Classical Methods of Control Systems

BLOCK DIAGRAM REPRESENTATION OPEN LOOP PLANT INPUT

x(t ) X (s)

G (s )

OUTPUT

y (t ) Y ( s)

Gc R

Y ( s) = G ( s) × X ( s) Y (s) G ( s) = X (s)

CONTROLLER

GP

X

PLANT

Error, E = R − Y = 0 is desired

Y

X = RGC Y = XGP ⇒ Y = RGC GP Now , E = R − RGC GP = 0 ⇒ R(1 − GC GP ) = 0 Q R ≠ 0, 1 − GC GP = 0 1 GC = GP

CLOSED LOOP CONTROLLER

R

E ─

C

PLANT

U

G1

MEASUREMENT

H

Y = UG1 U = EC E = R − YH

Y CG1 = R 1+ CG1H

Denominator : 1 + CG1 H = 1 + GH { G

Y

1 + GH → Characteristic polynomial 1 + GH = 0 → Characteristic equation (CE) Y G = R 1 + GH If CE is known, we can determine, • order of system • eigen values • stability • response characteristics

EXAMPLE Applying Conservation of Angular Momentum h& = τ − mgl sin θ

h = ml θ&

(

h& = ml 2θ&&

⇒

2

h = Iθ&

ml 2θ&& + mgl sin θ = τ 1 mgl & & θ + 2 sin θ = 2 τ ml ml

τ l

θ

)

m

ml 2θ&& = τ − mgl sin θ Need to linearize • Operating point • Static equilibrium point

θ = θ (t )

g 1 & & θ + sin θ = 2 τ l ml

τ = τ (t ) θ& = θ&& = 0

Static Equilibrium,

θ 0 ,τ 0

l

1 sin θ 0 = τ = τ 0 0 2 gml gml Assume τ 0 = 0 ⇒ ⇒

180o

sin θ 0 = 0

θ 0 = 0 or 180

© 2005 P. S. Shiakolas

o

o

0o

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

LINEARIZATION Æ Taylor’s Series Expansion

(

g 1 & & θ + sin θ − 2 τ = 0 = f θ&&, g , l , m,τ , θ l ml

(

)

)

∂f && && ∂f ∂f ( ( − θ0 + − l0 ) f − f0 = g − g0 ) + l θ 424 3 ∂l 123 ∂θ&& 0 ∂g 0 1 0 Δg =0 Δl =0 ∂f ∂f ∂f (1 ( ( + m − m0 ) + τ −τ 0 ) + θ − θ0 ) 3 ∂m ∂m 0 424 ∂θ 0 0 Δm = 0 ⎛ 1 ⎞ ⎛g ⎞ & & Δf = 0 = 1Δθ + ⎜⎜ − ⎟⎟ Δτ + ⎜ cos θ ⎟ Δθ ⎝l ⎠0 ⎝ ml 2 ⎠0 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

⎛ 1 ⎞ ⎛g ⎞ & & Δθ + ⎜ cos θ ⎟ Δθ = ⎜⎜ ⎟⎟ Δτ ⎠0 ⎝l ⎝ ml 2 ⎠ 0 If the equilibrium point is θ = 0o ⇒ 1 ⎛g ⎞ & & Δθ + ⎜ cos 0 ⎟Δθ = Δτ ⎝l ⎠ ml 2 g 1 & & Δθ + Δθ = Δτ l ml 2

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

If the equilibrium point is θ = 180o ⇒ 1 ⎞ ⎛g & & Δτ Δθ + ⎜ cos180 ⎟Δθ = ⎠ ⎝l ml 2 1 g & & Δθ − Δθ = Δτ l ml 2

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

LAPLACE TRANSFORM (Assume Zero ICs) g 1 & & Δθ + Δθ = Δτ ⇒ Laplace Transform 2 l ml

g 1 ΔΘ( s ) s + ΔΘ( s ) = ΔΓ l ml 2 2

1 ⎛ 2 g⎞ ⎜ s + ⎟ΔΘ = 2 ΔΓ l⎠ ⎝ ml ⎛ ⎞ ΔΘ 1 ⎜ 1 ⎟ ⎜ ⎟ = ΔΓ ml 2 ⎜ s 2 + g ⎟ ⎜ ⎟ l ⎠ ⎝ © 2005 P. S. Shiakolas

PLANT

ΔΘ

ΔΓ

or

⎛ ⎞ 1 ⎜ 1 ⎟ ⎜ ⎟ΔΓ ΔΘ = ml 2 ⎜⎜ s 2 + g ⎟⎟ l ⎠ ⎝

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

g s + = 0 ⇒ characteristic equation l g Im eigen values ⇒ s = ± i l 2

C.E. for θ 0 = 180o ⇒

Unstable region

g 2 g s − =0 ⇒ s=± l l If the eigen values have • positive real parts, then the system is unstable • negative real parts, then the system is stable • zero real part, then the system is marginally stable © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Re

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

g 1 & & Δθ + Δθ = Δτ l ml 2

State space ⇒ representation

Set of first order DFQs z1 = Δθ z 2 = z&1 = Δθ& ⇒ z&1 = z 2 Number of DFQs = Order of system g 1 z&2 + z1 = Δτ 2 l ml g 1 z&2 = − z1 + Δτ 2 l ml © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

z&1 = 0 z1 + 1z2 + 0Δτ g 1 Δτ z&2 = − z1 + 0 z2 + 2 l ml x& = Ax + Bu where ⎛ z&1 ⎞ ⎡ 0 ⎜ ⎟=⎢ g ⎜ z& ⎟ ⎢− ⎝ 2⎠ ⎣ l Output : © 2005 P. S. Shiakolas

A = coefficient matrix − square B = order of the system

1⎤ ⎛ z ⎞ ⎛ 0 ⎞ 1 ⎥ ⎜ ⎟ + ⎜ 1 ⎟Δτ 0⎥ ⎜ z ⎟ ⎜⎜ 2 ⎟⎟ ⎦ ⎝ 2 ⎠ ⎝ ml ⎠ y = Cx + Du ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

C.E. ⇒ det (SI − A) = 0 ⎡s det ⎢ ⎢0 ⎢⎣ s det g l

0 − g s − l

0

1⎤ ⎥=0 0⎥ ⎥⎦

−1 s

=0

g s + =0 l 2

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

PROJECT PENDULUM

g

mc

θ 0 = 180o

F

l

x

θ

© 2005 P. S. Shiakolas

mr

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

MAGLEV

electro magnet

g

sensor

object mass, m L

R

Magnet

V

I

Fmagnet

mass

Fmagnet = f (k , I , x )

weight © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

• Derive system dynamic equations • Linearize, if needed • Derive state space representation • Find transfer function • Find characteristic equation

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

LAPLACE TRANSFORMS Final Value Theorem (FVT)

f (∞) ← lim s F ( s ) s →0 Condition: All the poles of F(s) other than s = 0 must have negative non - zero real parts. Poles: Are points at which the f(t) or its derivatives → infinity Zeros: Are points at which the f(t) equals zero © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

