Transfer

# Transfer - ME375 Handouts Transfer Function Analysis •...

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Unformatted text preview: ME375 Handouts Transfer Function Analysis • • • • • Free & Forced Responses Transfer Function Poles & Zeros Static Gain Gain Obtain I/O Model using Transfer Functions School of Mechanical Engineering Purdue University ME375 Transfer Functions - 1 1 ME375 Handouts 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: 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: School of Mechanical Engineering Purdue University ME375 Transfer Functions - 2 2 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 • 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 3 ME375 Handouts Example Find the free and forced responses of the following I/O model where u ( 0 ) = 0 : 5 y + 30 y + 50 y = 2 u + u School of Mechanical Engineering Purdue University ME375 Transfer Functions - 4 4 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: forced – Taking the LT of the ODE: L ⎡ y ( n) ⎤ = s nY (s) ⎣ ⎦ ∴ an s nY ( s) + an −1s n −1Y ( s) + + a1sY ( s) + a0Y ( s) = bm s U ( s ) + bm−1s m−1U ( s ) + m + b1sU ( s) + b0U ( s) ⇒ ⎡ an s n + an −1s n −1 + ⎣ + a1s + a0 ⎤ ⋅ Y ( s) = ⎡bm s m + bm−1s m−1 + ⎦ ⎣ D( s ) N (s) ⇒ Y ( s) = + b1s + b0 ⎤ ⋅U ( s) ⎦ bm s m + bm−1s m−1 + + b1s + b0 N ( s) ⋅U ( s) = ⋅U ( s) = G( s) ⋅U ( s) n n −1 an s + an −1s + + a1s + a0 D( s ) G(s) Transfer Function School of Mechanical Engineering Purdue University ME375 Transfer Functions - 5 5 ME375 Handouts 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: – (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 6 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 The transfer function of the system is: bm s m + bm−1s m−1 + + b1s + b0 G( s) ≡ an s n + an −1s n −1 + + a1s + a0 – The transfer function can be interpreted as: u(t) Input Differential Equation y(t) U(s) Output Input G(s) Y(s) Output s - Domain Time Domain School of Mechanical Engineering Purdue University ME375 Transfer Functions - 7 7 ME375 Handouts Poles and Zeros Given a transfer function (TF) of a system: bm s m + bm −1 s m −1 + G (s) ≡ 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 n + an −1s n−1 + = an ( s − p1 )( s − p2 ) ⇒ p1 , p2 , G(s) ≡ + b1 s + b0 N ( s ) = + a1 s + 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 8 ME375 Handouts Static Gain • Static Gain ( G(0) ) The value of the transfer function when s = 0. If G( s ) ≡ bm s m + bm−1s m−1 + + b1s + b0 N (s) = an s n + an−1s n−1 + + 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 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 9 ME375 Handouts Example 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 ? steady School of Mechanical Engineering Purdue University ME375 Transfer Functions - 10 10 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: free – Taking the LT of the Homogeneous ODE: an y ( n ) + an −1 y ( n −1) + a n ⎡ s n Y ( s ) − s n −1 y (0) − ⎣ − y ( n −1) (0) ⎤ + a n −1 ⎡ s n −1Y ( s ) − s n − 2 y (0) − ⎦ ⎣ L ⎡ y(n) ⎤ ⎢ ⎥ ⎣ ⎦ + a1 y + a0 y = 0 L ⎡ y ( n −1) ⎤ ⎢ ⎥ ⎣ ⎦ + + a1 [ sY ( s ) − y (0) ] + a 0 Y ( s ) = 0 L⎡ y⎤ ⎣⎦ n n −1 ⎡ ⎣ a n s + a n −1 s + ⎤ + a1 s + a 0 ⎦ ⋅ Y F ree ( s ) ⇒ − y ( n − 2 ) (0) ⎤ ⎦ = ( a n s n − 1 + a n −1 s n − 2 + + a1 ) y (0) + L⎡ y⎤ ⎣⎦ + y ( n −1) (0) F (s) A Polynomial of s That depends on ICs ⇒ Y Free ( s ) = F (s) a n s + a n −1 s + n n −1 + a1 s + a 0 School of Mechanical Engineering Purdue University ME375 Transfer Functions - 11 11 ME375 Handouts Example Ex: Find the free response of the following system for y( 0 ) = 1 and y( 0 ) = 1 : 5 y + 30 y + 50 y = 2 u + u School of Mechanical Engineering Purdue University ME375 Transfer Functions - 12 12 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 + an −1s + + a1s + a0 an ( s − p1 )( s − p2 ) n n −1 ( s − pn ) An A1 A2 + ++ s − p1 s − p2 s − pn ≠ ≠ p n i .e. n distinct poles 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 + 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 + A n e p n ⋅t Img. i i i i Real i School of Mechanical Engineering Purdue University ME375 Transfer Functions - 13 13 ME375 Handouts Complete Response U(s) Input • G ( s) ≡ N ( s) D(s) ICs: y(0) , y(0) , Y(s) Output , y(n−1) (0) Complete Response 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 14 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. M1z1 + ( B1 + B2 ) z1 + ( K1 + K2 ) z1 − B2 z2 − K2 z2 = f (t) z2 M2 K2 M2 z2 + B2 z2 + K2z2 − B2 z1 − K2 z1 = 0 B2 z1 M1 K1 f(t) B1 Input: Output: School of Mechanical Engineering Purdue University ME375 Laplace - 15 15 ME375 Handouts Obtaining I/O Model Using TF School of Mechanical Engineering Purdue University ME375 Laplace - 16 16 ...
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