Control Systems Formula Sheet PDF

Preview

Summary

Control Systems Formula Sheet...

Description

University of Hertfordshire Faculty of Engineering & Information Sciences Control Systems Formula Sheet

1.

DAG 29/09/03

SYSTEMS MODELLING

Time Domain

Laplace Domain

f(t)

F(s) or L[f(t)]

a.f(t) + b.g(t)

(a and b constant)

a.F(s) + b.G(s) sF(s) − f ( 0)

df f ′( t ) = f& ( t ) = dt d2 f f ′′( t) = 2 dt n d f dt n

s2 F( s) − sf ( 0) − f ′(0) s n F( s) − s n−1 f ( 0) − s n− 2 f ′ (0)− ....− f ( n−1) (0) 1 F(s) s e-sT F(s)

t

∫ f ( t) dt 0

f(t-T)

(T is a time delay) δ(t)

1

1

1 s n!

tn e

sn+1 1 s+ a n!

-at

tn e-at

( s+ a) n+1 sin(bt)

b s + b2 s 2 s + b2 b 2

cos(bt) e-at sin(bt)

(s + a )

2

+ b2

s+a

e-at cos(bt)

(s + a )2 + b 2 Initial and Final Value Theorem f ( 0) =

lim t→0

lim

[f ( t )] = s →∞ [sF(s)]

f ( ∞) =

1

lim t→∞

lim

[ f( t)] = s→0 [ sF( s)]

Transfer Function Definition

G(s) =

The Laplace Transform of the output The Laplace Transform of the input

=

L [y(t) ]

Y(s) U(s)

=

L [ u(t) ]

Assuming all initial conditions are zero.

System

Transfer Function

Differential Equation

Gain

G(s) = K

y = Ku

Integrator

First Order

Second Order

System

G (s) =

G (s) =

G( s) =

1 s

y=

K 1 + sT

T

Kω n2

dy + y = Ku dt

d 2y dy + 2ζω n + ω n2y = Kω n2u 2 dt dt

s2 + 2ζω n s+ ω 2n

Unit Step Response (assuming all ic’s = 0)

−t   y = K 1 − e T   

First Order

Second Order

∫ u. dt

  ζω n t  sin (ω d t)   y = K1 − e −ζωn cos(ω d t ) + ωd     where ω d = ω n 1 − ζ 2 ,

−

overshoot = e

ωn =

πζ 1 − ζ2

2

,

ζ=

π t max 1− ζ 2

,

( ln(overshoot)) 2 2 π 2 + ( ln( overshoot) )

Rules for Block Diagram Manipulation

GA (s)

GB (s)

GA (s)GB(s)

GA (s) GA (s) ± GB (s)

+ ± GB (s)

G(s)

= G(s)

G(s)

1 G(s) G(s) G(s)

+

G(s)

+

G(s)

±

± G(s)

G(s)

+

+

G(s) ±

± 1 G(s)

+

G(s) G(s) 1 + G(s)H(s) H(s)

3

2.

SYSTEM PERFORMANCE

y ss =

Steady State Output

lim lim y( t ) = sY (s) t→∞ s→0

e ss = u - yss

Steady State Error

Steady State Error for Unity Feedback Systems Step u(t) = 1 Type 0

e ss =

1 1+ K p

Type 1

ess = 0

Type 2

ess = 0

Ramp u(t) = t

Parabolic u(t) = t2/2

ess = ∞

ess = ∞

ess =

ess = ∞

1 Kv

ess = 0

Position Error Constant

Velocity Error Constant

Acceleration Error Constant

ess =

Kp =

lim [ G o ( s)] s →0

Kv =

lim [ sGo( s) ] s →0

Ka =

4

lim s→0

[s G (s)] 2

o

1 Ka

3.

SIMPLE CONTROLLERS

Controller Gc(s) Proportional

K

K+

Proportional + Integral

Ki s

s +α  K or K   where α = i K  s 

K + K ds

Proportional + Derivative

or K (1 + sTd) where Td = K d K K+

Proportional + Integral + Derivative (Three Term Controller) K

Lead Controller or Lag Controller or

4.

Ki + Kd s s

s +a s+b

where a < b

1 + sT 1+ sα T s +a K s+b K

K

1 + sT 1+ sα T

where α < 1 where a > b

where α > 1

ROUTH STABILITY CRITERION

D(s) = ansn + an-1sn-1 + an-2sn-2 + an-3sn-3 + an-4sn-4 + an-5sn-5 + ... = 0 sn

an

an-2

an-4

n-1

s

an-1

an-3

an-5

sn-2 . . . s0

bn-1 . . . hn-1

bn-3 . . .

bn-5 . .

5

bn −1 =

a n −1 a n − 2 − a n a n −3 a n− 1

b n− 3 =

a n −1 a n −4 − a n a n −5 a n −1

5.

ROOT LOCUS

No

Drawing Rules

1

All loci start for K = 0 at the Open Loop Poles and finish for K = ∞ at either Open Loop Zeros or s = ∞.

2

There will always be a locus on the real axis to the LEFT of an ODD number of Open Loop Poles and Zeros.

