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Transfer Func filled - ME375 Handouts Transfer Function...

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Unformatted text preview: ME375 Handouts Transfer Function Analysis • Free & Forced Responses • Transfer Function • System Stability School of Mechanical Engineering Purdue University ME375 Transfer Functions - 1 Free & Forced Responses Ex: Let’s look at a stable first order system: Let’ τ y + y = Ku – Take LT of the I/O model and remember to keep tracks of the ICs: L [τ y + y ] = L [ Ku ] ⇒ τ( )+ =K⋅ – Rearrange terms s.t. the output Y(s) terms are on one side and the input U(s) and IC terms are on the other: ( ) ⋅ Y (s) = ( ) ⋅U (s) + ( ) ⋅ y(0) – Factor out the output side of the equation: ⎛ Y (s) = ⎜ ⎝ ⎞ ⎛ ⎟ ⋅U ( s ) + ⎜ ⎠ ⎝ School of Mechanical Engineering Purdue University ⎞ ⎟ ⋅ y (0) ⎠ ME375 Transfer Functions - 2 1 ME375 Handouts Free & Forced Responses • Free Response (u(t) = 0 & nonzero ICs) – The response of a system to zero input and nonzero initial conditions. 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 conditions. conditions. – Can be obtained by • Assume zero ICs and use LT and ILT to solve for the forced response response (replace differentiation with s in the I/O ODE). School of Mechanical Engineering Purdue University ME375 Transfer Functions - 3 In Class Exercise Find the free and forced responses of the following I/O model: y + 4 y − y + 5 y = 2u − u + u School of Mechanical Engineering Purdue University ME375 Transfer Functions - 4 2 ME375 Handouts Transfer Function Given a general nth order system model: a n y ( n ) + a n −1 y ( n −1) + + a1 y + a 0 y = bm u ( m ) + bm −1 u ( m −1) + + b1 u + b0 u The forced response (zero ICs) of the system due to input u(t) is: – Taking the LT of the ODE: ( WHY? ) L ⎡ y ( n ) ⎤ = s nY ( s) ⎣ ⎦ n −1 an s Y ( s) + an −1s Y (s) + + a1sY (s) + a0Y ( s) = bm s mU ( s) + bm−1s m−1U ( s) + n ∴ ⇒ ⎡ an s + an−1s ⎣ n n −1 + + a1s + a0 ⎤ ⋅ Y ( s) = ⎡bm s + bm−1s ⎦ ⎣ m D( s ) ⇒ Y (s) = + b1sU (s) + b0U (s) m −1 + + b1s + b0 ⎤ ⋅U ( s) ⎦ N (s) m −1 bm s + bm−1s + + b1s + b0 N (s) ⋅U (s) = ⋅U ( s) = G ( s ) ⋅U ( s) n n −1 an s + an−1s + + a1s + a0 D( s ) m G(s) Transfer Function School of Mechanical Engineering Purdue University ME375 Transfer Functions - 5 Examples Find the transfer function of the system. (2) For the following 2nd order system: y + 2ζω n y + ω n2 y = K ω n2 u Find the transfer function of the system. – Taking LT of the ODE: ODE: – Taking LT of the ODE: ODE: (1) Recall the first order system: τ y + y = K u School of Mechanical Engineering Purdue University ME375 Transfer Functions - 6 3 ME375 Handouts Transfer Function Given a general nth order system: an y(n) + an−1 y(n−1) + + a1 y + a0 y = bm u(m) + bm−1u(m−1) + + b1u + b0 u The transfer function of the system is: b s m + b s m−1 + + b1s + b0 G ( s) ≡ m n m−1 n −1 an s + an −1s + + a1s + a0 – The transfer function can be interpreted as: u(t) Differential Equation Input y(t) ⇒ U(s) Output ⇐ Input Time Domain G(s) Y(s) Output s - Domain School of