Control System Part 3

7/21/2019 Control System Part 3

http://slidepdf.com/reader/full/control-system-part-3 1/29

Lecture slides for ELEN90055 prepared by Michael Cantoni (c) 2011, 2012, 2013

ELEN90055 Control Systems: Part IIIM.Cantoni (c) 2011, 2012, 2013




C s

E sR s Y s

Di(s)

U s

Do(s)

Dm sY m

s

C (s)

G0 s

Y (s)

Do(s) =

1

1 + G0(s)C (s)Y (s)

R(s) =

G0(s)C (s)

1 + G0(s)C (s)

wit on y o = 0wit on y r = 0

G0(s) = B0(s)

A0(s) nominal plant transfer function, C (s) =

P (s)

L(s) controller transfer function

B0(s), A0(s)} and P (s), L(s)} coprime (no common roots) polynomial pairs

Di(s), Do(s), Dm(s) plant input, plant output and measurement noise uncertainty

R(s), U (s), Y (s) reference, control, and plant output signals

Y (s) = D0(s) + G0(s)U (s) + Di(s)

U (s) = C (s)R(s)−Dm(s)− Y (s)


E sR s Y s

Di s

U s

Do s

Dm sY m s

80

60

40

20

0

20

M a g n i t u d e ( d B )

100

102

104

106

90

45

0

45

90

P h

a s e ( d e g )

Frequency response

Frequency (rad/sec)

K 100

0.01 s + 1

K = 1

K = 10

K = 1K = 10

1

1 + G0C

G0C

1 + G0C

G0(s) = 100

0.01 s + 1 C (s) = K

K = 1

K = 10

K = 1K = 10

K K




C s

E sR s Y s

Di(s)

U s

Do(s)

Dm sY m

s

G0 sH s

Y (s) = G0(s)C (s)H (s)

1 + G0(s)C (s) R(s) +

1

1 + G0(s)C (s)Do(s) +

G0(s)

1 + G0(s)C (s)Di(s)−

G0(s)C (s)

1 + G0(s)C (s)Dm(s)

C s

U (s) = C (s)H (s)

1 + G0(s)C (s)R(s)−

C (s)

1 + G0(s)C (s)Do(s)−

G0(s)C (s)

1 + G0(s)C (s)Di(s)−

C (s)

1 + G0(s)C (s)Dm(s)

H (s) is a stable transfer function (reference filter)


S 0 s + T 0 s = 1, S i0 s = G0 s S 0 s = T 0 s /C s , S u0 s = S 0 s C s = T 0 s /G0 s

Y (s) = T 0(s)H (s)R(s) + S 0(s)Do(s) + S i0(s)Di(s)− T 0(s)Dm(s)

U (s) = S u0

H (s)R(s)−Dm(s)−Do(s)−G0(s)Di(s)

For a nominal plant G0(s) = B0(s)A0(s)

and controller C (s) = P (s)L(s)

:

T 0(s) .=

G0(s)C (s)

1 + G0(s)C (s) =

B0(s)P (s)

A0(s)L(s) + B0(s)P (s) complementary sensitivity

S 0(s) .=

1

1 + G0(s)C (s) =

A0(s)L(s)

A0(s)L(s) + B0(s)P (s) (output) sensitivity

S i0(s) .=

G0(s)

1 + G0(s)C (s) =

B0(s)L(s)

A0(s)L(s) + B0(s)P (s) input-disturbance sensitivity

S u0(s) .=

C (s)

1 + G0(s)C (s) =

A0(s)P (s)

A0(s)L(s) + B0(s)P (s) control sensitivity




The problem spec. translates to |S 0 jω = 1−T 0 jω | << 1 for ω ∈ [0, 4 rad sec

Given a

nominal plant model G0(s) = 4(s+1)(s+2)2 and

output disturbance model do(t) = k + dv(t), where k is a constant and dv(t)has energy confined to the frequency band (0, 4) rad/sec,

design a feedback controller that desensitizes the plant output to the disturbance