3

If there are n Open Loop Poles and m Open Loop Zeros, there will be n-m loci ending at infinity on asymptotes at angles to the real axis of

±

4

180° , n −m

±

540° , n−m

±

900° , K n− m

The asymptotes meet on the real axis at m

n

∑ p i - ∑ zi i =1

σ =

i =1

n - m

where pi is the position of the i’th open loop pole and zi is the position of the i’th open loop zero.

5

Where two real loci meet on the real axis they “breakaway” from the real axis at ±90° to form two complex loci, symmetrical about the real axis. Where two complex loci meet on the real axis they “break-in” to form two real loci, moving in opposite directions along the real axis. The “breakaway” and “break-in” points are given by the roots of

6

dK = 0 ds

The points of intersection of a locus with the imaginary axis can be determined by solving the equation

D ( j ω)

+

KN ( j ω )

=

0

Remember both the real part and the imaginary parts must be satisfied in this equation. Hence this will give two simultaneous equations one that will give values of ω while the other gives the K value at the crossing point.

7

The angle of departure of a locus from a complex open loop pole is given by;

φ d = 180° −

n

∑φ i =1 i≠ d

m

i

+

∑ψ

i

i =1

where φi is the angle from the i’th open loop pole and ψi is the angle from the i’th zero. The angle of arrival of a locus at a complex zero is given by; m

n

ψa = 180 °

+

∑ φi i= 1

6

−

∑ψ i= 1 i≠ a

i

n

Magnitude Condition

∏P

i

K=

i=1 m

∏Z

i

i =1

m

n

180° =

Angle Condition

∑φ

i

∑ψ

-

i =1

ω=±

Lines of constant damping

1 - ζ2 ζ

i

i =1

σ

where s = σ + jω

NB Straight line through the origin of the s plane. Also lines makes an angle cos-1 ζ with the negative real axis

Lines of constant undamped natural frequency

6.

σ 2 + ω 2 = ω 2n NB Circle with centre on the origin of the s plane and radius ωn

FREQUENCY RESPONSE METHODS

y(t) = R.sin(ωt + φ)

u(t) = sin(ωt)

G(s) R = G ( jω ) = a 2 + b 2

G(jω ) = a(ω ) + jb(ω )

 b φ = ∠G( jω) = tan − 1    a

g = 20 log 10 R y(t) = R.sin(ωt + φ)

u(t) = sin(ωt) G1(s)

R = R 1R 2R 3

G2 (s)

G3 (s)

g = g1 + g 2 + g3

7

φ = φ1 + φ 2 + φ3

Nyquist Diagrams of Common System Elements

R

φ

K

0°

G(s)

Nyquist Diagram

Gain Term

K

K

Integrator 1 s

1 ω

-90°

ω

90°

Differentiator

s

First Order “Lag” 1 − tan − (ωT )

1 1 1 + sT

1

1 + ω 2T 2

First Order “Lead” tan −1( ωT)

1 + ω 2T 2

1 + sT

Second Order Term ω 2n s2 + 2 ζωn s + ω2n

 2ζω ω  − tan− 1 2 n 2   ωn − ω 

ω2n

(ω

2 n

− ω2

)

2

+ ( 2ζω nω )

2

Pure Time Delay

e − sTD

−ωTD

1

(in radians)

8

1

Bode Plots of Common System Elements

g (dB) G(s)

Phase Plot

Gain Plot

φ(deg)

Gain Term 20log10 K

2 0 lo g 1 0 K

0

K 0°  1 20 log10   ω

Integrator 1 s

1 0

0

-90° -20 dB/dec

20 log10 ( ω )

Differentiator s

20 dB/dec

-90

90

90° 0

0 1

First Order “Lag” 1 1 + sT

First Order “Lead” 1 + sT

 1 20log10   1 + ω2 T2

   

1/T 0

-3dB

ω n2 s + 2 ζω ns + ω 2n

2 n

− ω2

)

2

1/T

10/T

20 dB/dec 90

3dB

45

0

   2 + (2ζω n ω )  

0

-20 log10(2ζ)

ω n2

(ω

0.1/T

-90

1/T

2

10/T

-20 dB/dec

2 2 20log10  1 + ω T   

  20 log 10   

1/T

-45

− tan −1 (ωT )

tan −1( ωT)

Second Order Term

0.1/T 0

ωn 0

0

ωn -90

 2ζω ω  − tan −1 2 n 2   ωn − ω 

-40 dB/dec

9

180

7.

DIGITAL CONTROL

z Domain (T = sample time period) F(z) or Z[f(k)]

Discrete Time Domain f(k)

Laplace Domain F(s)

a.f(k) + b.g(k)

a.F(z) + b.G(z)

a.F(s) + b.G(s)

f(k+1)

zF(z) - zf(0)

esT F(s)

f(k-1)

z-1F(z)

e-sT F(s)

δ(k)

1

1

δ(k-n)

z-n

e-snT

1

z z−1 Tz

1 s 1 s2

k

( z − 1) 2 1 2 k 2!