Mechanical Engineering Purdue University ME375 Transfer Functions - 7 Poles and Zeros Given a transfer function (TF) of a system: G (s) ≡ bm s m + bm −1 s m −1 + an s n + an −1 s n −1 + • Poles The roots of the denominator of the TF, i.e. the roots of the characteristic equation. D ( s) = an s + an−1s n n −1 + = an ( s − p1 )( s − p2 ) ⇒ p1 , p2 , G ( s) ≡ + b1s + b0 N ( s ) = + a1s + a0 D ( s ) • Zeros The roots of the numerator of the TF. N ( s ) = bm s m + bm−1s m−1 + = bm ( s − z1 )( s − z2 ) + a1s + a0 ( s − pn ) = 0 ⇒ z1 , z2 , + b1s + b0 ( s − zm ) = 0 , zm : m zeros of the TF , pn : n poles of the TF bm s m + bm−1s m−1 + + b1s + b0 N ( s) = = an s n + an −1s n−1 + + a1s + a0 D( s) School of Mechanical Engineering Purdue University ME375 Transfer Functions - 8 4 ME375 Handouts Static Gain • Static Gain ( G(0) ) The value of the transfer function when s = 0. If b s m + b s m−1 + + b1s + b0 N ( s) G( s) ≡ m n m−1 n−1 = an s + an−1s + + a1s + a0 D(s) ⇒ K S = G(0) = Ex: For a second order system: y + 2ζω n y + ω n 2 y = K sω n 2 u Find the transfer function and the static gain. N (0) b0 = D(0) a0 The static gain KS can be interpreted as the steady state value of the unit step response. Ex: Find the steady state value of the system y + 3 y + 5 y + 7 y = u + 2u + u to a step input of magnitude 2. School of Mechanical Engineering Purdue University ME375 Transfer Functions - 9 In Class Exercise Given an I/O model of a 2nd order system: 5y + 2y + 4y = 6 u (1) Find the transfer function of the system (2) Find the poles and zeros of the system (3) What is the system forced response Y(s) to a unit step input u(t) = 1? (4) As time goes to infinity, what is the steady state value of the unit step response ? School of Mechanical Engineering Purdue University ME375 Transfer Functions - 10 5 ME375 Handouts A Closer Look at Free Response Given a general nth order system model: an y( n) + an−1 y( n−1) + + a1 y + a0 y = bm u( m) + bm−1 u( m−1) + + b1 u + b0 u The free response (zero input) of the system due to ICs is: – Taking the LT of the Homogeneous ODE: an y ( n ) + an −1 y ( n −1) + an ⎡ s Y ( s ) − s ⎣ n n −1 y (0) − −y ( n −1) (0) ⎤ + a n −1 ⎡ s ⎦ ⎣ n −1 Y (s) − s n−2 y (0) − + a1 y + a0 y = 0 − y ( n − 2 ) (0) ⎤ ⎦ L ⎡ y ( n −1) ⎤ ⎢ ⎥ ⎣ ⎦ L ⎡ y(n) ⎤ ⎢ ⎥ ⎣ ⎦ + + a1 [ sY ( s ) − y (0) ] + a 0 Y ( s ) = 0 L⎡ y⎤ ⎣ ⎦ L⎡ y⎤ ⎣ ⎦ ⎡ a n s n + a n −1 s n −1 + ⎣ + a1 s + a 0 ⎤ ⋅ YF ree ( s ) ⎦ ⇒ = ( a n s n −1 + a n −1 s n − 2 + + a1 ) y (0) + + y ( n −1) (0) F (s) A Polynomial of s T hat depends on ICs ⇒ YFree ( s ) = F (s) a n s n + a n −1 s n −1 + + a1 s + a 0 School of Mechanical Engineering Purdue University ME375 Transfer Functions - 11 A Closer Look at Free Response Ex: Find the free response of the following system: 5 y + 30 y + 50 y = 2 u + u Ex: Perform partial fraction expansion (PFE) of the above free response when: response y( 0 ) = 1 and y( 0 ) = 1 School of Mechanical Engineering Purdue University ME375 Transfer Functions - 12 6 ME375 Handouts Free Response and Pole Position The free response of a system can be represented by: YFree ( s ) = = Assume ⇒ p1 ≠ p 2 F (s) F (s) = an s n + an −1s n −1 + + a1s + a0 an ( s − p1 )( s − p2 ) ( s − pn ) An A1 A2 + + + s − p1 s − p2 s − pn ≠ ≠ p n i.e. n d istinct p oles y F r e e ( t ) = L − 1 Y F r e e ( s ) = A1 e p 1 ⋅t + A 2 e p 2 ⋅t + + A n e p n ⋅t R |p = 0 ⇒ | p is real R p < 0 ⇒ S | |p > 0 ⇒ | T | S | | p = σ + jβ Rσ < 0 ⇒ |σ = 0 ⇒ | S | |σ > 0 ⇒ T T Img. i i i i Real i School of Mechanical Engineering Purdue University ME375 Transfer Functions - 13 Complete Response U(s) Input • Complete Response G ( s) ≡ Y(s) N ( s) D( s) ICs: y(0) , y(0) , Output (n−1) ,y (0) Y ( s ) = YF orced ( s ) + YF ree ( s ) = G ( s ) ⋅ U ( s ) + YF ree ( s ) = ⋅ U (s) + Q: What part of the system affects both the free and forced response ? response Q: If U(s) = 0 and there are non-zero ICs, what will guarantee that y(t) → 0 ? U( nony( School of Mechanical Engineering Purdue University ME375 Transfer Functions - 14 7 ME375 Handouts Stability • Stability Concept Describes the ability of a system to stay at its equilibrium position (for linear position systems: all state variables = 0 or y(t) = 0) in the absence of any inputs. – A linear time invariant (LTI) system is stable if and only if (iff) its free response converges to zero. Ex: Pendulum Ball on curved surface School of Mechanical Engineering Purdue University ME375 Transfer Functions - 15 Stability of LTI Systems • Stability Criterion for LTI Systems an y(n) + an−1 y(n−1) + Stable + a1 y + a0 y = bm u(m) + bm−1 u(m−1) + ⇐⇒ all poles of D(s) = an sn + an−1sn−1 + + b1 u + b0 u + a1s + a0 lie in the left-half complex plane Characteristic Polynomial LHP • Comments on LTI Stability – Stability of an LTI system does not depend on the input. (why?) (why?) – For 1st and 2nd order systems, stability is guaranteed if all the coefficients of the the characteristic polynomial are positive. D( s ) = a1s + a0 : Stable ⇔ ai > 0 ∀ i or ai < 0 ∀ i D( s ) = a2 s + a1s + a0 : Stable ⇔ ai > 0 ∀ i or ai < 0 ∀ i 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 transient response of the system. School of Mechanical Engineering Purdue University ME375 Transfer Functions - 16 8 ME375 Handouts In Class Exercises (1) Find the transfer function of the following I/O equation: − y − 2 y − 5y = 2u + u (2) Determine the system’s stability. system’ (3) Plot the poles and zeros of the system on the complex plane. (1) Find the transfer function of the following I/O equation: y + y + 6 y = u − 3u + 4 u (2) Determine the system’s stability. system’ (3) Plot the poles and zeros of the system on the complex plane. School of Mechanical Engineering Purdue University ME375 Transfer Functions - 17 Example (Inverted) Pendulum (1) Derive a mathematical model for a pendulum. (2) Find the equilibrium positions. (3) Discuss the stability of the equilibrium positions. School of Mechanical Engineering Purdue University ME375 Transfer Functions - 18 9 ME375 Handouts Example (Inverted Pendulum) School of Mechanical Engineering Purdue University ME375 Transfer Functions - 19 10 ...
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