Choosing

T 0(s) = G0(s)C (s)

1 + G0(s)C (s) =

1

(τ 1s + 1)2(τ 2s + 1)2

with τ 1 = 0.1 and τ 2 = 0.01, achieves |S 0( jω)| << 1 for ω ∈ [0, 4) and the

corresponding controller is

C (s) = T 0(s)

(1 − T 0(s)) ·

1

G0(s) =

106(s + 1)(s + 2)2

4((s + 10)2(s + 100)2 − 106)


10

100

10

104

0

100


Bode Diagram

Frequency (rad/s)

10

100

10

104

0

100


Bode Diagram

Frequency (rad/s)

10

100

10

104

0

100

M a

g n i t u d e ( d B )

Bode Diagram

Frequency (rad/s)

10

100

10

104

0

100

M a

g n i t u d e ( d B )

Bode Diagram

Frequency (rad/s)

C jω

G0( jω)

S 0( jω)S i0 jω

S u0 jω

T 0 jω

S u0(s) = T 0(s)G0(s)

= (s+1)(s+2)2

4(0.1s+1)2(0.01s+1)2 =⇒ |S u0( j3)| ≈ 20 dB!




|Λ0( jω)|

|T 0( jω)| S 0( jω)|

good tracking of low-

frequency reference

low sensitivity to high-frequency

measurement noise

good rejection of low-frequency

output disturbance

|Λ0 jω.= G0 jω C jω |

large low-frequency loop-gain

small high-frequency loop-gain


Frequency response

Frequency (rad/sec)

NOTE: T 0( jω) = Λ0( jω)

1 + Λ0( jω); S 0( jω) =

1

1 + Λ0( jω)

loop-gain bandwidth ωb : |Λ0( jωb)| = 1


If |Λ0( jω)| ≈ 1, then both |S 0( jω)| and |T 0( jω)| could be large – all depends on

∠Λ0( jω) at these frequencies

Similarly, using 1− |Λ0| ≤ |1 + Λ0( jω)|, for frequencies where |Λ0( jω)| << 1,

|S 0( jω)| ≈ 1, |T 0( jω)| << 1, |S i0( jω)| ≈ |G0( jω)|, |S u0( jω)| ≈ |C ( jω)|.

Consider the inequalities |Λ0( jω)| − 1 ≤ |1 + Λ0( jω)| ≤ 1 + |Λ0( jω)|.For any frequency such that |Λ0( jω)| > 1,

1|Λ0(jω)|+1

≤ |S 0( jω)| ≤ 1|Λ0(jω)|−1

|Λ0(jω)||Λ0(jω)|+1 ≤ |T 0( jω)| ≤ |Λ0(jω)|

|Λ0(jω)|−1

1|C (jω)| ·

|Λ0(jω)||Λ0(jω)|+1 ≤ |S i0( jω)| ≤ 1

|C (jω)| · |Λ0(jω)||Λ0(jω)|−1

1

|G0(jω)| ·

|Λ0(jω)|

|Λ0(jω)|+1 ≤

|S u0( jω

)|≤ 1

|G0(jω)| ·

|Λ0(jω)|

|Λ0(jω)|−1 .

As such, for frequencies where |Λ0( jω)| >> 1,

|S 0( jω)| << 1, |T 0( jω)| ≈ 1, |S i0( jω)| ≈1

|C ( jω)|, |S u0( jω)| ≈

1

|G0( jω)|.




C s

E sR s Y s

Di(s)

U s

Do(s)

Dm sY m

s

G0 s

ess.