( s + 2 )(s + 10 ) G ( s) = s(s + 1)(s + 7 )(s + 17 )2 2 zeros, 5 poles Im

Re -17

© 2005 P. S. Shiakolas

-10

-7

-2 -1 0

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

TRANSFER FUNCTIONS (T.F.) L[output ] T.F. = G ( s ) = L[input ] Zero I.C.'s Y ( s ) bm s m + bm −1s m −1 + ... + b1s + b0 = , n≥m X (s) s n + an −1s n −1 + ... + a1s + a0 n → Order of the system • Applicable only for Linear Time Invariant (LTI) Systems • Independent of input (magnitude and time) © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

BLOCK DIAGRAMS Output

Input

a1 a2

Description of component or subsystem

Σ

Branching point

b1

─

B

A

Signal

a3

Summing JCT: a1 + a2 − a3 = b1 © 2005 P. S. Shiakolas

C

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

D E

R

─

C

P

G H

PRINCIPLE OF SUPERPOSITION (LTI)

C = C R, D = 0 + R

─

CR ─

P

G H

D P

G H

GP CR = R 1 + GPH © 2005 P. S. Shiakolas

C D, R = 0

P CD = D 1 + PHG

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

CD

GP P C= R+ D 1 + GPH 1 + GPH 1 + GPH → Characteristic polynomial

STATE SPACE REPRESENTATION System of linear DFQ’s Æ System of First Order linear DFQ’s

x

k m

f (t )

b © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

m&x& + bx& + kx = f (t )

2 nd order linear DFQ ⇒ 2 first order linear DFQ in state space State space variables : z1, z2 z1 = x

z2 = z&1 = x& z&2 = &z&1 = &x& m z&2 + bz2 + kz1 = f 1 z&2 = (− kz1 − bz2 + f ) m & z&1 = z2

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

x& = Ax + Bu y = Cx + Du Y X

( x, y ) 2 variables l 2 = x 2 + y 2 Constraint Equation

l

θ

Only one variable for system state x& = A x + Bu L ⇒ sIX − x0 = AX − BU

(sI − A)X = BU + x0 X = (sI − A)−1{BU + x0 } © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

y = C x + Du L ⇒ Y = CX + DU

[

]

Y = C (sI − A)−1{BU + x0 } + DU

{

}

Y = C (sI − A)−1 B + D U + C (sI − A)−1 x0

Denominator of the above eq. sI − A Therefore, sI − A is the Characteristic Polynomial & sI − A = 0 is the Characteristic Equation of the system. © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

• Time Step for Numerical Integration • Dynamics Æ Differential Equations ¾ Linearize ¾ State Space • Block Diagram Æ State Space form

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

INTEGRATOR

x&

1

x1

x

s

x2

1 x1 = x2 s

x1 = x&2 State variables at output of integrators

1 © 2005 P. S. Shiakolas

f ( s)

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Example: U

10 s ( s + 2)

─

Z

1 s +1

U

1 s

─ X3

© 2005 P. S. Shiakolas

X2

10 s ( s + 2)

1 s +1 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

X1

Z

Y FF CE : = U 1 + FF × FB 10 1 Denominator : 1 + =0 s ( s + 2) s + 1

(s 2 + 2s )(s + 1) + 10 = 0 ← 3rd Order

10 X2 = X 1 ⇒ 10 x2 = x&1 + 2 x1 s+2 1 X1 x1 = x&3 + x3 = X3 ⇒ s +1 1 (U − X 3 ) = X 2 ⇒ u − x3 = x&2 s © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

x&1 = −2 x1 + 10 x2 + 0 x3 + 0u x&2 = 0 x1 + 0 x2 − 1x3 + 1u x&3 = 1x1 + 0 x2 − 1x3 + 0u ⎛ x&1 ⎞ ⎡− 2 ⎜ ⎟ ⎢ ⎜ x& ⎟ = ⎢ 0 ⎜ 2⎟ ⎢ ⎜ x& ⎟ ⎢ 1 ⎝ 3⎠ ⎣

10 0 0

0 ⎤ ⎛ x1 ⎞ ⎡0⎤ ⎥⎜ ⎟ ⎢ ⎥ − 1⎥ ⎜ x2 ⎟ + ⎢1⎥ u ⎥⎜ ⎟ ⎢ ⎥ 1 ⎥⎦ ⎜⎝ x3 ⎟⎠ ⎢⎣0⎥⎦

Z = 1x1 + 0 x2 + 0 x3 + 0u Y = [1 © 2005 P. S. Shiakolas

0

0] x + 0u ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Transfer Function has zeros or derivatives of input

Y f1 ( s ) = U f 2 (s) X s+a = U s 2 + 2ξωn s + ωn2 &x& + 2ξωn x& + ωn2 x = u& + au Converting into state space z1 = x z2 = x& = z&1 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

z&1 = z2 z&2 = −ωn2 z1 − 2ξωn z2 + au + u&

U

X

TF

X X Z TF = = U ZU

U

© 2005 P. S. Shiakolas

1 DEN

Z

NUM

X

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

State space representation w.r.t. variable z

X s+a = U s 2 + 2ξωn s + ωn2 From the block diagram, Z X 1 = = s+a & 2 2 U s + 2ξωn s + ωn Z Z ⇒ &z& + 2ξωn z& + ωn2 z = u U Converting into state space, φ =z φ = φ& = z& 1

© 2005 P. S. Shiakolas

2

2

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

φ&1 = 0φ1 + 1φ2 + 0u φ&2 = −ωn2φ1 − 2ξωnφ2 − 1u ⎛ φ&1 ⎞ ⎡ 0 ⎜ ⎟=⎢ ⎜ φ& ⎟ ⎢− ω 2 ⎝ 2⎠ ⎣ n

⎤ ⎛ φ1 ⎞ ⎛ 0 ⎞ ⎥⎜ ⎟+ ⎜ ⎟u − 2ξωn ⎥⎦ ⎜⎝ φ2 ⎟⎠ ⎜⎝ − 1⎟⎠ 1

X ⇒ x = z& + az Z x = aφ1 + 1φ2 + 0u © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

FEEDBACK CONTROL SYSTEM CHARACTERISTICS

STATE SPACE

BLOCK DIAGRAM

TRANSFER FUNCTION

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

PLANT

CONTROLLER

R

E ─

C

U

Y

G1

MEASUREMENT

H

F = EG

R

Φ

X GP = R 1 + GPH

─

CE :1 + GPH = 0

Find Φ such that the block diagrams are equivalent.

© 2005 P. S. Shiakolas

Unity Feedback

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

X

EFFECTS OF FEEDBACK • Affects • closed loop behavior • stability • bandwidth • overall gain • Reduces sensitivity of the output to disturbances and system/plant parameter changes. • Improve the steady state error • Ease control of transient response © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

GENERAL PROPERTIES OF F.B.C.S. T R

E ─

P

G H

GP P X= R+ T 1 + GPH 1 + GPH CE :1 + GPH = 0 E = Reference Signal − Output Signal E = R − XH © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

X

DEP

SENSITIVITY

L2 L1

Output , C or X C = f (G, P, H , T , R ) Linearize ⇒ ∂C ∂C ∂C ΔG + ΔP + ΔH ΔC = ∂H 0 ∂G 0 ∂P 0 ∂C ∂C + ΔR + ΔT ∂R 0 ∂T 0 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