T2 2!

e-ak

 z (z + 1)   3   ( z − 1)  z

1 s

1 s+a 1

z − e − aT Tze −aT

ke-ak

(z − e ) −aT

1 - e-ak -ak

k - (1 - e )/a

(s + a )2

2

z(1 − e − aT )

[

( z − 1) (z − e

− aT

a s (s + a )

)

z z (aT − 1 + e −aT ) + (1 − e −aT − aTe −aT a( z − 1) ( z − e

− aT

2

3

)

]

a s (s + a ) 2

sin(ak)

z sin( aT)

a

cos(ak)

z − 2 z cos( aT) + 1 z(z − cos( aT))

s + a2 s s2 + a 2

2

z 2 − 2z cos( aT) + 1 e-ak sin(bk) e-ak cos(bk)

z e− aT sin( bT) z 2 − 2ze−aT cos( bT) + e− 2aT

( s + a) 2 + b2

z(z − e− aT cos( bT))

s+ a

z 2 − 2ze −aT cos( bT) + e −2 aT Zero Order Hold (A/D and D/A converters)

2

GH (z)

10

b

(s + a )

2

+ b2

1− e − sT s

8.

STATE SPACE METHODS

Standard Form for State Space Model

Digital

x& = Ax + Bu

x (k + 1) = Fx (k ) + G u (k )

y = Cx + Du

y(k ) = C x (k ) + Du ( k )

G (s) = C (sI − A ) B + D

G (z ) = C( zI − F) G + D

det( sI − A) = 0

det ( zI − F ) = 0

Characteristic Equation

t   x( t) = e  x(0) + e− Aτ Bu(τ ) dτ   0

∫

At

[

Φ( t) = eAt = L-1 ( sI − A )− 1

Transition Matrix

Φ ( t ) = e At = I + At + State Feedback Equation

Characteristic Equation with state feedback

x (k + 1) = F k+ 1x (0) +

]

G = A −1 [F − I ]B

x& = Ax + B( u − Kx)

x (k + 1) = Fx (k ) + G( u (k ) − K x ( k ))

det (sI − A + BK) = 0

det ( zI − F + GK ) = 0

]

y =

0

1

0

L

0

0

1

L

M

M

M

O

0

0 − a1

0

L

− a2

L

− a0

[b 0

∑ FiGu (k − i )

F = eAT

A 2 t 2 A3 t3 + +L 3! 2!

[

   x& c =     

k

i= 0

2 n −1 M c = BM ABMA BMLM A B

Controllability Matrix Controllable Canonical Form

−1

−1

Transfer Function Matrix

State Time Response

Continuous

b 1 L L b n− 2

0  0   M x c  1  − a n −1

Mc

 0  0   + M u    0  1  

 0  0  xc (k + 1) =  M   0  − a0 y(k ) =[ b0

b n− 1]xc

11

[

]

= GM FGM F2 GMLMFn − 1G

1

0

0

1

L L

M

M

O

0

0

L

− a1

− a2

L

b1 L L b n− 2

b n − 1]x c (k)

0   0  0 0     M  xc (k) +  M u (k)    1   0 − an −1   1

Ackermann’s Formula for State Feedback

If desired CE is s n + α n− 1s n− 1 + L + α 1s + α 0 = 0

K = [ 0 0 K 1] M−c 1φ ( A)

K = [ 0 0 K 1] M −c1 φ( F)

where φ(A ) = α 0I + α 1A + L + α n − 1A n−1 + A n

State Estimator Equation

Characteristic Equation for a State Estimator

y (k ) − yˆ(k )) x (k + 1) = Fxˆ(k ) + Gu(k ) + P ( ~

det( sI − A + PC) = 0

det (zI − F + PC) = 0

 C    CA  =  CA 2     L  n −1  CA 

Mo

 0  1  = x& o  0   L  0

0

L

0

0

L

0

1

M

0

L

O

L

0

L

1

− a0   b0   b  −a 1   1  −a 2 x o +  b 2  u    M   M    −a n − 1  b n − 1

If desired CE is s n + α n− 1s n− 1 + L + α 1s + α 0 = 0

[

P = φ(A )M −o1 [ 0 0 K 1]

 0   1 xo (k + 1) =  0   L  0  y(k )= [0 0 0 L

y = [0 0 0 L 0 1 ]x o

Ackermann’s Formula for State Estimators

where φ( F ) = α 0I + α 1F + L + α n− 1F n− 1 + F n

x&$ = Ax$ + Bu + P( ~ y − y$ )

Observability Matrix

Observable Canonical Form

If desired CE is z n + α n −1 z n −1 + L + α1 z + α 0 = 0

]T

Mo

 C   CF  = CF 2     L  n− 1  CF 

0

L

0

0 1

L

0 0

L

M O

L

0

L

1

 b0       b1    xo (k) +  b 2  u (k )    M   M     − a n− 1 bn −1  − a0 − a1 − a2

0 1]xo (k)

If desired CE is z n + α n −1 z n −1 + L + α1 z + α 0 = 0

[

]T

P = φ (F )M −o1 [0 0 K 1 ]

where φ(A ) = α 0I + α 1A + L + α n − 1A n−1 + A n

12

where φ( F ) = α 0I + α 1F + L + α n− 1F n−1 + F n...