= limt→∞

e(t) = lims→0

sE (s) = lims→0

sS 0(s)R(s) = lims→0

s 1

1 + Λ0(s)R(s) = 0

ess = 0

ess = lims→0 s

1

1 + Λ0(s)

1

s =

1

1 + Λ0(0) = S 0(0)

S 0 s

Λ0(0) =∞ =⇒ S 0(0) = 0


r(t), di(t), do(t), dm(t)

G0(s) = 1s

, C (s) = K =⇒ T 0(s) = K s + K

, S 0(s) = ss + K

,

S i0(s) = 1

s + K , S u0(s) =

Ks

s + K

Λ0(s) = G0(s)C (s)

e t , u t , y t , ym t




G0(s) = s− 1

s + 3 , C (s) = 1

s− 1 =⇒ S 0(s) = s + 3

s + 4 , T 0(s) = 1

s + 4 ,

S i0(s) = s− 1

s + 4, S u0(s) =

s + 3

(s− 1)(s + 4)

A0(s)L(s) + B0(s)P (s) = (s + 3)(s− 1) + (s− 1) = (s− 1)(s + 4)

{A0, B0}, {L,P }

Given a strictly proper nominal plant model G0(s) = B0(s)A0(s)

and proper controller

C (s) = P (s)L(s) the closed-loop is internally stable if, and only if, all zeros of the

nominal characteristic polynomial

A0(s

)L

(s

) +B

0(s

)P

(s

)

have strictly negative real part.


S 0.

=1

1 + G0C ; T 0

.=

G0C

1 + G0C ; S i0

.=

G0

1 + G0C ; S u0

.=

C

1 + G0C

Λ0

.= G0C

G0(s)

C s




as2+ bs+ c has zeros at s =

−b±√ b2−4ac

2a

n = 2

an = 1

for p(s) = s2 + ks + (1 − k)

if [ p(s0) = 0 =⇒ (s0) < 0 ]

then 0 < k < 1

s3 + s2 + s+ 1

has zeros at

s = −1,± j

given a degree n > 0 polynomial p(s) = sn + an−1sn−1 + · · · + a1s + a0, if the

zeros of p(s) all have negative real part, then ai > 0 for i = 0, 1, . . . , n

follows by factorizing p(s) into product of first and second order polynomials with

positive coefficients, which is possible because all zeros have negative real part

so if any of the polynomial co-efficients is zero or negative we know there must be

a zero with non-negative real partthe condition is sufficient when n = 1, 2, but not in general


G0(s) = 1s(s+1.5)

and C (s) = K + s

0.1s+1

0.1s3 + 1.15s2 + (0.1K + 2.5)s + K

K

K > 0 and (0.1K + 2.5) > 0 (i.e. K > 0)

so stable if K > 0 and 3K

230 + 2.5 > 0

(i.e. for all K > 0)

s3 0.1 (0.1K + 2.5)

s2 1.15 K

s1 3K

230 + 2.5 0

s0

K

For p(s) =4

k=0

aksk, a4 > 0,

s4 a4 a2 a0s3 a3 a1 0

s2

b0.

=−

a4a1−a2a3

a3 b1.

=−

a4·0−a0a3

a3 = a0 0s1 c

.= −

a3b1−a1b0

b00

s0 −

b0·0−b1c

c = b1




G0(s) = 1s(s+1.5)

and C (s) = K + s

0.1s+1

0.1s3 + 1.15s2 + (0.1K + 2.5)s + K

K

w = s+ 1

w3 0.1 0.1K + 0.5

w2 0.85 0.9K − 1.45

w1 1

170(−K + 114)

w0 0.9K −

1.45

0.1(w − 1)3 + 1.15(w − 1)2 + (0.1K + 2.5)(w − 1) + K

= 0.1w3 + 0.85w2 + (0.1K + 0.5)w + (0.9K − 1.45)

so all closed-loop poles have real part less than −1 if 2918

< K < 114

so zeros lie in left-half plane if 0.9K − 1.45 > 0 and 1

170(114−K ) > 0

−1

(w) < 0 ⇔ (s) < −1


K

K > 0

1 + K · F s0 = 0

F (s) :

s0

K =1

|F (s0)|

1 + K · F (s) = 0 where F (s) = M (s)

