{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

5 - TransferFunctions5 filled

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online