Transfer Function

# Transfer Function - 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: & τ y + y = Ku – Take LT of the I/O model and remember to keep tracks of the ICs: 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: out the output side of the equation: School of Mechanical Engineering Purdue University 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. zero nonzero – Can be obtained by 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. nonzero zero – 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 ME375 Transfer Functions - 3 In Class Exercise Find the free and forced responses of the following I/O model: &&& + 4 && − y + 5 y = 2u − u + u && & y y& 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) + 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 The forced response (zero ICs) of the system due to input u(t) is: is: – Taking the LT of the ODE: ( WHY? ) L ⎡ y ( n ) ⎤ = s nY ( s) ⎣ ⎦ n −1 an 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 ∴ ⎡ ⎤ ⎡ ⎤ ⇒ ⎣ an 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 44444 2 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 an s + an−1s + L + a1s + a0 D( s ) 14444 244444 4 3 m G(s) Transfer Function School of Mechanical Engineering Purdue University ME375 Transfer Functions - 5 Examples Find the transfer function of the system. the transfer function of the system (2) For the following 2nd order system: && + 2 ζ ω n y + ω n 2 y = K ω n 2 u & y Find the transfer function of the system. the transfer function of the system – Taking LT of the ODE: – (1) Recall the first order system: & τ y+ y= Ku Taking LT of the ODE: 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) + L+ a1 y + a0 y = bm u(m) + bm−1u(m−1) + L+ b1u + b0 u The transfer function of the system is: b s m + b s m−1 + L + b1s + b0 G ( s) ≡ m n m−1 n −1 an s + an −1s + L + 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 + L + b1s + b0 N ( s ) = an s n + an −1 s n −1 + L + a1s + a0 D ( s ) • 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 • Zeros The roots of the numerator of the TF. N ( s ) = bm s m + bm−1s m−1 + L + b1s + b0 = bm ( s − z1 )( s − z2 )L ( s − zm ) = 0 + L + a1s + a0 = an ( s − p1 )( s − p2 )L ( s − pn ) = 0 ⇒ z1 , z2 , L , zm : m zeros of the TF ⇒ p1 , p2 , L , pn : n poles of the TF G ( s) ≡ bm s m + bm−1s m−1 + L + b1s + b0 N ( s) = = an s n + an −1s n −1 + L + 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 when s = 0. If hen b s m + b s m−1 + L + b1s + b0 N ( s) G( s) ≡ m n m−1 n−1 = an s + an−1s + L + a1s + a0 D(s) ⇒ K S = G(0) = Ex: For a second order system: && + 2ζω n y + ω n 2 y = K sω n 2 u & y 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 step response. Ex: Find the steady state value of the system && & &&& + 3 && + 5 y + 7 y = u + 2 u + u & y y 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: & 5 && + 2 y + 4 y = 6 u y (1) Find the transfer function of the system (2) Find the poles and zeros of the system (3) (3) What is the system forced response Y(s) to a unit step input u(t) = 1? syste to step (4) As time goes to infinity, what is the steady state value of the unit step response ? steady 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) + L+ a1 y + a0 y = bm u( m) + bm−1 u( m−1) + L+ b1 u + b0 u free 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) + L + a1 y + a0 y = 0 & a n ⎡ s n Y ( s ) − s n −1 y (0) − L − y ( n −1) (0) ⎤ + a n −1 ⎡ s n −1Y ( s ) − s n − 2 y (0) − L − y ( n − 2 ) (0) ⎤ ⎣ ⎣ 3 14 4 4 4 442 4 4 4 4 4 4 ⎦ 3 14 4 4 4 4 2 4 4 4 4 4 4⎦ 4 L ⎡ y ( n −1) ⎤ ⎢ ⎥ ⎣ ⎦ L ⎡ y(n) ⎤ ⎢ ⎥ ⎣ ⎦ + L + a1 [ sY ( s ) − y (0) ] + a 0 Y ( s ) = 0 { 14 424 4 3 L⎡ y⎤ ⎣⎦ L⎡ y⎤ ⎣ &⎦ n n −1 ⎡ ⎣ a n s + a n −1 s + L + a1 s + a 0 ⎤ ⋅ YF ree ( s ) ⎦ = ( a n s n −1 + a n −1 s n − 2 + L + a1 ) y ( 0) + L + y ( n −1) (0) 14 4 4 4 4 4 4 424 4 4 4 4 4 4 43 4 ⇒ 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 + L + 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 && + 3 0 y + 5 0 y = 2 u + u y Ex: Perform partial fraction expansion (PFE) of the above free response when: & 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 + L + a1s + a0 an ( 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 y F re e ( t ) = L − 1 Y F re e ( s ) = A1 e p1 ⋅t + A 2 e p 2 ⋅t + L + 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) ≡ N ( s) D( s) Y(s) Output ICs: y(0) , y(0) , L , y(n−1) (0) (0) &(0) (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 ? Q: If U(s) = 0 and there are non-zero ICs, what will guarantee that y(t) → 0 ? and non- School of Mechanical Engineering Purdue University ME375 Transfer Functions - 14 7 ME375 Handouts Obtaining I/O Model Using TF Obtain the I/O model for the vibration absorber shown below. The input is the below. force f(t) and the output is the displacement of mass m2. & & M1&&1 + ( B1 + B2 ) z1 + ( K1 + K2 ) z1 − B2z2 − K2z2 = f (t) z z2 M2 K2 & & M2&&2 + B2z2 + K2 z2 − B2 z1 − K2z1 = 0 z B2 z1 M1 K1 f(t) B1 Input: Output: School of Mechanical Engineering Purdue University ME375 Laplace - 15 Obtaining I/O Model Using TF School of Mechanical Engineering Purdue University ME375 Laplace - 16 8 ...
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## This note was uploaded on 12/23/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue.

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