IND

OPEN LOOP

H =0

CLOSED LOOP

GP P R+ T C= 1 + GPH 1 + GPH

C = GPR + PT SENSITIVITY TO DISTURBANCES ΔR = 0 ; ΔT ≠ 0 ΔG = ΔP = ΔH = 0 P ⎛ ⎞ ∂C ΔC = ⎜ ⎟ΔT ∂C = ΔT 1 + GPH ⎠ ⎝ ∂T 0

ΔC = PΔT © 2005 P. S. Shiakolas

P ΔC ΔT 1 + GPH if R = 0, = P C T 1 + GPH ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

SENSITIVITY TO SENSOR GAIN

GP C= R 1 + GPH − GPGPR ∂C ⇒ ΔC = ΔH 2 ∂H (1 + GPH ) ΔC ⇒ Normalize C ΔC − G 2 P 2 R (1 + GPH )2 = ΔH C GP 1 + GPH R © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ΔC − GP = ΔH C 1 + GPH if 1 + GPH >> 0 ⇒ ΔC ΔH GP ≅− ΔH ≅ − C GPH H ΔC ΔH ΔC ΔH ≅ − ⇒ Abs Value = C H C H One − to − one mapping

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

EXAMPLE

G=

R unit step

ωn2

s(s + 2ζω n )

─

C

G

ζ = 0.6 & ωn = 5 rad/sec Find the time domain performance specifications.

π −β Rise time → t r = ωd t r = 0.554 sec

ωd = ωn 1 − ζ 2 ωd = 4 rad/sec β = atan

1− ζ

ζ

β = 0.927 rad © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

2

π Peak time → t r = = 0.785 sec ωd Settling time → 2% t s = 5% t s =

4

ζω n 3

ζω n −

Max overshoot → M p = e

© 2005 P. S. Shiakolas

= 1.333 sec = 1.000 sec ζ

1−ζ

2

π

= 0.095

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

EXAMPLE k G= s(s + 1)

R

C

G

─

H

H = 1 + kh s

k

Determine k & k h s.t.

1 s (s + 1)

max. overshoot = 0.2 units peak time = 1.0 sec R → unit step Obtain tr , t s © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

−

Mp =e

ζ 1−ζ

2

π

= 0.456

π tp = ⇒ ωd = 3.14 ωd ωd = ωn 1 − ζ 2 ⇒ ωn = 3.53 CE : 1 + GH = 0 k (1 + kh s ) = 0 1+ s (s + 1) s 2 + (1 + kkh )s + k = 0 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Compare with standard 2 nd order s 2 + 2ζω n s + ωn2 = 0

ωn2 = k ⇒ k = 12.5 2ζω n − 1 2ζω n = 1 + kkh ⇒ k h = = 0.178 k

π −β tr = = 0.65 sec ωd t s = (2% )

4

ζω n

© 2005 P. S. Shiakolas

= 2.48 sec ; (5% )

3

ζω n

= 1.86 sec

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

EXAMPLE

Negative Feedback C.L.

k (s + 2 ) GH = s (s − 2 )

a) k such that ζ of C.L. poles = 0.707 b) k such that C.L. poles are on Im. axis 1 + GH = 0 ⇒ s 2 + (k − 2 )s + 2k = 0

ωn2 = 2k ⇒ ωn = 2k 2ζω n = k − 2 ⇒ 2ζ 2k = k − 2 a)

ζ = 0.707 ⇒ 2(0.707 ) 2k = k − 2

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

k 2 − 8k + 4 = 0

Im

Re

b) C.L. poles on Im − axis

Re − part of eigen value = 0 s1,2 = ±ωj ⇒ (s − s1 )(s − s2 ) = 0 s 2 + ω 2 = 0 ⇒ 2ζω n = 0 = k − 2 ⇒k = 2 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Im

constant ζ − line

k =2

k = 0.53

jω d

β

Re

− ζω n

⎛ ω 1− ζ 2 ⎛ ωd ⎞ ⎟⎟ = atan⎜ n β = atan⎜⎜ ⎜⎜ ζω ζω n ⎝ n⎠ ⎝ © 2005 P. S. Shiakolas

⎞ ⎛ 1− ζ 2 ⎟ = atan⎜ ⎟⎟ ⎜⎜ ζ ⎠ ⎝

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

⎞ ⎟ ⎟⎟ ⎠

STANDARD CONTROLLERS R

─

GC

GP H

ONE TERM CONTROLLER

Gc = k H =kf

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

C

TWO TERM CONTROLLERS a) PI: Proportional Integral ⎛ 1 ⎞ kc ⎛ τ i s + 1 ⎞ ⎟⎟ ⎟⎟ = ⎜⎜ Gc = kc ⎜⎜1 + ⎝ τis ⎠ τi ⎝ τis ⎠ b) PD: Proportional Derivative

Gc = k p + τs = kc (τ d s + 1)

THREE TERM CONTROLLERS PID: Proportional Integral Derivative

⎛ 1 ⎞ ⎟⎟ Gc = kc ⎜⎜1 + τ d s + τis ⎠ ⎝ © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

R

─

GC

GP H

1 Gp = s (τs + 1) No Controller

1 C τ = R s2 + 1 s + 1

τ

© 2005 P. S. Shiakolas

τ

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

C

PD Controller

PD ⇒ Gc = kc (1 + τ d s )

kc (1 + τ d s )

GcG p C = = R 1 + GcG p 2 ⎛ 1 + kcτ d s +⎜ ⎝ τ PI Controller

τ

⎞ kc ⎟s + τ ⎠

⎛ 1 ⎞ ⎟⎟ PI ⇒ Gc = kc ⎜⎜1 + ⎝ τis ⎠

kc (1 + τ i s )

( τ iτ ) C = R s 3 + 1 s 2 + kc s + kc τ

© 2005 P. S. Shiakolas

τ

τ iτ

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Two Term – Velocity Feedback

H = 1+ k f s

R

─

kc

KC

K cG p C τ = = R 1 + K cG p 2 ⎛⎜ 1 + kc k f s +⎜ ⎝ τ R

─

─

kc τs + 1

GP H

⎞ kc ⎟⎟ s + τ ⎠ GP

kf H © 2005 P. S. Shiakolas

C

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

C

STEADY STATE ERROR TABLE Type 0 1

© 2005 P. S. Shiakolas

Step A 1+ k p 0

INPUT Ramp Parabolic

∞ A kv

∞ ∞

2

0

0

A ka

3

0

0

0

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

EXAMPLE Design a controller that would satisfy :

⋅ ess for ramp imput < 10% of input ⋅ ζ of dominant poles ≥ 0.707 ⋅ settling time (2% ) ≤ 3 sec

k1 G= s (s + 2 ) H = 1 + k2 s R=

A s2

© 2005 P. S. Shiakolas

R ─

G H

C k1 = R s 2 + (2 + k1k 2 )s + k1 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

C

Type of system = Type 1 GH =

k1

s1 (s + 2 )

(1 + k2 s )

Assume FVT applicable ess = lim sE ( s ) = lim s(1 − TF )R s →0

s →0

⎛ ⎞ + + s k k 2 1 2 ⎟ = lim A⎜ s → 0 ⎜⎝ s 2 + (2 + k1k 2 )s + k1 ⎟⎠ A(2 + k1k 2 ) 10 ess = < A k1 100 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

CE : s + (2 + k1k 2 )s + k1 = 0 2

∴ ωn2 = k1 & 2ζω n = 2 + k1k 2

ωn = k1 2ζω n 2 + k1k 2 < 0.1 ⇒ < 0.1 k1 k1 2(0.707)ωn Let ζ = 0.707 ⇒ < 0.1 k1 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

2 ωn > 14.14 ⇒ k1 = ωn = 199.94 ≈ 200

2ζω n = 2 + k1k 2 ⇒ k 2 = 0.09 Check third requirement 4 t s (2% ) = = = 0.4 < 3 sec ζω n (0.707)(14.14) 4

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

STABILITY

STABLE t

Im

NEUTRAL Re

t

UNSTABLE t © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

RUTH-HURWITZ CRITERION CE : an s n + an −1s n −1 + an − 2 s n − 2 + ... + a1s + a0 = 0 TESTS FOR APPLYING R.H.

