Control System Part 2

7/23/2019 Control System Part 2

http://slidepdf.com/reader/full/control-system-part-2 1/24

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

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




t

tt


di do

ye u




y(t) = k(u(t))

y(t) =m

t

0

· · ·

t

0

gm(t, τ 1, . . . , τ n)u(τ 1) · · ·u(τ n)dτ 1 . . . dτ n

Y σ + jω = G σ + jω U σ + jωy(t) =

t

0

g(t− τ )u(τ )dτ

y t , y t , y t , . . . , u t , u t , . . . = 0





LC R

u V

i A

M

kg

k (Nm−1)

u t y t

LC y(t) + L

R y(t) + y(t) = u(t)

y V

u (m3s−1)

y (m)

u N

y m

u t y t

k√ y (m3s−1)

u t y t

A

Ay(t) + k

y(t) = u(t− τ )

τ

v(t) = dv

dt (t)

M y t + ky t = u t


x

2 n + 1

dxdt

(t) = f (x(t), u(t)); y(t) = g(x(t), u(t))

(dny

dtn(t), . . . ,

dy

dt (t), y(t),

dnu

dtn(t), . . . ,

du

dt (t), u(t)) = 0

t

yk, uk k = 0, . . . , n

yn, . . . , y0, un, . . . , u0 , which satisfies yn, 0, . . . , 0 ≡ 0,




d

dt

x1

x2

=

−

βα 1

−

γ α 0

x1

x2

+

−

βαδ

η − γ αδ

u

y = 1

α 0

x1

x2

+

δ

αu

αd2y

dt2 + β

dy

dt + γ y

−

δ

d2u

dt2 −

du

dt −

ηu= 0 (α = M,β = 0, γ = k, δ = 0, = 0, η = 1)

d2

dt2( x1 = αy − δ u )

d

dt( x2 =

dx1

dt + β y − u )

0 = dx2

dt + γ y − ηu


2(n + 1)

(0, . . . , 0, y, 0, . . . , 0, u) = 0

yn = dny

dtn (t), . . . , y1 = dy

dt (t), y0 = y(t), un = dnu

dtn (t), . . . , u1 = du

dt (t), u0 = u(t)

(dny

dtn(t), . . . ,

dy

dt (t), y(t),

dnu

dtn(t), . . . ,

du

dt (t), u(t)) = 0 ()

u · , y ·

u · , y ·

dky

dtk (t) = 0,

dku

dtk (t) = 0 for k = 1, . . . , n and all times t

(u, y)




(0, . . . , 0, y, 0, . . . , 0, u) = 0

yn

= , . . . , y1 = , y0 = y, un

= , . . . , u1 = , u0 = u

0

δ u = u− u → δ y = y − y

lin(dnδ y

dtn, . . . ,

dδ y

dt , (δ y + y),

dnδ u

dtn, . . . ,

dδ u

dt , (δ u + u)) = 0

u

yn

, . . . , y0 − y , un

, . . . , u0 − u

dkδydtk

=dky

dtk

dkδudtk

=dkudtk

u, y

lin(yn, . . . , y1, y0, un, . . . , u1, u0).

= (0, . . . , 0, y, 0, . . . , 0, u) + ∂

∂ yn

(0,...,0,y,0,...,0,u)

(yn − 0) + . . . + ∂

∂ y0

(0,...,0,y,0,...,0,u)

(y0 − y)

+ ∂

∂ un

(0,...,0,y,0,...,0,u)

(un − 0) + . . . + ∂

∂ u0

(0,...,0,y,0,...,0,u)

(u0 − u)


u

y

uaub

yb

ya

y, u = 0

u

y




θ

F a

MgM

F a cos θ

l − l cos θl sin θ

u

F a sin + u

u

sin θ ( F a cos θ −M g = M d2

dt2(l − l cos θ) = M l

d

dt( θ sin θ) = M l (θ sin θ + θ2 cos θ) )

cos θ ( − F a sin θ − u = M d2

dt2(l sin θ) = M l

d

dt( θ cos θ) = M l (θ cos θ − θ2 sin θ) )+

θ · , u · , F a ·

M l θ + u cos θ + M g sin θ = 0

F a ·

θ = 0 u = 0

is one possible equilibrium

operating point


θF a

Mg

M

F a cos θ

u

F a sin + u

u

(θ, θ, θ, u) = M l θ + u cos θ + M g sin θ = 0

(θ2, θ1, θ0, u0) = M lθ2 + u0 cos θ0 + Mg sin θ0

≈ (0, 0, θ, u) + ∂

∂θ2

(0,0,θ,u)