D(s) =

m

k=1(s− β k)

n

k=1(s− αk)

e.g. 0.1s3 + 1.15s2 + (0.1K + 2.5)s + K = 0⇔ 1 + K 0.1s+1

s(0.1s2+1.15s+2.5) = 0

lim|ω|→∞

∠F ( jω) = (m− n)π

2

if this is not the case

(e.g. F (s) = −1/s(s + 1)) then

analysis must be modified to

handle additional π rad phase

(2l + 1)π = ∠F (s0) =m

k=1∠(s0−β k)−

n

k=1∠(s0−αk) for l = 0,±1,±2, . . .

magnitude condition: |K · F s0 | = 1

phase condition: ∠K · F (s0) = (2l + 1)π for l = 0, ±1, ±2, . . .




K

K

F s

F s

1 + K · F (s) = D(s) + KM (s)

D(s)

1 + K · F (s) = K ·

1

K D(s) + M (s)

D(s)

s s s

s− γ

(s− γ )

∠(s− γ ) =

−∠(s− γ )

γ ∠ s− γ = 0

γ

γ

s− γ s− γ

∠ s− γ = π

γ


n m

K →∞

K → 0

n−m

m− n

∞

∞

s = σ ηk

ηk

n > m

s = −σ

σ .=

n

k=1αk −

m

k=1β k

n−m and ηk

.=

(2k − 1)π

n−m for k = 1, . . . , n−m




K

F (s) n > m

|F s |→ 0 |s|→∞ n > m

|s|

K 1 + KF s 1 + K 1

sn−m

σ

s0

σ = −

(an−1 − bm−1)

n−m

=

n

k=1αk −

m

k=1β k

n−m

F (s) = sm + bm−1s

m−1 + · · · + b1s + b0

sn + an−1sn−1 + · · ·a1s + a0

=

m

k=1(s− β k)

n

k=1(s− αk)

= 1

sn−m + (an−1 − bm−1)sn−m−1 + · · · + d1s + d0 + cm−1sm−1+···+c1s+c0

sm+bm−1sm−1+···+b1s+b0

≈

1

(s− σ)n−m =

1

sn−m − σ(n−m)sn−m−1 + · · · + (n−m)(−σ)n−m−1s + (−σ)n−m

(n−m)∠ 1

s0= ∠− 1 = (2l + 1)π, l = 0,±1, . . . ,

∠s0 = 2k − 1 π/ n−m , k = 1, 2, . . . , n−m


0.1s3 + 1.15s2 + (0.1K + 2.5)s + K = 0 ⇔ 1 + K 0.1s+1

s(0.1s2+1.15s+2.5) = 0

i.e. F (s) = (s+10)s(s+8.5895)(s+2.9105)

Root Locus

Real Axis

I m a g i n a r y A x i s

>> z = [-10];

>> r = roots([0.1,1.15,2.5])

r =

-8.5895

-2.9105

>> p = [0 r(1) r(2)];

>> g = 1;

>> F = zpk(z,p,g)

Zero/pole/gain:

(s+10)

---------------------

s (s+8.589) (s+2.911)

>> rlocus(F);

as K →∞ two (n−m = 2)

roots approach ∞ along asymptotes

that intersect the real axis at

σ = 0− 8.5895− 2.9105 + 10

2

with angles η1 = π

2, η2 =

3π

2

as K →∞ one root approac es

the zero of F (s)