⋅ a0 ≠ 0 ; All ai Real ⋅ If any of the coefficients are zero or change sign, then system is unstable

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ROUTH TABLE

Power

Coefficients

sn

an

an − 2

an − 4

s n −1

an −1 an −3

an − 5

sn−2

b1

b2

s n −3

c1

c2

b3

M s0 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

− 1 an b1 = an −1 an −1

an − 2

− 1 an b2 = an −1 an −1

an − 4

an − 3

an − 5

− 1 an −1 c1 = b1 b1

an − 3

− 1 an −1 c2 = b1 b1

an − 5

© 2005 P. S. Shiakolas

b2

b3 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

RH States – The number of poles of CE with positive real parts (unstable poles) is equal to the number of sign changes in the first column of the Routh array. Case 1 2 nd Order System : a2 s 2 + a1s + a0 = 0

s

2

1

a2

a0

s

a1

0

s0

b1 = a0

− 1 a2 b1 = a1 a1

a0 0

−1 (− a1a0 ) b1 = a1 b1 = a0

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

3

2

CE : a3 s + a2 s + a1s + a0 = 0 a3 a1 3 1 − s a3 a1 b1 = a a0 2 a2 2 s a2 a0 −1 (a0a3 − a1a2 ) b1 = s1 b1 b2 a2 s0

c1 = a0

− 1 a2 c1 = b1 b1 © 2005 P. S. Shiakolas

a0 0

1 (a1a2 − a0a3 ) = a2

= a0

For stable system, b1 > 0 ⇒ a1a2 > a0 a3

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

CE : s 3 + s 2 + 2 s + 24 = 0 2 −1 1 3 s 1 2 b1 = = −22 1 1 24 s2 1 24 24 −1 1 = 24 s1 b1 = −22 c1 = b 0 1 b1

s0

c1 = 24

System is unstable as there is a sign change in the first column

# of unstable poles = # of sign changes in the first column

Eigen values : s1,2 = +1 ± j 7 & s3 = −3 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

CL feedback system with negative feedback

Gp =

1 a3s 3 + a2 s 2 + a1s + a0

Gc = kc

H =1

• Will this system be stable or unstable? • Is there a limiting value for kc for stability?

CE : 1 + Gc G p = 0 a3s 3 + a2 s 2 + a1s + a0 + kc = 0 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Applicable ? All coefficients > 0

a0 + kc > 0 ⇒ kc > − a0 a1a2 − a3a0 − a3kc b1 = >0 a2 c1 = a0 + kc ⇒ kc > − a0 a1a2 − a3a0 − a3kc a3a0 − a1a2 > ⇒ kc < a3 a2 a2 a1a2 − a3a0 − a0 < kc < a3 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

CE : s 4 + ks 3 + s 2 + s + 1 = 0 • Determine the range of k for stability 4

1

1

s3

k

1

s

s2

b1

s1

c1

s0

d1

b2

1

−1 1 b1 = k k

1

−1 1 b2 = k k

1

1 = (k − 1) 1 k =1

0

k2 c1 = 1 − k −1 d1 = 1

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

1 b1 = (k − 1) > 0 ⇒ k > 1 k k2 k −1− k 2 c1 = 1 − >0 >0 ⇒ k −1 k −1 negative − 1 + k (1 − k ) >0 k −1 positive The above range of inequality is false and therefore there is no range of k that yields us a stable system. Hence, system is unstable. © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Case 2 Zero in first column while all other elements are non-zero 5

4

3

2

CE: s + 2 s + 2 s + 4 s + 11s + 10 = 0

s5

1

2

11

s4

2

4

10

s

3

s2 1

s s

0

b1 c1 d1 e1

© 2005 P. S. Shiakolas

b2

ε −

12

ε

−1 1 b1 = 2 2

2

−1 1 b2 = 2 2

11

−1 1 c1 = b1 b1 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

=0

4

10 11 b2

=6

Case 3 Whole Row full of zeros 3

2

CE: s + s + 2 s + 2 = 0 ⋅ s1 row is all zeros

s3

1

2

s2

1

2

s1

b1 = 0

s0

c1

⋅ Form the auxillary polynomial

1s 2 + 2 s 0 ⇒ s 2 + 2 Eigen values of s 2 + 2 is ± 2 j which are symmetric about the real axis.

Differentiate auxillary polynomial ⇒ 2 s © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

CE : s 4 + s 3 + 3s 2 + 2 s + 2 = 0

s4 s

3

1

3

k

2

s2

b1

b2

s1

c1 = 2

s0

d1

© 2005 P. S. Shiakolas

2

b1 = 1 b2 = 2 c1 = 0 Q = b1s 2 + b2 = s 2 + 2 dQ = 2s ds − 1 b1 d1 = c1 c1

b2 0

= b2

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

5

4

3

2

CE: s + 2 s + 24 s + 48s − 25s − 50 = 0 s5

1

24

− 25

b1 = b2 = 0

− 50

Q = 2 s 4 + 48s 2 − 50

4

2

48

s3

b1

b2

s2

c1

∴b1 = 8 & b2 = 96

s1

d1

c1 = 24 & c2 = 50

s0

e1

d1 = 112.7 & e1 = −50

s

dQ = 8s 3 + 96 s ds

∴ System is unstable © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

CE : s 3 + 2 s 2 + 4 s + k = 0 • Determine the range of k for stability

s3

1

4

s2

2

k

1

s s

0

b1 c1

−1 1 b1 = 2 2

4

8−k = >0 2 k

−1 2 c1 = b1 b1

k

=k >0

0

8−k > 0 ⇒ k < 8 0 0 ⇒ k > −1 18 − 2k 3 b1 = >0 s 1 5 4 c1 = 2 + 2k

s2

4

s1

b1

s0

18 − 2k >0 ⇒ k 0 ⇒ k p > −1 kI > 0

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

s4

1

s3

4 2 + 2k p

s2

b1

s1

c1

s0

d1

© 2005 P. S. Shiakolas

5

b2

2k I

b1 =

18 − 2k p 4

>0

⇒ kp < 9 b2 = 2k I d1 = 2k I > 0

[(

)(

)

4 c1 = 1 + k p 9 − k p − 8k I 18 − 2k p

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

]

ROOT LOCUS (RL) • Graphical technique that will allow us to plot the path or locus of the eigen values as one parameter changes (0 to ∞) • RL has the ability to • Determine closed loop system behavior from open loop conditions • Determine the effects of one parameter qualitatively

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Complex Algebra Review

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ROOT LOCUS (RL) • Graphical technique that will allow us to plot the path or locus of the eigen values as one parameter changes (0 to ∞) • RL has the ability to • Determine closed loop system behavior from open loop conditions • Determine the effects of one parameter qualitatively

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Complex Algebra Review

A = α + β j = A e jθ

Im

⎛β ⎞ θ = atan2⎜ ⎟ = ∠A ⎝α ⎠

A

θ

A = α2 + β2

A1 e jθ1

A1 A1 j (θ1 ±θ 2 ) = = e j θ A2 A2 e 2 A2 A1 ± A2 = (α1 + β1 j ) ± (α 2 + β 2 j ) = (α1 ± α 2 ) + (β1 ± β 2 ) j © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Re

Background on RL

CLTF

C G = R 1 + GH

R ─

CE : 1 + GH = 0

G H

GH: Ratio of polynomials in s Where s: complex quantity

s = σ ± jω 1 + GH s =σ ± jω = 0 GH s =σ ± jω = −1 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

C

Magnitude GH s =σ ± jω = − 1 Angle

∠GH s =σ ± jω = ∠(− 1 + 0 j ) = ±180(2n + 1) n = 0,1, 2...

