week5notes(1)

8/3/2019 week5notes(1)

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UNIVERSITY OF NORTH FLORIDA, SCHOOL OF ENGINEERING

EEL 4657C: Linear Control Systems - Fall 2011

Lecture Notes (Week 5) 1

O. Patrick Kreidl, Assistant Professor

Topic 1: Mathematical Background (Some of what was covered in weeks 1 to 4)

Topic 1.1: Continuous-Time LTI Dynamic Systems

Topic 1.2: Unilateral Laplace Transform

Topic 2: Transfer Functions (Some of what will be covered in week 5)

Topic 2.1: Stability

Topic 2.2: Frequency Response

1

Adapted from the Spring 1997 course notes generated by Munther A. Dahleh and O. Patrick Kreidl for6.003 Signals and Systems, MIT Department of Electrical Engineering and Computer Science.


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UNF-SE EEL 4657C: Fall 2011 Lecture Notes (Week 5) Kreidl2

1 Mathematical Background

1. Continuous-Time LTI Dynamic Systems

Differential Equation vs Input-Output Description

Linearity and Time-Invariance (LTI)

Solving ODEs with Constant Coefficients2. Unilateral Laplace Transform

Definition

Examples

Properties

Impedance Method


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Continuous-Time Dynamic Systems

The term dynamic refers to phenomena that produce

time-changing patterns (signals), the characteristics ofthe pattern at one time being interrelated with thoseat other times.

The term continuous-time refers to signals that de-pend on a time variable t which can have any value

on the continuous real-line; that is, t (,).

Commonly, we model such phenomena using differen-tial equations.

Many, many examples: from real processes to man-made systems to computer algorithms.

Signals and Systems: Development of tools for anal-ysis and synthesis of such systems.


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Dynamic Systems: Circuits

CL

i 1 i 2 i 3

v(t)

+

-

Ri(t)

v = R i1 v = Ldi2dt

i3 = Cdv

dt

(KCL) i(t) = i1(t) + i2(t) + i3(t)

=v(t)

R+ i2(t) + C

dv(t)

dtRe-arranging terms,

dv(t)

dt= 1

RCv(t) 1

Ci2(t) +

1

Ci(t)

di2(t)

dt=

1

Lv(t)

= Cd2v(t)

dt2+

1

R

dv(t)

dt+

1

Lv(t) = i(t)

2nd order differential equation driven by i(t)


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Diff Eqn vs Input-Output

Diff Eqn:

dny(t)

dtn+ an1

dn1y(t)dtn1

+ + a0 y(t) =

b0 u(t) + b1du(t)

dt+ + bm1du

m1(t)dtm1

Input-Output:

System

signal in

u y

signal out

Diff Eqn gives information about latent variables ofthe system

Input-Output gives a description of the behavior ofthe system

The two descriptions are strongly related!


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Linearity and Time-Invariance

Component Level:i

+ v -

f(i, v) = 0 is called a constitutive relation

Linearity: Iff(i1, v1) = 0 and f(i2, v2) = 0

f(i1 + i2, v1 + v2) = 0for any constants ,

Time-Invariance: Iff(i(t), v(t)) = 0

f(i(t

), v(t

)) = 0

for all

Example: Capacitor Component

i(t) = Cdv(t)

dt

linear, time-invariant

i(t) = C(t)dv(t)

dtlinear but not time-invariant

i(t) = Cd(v(t) + 2v2(t))

dtnot linear but time-invariant


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Linearity and Time-Invariance

Diff Eqn

A circuit consisting of linear time-invariant compo-nents is an example of a LTI system.

In general, LTI systems take the formdny(t)

dtn

+ an

1dn1y(t)

dtn

1+

+ a0 y(t) =

b0 u(t) + b1du(t)

dt+ + bm1du

m1(t)dtm1

where ai, bi are constants

Input-Output Superposition:

ify1 is response for u1 and y2 is response for u2,then 1y1 + 2y2 is response for 1u1 + 2u2

Time Invariance:ify1(t) is response for u1(t),then y1(t ) is response for u1(t )

(Q: what about initial conditions?!)