(θ2 − 0) + ∂

∂θ1

(0,0,θ,u)

(θ1 − 0)

+ ∂

∂θ0

(0,0,θ,u)

(θ0 − θ) + ∂

∂ u0

(0,0,θ,u)

(u0 − u)

= 0 + Ml(θ2 − 0) + 0(θ1 − 0) + (−u sin θ + Mg cos θ)(θ0 − θ) + cos θ (u0 − u)

= lin(θ2, θ1, θ0, u0)

0

For small δ u = u−

u and δ θ .= θ−

θ we have δ θ = θ, δ θ = θ and the approximate linear model

M l δ θ + (M g cos θ − u sin θ) δ θ + cos θ δ u = 0

(u, θ)




eAt .= I +

∞

n=1

1

n!(At)n

dx

dt (t) = Ax(t) +Bu(t), y(t) = Cx(t) +Du(t),

x(t) = eAtx(0) +

t

0

eA(t−τ )Bu(τ )dτ ; y(t) = Cx(t) +Du(t)

u

·

y ·

x 0

A,B,C ,D


φ a1u1 + a2u2 = a1φ u1 + a2φ u2 a1, a2 u1 · , u2 ·

φ(uτ ) = yτ for all τ > 0 where sτ (t) =

s(t− τ ) if t ≥ τ ;0 otherwise

φ = u · → y ·

s

sτ

τ

x 0 = 0A,B,C ,D

x(t) =

t

0

eA(t−τ )Bu(τ )dτ ; y(t) = Cx(t) +Du(t)

y(t) u(τ ) τ > t

with y = φ(u) note that




u

t

t1 t2 t3

u(t1)

u(t0)

u(t2)

u(t3)u(t1)− u(t0)

t0 =

u(t) = u(t0)ς (t) +n>0

[u(tn)− u(tn−1)] ς (t− tn)

where the unit step ς (t) .=

1 if t ≥ 00 otherwise

φ = u · → y ·


φ = u(·) → y(·)

h .= φ ς

hτ (t)

.=

h(t− τ ) if t ≥ τ

0 otherwise t0 = < t1 < t2, . . .

y.

= φ uu(t) .

= u(t0)ς (t) +n>0

[u(tn)−

u(tn−1)] ς (t−

tn)

y(t) = u(t0)h(t) +

n>0

[u(tn)− u(tn−1)] htn

(t) =

maxn:tn<t

n=0

htn

(t)− htn+1

(t)u(tn)

y(t) → y(t) =

t

0

g(t− τ )u(τ )dτ

˜(t

)→ u

(t

)

h(t− tn+1 + (tn+1 − tn))− h(t− tn+1)

tn+1 − tn→

dh

dt (t−

tn+1)

.

= g(t−

tn+1)




r s

−M u

y∞ +M p

y∞y∞ + δ

y∞ −

y(t) = (g ∗ u)(t) .=

t

0

g(t − τ )u(τ )dτ

u(·) → y(·)

g

g

y∞ steady-state final value(sign normalized to 1 here)

tr rise time to within specified

fraction of y∞ (here y∞−δ

y∞)

ts settling time after which responseremains within δ of y∞

M p overshoot - maxiumum amount the(normalized) response exceeds y∞(%)

M u undershoot - maximum amount the(normalised) response fall below 0(%)

0 2 4 6 8 10 12 14 16-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

A m

p l i t u d e


1 = ej0

−1 = e±jπ

− j = e−j π2

j = ej 2

re−jφ

(r < 1)

Rej(π−φ)

(R > 1)

z1 + z2 = z1 + z2 + j z1 + z2

z1 · z2 = |z1||z2|ej(∠z1+∠z2)

j sin

cos

z = z + j z

= ( (z)2 +(z)2) ej arctan((z)/(z))

= |z|ej∠z

ez = e(z) ej(z) = e(z)(cos(z) + j sin(z))

ejφ= e

j(φ±2πn)

(n = 0, 1, 2, . . .)

z

z




DEFINITION: The Laplace transform of a time-domain signal y(·) is denoted

Y (s) = L[y](s) .=

∞

0

y(t)e−st dt

defined for complex s = σ + jω that satisfy σ < σ = (s) (i.e. s ∈ ROC(Y )), where

