Transfer function and linearization -...

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Transfer functionLinearization

Transfer function and linearization

Daniele Carnevale

Dipartimento di Ing. Civile ed Ing. Informatica (DICII),

University of Rome “Tor Vergata”

Corso di Controlli Automatici, A.A. 2014-2015

Testo del corso: Fondamenti di Controlli Automatici,

P. Bolzern, R. Scattolini, N. Schiavoni.

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Example

Space state and transfer function

Consider Single Input Single Output (SISO) Linear Time-invariant System(LTI) described by the differential equation (continuous time)

x = Ax+Bu, y = Cx+Du, (1)

with x ∈ Rn, u ∈ R, y ∈ R.The solution ϕ(t, x0, u(·)) := x(t) of (1) with initial condition x0 is

x(t) = eAtx0 +

eA(t−τ)Bu(τ)d τ, [proof: substitute in (1)] (2)

the system output is then

y(t) = CeAtx0︸︷︷︸free response

eA(t−τ)Bu(τ)d τ +Du(t)︸︷︷︸forced response

The free response is characterized by system modes of the type eλi tthi/hi!,where λi ∈ σ{A} ⊂ C and 0 ≤ hi ≤ n is the dimension of the largest Jordanblock associated to λi. The forced response contains the modes of the inputand the output eαj ttqj/qj !. There might be higher order (tqj+1) polynomialsin the response if αj = λi for some i and j (resonance).

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Example

Laplace transform: the transfer function

Applying Laplace transform to (1) yields

L[x(t)](s) := x(s) = (sI −A)−1x0 + (sI −A)−1Bu(s) (4)

L[y(t)](s) := y(s) = C(sI −A)−1x0 +(C(sI −A)−1B +D

)u(s). (5)

Definition (Transfer function)

Assuming x0 = 0, then

FDT (s) :=y(s)

u(s)= C(sI −A)−1B +D =

Γmi=0(s− zi)Γni=0(s− pi)

is the system transfer function and is a proper rational function (ratio ofpolynomials of s where m ≤ n ≤ n). Furthermore, if every pole (root of thedenominator) pi of the FDT has non-positive real part (pi ∈ C−0 ) then ifu(t) = E cos(ωt+ θ) it holds

limt→∞

y(t) = ρ(ω)E cos(ωt+ θ + ϕ(ω)), (7)

where ρ(ω) := |FDT (jω)| and ϕ(ω) := FDT (jω).

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Example

Transient response

The forced response, using inverse Laplace transform and residuals can bewritten as

y(t) = L−1[C(sI −A)−1x0](t) + L−1[FDT (s)u(s)](t)

= |x0=0 L−1

Ri,j(s− pi)j

Ri,j(s− αi)j

Ri,j tj−1epi t

(j − 1)!︸︷︷︸transient response

Ri,j tj−1eαi t

(j − 1)!︸︷︷︸regime response

If pi ∈ C− then

limt→∞

Ri,j tj−1epi t

(j − 1)!︸︷︷︸transient response

What happen if pi ∈ C−0 ? Does exist an initial condition x0 such thaty(t) = ρ(ω)E cos(ωt+ θ + ϕ(ω)) for all t ≥ 0?

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Example

First order rational functions (not all of them are proper transfer functions)

P1(s) =1

s+ 10, P2(s) =

1, P3(s) =

0.1s+ 1, P4(s) =

−0.1s+ 1

P5(s) =1

s− 10, P6(s) =

s+ 100

s+ 10, P7(s) = 2

s+ 100.

−100

10−1 100 101 102 103 104−180

Bode Diagram

Frequency (rad/s)

P1P2P3P4P5P6P7

Figure : Bode plots - first order functions. 5 / 13

Example

P1 = t f ( 1 , [ 1 1 0 ] ) ; P2 = t f ( [ 1 1 0 ] , 1 ) ; P3 = t f ( 1 , [ 0 . 1 1 ] ) ;P4 = t f ( [−0.1 1 ] , [ 1 ] ) ; P5 = t f ( 1 , [ 1 −10]) ;P6 = t f ( [ 1 1 0 0 ] , [ 1 1 0 ] ) ; P7 = zpk ( [ −1 0 ] , [ −1 0 0 ] , 2 ) ;

f i g u r e ( 1 )bode ( P1 , P2 , P3 , P4 , P5 , P6 , P7 ) ;legend ( ’ P 1 ’ , ’ P 2 ’ , ’ P 3 ’ , ’ P 4 ’ , ’ P 5 ’ , ’ P 6 ’ , ’ P 7 ’ )set ( f i n d a l l ( gcf , ’ t y p e ’ , ’ l i n e ’ ) , ’ l i n e w i d t h ’ , 3 )hold ong r i d on

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Example

Step response: first order rational functions

P1(s) =1

s+ 10, P2(s) =

1, P3(s) =

0.1s+ 1, P4(s) =

−0.1s+ 1

P5(s) =1

s− 10, P6(s) =

s+ 100

s+ 10, P7(s) = 2

s+ 100.