EXAMPLE Im

k GH = s +α GH =

k

(σ + α ) + jω

© 2005 P. S. Shiakolas

−α

s = σ ± jω

θ

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Re

GH =

k

(σ + α )2 + ω 2

∠GH = θ = −atan

ω σ +α

= ±180(2n + 1)

k +1 = 0 GH + 1 = 0 ⇒ s +α

s + (k + α ) = 0 ⇒ s = −(k + α )

k =∞ x ∞ © 2005 P. S. Shiakolas

Im

k =0 x −α ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Re

Properties of RL

GH = −1 where G = k g kg

Ng Dg

kf

Nf Df

Ng Dg

& H =kf

= −1 ⇒ k = k g k f = −

Nf Df

Ng N f Dg D f

Dg D f = 0 ⎫ ⎪ if k = 0 ⇒ ⎬ Open loop poles N g N f = ∞ ⎪⎭ Dg D f = ∞ ⎫ ⎪ if k = ∞ ⇒ ⎬ Open loop zeros N g N f = ∞ ⎪⎭ © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

OLTF :

CLTF :

k

Ng N f Dg D f

C = R

kg

Ng

kg

Ng

Dg Dg = Dg D f + kN g N f Ng N f 1+ k Dg D f Dg D f

(

k → 0 : CL poles approaches OL poles k → ∞ : CL poles approaches OL zeros

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

)

EXAMPLE

k GH = s +1

0≤k ≤∞

CE : s + (1 + k ) = 0 OL poles : s = −1 OL zeros : ∞

k =∞ x ∞ © 2005 P. S. Shiakolas

Im

k =0 x -1 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Re

EXAMPLE

k (0.4 s + 1) OL pole : s = −1 GH = s +1 OL zero : s = −2.5 Im

TP2

o

TP3

TP1

φ TP5

x

α

-1

Re

TP4

∠GH = ∠k + ∠0.4 s + 1 − ∠s + 1 ∠GH = 0 + φ − α = ±180(2n + 1) © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

CE : (1 + 0.4k )s + 1 + k = 0 1+ k s=− 1 + 0.4k k = 0 ⇒ s = −1 k = ∞ ⇒ s = −2.5 k = 5 ⇒ s = −2 101 k = 100 ⇒ s = − 41 1001 k =0 ⇒ s=− 401 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

GH s = −1.5 = 1 k (0.4 s + 1) =1 s + 1 s = −1.5 − 0.8k = 1 k = 1.25

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ROOT LOCUS (Cont…) Starts at O.L. poles k varies from 0 to ∞ Ends at O.L. zeros.

1 + kGH = 0 where kGH is OLTF k OLTF: GH = s (s + 1)

(

)

OL Poles :

s = −1 OL Zeros : none

CE : s 2 + s + k = 0 1 1 − 1 ± 1 − 4k 1 − 4k s1,2 = =− ± 2 2 2 © 2005 P. S. Shiakolas

s=0

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

∞

x

θ2

-1

-1/2

Im

x

θ1 0

−∞ © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Re

k

CL poles

0

0

−1

1 4

1 − 2

1 − 2

1 2

1 1 − + j 2 2

10

1 39 − + j 2 2

1000 © 2005 P. S. Shiakolas

1 3999 − + j 2 2

1 1 − − j 2 2 1 39 − − j 2 2 1 3999 − − j 2 2

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Identify controller gain k to yield eigen value at 1 s = − ±3j 2 GH for all s = − 1 = 1 k =1 s (s + 1) s = − 1 + 3 j 2

find k .

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

1 0 < k < , the eigen values are on the the Re - axis 4 i.e. they behave as the first order system.

(s + α )(s + β ), ζ

> 1.

1 k = , the eigen values are on the Re - axis 4 & repeating. (s + γ )2 , ζ = 1. 1 k > , the eigen values are complex conjugates. 4

ζ < 1. © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Im

s 2 + 2ζω n s + ωn2 = 0

x

-1

− ζω n

ωd

β

x

-1/2

0

ωd ωn 1 − ζ 2 ⎛⎜ 1 − ζ 2 tan β = = =⎜ ζω n ωnζ ⎜ ζ ⎝

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Re

⎞ ⎟ ⎟⎟ ⎠

EXAMPLE ζ = 0.6 (desired damping)

1 − 0.6 2 tan β = = 1.333 ⇒ β = 53.1o 0.6 1 − ζω n = − ⇒ ωn = 0.833 2

ωd = ωn 1 − ζ 2 = 0.6664 Point of interest ⇒ Desired eigen value 1 Find k to yield s = − ± 0.6664 j 2 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

RULES FOR DRAWING ROOT LOCUS Example k OLTF : GH = (s + 1)(s + 2)(s + 3)

• Find OL poles : 3 (−1, − 2, − 3) OL zeros : 0 • Draw OL poles & zeros • Start OL poles (k = 0 ) End OL zeros (k → ∞ ) • # Loci = 3 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

• Symmetry about Re − axis • n → # OL poles = 3 m → # OL zeros = 0 • # asymptotes = na = n − m = 3 (loci going to ∞) • Angle of asymptotes with the Re − axis ± 180(2q + 1) θq = n−m

± 180(2q + 1) = = ± 60(2q + 1) 3

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

q

θq

0

± 60

1 ± 180 2 ± 300

• Intersection of asymptotes with Re − axis

(Centroid of

poles & zeros )

∑ poles − ∑ zeros σa = n−m

( - 1 - 2 - 3) − 0 = = −2 3

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

• Loci on Re − axis : Loci lie on Re - axis in regions where there is an odd number of poles and zeros (on Re - axis) to the right of it. • Intersection of RL with Im - axis will yield limits of gain for stability − Routh - Hurwitz − On Im - axis s = jw Substitute s = jw in CE CE : s 3 + 6 s 2 + 11s + 6 + k = 0 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

⇒ ( jw)3 + 6( jw)2 + 11( jw) + 6 + k = 0

( ) Im ⇒ (− ω 3 + 11ω ) j = 0 ω (11 − ω 2 ) = 0 ⇒ ω = 0

Re ⇒ − 6ω 2 + 6 + k = 0

ω = ± 11

( )

2

− 6 11 + 6 + k = 0 ⇒ k = 60 • Breakin/Breakaway point CE :1 + kGH = 0 = f ( s ) © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

A + kB = 0 = f ( s )

( A = A( s), B = B( s) )

A k =− B dk d ⎛ A⎞ min/ max ⇒ =− ⎜ ⎟=0 ds ds ⎝ B ⎠ The value of s is breakin/away pts if k for that s is positive.