σ

.= inf σ : |y(t)|e−σt < k for all t ≥ 0 and some real constant k

Find Y a(s) = L[ya](s), where ya(t) .= e−at and a is a real constant :

Y a(s) =

∞

0

e−ate−st dt =

e−(s+a)t

−(s + a)

∞0

= 1

(s + a) for (s) > −a

Find Y n(s) = L[yn](s), where yn(t) .= tn

n! =

t0

yn−1(τ )dτ and n ≥ 0 is an integer :

Y n(s) =

∞

0

tn

n!e−st dt =

tn

n!

e−st

−s

∞

0

+

∞

0

t(n−1)

(n− 1)!

e−st

s dt

= 1

sY n−1(s) =

1

snY 0(s) =

1

s(n+1), for (s) > 0


TRANSFORM POST INTEGRATION: With yint(t) .=

t

0 y(τ )dτ ,

Y int(s) =

∞

0

t

0

y(τ )dτ

e−stdt

=

t

0

y(τ )dτ

e−st

−s

∞0

+ ∞

0

y(t)e−st

s dt

= 1

sY (s) (s ∈ z : (z) > 0 ∩ ROC(Y ))

TRANSFORM POST SHIFTING: With yτ (t) = y(t− τ ) if t ≥ τ ; 0 otherwise,

Y τ (s) =

∞

τ

y(t− τ )e−st dt =

∞

0

y(ν )e−s(ν +τ ) dν = e−sτ Y (s) (s ∈ ROC(Y ))

TRANSFORM POST EXPONENTIAL WEIGHTING: With yz(t) = ezty(t),

Y z(s) =

∞

0

y(t)e−(s−z)t dt = Y (s− z) (s ∈ (z) + w : w ∈ ROC(Y ))




an = 0

General (proper) ODE model form:

nk=0

akdky

dtk =

nk=0

bkdku

dtk .

Assuming zero initial conditions:

nk=0

aksk

Y (s) =

nk=0

bksk

U (s).

This yields the model Y (s) = G(s)U (s) where G(s) = B(s)

A(s)

TRANSFORM POST DIFFERENTIATION:

L[ y](s) =

∞

0

dy

dt e−st dt =

y(t)e−st

∞

0 + s

∞

0

y(t)e−st dt;

= sY (s)−

y(0)

L[dny

dtn ](s) = snY (s)− sn−1y(0)− sn−2y(0)− . . .−

dn−1y

dtn−1 (0)

LINEARITY OF THE TRANSFORM:

L[a1y1 + a2y2](s) = a1Y 1(s) + a2Y 2(s) (because integration is linear)


0

∞

Using the standing causality assumption g(t) = 0 for t < 0, it follows that

L[g ∗ u](s) =

∞

0

e−st

t

0

g(t − τ )u(τ ) dτ dt

= ∞

0 ∞

0

e−stg(t − τ )u(τ ) dt dτ

=

∞

0

∞

−τ

e−s(T +τ )g(T )u(τ ) dT dτ (T = t − τ )

=

∞

0

u(τ )e−sτ

∞

−τ

g(T )e−sT dT

dτ

=

∞

0

u(τ )e−sτ

∞

0

g(T )e−sT dT

dτ = G(s) · U (s)




ith Y (s) = L[y](s), the following hold whenever the limits exist.

FINAL VALUE: limt→∞

y(t) = limσ→0+

(s Y (s))s=σ+j0

INITIAL VALUE: limt→0

y(t) = limσ→∞

(s Y (s))s=σ+j0

Suppose there exist constants K, a > 0 and y∞ such that |y(t) − y∞| < Ke−at.

Then y∞ = limt→∞ y(t), Y (s) is defined for (s) > 0, and

limσ→0+

σ ∞

0

y(t)e−σtdt − y∞

= lims→0+

σ

∞

0

(y(t) − y∞)e−σtdt

≤ lim

σ→0+

σ ∞

0

Ke−ate−σtdt ≤ limσ→0+

σ

K

σ + a

= 0.

