 y + a0 y = bm u( m) + bm−1 u( m−1) + L+ b1 u + b0 u The free response (zero input) of the system due to ICs is: – Taking the LT with zero input, i.e., & y ( n ) + an−1 y ( n −1) + L + a1 y + a0 y = 0 ⎡ s n Y ( s ) − s n −1 y (0) − L − y ( n −1) (0) ⎤ + L + a1 [ sY ( s ) − y (0) ] + a 0 Y ( s ) = 0 ⎣ { 14 424 4 3 14 4 4 4 4 424 4 4 4 4 4 ⎦ 3 L⎡ y⎤ L ⎡ y(n) ⎤ ⎢ ⎥ ⎣ ⎦ ⇒ ⇒ L⎡ y⎤ ⎣ &⎦ ⎣⎦ ⎡ s n + a n −1 s n −1 + L + a1 s + a 0 ⎤ ⋅ YFree ( s ) = ( s n −1 + a n −1 s n − 2 + L + a1 ) y (0) + L + y ( n −1) (0) ⎣ ⎦ 144 44444 24444 4444 4 3 F (s) A Polynom ial of s T hat dep ends on IC s Y Free ( s ) = ME575 Session 5 – Transfer Functions F (s) s n + a n − 1 s n − 1 + L + a1 s + a 0 School of Mechanical Engineering Purdue University Slide 10 5 ME 575 Handouts Free Response and Pole Locations The free response of a system can be represented by: YFree ( s ) = F (s) F (s) = s n + an −1s n −1 + L + a1s + a0 ( s − p1 )( s − p2 ) L ( s − pn ) An A1 A2 + +L+ s − p1 s − p2 s − pn ≠ L ≠ p n i .e. n d istinct p oles = Assume p1 ≠ p 2 ⇒ yFree (t ) = L−1 [YFree ( s )] = A1e p1⋅t + A2 e p2 ⋅t + L + An e pn ⋅t ⎧ ⎪ ⎪ pi is real ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ pi = σ + j β ⎪ ⎩ ⎧ pi < 0 ⇒ ⎪ ⎨ pi = 0 ⇒ ⎪ ⎩ pi > 0 ⇒ Img. Real ⎧σ < 0 ⇒ ⎪ ⎨σ = 0 ⇒ ⎪σ > 0 ⇒ ⎩ ME575 Session 5 – Transfer Functions School of Mechanical Engineering Purdue University Slide 11 Transfer Function of State Space Model • Linear Time-Invariant (LTI) State Space Model: Time& x(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) • Dynamic Response in s-Domain s- • Transfer Function ME575 Session 5 – Transfer Functions School of Mechanical Engineering Purdue University Slide 12 6 ME 575 Handouts Transfer Function of Time Delay • Mathematical Model in Time-Domain Timet < Td ⎧0 z (t ) = ⎨ ⎩ x(t − Td ) t > Td • Mathematical Model in s-Domain s∞ ∞ Z ( s ) = ∫ − z (t )e − st dt = ∫ x(t − Td )e − st dt = 0 Td • Transfer Function for Pure Delay of Td sec ME575 Session 5 – Transfer Functions School of Mechanical Engineering Purdue University Slide 13 Stability • Stability Concept Describes the ability of a system to stay at its equilibrium position in the absence of any inputs. – A linear time invariant (LTI) system is stable if and if only if (iff) its free response converges to zero for all ICs. Ex: Pendulum ME575 Session 5 – Transfer Functions School of Mechanical Engineering Purdue University Slide 14 7 ME 575 Handouts Stability of LTI Systems • Stability Criterion for LTI Systems & & y(n) + an−1 y(n−1) + L+ a1 y + a0 y = bm u(m) + bm−1 u(m−1) + L+ b1 u + b0 u Stable ⇐⇒ all poles lie in the (open) left-half plane 14 244 4 3 LHP ⇐⇒ all roots of D(s) = sn + an−1sn−1 + L+ a1s + a0 lie in the (open)left-half plane 14 244 4 3 14444 24444 3 Characteristic Polynomial LHP School of Mechanical Engineering Purdue University ME575 Session 5 – Transfer Functions Slide 15 Stability of LTI Systems • Comments on LTI Stability – Stability of an LTI system does not depend on the input. (why?) – For 1st and 2nd order systems, stability is guaranteed if all the coefficients the of the characteristic polynomial are positive: D( s) = s + a0 : Stable ⇔ a0 > 0 D( s) = s + a1s + a0 : Stable ⇔ a1 > 0 and a0 > 0 2 – Effect of Poles and Zeros on Stability • Stability of a system depends only on its poles. • Zeros do not affect system stability. • Zeros affect the dynamic response of the system. ME575 Session 5 – Transfer Functions School of Mechanical Engineering Purdue University Slide 16 8 ...
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This note was uploaded on 12/09/2011 for the course ME 575 taught by Professor Meckl during the Fall '10 term at Purdue University-West Lafayette.

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