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

RULES FOR DRAWING ROOT LOCUS (Cont..)

A k=− B

A = (s + 1)(s + 2)(s + 3)

B =1

dk d ⎛⎜ s 3 + 6 s 2 + 11s + 6 ⎞⎟ =0 =− ⎟ 1 ds ds ⎜⎝ ⎠

(

)

= − 3s 2 + 12 s + 11 = 0 s1 = −2.6 & s2 = −1.4 1. Evaluate k ⇒ breakin/away. Reject if k < 0 2. Examine the Re − axis © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ROOT LOCUS Root Locus 5 4

Constant ζ line

3

IMAGINARY AXIS Imaginary Axis

ω = 11, k = 60

2

β

1 0 -1

s = −1.4 k = 0.384

-2 -3 -4 -5 -8

-7

-6

-5

-4

-3

-2

-1

Real Axis

REAL AXIS © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

0

1

Example GH =

10(s + α ) 2

s + 5s + 6

0 0

Root Locus 20

n=3 na = 3

θ a = ±60(2q + 1) σ a = −5.33

10 Imaginary Axis

m=0

15

5 0 -5 -10 -15 -20 -30

© 2005 P. S. Shiakolas

-25

-20

-15

-10 -5 Real Axis

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

0

5

10

At intersection with Im − Axis s = jω CE : ( jω )3 + 16( jω )2 + 44( jω ) − 160 + k = 0 − jω 3 − 16ω 2 + 44 jω − 160 + k = 0

(−16ω 2 −160 + k )+ (− ω 3 + 44ω )j = 0 ω (− ω 2 + 44) = 0 ⇒ ω = 0 or ω = 44 − 16ω 2 − 160 + k = 0 ⇒ − 16(44 ) − 160 + k = 0 k = 864 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

RH : Let α = k − 160 s 3 + 16 s 2 + 44s + α = 0

1 1 b1 = − 16 16

1 = − (α − 704) α 16

44

s3

1

44

s2

1

α

s1

b1

k < 864

s0

c1

c1 = α = k − 160 > 0 ⇒ k > 160

b1 > 0 ⇒ α < 704 ⇒ k − 160 < 704

160 < k < 864

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

FREQUENCY RESPONSE METHODS Frequency Response: It is the steady state response of a system due to a sinusoidal input for a stable or marginally stable system. Why use frequency response methods? For LTI system, the response is of the same form as the input. If input is sinusoidal (w, A), output will also be sinusoidal (w, B, φ). © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

INPUT/OUTPUT RELATIONSHIP FOR F.R. y (t ) x(t ) STABLE G (s ) L.T.I. Y (s) X (s)

L Let : x(t ) = X sin ωt ⎯⎯→ X ( s ) =

Xs s2 + ω 2

p( s) p(s) = G ( s) = q ( s ) (s + s1 )(s + s2 )......(s + sn ) Y ( s) = G (s) X ( s) p(s) p(s) Xs = X (s) = (s + s1 )(s + s2 )......(s + sn ) s 2 + ω 2 q(s) © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Partial Fraction Expansion bn b1 b2 a a Y ( s) = + + ... + + + s + s1 s + s2 s + s n s + jω s − j ω a = conjugate of a L- 1 ⎯⎯ ⎯→ y (t ) = b1e − s1t + b2e − s2t + ... + bn e − sn t + ae − jωt + a e jωt 1444442444443 = 0 for steady state (t ≥ 4τ ) yss (t ) = ae − jωt + a e jωt © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Residue theorem ⇒ a & a a = Y ( s ) ( s + jω ) s = − j ω

ωX (s + j ω ) s = − j ω a = G( s) (s + jω )(s − jω ) ωX a = G ( − jω ) − 2 jω

ωX & a = G ( jω ) 2 jω

G ( j ω ) = G ( jω ) e j φ

φ = ∠G ( jω )

G ( − j ω ) = G ( jω ) e − j φ © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

a=

G ( jω ) Xe − jφ

y ss (t ) =

−2j

& a=

X G ( j ω ) e − j φ e − jω t −2j

G ( jω ) Xe jφ 2j +

X G ( j ω ) e jφ e jω t 2j

⎧⎪ e − j (φ +ωt ) e j (φ +ωt ) ⎫⎪ + y ss (t ) = X G ( jω ) ⎨ ⎬ 2 j ⎪⎭ ⎪⎩ − 2 j e iα − e − iα We know , Euler' s identity sin α = 2j © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

y ss (t ) = X G ( jω ) sin(ωt + φ )

φ > 0, phase lead φ < 0, phase lag EXAMPLE

x(t )

G (s )

k G(s) = τ s +1

y (t )

x(t ) = X sin ωt

y ss (t ) = X G ( jω ) sin(ωt + φ ) © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

k k ⇒ G ( jω ) = G ( jω ) = 1 + τωj 1 + τ 2ω 2

φ = ∠G ( jω ) = ∠k − ∠(1 + τωj ) ⎛ τω ⎞ = −atan⎜ ⎟ ⎝ 1 ⎠ yss (t ) = X

© 2005 P. S. Shiakolas

k 1 + τ 2ω 2

sin (ωt − atan (τω ) )

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

OBSERVATIONS :

τω > 1 (ω very large) Magnitude of s.s. is approximately 1

ω

Phase of s.s. is approximately − 90o

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

BODE PLOT Amplitude & Phase Linear

MAGNITUDE

Logarithmic w

Linear

PHASE

Logarithmic winput © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Bode Plot: It is a graphical representation of the steady state output due to a sinusoidal input for a range of values of the input frequency. Standard Representation for Magnitude of G(jw) is logarithmic and is given as

20 log10 G ( jω ) db (decibels)

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

How to Draw a Bode Plot

In General, a T.F. G ( s) =

k g (s + a )

2 2 ( ( ) + + + 2 s s b s ζω n s ωn )

Normalizing, ⎛1 ⎞ k g a⎜ s + 1⎟ a ⎝ ⎠ G ( s) = 2 ⎛ ⎞ ⎛ ⎞ 2ζ ⎛1 ⎞ 2 ⎜⎜ s ⎟ ⎟ 1 s b⎜ s + 1⎟ ωn ⎜ ⎜ s + + ⎟ ⎟ b ω ω ⎝ ⎠ n ⎝⎝ n ⎠ ⎠ © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

s = jω k (τ aωj + 1) G ( jω ) = ⎛ ⎛ jω ⎞ 2 2ζω ⎞ ⎟⎟ + jω (τ bωj + 1)⎜⎜ ⎜⎜ j + 1⎟⎟ ωn ⎠ ωn ⎝ ⎝ ⎠ G ( jω ) =

k (1 + τ aωj ) ⎛ ⎛ ω ⎞ 2 2ζω ωj (1 + τ bωj ) ⎜⎜1 − ⎜⎜ ⎟⎟ + ωn ⎠ ωn ⎝ ⎝

Plot Magnitude in db © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

⎞ j ⎟⎟ ⎠

20 log G ( jω ) = 20 log k + 20 log (1 + τ aωj )

− 20 log ωj − 20 log (1 + τ bωj )