Now suppose there exist constants K > 0 and b such that | y(t)| ≤ Kebt, whereby

L[ y](s) = sL[y](s) − y(0) is defined for (s) > b. Then

limσ→∞

|σY (σ) − y(0)| = limσ→∞

∞

0

y(t)e−σtdt

≤ limσ→∞

∞

0

Kebte−σtdt = limσ→∞

K

σ − b = 0


LR

i A

θ (rad)

J

u V

B

K 1 θ (V)

linearised motor equation: τ (t) = K 2i(t) L↔ T (s) = K 2I (s)

Newton’s second law: J θ(t) = τ (t)−B θ(t) L↔ J (s2Θ(s)− sθ(0)− θ(0))

= T (s)−B(sΘ(s)− θ(0))

Kirchoff ’s voltage law: u(t) = Ri(t) + Ldi

dt(t) + K 1

θ(t)

L↔ U (s) = R I (s) + L(s I (s)− i(0)) + K 1(sΘ(s)− θ(0))

u (V) θ (rad)

T 1 and T 2

Θ(s)

U (s) =

K 2

s

(sL + R)(s J + B) + K 1K 2

= K

s ·

1

(T 1s + 1)(T 2s + 1) RJ + BL)2 >

4(K 1K 2 + RB)LJ

τ (Nm)




K ≈ 2.2; T 1 ≈ 12.5 (ms); T 2 = 1 (ms).

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.02.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5Impulse Response

Time (sec)

A m

p l i t u d e

A3

A2

A1

K = 2.2;

T1 = 12.5e-3;

T2 = 1e-3;

T = T1*4;

A1 = K;

A2 = (-K*T1)*1/(-T2/T1 + 1);

A3 = (-K*T2)*1/(-T1/T2 + 1);

sys1 = A1*tf(1,[1 0]);

sys2 = A2*tf(1,[T1,1]);

sys3 = A3*tf(1,[T2,1]);

sys = sys1 + sys2 + sys3;

impulse(sys,’r’,sys1,’y’,sys2,’g’, ...

sys3,’b’, [0:0.0001:T])

Θ(s)

U (s) =

K

s ·

1

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

A1

s +

A2

T 1s + 1 +

A3

T 2s + 1

L↔ (A1 +

A2

T 1e−

t

T 1 + A3

T 2e−

t

T 2 )ς (t)

here A1 = K, A2 = K −1/T 1

1(−T 2/T 1+1) , A3 = K −1/T 2

1(−T 1/T 2+1)


A1

A2

A1

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.00.5

0

0.5

1

1.5

2

2.5Step Response

Time (sec)

A m p l i t u d e

K = 2.2;

T1 = 12.5e-3;

T2 = 1e-3;

T = T1*4;

A1 = K/(-T2/T1 + 1);

A2 = K/(-T1/T2 + 1);

sys1 = A1*tf(1,[T1,1]);

sys2 = A2*tf(1,[T2,1]);

sys = sys1 + sys2;

step(sys,’r’,sys1,’g’, ...

sys2,’b’,[0:0.0001:T])

Ω(s) = sΘ(s)− θ(0) L↔ ω(t) = θ(t)

Ω(s)

U (s) =

K

(T 1s + 1)(T 2s + 1)

For a unit step input: Ω(s) = K

(T 1s + 1)(T 2s + 1) ·

1

s

= A1

T 1s + 1 ·

1

s +

A2

T 2s + 1 ·

1

s

where A1 = K (−T 2/T 1 + 1)

and A2 = K (−T 1/T 2 + 1)




u (m3s−1)

y m

k√ y (m

3

s

−1

)

u t y t

A

Ay(t) + k

y(t) = u(t− τ )

τ

Linearizing about the equilibrium (uq, yq.

= u2q/k2), with δ y

.= y − yq and δ u

.= u − uq :

Aδ y(t) + k

2√ yq

δ y(t) = δ u(t− τ )

L

G(s) = e−sτ

As + k2√ yq

A rational (Pade) approximation of the delay: e−sτ = e−sτ /2

esτ /2 ≈

1 − (τ /2) s

1 + (τ /2) s


F s

F s

C (s)

H s

F s · G s

F s + G s

G(s) · C (s)

1 + G(s) · C (s)

G s)

1 + G(s) · H (s)

U s Y s

U s

Y (s)

U s Y s

U s Y s

U s Y s

U s Y (s)

U (s) Y (s)

U s Y s

G s

G s

G(s)

G s





ak, bk αk, β k

σk ± jωk

k

αk

G(s) = bms

m + · · · + b1s + b0

ansn + · · · + a1s + a0

= K (s− β 1) · · · (s− β m)

(s− α1) · · · (s− αn) = K

m

k=1(s− β k)

n

k=1(s− αk)

ith relative degree nr = n−m≥ 0.