at s = −10

as K → 0 the locus of roots

approaches the poles of F (s)at s = 0,−2.9105,−8.5895




Λ0 jω = G0 jω C jω

2

Real Axis


Real Axis


s

s

→

2π

s + 1


0 3 4 5 6 7 8

0

3

4

5

Real Axis


Real Axis


Real Axis


Real Axis


s → (s + 1)2

two clockwise

encirclements

of the origin

when two zerosare encircled

s →

1

s + 1

one anti -clockwise

encirclement

of the origin

when one pole

is encircled




Real Axis


0

0

Real Axis


s →

s + 1

(s + 1.25)2

N = Z − P

N = number of clockwise encirclements of the origin made by

F (s) as s traverses a closed contour in clockwise direction

Z = number of zeros of F (s) inside the contour

P = number of poles of F (s) inside the contour


Λ0 jω = −1 for −∞ < ω <∞

s

s

ε

→∞

→ 0

|Λ0 jω p | = ∞

jω p

− jω p

x

x

1 + Λ0(s) = 1 + G0(s)C (s)

# encirclements of 0 made by 1 +Λ0 s = # encirclements of −1 made by Λ0 s

Nyquist stability criterion:

By Cauchy’s principle of the argument

the closed-loop has Z = N + P

unstable poles where

P = # RHP poles of Λ0(s)

N = # clockwise encirclements of

− 1 made by Λ0(s) as s traverses N ;

The closed-loop is stable if, and only if,

N = −P (i.e. Λ0(s) makes P anti-

clockwise encirclements of − 1.)

N (, ε)




limω→0

Λ0 jω

lim→∞

Λ0(ejθ)s

0 ≤ ω <∞

Λ0 s

ε→ 0

−nπΛ0(s)

n

2

−1 + j0

{Λ0(s) | s ∈ N (, ε) for sufficiently large and sufficiently small ε}

lim|s|→∞

Λ0(s) = 0


1

Nyquist Diagram

Real Axis


2 3

2

Nyquist Diagram

Real Axis


Λ0(s) = 1

(s + 1)2

Λ0(s) = 20

(s + 1)(s + 2)(s + 3)

ω = 0ω = +∞

ω = −∞

ω = 0

ω = +∞

ω = −∞

Λ0( jω) = 1 − ω

2

(1 + ω2)2

− j2ω

(1 + ω2)2

ω = 1

ω = −1

Λ0(s)

>> lambda = tf(1,[1 2 1]);

>> nyquist(lambda);




Nyquist Diagram

Real Axis

I m a g

i n a r y A x i s

Nyquist Diagram

Real Axis

I m a g i n a r y A

x i s

Λ0(s) = 5

(s + 1)(s− 2)(s + 3)

Λ0(s) = 1

s(s + 1)

ω = 0ω = +∞

ω = −∞

ω = −∞

ω = +∞

Λ0(s)

ω = 0+

ω = 0−

s = 0


Nyquist Diagram

Real Axis


Λ0(s) = 1.5(s + 1)

s(s− 4)

Λ0(s) = e−2s

(s + 1)2

ω = +∞

ω = −∞

ω = 0

ω = 0+

ω = 0−

Λ0(s)

Nyquist Diagram

Real Axis


Λ0(s)





1 + j0

−1 + j0

Nyquist Diagram

Real Axis


0 5 10 15 20 25 30 35 40 45 500

0.1

0.2

0.3

0.4

0.5

0.6

0.7Step Response

Time (sec)

A m p l i t u d e

L−1[

Λ0

1 + Λ0

· 1

s](t)

ω = 0

ω = +∞

closed-loop poles: s = −0.9393;

s = −0.0304± j1.3936

Λ0(s) = 0.825

s3 + s

2 + 2s + 1




Λ0(s) = G0(s)C (s)

Nyquist Diagram

Real Axis

I m a g i n a r

y A x i s

|a|

η

Λ0(s) = 30

(s + 1)(s + 2)(s + 3)

|1 + Λ0| = dist. from − 1 + j0 to Λ0

cross-over frequency

|Λ0( jω)| = 1

> 30◦

good

> 15 goo

< 4 goo

PM: M f = φ (degrees or radians)

GM: M g = 20 log10

( 1

|a|) = −20log

10 |a| (dB)