⎛ ⎛ ω ⎞ 2 2ζω ⎟⎟ + − 20 log ⎜⎜1 − ⎜⎜ ωn ⎜ ⎝ ωn ⎠ ⎝

(

⎞ j ⎟⎟ ⎟ ⎠

20 log G ( jω ) = 20 log k + 20 log 1 + (τ aω )2 − 20 log ω − 20 log 2

© 2005 P. S. Shiakolas

)

(1 + (τ bω )2 )

⎛⎧ 2 ⎫2 2⎞ ⎜ ⎪ ⎛ ω ⎞ ⎪ ⎛ 2ζω ⎞ ⎟ ⎟⎟ ⎟ ⎟⎟ ⎬ + ⎜⎜ − 20 log⎜ ⎨1 − ⎜⎜ ⎜ ⎪⎩ ⎝ ωn ⎠ ⎪⎭ ⎝ ωn ⎠ ⎟ ⎝ ⎠ ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Phase

∠G ( jω ) = ∠k + ∠(1 + τ aωj ) − ∠jω ⎛ ⎛ ω ⎞ 2 2ζω ⎟⎟ + − ∠(1 + τ bωj ) − ∠⎜⎜1 − ⎜⎜ ωn ⎜ ⎝ ωn ⎠ ⎝ ⎛ τ aω ⎞ ⎛ω ⎞ = φ + a tan⎜ ⎟ − atan⎜ ⎟ ⎝0⎠ ⎝ 1 ⎠ ⎛ 2ζω ⎞ ⎜ ⎟ ωn ⎟ ⎛ τ bω ⎞ ⎜ − a tan⎜ ⎟ − a tan⎜ 2⎟ ⎝ 1 ⎠ ⎜⎜ 1 − ⎛⎜ ω ⎞⎟ ⎟⎟ ⎝ ⎝ ωn ⎠ ⎠ © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

⎞ j ⎟⎟ ⎟ ⎠

Constant

20 log k ⇒ constant in db

∠k = 0

Linear Axis

| | db

0.1

1

10

ω

o 0.1 © 2005 P. S. Shiakolas

1

10

ω

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Pole at Zero or Origin 1 1 → Pole at origin s jω

1 = −20 log jω = −20 log ω 2 20log jω = −20 log ω db

ω 1 ∠ = −atan = −90o 0 jω ω

© 2005 P. S. Shiakolas

(Constant )

0.1

20 log ω − 20

− 20 log ω 20

1 10

0 20

0 − 20

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

40 Passes Through ω=1

20

| | db

0.1

1

10

100

ω (log)

-20

Slope -20 db/dec

-40

⎛⎜ 1 ⎞⎟ 2 ⎛⎜ 1 ⎞⎟ ⎝ jω ⎠ ⎝ jω ⎠

Slope -40 db/dec

o 0.1 -90 -180 © 2005 P. S. Shiakolas

1

10

100

⎛⎜ 1 ⎞⎟ ⎝ jω ⎠ 2 ⎛⎜ 1 ⎞⎟ ⎝ jω ⎠

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ω (log)

Dual Pole at Origin

20 log −

1

ω

2

= 20 log

1

ω

2

( )

= −20 log ω

2 2

= −20 log ω 2 = 2(− 20 log ω ) ⎛ 0 ⎞ ∠− = ∠1 − ∠ − ω = 0 − atan⎜⎜ ⎟⎟ ω2 ⎝ −ω2 ⎠ 1

2

( )

= −180o = 2 90o

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

⎛ ⎛ 1 ⎞⎞ ⎟⎟ ⎟⎟ ⇒ N ⎜⎜ − 20 log⎜⎜ SN ⎝ jω ⎠ ⎠ ⎝ 1

where N → slope − 20 × N db/dec ⇒ N (− 90 ) S ± N ⇒ Mag passes through ω = 1 Slope ± N × 20 db/dec ⇒ Phase ± N × 90o © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Simple Pole or Zero

1 1 + τωj

(1 ± τωj )

±N

Break Frequency Corner Frequency

1

τ

1 2 2 20 log = −20 log 1 + τ ω 1 + τωj

(

= −10 log 1 + τ 2ω 2 when ω =

1

τ

© 2005 P. S. Shiakolas

)

⇒ τω = 1⇒ τ 2ω 2 = 1 = −10 log(1 + 1) = −10 log 2 = −3.01 db ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ωτ 1

⇒ Mag ≈ − 10 log 1 = 0 db ∠ ≈ 0o ⇒ Mag ≈ − 3.01 db ∠ ≈ −45o

(

⇒ Mag ≈ − 10 log τ 2ω 2 ≈ − 20 log(τω )

)

≈ − 20 db/dec ∠ ≈ −90o © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

0.1

τ

| | db

1

τ

10 1

τ

100 1

τ

ω (log)

-20 Slope -20 db/dec

-40

0.1

o

τ

1

τ

10 1

τ

100 1

τ

-45

-90

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ω (log)

(1 + τωj )± N Mag : ωτ > 1 ⇒ 20 × (± N ) db/dec Phase : ωτ > 1 ⇒ 90 × (± N )

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Quadratic Poles & Zeros Quadratic Poles 1 G ( s) = s 2 + 2ζω n s + ωn2

G ( s) =

1

1

ωn2 ⎛ s ⎞ 2 s ⎜⎜ ⎟⎟ + 2ζ +1 ωn ⎝ ωn ⎠

G ( jω ) =

1 2

⎛ω ⎞ ω ⎟⎟ + 2ζ 1 − ⎜⎜ j ωn ⎝ ωn ⎠

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

2

⎛ω ⎞ ω ⎟⎟ + 2ζ j Mag : 2 − log G ( jω ) = −20 log 1 − ⎜⎜ ωn ⎝ ωn ⎠ ⎡⎧ 2 ⎫2 2⎤ 1 ω⎫ ⎥ ⎢⎪ ⎛ ω ⎞ ⎪ ⎧ ⎟⎟ ⎬ + ⎨2ζ = −20 × log ⎢⎨1 − ⎜⎜ ⎬ ⎥ 2 ωn ⎠ ⎪ ⎩ ωn ⎭ ⎝ ⎪ ⎢⎩ ⎥ ⎭ ⎣ ⎦

ω Let =u ωn

(

)

⎡ 2⎤ 2 2 = −10 log ⎢ 1 − u + (2ζu ) ⎥ ⎦ ⎣ © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

⎛ 2ζu ⎞ Phase ∠G ( jω ) = −atan⎜⎜ ⎟ 2⎟ ⎝1− u ⎠ if u > 1, Mag ≈ −10 log u 4 = −40 log u slope = −40 db/dec ⎛ 2ζu ⎞ ⎛ 1 ⎞ ⎟ ≈ −atan⎜ Phase ≈ -atan⎜⎜ ⎟ 2⎟ ⎝ −u ⎠ ⎝ −u ⎠ ≈ −180o © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

if u = 1, Mag = −10 log(2ζ )2 = −20 log(2ζ ) ⎛ 2ζ Phase = −atan⎜ ⎝ 0 | | db -20

0.1u

u =1

⎞ o 90 = − ⎟ ⎠ 10u

-40

o

ω (log) Slope -40 db/dec

0.1u

u =1

10u

ω (log)

-90 -180 © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Max Amplitude occurs at ωr (resonant frequency)