s− 1

s2 + 3s + 2 =

s− 1

(s + 1)(s + 2)

s2 + 2s + 2

s3 + 3s2 + 3s + 1 =

(s + 1 + j)(s + 1− j)

(s + 1)3

s

s

s

s

L

G(s) = bmsm + · · · + b1s + b0

ansn + · · · + a1s + a0= K

m

k=1(s− β k)

n

k=1(s− αk)

=n

k=1

Bk

s− αk

where m < n and the residues Bk = lims→αk

(s−

αk)G(s)

g(t) =

n

k=1 Bke

αk for t ≥ 0


0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3

3.5

4

4.5Single real pole

Time (sec)

0 0.5 1 1.5 2 2.55

4

3

2

1

0

1

2

3

4

5Complex conjugate poles

Time (sec)

α = ±0.5

eαt

for t ≥ 0

1/α

α = ± 0.5± j10

e(α)t

2π/(α)

= 2π/ω

1/ α = 1/σ

e(α)t

(α) = 0

en α = σ + jω we on y cons er

(eαt) = (eσtejωt) = eσt cos(ωt)

because there is a conjugate pole

that contributes to cancel the

imaginary part and reinforce the

real part as the corresponding

residue is also conjugate ...

note that the angle of the residue

aff ects the sinusoidal term

(|B|ejφeαt) = |B|eσt cos(ωt + φ)

tneαt

and eσt

tn cos(ωt + φ)

1(s− α)n+1




y(t) + 2ψωn y(t) + ω2

ny(t) = ω2

nu(t)

L↔ G(s) =

Y (s)

U (s) =

ω2

n

s2 + 2ψωns + ω2n

ψ = 0.9

ψ = 0.1

(s)

s

σ = −ψωn

ω = ωn

1− ψ2

os−1 ψ

0 1 2 3 4 5 61

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1Canonical secondorder impulse response

Time (sec)

ωn

or 0 < ψ < 1 the poles are −2ψωn ±

(2ψωn)2 − 4ω2

n

2 = −ψωn ± jωn

1− ψ2

ωn = 2 (rad/sec)


eαt for each isolated real pole s = α; tneαt for each n-times repeated real pole s = α;

eσt cos(ωt + φ) for each conjugate pair of complex poles s = σ ± jω; and

eσttn cos(ωt + φ) for each n-times repeated complex conjugate pair s = σ ± jω




g(t)

G s = L[g] s

G s

G s

T

0 |g(t)|dt<a+bT

BOUNDED INPUTS map to BOUNDED OUTPUTS

ALL poles lie in s : (s) < 0 (i.e. strictly in the left-half plane)

For any input satisfying u = maxt≥0

|u t | < ∞,

|y(t)| = |(g ∗ u)(t)| ≤ t0

|g(t− τ )| · |u(τ )|dτ = t0

|g(ν )| · |u(t− ν )|dν ≤ u ∞0

|g(ν )|dν .

BOUNDED INPUTS map to BOUNDED OUTPUTS ⇐⇒

∞0

|g(t)|dt < ∞


αk (k = 1, . . . , n)

’transient’ decays to 0

u(t) = ejωt = cos(ωt) + j sin(ωt) L↔ U (s) = 1

s−jω

=

G(0)

Bk = lims→αk

(s− αk)G(s)

Note αk = jω

since (αk) < 0

’steady-state’ response

Y (s) = G(s)U (s) = B(s)

A(s)

1

s − jω=

n

k=1

Bk/(αk − jω)

(s − αk) +

G( jω)

(s − jω)

L

y(t) =

n

k=1

Bk

(αk − jω)eαkt + G( jω)ejωt




0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 56

4

2

0

2

4

6Step Response

Time (sec)

A m p l i t u d e c = −10

c = 10

c = −0.25

c = 0.25

c = 0.1

c = −0.1

s = c

G(s) = −(s− c)

c(s + 1)(0.5s + 1)

e−t; e

−t/0.5; 1 for t ≥ 0

c = 10

s =−1


G s G 0 = 1

c > 0

G c = 0

1 − δ ≤ |y t | ≤ 1 + δ for all t ≥ ts δ small ∼ 0.01

y t

M u ≥1− δ

ects − 1≈

1

c tsfor small c ts and δ

okay because

c ∈ ROC(V )

= s :(s)>0

Define v(t) = 1 − y(t) so that maxt≥0

v(t) = 1 + M u > 0 and v(t) ≤ δ for t ≥ ts.

We have V (s) =

1 −G(s)1

s and V (c) =

1

c =

∞0

v(t)e−ct dt.