SP: 1

η =

1

minω |1 + Λ0( jω)| = max

ω

|S 0( jω)|


maxω

|Λ0 jω | < 1


P h a s e ( d

e g )

Bode Diagram

Frequency (rad/sec)

M f

M g

Λ0(s) =

(s + 1)(s + 2)(s + 3) S 0 =1

1 + Λ0

20log10

1

minω |1 + Λ0( jω)|

2π

Λ0(s)




G s = G0 s + Gε s

G0 s

G0 s

G0(s)

G∆ s

Gε(s)

G s = G0 s 1 + G∆ s

Note: Gε s = G∆ s G0 s


G0(s) = 1

s(s2 + s + 1)(0.3s + 1)

G1(s) = 1s(s2 + s + 1)(0.2s + 1)

G2(s) = 1

s(s2 + s + 1)(0.5s + 1)

Gε2

G∆2

Gε1

G∆1

G0

10

100

101

0

10


Frequency (rad/sec)

G0

G0(s) =s(0.5s + 1)

Gε2

Gε1

G∆2

G∆1

G0

G0(s) = 11

(10s + 1)(s2 + s + 1)

10

10

100

101

0

10


Frequency (rad/sec)

10

100

101

102

0

10

20


Frequency (rad/sec)

Gε2

G∆2

Gε1

G∆1

G2(s) = e−0

.2s

s(0.5s + 1)

G1(s) = 11

(20s + 1)(s2 + s + 1)

G2(s) =(5s + 1)(s2 + s + 1)

G1(s) = e−0

.4s

s(0.5s + 1)

Note: perturbations

Gε and G∆

are all stable for the

examples shown; may

not always be the case




Suppose G0(s) is the nominal transfer function model of a transfer function G(s)

Suppose C (s) is a strictly proper feedback controller transfer function that internally

stabilizes G0(s)

Suppose Λ0

.= G0C and Λ

.= GC have same number of RHP poles

e.g. G = G0 + Gε or G = G0(1 + G∆) with Gε, G∆ stable

Nyquist Diagram

Real Axis


|1 + Λ0( jω)|

Λ0 jωΛ jω

|GεC jω | = |G∆Λ0 jω |

|G( jω)| · |C ( jω)|

|1 + Λ0( jω)| < 1 or |G∆( jω)| ·

|Λ0( jω)|

|1 + Λ0( jω)| < 1 for all frequencies

Then C (s) also internally stabilizes G(s) if

The condition ensures the number of encirclements

of the critical point does not change

The condition is only sufficient (i.e. potentially conservative)

In practice may only have frequency dependent bound on

|G∆( jω)| with no phase information and so hard to do better

Result can be extended to unstable perturbations - just

need to account for appropriate number of encirclements

as required to satisfy Nyquist stability criterion


Suppose G0(s) is the nominal transfer function model of a transfer function G(s)

Let G∆(s) = G(s)−G0(s)G0(s)

be the multiplicative modelling error

Nominal and achieved performance similar if S ∆ jω ≈ 1 + j0 for all frequencies

his is roughly the case provided |T 0( jω)| is sufficiently small where |G∆( jω)| be-comes significant (often high frequency) so that |T 0( jω)G∆( jω)| << 1.

For a controller C (s) that achieves the nominal closed-loop sensitivity functions

S 0(s), T 0(s), S i0(s) and S u0(s)

we have by direct calculation the following perturbed sensitivity functions:

S (s) = S 0(s)S ∆(s)

T (s) = T 0(s)(1 + G∆(s))S ∆(s)

S i(s) = S i0(s)(1 + G∆(s))S ∆(s)

S u(s) = S u0(s)S ∆(s)

where S ∆(s) .= 1

1+T 0(s)G∆(s)

Model uncertainty typically limits loop-gain bandwidth





2

3

P h a s e ( d e g )

Frequency (rad/sec)