ωr = ωn 1 − 2ζ 2

ζ < 0.707

M Pω r = G ( jωr ) = ⎧⎨2ζ 1 − ζ 2 ⎫⎬ ⎩ ⎭

(

−1

⎛ Mag ⎞ ⎟⎟ × G ( jω ) × sin ωt + 0o Output = ⎜⎜ ⎝ Input ⎠

© 2005 P. S. Shiakolas

)

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

First Order Simple Pole at ω = 0.1 ⇒ 0 db

τ

20 log G ( jω ) = 0 db ⇒ G ( jω ) = 1 ⎛ mag ⎞ ⎟⎟ × 1× sin(ωt + 0o ) output = ⎜⎜ ⎝ input ⎠ at ω = 100

τ

20 log G ( jω ) = −40 db ⇒ G ( jω ) = 0.01 ∠ ≈ −90

o

© 2005 P. S. Shiakolas

⎛ mag ⎞ ⎟⎟ × 0.01× sin(ωt − 90o ) output = ⎜⎜ ⎝ input ⎠ ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Example

G ( s) =

10 ( s + 3) s ( s + 2) ( s 2 + s + 2)

Normalize ⎛1 ⎞ 10 × 3 ⎜ s + 1⎟ ⎝3 ⎠ G (s) = 2 ⎞ ⎛ 1 ⎛1 ⎞ ⎜⎛ s ⎞ s × 2⎜ s + 1⎟ × 2 ⎜ ⎟ + s + 1⎟ ⎟ 2 ⎝2 ⎠ ⎜⎝ ⎝ 2 ⎠ ⎠ ⎛ jω ⎞ 7.5 ⎜ + 1⎟ 3 ⎝ ⎠ G ( jω ) = 2 ⎛ jω ⎞⎟ ⎛ jω ⎞ ⎜ ⎛ jω ⎞ jω ⎜ +1 + 1⎟ ⎜ ⎟ + ⎟ 2 ⎝ 2 ⎠ ⎜⎝ ⎝ 2 ⎠ ⎠ © 2005 P. S. Shiakolas ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

FACTORS Break Freq.

Slopes db/dec

Angle deg.

7.5

j

ω 3

+1

( jω )

−1

⎛ ω ⎞ ⎜ j + 1⎟ ⎝ 2 ⎠

−1

(Quad )−1

*

3

1

2

2

Low Freq.

*

0

-20

0

0

High Freq.

*

20

-20

-20

-40

Low Freq.

0

0

-90

0

0

-45

-90

-90

-180

Break Freq. High Freq.

© 2005 P. S. Shiakolas

45 0

90

-90

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Final Slope (ω >> 1) = − 60 db/dec Final Phase (ω >> 1) = −270o

Constant : 20 log(Constant) db = 20 log(7.5) = 17.5 db Quadratic : Find ωr , ζ , M pω r from formulae. © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Discussion/review from last example

ω1

| | db -20 -40

© 2005 P. S. Shiakolas

Simple Zero Slope = 20 db/dec

ω2

Constant

ωr = M p

ω3

Simple Pole Slope = -20 db/dec

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ω (log)

Quadratic Slope = -40 db/dec

Magnitude

ω1

| | db

ω2

ω3

-20 -40

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ω (log)

Phase 90 45

ω (log)

o -45 -90 -135 -180 -270

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

MINIMUM PHASE & NON-MINIMUM PHASE SYSTEMS • Poles & Zeros in left-hand s-plane Æ Minimum Phase • Poles or Zeros in right-hand s-plane Æ Non-Minimum Phase 1 + j ωT 1 − j ωT G1 ( jω ) = G2 ( jω ) = 1 + jωT1 1 + jωT1

⎛ ωT ⎞ ⎛ ωT1 ⎞ ∠G1 = atan⎜ ⎟ − atan⎜ ⎟ ⎝ 1 ⎠ ⎝ 1 ⎠ ⎛ − ωT ⎞ ⎛ ωT1 ⎞ ∠G2 = atan⎜ ⎟ − atan⎜ ⎟ ⎝ 1 ⎠ ⎝ 1 ⎠ © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

1 1 > T T1 o o

1 T1

1 T1

1 T

1 T

CUT-OFF FREQUENCY OR BANDWIDTH Cut-off frequency is the frequency at which the magnitude of the closed loop frequency is at 3 db below the zero frequency value. © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

| | db

ωb

0

ω (log)

-3 db Bandwidth

| | db

ω (log)

0 -3 db Bandwidth

Find the attenuation factor at -3 db magnitude. (Find real # corresponding to magnitude of -3 db) © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

e − jω L G ( jω ) = 1 + j ωT

L = 0.5 T = 1.0

Magnitude (db) = 20 log G ( jω ) = 20 log e − jωL − 20 log 1 + jωT Phase ∠ = ∠e − jωL − ∠1 + jωT ⎛ ωT ⎞ = − ωL − a tan⎜ ⎟ { ⎝ 1 ⎠ 57.3ωL deg

φ = −ωL or − 57.3Lω © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

STABILITY ANALYSIS IN FREQUENCY DOMAIN USING BODE PLOTS

ωcg

| | db 0 o -180

ω (log) Positive Gain Margin (G.M.)

Positive Phase Margin (P.M.)

© 2005 P. S. Shiakolas

ωcp

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ω (log)

Phase Margin: It is the amount of additional phase lag at ωcg required to bring the system to the verge of instability. γ = 180 + φ ω cg

( ) if γ < 0 (φ < −180o ) , System is unstable if γ > 0 φ > −180o , System is stable

Gain Margin: It is the reciprocal of the magnitude |G(jω)| at ωcp. GM indicates how much the gain can be increased before the system becomes unstable OR how much it must be decreased for the system to become stable. © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

kg =

1 G ( jω ) ω

cp

k g db = 20 log k g = −20 log G ( jω ) ω

cp

if k g (db) > 0, system is stable if k g (db) < 0, system is unstable For satisfactory performance and to guard against variations in performance of system components,

30o < PM < 60o GM > 6 db © 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

Example

k G (s) = s (s + 1)(s + 5)

Find k for stability and draw BP for k = 1,10,100 Root Locus 10

n=3

k = 30

m=0

ω= 5

na = n − m = 3

θ a = ±60(2q + 1) −6 σa = = −2 3

Imaginary Axis

5

0

-0.472 -5

-10 -14

© 2005 P. S. Shiakolas

-12

-10

-8

-6 -4 Real Axis

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

-2

0

2

4

STATIC ERROR CONSTANTS Slope of magnitude curve at low frequencies

G( s) = Type 0

k (T1s + 1)(T2 s + 1)....

s N (Ta s + 1)(Tb s + 1)....

(s = jω )

Static position

At low frequencies, the magnitude of G(jω) equals to k or kp

lim G ( jω ) = k = k p

ω →0

© 2005 P. S. Shiakolas

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

kp db

| | db

-20 db/dec -40 db/dec

ω (log)

Type 1 Static Velocity

k (s + 2 ) G( s) = s (s + 5)(s + 8)

| | db

k ( jω + 2 ) G ( jω ) = jω ( jω + 5)( jω + 8) © 2005 P. S. Shiakolas

1

ME 5303 Classical Methods of Control Systems – Analysis & Synthesis

ω (log)

At low frequencies, kv , ω