As such, 1

c = V (c) =

ts

0

v(t)e−ct dt +

∞ts

v(t)e−ct dt

≤ (1 + M u) ts

0

e−ct dt + δ ∞

ts

e−ct dt

= (1 + M u)1 − e−cts

c + δ

e−cts

c .




M u > 1

c ts=⇒ M u large when ts small

Suppose G(s) is a stable transfer function, with G(0) = 1, G(c) = 0 for a

real valued c < 0 and a dominant pole having real part − p < 0, such that

(i) η .=

c

p

<< 1 and (ii) the unit step response y(t) satisfies

|1 − y(t)| < Ke− pt for t ≥ ts and some K > 0,

where ts is the settling time for a suitable δ .

Then the step response has a percentage overshoot

M p ≥1

e−c ts − 1

1 −

K η

1 − η

.


u(t) = ejω

t

|G jω |

∠G jω

t→∞

yss(t) = G( jω)ejωt = |G( jω)|ej(ωt+∠G(jω))

‘frequency response’




0 1 2 3 4 5 6 7 8 9

0

1Sinusoidal Response

Time (sec)

A m p l i t u d e

2π/52π/5

|G( j5)|

= 4

√ 102 + 1

G(s) = 4

2s + 1 with u(t) = sin(5t)

L↔

5

s2 + 25

y(t) = L−1

4

2s + 1 ·

5

s2 + 25

(t) = ytr(t) + |G( j5)| sin(5t + ∠G( j5))

∠G( j5)

5 = 0.2 · arctan

(G( j5))

(G( j5))= −0.294

transient decays to 0 as t→∞


1

Bode Diagram

Frequency (rad/sec)

Magnitude(dB)

1

Phase(deg)

Bode Diagram

Frequency (rad/sec)

1

Bode Diagram

Frequency (rad/sec)

Magnitude(dB)

1

Phase

(deg)

Bode Diagram

Frequency (rad/sec)

Magnitude(dB)

Phase(deg)

Magnitude(dB)

Ph

ase(deg)

K

s + 1

1

s2 + 2ψs + 1

as + 1

(s + 1)2−as + 1

(s + 1)2

= 1

= 100

ψ = 0.05

= 0.95

ψ = 0.05

ψ = 0.95

a = 10

a = 0.1

a = 10

a = 0.1

a = 10

a = 0.1

a = 10

a = 0.1




0

10

20

30

40

50

60

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

10

10

100

101

102

103

0

45

90

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)


P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

s + 1

0.1s + 120dB

1 decade

20dB

1 decade

1

0.1s + 1

1s + 1

ω1

jω2

jω3

jω3 + 1

1

for large ω we have:

| jω + 1| ≈ |ω| = 20 log10

|ω| dB

and ∠( jω + 1) = π/2 rad

for small ω we have:

| jω + 1| ≈ 1 = 0 dB and

∠( jω + 1) = 0 rad


20dB

1 decade

20dB

1 decade

jω1

jω2

jω3

jω3 − 1

1

0

10

20

30

40

50

60


10

10

100

101

102

103

90

135

180

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

10

10

100

101

10

103

P h a s

e

( d e g )

Bode Diagram

Frequency (rad/sec)

0

20dB

1 decade

s− 1

0.1s− 1

1

0.1s− 1

1

s− 1

for large ω we have:

| jω − 1| ≈ |ω| = 20 log10

|ω| dB

and ∠( jω − 1) = π/2 rad

for small ω we have:

| jω − 1| ≈ 1 = 0 dB and

∠( jω − 1) = π rad





1

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

(s + 1)(s− 50)

s(s− 0.1)(s + 10)=

50(s + 1)(0.02s− 1)

s(10s− 1)(0.1s + 1)

(s + 1)

1

(0.1s + 1)

1

(10s− 1)

1

s

20 log10 50 = 34 dB

∠50 = 0 ra

1

s= lim

k→∞

k

ks + 1

By exploiting the complex number properites |A · B| = |A| · |B|,

log10

(|A| · |B|) = log10

(|A|) + log10

(|B|), log10

( 1

|A|) = − log

10(|A|),

∠A · B = ∠A + ∠B and ∠1

A = −∠A, it follows that both the

log-log magnitude and lin-log phase plots are simply the sum of

the respective magnitude and phase plots of component factors

(0.02s− 1)

Control System Part 2

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Transcript of Control System Part 2