G0S i0

T 0

C Λ0

Consider a feedback control loop with:

nominal plant model G0(s) = 2

(s + 1)(s + 2), unit step reference r(t) = ς (t),

input disturbance di(t) = A sin(t + φ), stable MME s.t. |G∆( jω)| ≤ ω√ ω2 + 400

and controller C (s) = 600 (s + 1)(s + 2)(s + 3)(s + 5)s(s2 + 1)(s + 100)

T 0 jω ≈ 1 over the range 0 ≤ ω ≤ 1,

by virtue of high gain of controller; as such,

reference tracking performance is good

gain of input sensitivity function is 0 at

1 rad/sec, by virtue of controller poles at ± j;

as such, input disturbance rejection is achieved

the LHP zeros of controller add phase to achieve

good phase margin; sensitivity peak also small

loop-gain bandwidth is maximized for fast response

while accounting for modelling uncertainty which

becomes of size 1√

2≈ 0.7 at 20 rad/sec

high gain of controller at high frequency could

be problem in the presence of measurement noise


1 + j0




ωb : |Λ0 jωb | = 1


S

0 s = 1− e

−τ s

T

0 (s) = e

−τ s1

10

100

101

102

0

10


Bode Diagram

Frequency (rad/sec)

T

0 (s)

S

0 s

Given a plant model G(s) = e−τ s

G0(s)

the muliplicative modelling error is

G∆(s) = e−τ s

− 1

which has same magnitude plot as S 0(s)

1

τ =

1

0.02 = 5

e−τ s

e−τ s

R

Do

Y

−

+ ++

+

+

1

1− e−τ s




Given a nominal plant G0(s) = B0(s)A0(s)

and any controller C (s) = P (s)L(s) , suppose

s = z is a zero and s = p a pole of Λ0(s) = G0(s)C (s) (i.e. z is an uncancelledero and p an uncancelled pole of the plant or controller):

T 0(z) = B0(z)P (z)A0(z)L(z) + B0(z)P (z)

= 0 S 0(z) = 1 − T 0(z) = 1

S 0( p) = A0( p)L( p)

A0( p)L( p) + B0( p)P ( p) = 0 T 0( p) = 1− S 0( p) = 1

If s = p is an uncancelled pole of C (s) then

S i0( p) = B0( p)L( p)

A0( p)L( p) + B0( p)P ( p) = 0


−

α > 0

z p z > −α and p > −α

z and p

∞

0

h(t)e−ztdt = lim

s→z

H (s) for any z in region of convergence for H (s) .= L[h](s)

S 0 z = 1, T 0 z = 0, S 0 p = 0, T 0 p = 1, S 0 0 = 0, T 0 0 = 1

s = 0

E (s) = S 0(s)(R(s)−D(s))

For R(s) = 1

s and D(s) = 0 OR R(s) = 0 and D(s) = −

1

s

(i) ∞

0

e(t)e−zt dt = E (z) = S 0(z)1

z

= 1

z

and

(ii)

∞

0

e(t)e− pt dt = E ( p) = S 0( p)1

p = 0




z

p α

s = 0

s = p

ii

i

y t > r t

s = z


ln |S 0 s | = − ln |1 + Λ0 s |

∞

0

ln |S 0( jω)|dω = 0




X (s) = X (s)Γ(s)

where X (s) is an order q − 1 polynomial depending on the initial

conditions and Γ(s) = sq +

q−1

n=0

cnsn

Γ s

x = 0 with x(0) = x; x(t) = x for t ≥ 0

x = 0 with x(0) = 0, x(0) = x; x(t) = xt for t ≥ 0

x + ω2x = 0 with x(0) = x, x(0) = 0; x(t) = x cos(ωt) for t ≥ 0

x + ω2x = 0 with x(0) = 0, x(0) = x; x(t) = x sin(ωt) for t ≥ 0

and more generally dqx

dtq+

q−1

n=0

cndnx

dtn= 0 with

dnx

dtn= xn for n = 0, . . . , q − 1


Λ0 s = G0 s C s

Γ(s) =

q−1

n=0

(s− γ n) (with (γ n) ≥ 0)