0 1 2 3 4 5 60

Step Response

Time (seconds)

P1P3P5P6P7

Figure : Step response - first order functions. 7 / 13

Example

Step response: first order rational functions

P1(s) =1

s+ 10, P2(s) =

1, P3(s) =

0.1s+ 1, P4(s) =

−0.1s+ 1

P5(s) =1

s− 10, P6(s) =

s+ 100

s+ 10, P7(s) = 2

s+ 100.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−10

time [s ]

P1P3P5P6P7

0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58

time [s ]

P1P3P5P6P7

Figure : Input response : u(t) = 2 sin(t · 2π · 30). 8 / 13

Example

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’ b−− ’ , ’ g−− ’ , ’ k−− ’ , ’ c−− ’ , ’ y−− ’ , ’m−− ’ , ’ r : ’ , ’ b : ’ , ’ g : ’ , . . .’ k : ’ , ’ c : ’ , ’ y : ’ , ’m: ’ , ’ r .− ’ , ’ b.− ’ , ’ g.− ’ , ’ k.− ’ , ’ c .− ’ , . . .’ y .− ’ , ’m.− ’ , ’ r−o ’ , ’ b−o ’ , ’ g−o ’ , ’ k−o ’ , ’ c−o ’ , ’ y−o ’ , ’m−o ’ , . . .’ r−x ’ , ’ b−x ’ , ’ g−x ’ , ’ k−x ’ , ’ c−x ’ , ’ y−x ’ , ’m−x ’ } ;

myFontSize = 1 6 ;

f i g u r e ( 1 )f o r i = 1 : 7

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endendho ld o f fl egend ( ’ P 1 ’ , ’ P 3 ’ , ’ P 5 ’ , ’ P 6 ’ , ’ P 7 ’ , ’ L o c a t i o n ’ , ’ NorthWest ’ )x l a b e l ( ’ t ime [ s ] ’ , ’ F o n t S i z e ’ , myFontSize , ’ I n t e r p r e t e r ’ , ’ Latex ’ )y l a b e l ( ’ y ( t ) ’ , ’ F o n t S i z e ’ , myFontSize , ’ I n t e r p r e t e r ’ , ’ Latex ’ )ho ld ong r i d on 9 / 13

Example

Second order rational functions

P1(s) =ω2n

s2 + 2ζωns+ ω2n

P2(s) =1 + 2s

s2 + 1s+ 1, P3(s) =

1−2ss2 + 1s+ 1

−100

10−2 10−1 100 101 102−180

−135

Bode Diagram

Frequency (rad/s)

10−2 10−1 100 101 102−90

Bode Diagram

Frequency (rad/s)

Figure : Left: P1(s) with ζ ∈ [0, 1]. Right: P2(s) and P3(s). 10 / 13

Example

Second order rational functions

P1(s) =ω2n

s2 + 2ζωns+ ω2n

P2(s) =1 + 2s

s2 + 1s+ 1, P3(s) =

1−2ss2 + 1s+ 1

, [Non-minum phase plants: zeros ∈ C+]

0 5 10 150

time [s ]

0 5 10 15−1

−0.5

time [s ]

Figure : Step response of P1(s) with ζ ∈ [0, 1]. Right: step response of P2(s) (RED)and P3(s) (BLUE).

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Linearization

Consider the nonlinear differential equation

x = f(x, u, d),

where the system state is x ∈ Rn with input u ∈ Rp and disturbance d ∈ Rq.Let (xe, ue, de) be the equilibrium triple such that f(xe, ue, de) = 0, then thelinearization of the system dynamics around such point is

˙x ≈ ∂f(x, u, d)

∣∣∣∣(xe,ue,de)

x+∂f(x, u, d)

u+∂f(x, u, d)

≈ Ax+Bu+Md, (9)

where x = x− xe, u = u− ue and d = d− de.

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Linearization: cart-pendulum

Figure : Inverted pendulum with cart (Matlab).

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