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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 nonzero 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 lefthalf 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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 Spring '10
 Meckle
 Mechanical Engineering

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