γ n

S 0 γ n = 0 and T 0 γ n = 1

limt→∞

e(t) = 0

Y (s) = T 0(s)R(s)

E (s) = R(s)− Y (s) = (1− T 0(s))R(s) = S 0(s)R(s)

U (s) = S u0(s)R(s)

E s = S 0 s R s




C s

E sR s Y sU s

D(s)

Gi0 s Go0 s

Gi0(s) = 1 and Go0(s) = G0(s)

G0 = Go0Gi0

Γ(s) = s

2

+ 1

Γ(s) =

q−1

n=0

(s− γ n) (with (γ n) ≥ 0)

γ n Gi0 s

Y s

Y (s) = S 0(s)Go0(s)D(s) U (s) = −S u0(s)Go0(s)D(s) = −

T 0(s)

Gi0(s)D(s)

Gi0 s = G0 s and Go0 s = 1


Gi0(s) = 1

S = S 0S ∆

S 0 =1

1+G0C

T 0 γ n = 1




H (s) T 0(s)

T 0 s H s = 1

H s

s = γ n

Y s = T 0 s H s R s ; E s = 1− T 0 s H s R s ; U s = S u0 s H s R s


r t = K 1 sin t + K 2 + ra t

K 1 K 2 ra t

[0, 5]rad/sec 3 rad/sec

T 0 s 3 rad sec

T 0 s H s 5 rad sec

s = 0 s = ± j

C (s) = (s2+3s+2)(5s2+3s+2)

s(s2+1)(s+5)

T 0(s) = 10s2+6s+4(s2+3s+4)(s+1)2

G(s) = 2

s2+3s+2

H s

0 s 2 T 0 s H s

H (s) = (s2+3s+4)(s+1)2

(10s2+6s+4)(0.01s+1)2 T 0(s)H (s) = 1(0.01s+1)2

ra(t)

5 rad sec




10

10

100

101

10

103

0

60


Bode Diagram

Frequency (rad/sec)

H (s)

T 0 s

S 0 s

T 0 s H s


C (s)E sR s Y sU s

D s

Gi0(s) Go0(s)

G0 = Go0Gi0

Gf s

Gf (s) = −Gi0(s)−1

Gi0

Y (s) = T 0(s)R(s) + S 0(s)Go0(s)(1 + Gi0(s)Gf (s))D(s)

U (s) = S u0(s)R(s)− (S u0(s)Go0(s)− S 0(s)Gf (s))D(s)




G0 = Go0Gi0 Gi0(s) = 2

s− 2 Go0(s) =

1

s + 1

C (s) = 3(s2 + 2s + 2)

s(0.1s + 1)(0.05s + 1)

10

100

101

102

103

0

10

20

30

M a g

n i t u d e ( d B )

Frequency (rad/sec)

Gi0 Go0

C s

S 0 s

0(s)

Λ0 s

Go0 s S 0 s

peaking of sensitivity is a result of limiting

loop-gain bandwidth to be not much more

than the bandwidth of the unstable pole

for robustness to MME above 10 rad/sec

correspondingly, injected disturbances with

frequencies in the range [2, 7] rad/sec

are not attenuated by much at plant output

if these components of the disturbance were

measurable, then feedforward could be used

to improve the level of attenuation ...


10

100

101

102

103

0


Frequency (rad/sec)

Go0(s)S 0(s)

Go0(s)S 0(s)(1 + Gi0(s)Gf (s))

1 + Gi0(s)Gf (s)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

0

0.5

1

1.5

2

2.5

Time (sec)

Gf (s) = −

s− 2

2(0.01s + 1) ≈ −Gi0(s)−1




−1 + j0

Documents

Control System Part 3