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Unforced Systems: Solution

dnx(t)

dtn + an1dn1x(t)

dtn1 + + a0x(t) = 0x(0), x(0), . . . , xn1(0) are given

If x(t) = Aest, then

A (sn + an1

sn

1 +

+ a0) est

0

s0, s1, . . . , sn1 are n roots of the polynomialsn + an1sn1 + + a0

If we assume all distict roots, the general solution has the

formx(t) = A0e

s0t + A1es1t + + An1esn1t

with A0, . . . , An1 are determined from initial conditions

For any root si repeated k times, there k terms in the

general solution of the formB1t

k1 + B2tk2 + + Bk

esit

with B1, . . . , Bk determined (alongside the Ais associ-ated to the distinct roots) from the initial conditions


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The Response to an Exponential Input

Example: Givend2y(t)

dt2+ 3

dy(t)

dt+ 2 y(t) = i(t) and

i(t)

t

i(t) = e - 1, t > 0t

y(0) = 1, y (0) = 0

Homogeneous solution: roots of 2 + 3 + 2

s1 = 2, s2 = 1 yh(t) = A1 e2t + A2 et

Particular solution: Assume yp(t) = A3et, thenA3e

t + 3A3et + 2A3e

t = et 6A3 = 1

yp(t) = (1/6)e

t

Total solution: Apply initial conditions toy(t) = yh(t) + yp(t) = A1e

2t + A2et + (1/6)et

y(0) = A1 + A2 + 1/6 = 1y(0) = 2A1 A2 + 1/6 = 0

A1 = 2/3A2 = 3/2

y(t) = (2/3)e2t + (3/2)et + (1/6)et

This procedure solves any o.d.e. with constant coeffs: higher-order characteristic equation means there

are more terms in homogeneous solution

different input means a different particular solution


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Examples

a)

f(t) =

1, t 00, otherwise

F(s) = 0 est dt = 1sR.O.C. Re[s] > 0

b)

f(t) = eat , t 00 , t < 0 F(s) =

0 e

(sa)t dt = 1s a

R.O.C. Re[s] > a

c)

f(t) = sin(t) , t 00 , t < 0 =

1

2j (ejt

ejt

) , t 00 , t < 0

F(s) = 12j

1

sj 1

s +j

= 1

s2 + 1

R.O.C. Re[s] > 0

d)

f(t) =

1, t T0, otherwise

F(s) = T est dt =esT

s

R.O.C. Re[s] > 0


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Properties of Laplace Transform

Linearity

a1f1(t) + a2f2(t) L a1F1(s) + a2F2(s) Uniqueness Theorem: If F1(s) = L[f1(t)] and

F2(s) = L[f2(t)] are equal in some region, then F1 =F2 and f1 = f2.

Inverse:f(t) = L1[F(s)] = 1

2j

C F(s)e

st ds

- Not interesting

- Use table look-up!

- Partial fraction expansion

Example:

F(s) =s + 2

s2 + 5s + 4

write=s + 2

(s + 1)(s + 4)

=1/3

s + 1+

2/3

s + 4 f(t) = 1

3et +

2

3e4t, t 0


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Properties (continued)

General Case; distinct poles

F(s) =bms

m + bm1sm1 + + b0sn + an1sn1 + + a0 n > m

=bms

m + bm1sm1 + + b0(s

p1)(s

p2)

(s

pn)

=K1

sp1 +K2

sp2 + +Kn

spnKi = (spi)F(s)|s=pi

f(t) = K1ep1t + + Knepnt t 0

Differentiation in time multiplication ins-domain

If F(s) = L[f(t)], then L[f(t)] = sF(s) f(0)(Verify: Integration by parts!)

Theorem useful for solving differential equations!


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Circuit Example (revisited)

y(t) + 3y(t) + 2y(t) = i(t) , y(0) = 1, y(0) = 0

i(t) = et 1, t 0 Take Laplace Transform of both sides (use Diff Thm)

y(t) L Y(s)y(t) L sY(s) y(0) = Y(s)y(t) L sY(s) y(0) = s2Y(s) sy(0) y(0)

Plug in; let i(0) = 0s2Y(s)sy(0)y(0)+3sY(s)3y(0)+2Y(s) = sI(s)i(0)

(s2 + 3s + 2)Y(s) (s + 3) = sI(s)

Y(s) =s + 3

s2 + 3s + 2 zero-input response

+sI(s)

s2 + 3s + 2 zero-state response

ZIR captures effect of initial conditions

ZSR captures effect of the input

In example, I(s) = 1s 1

1

s=

1

s 1

Perform Partial Fraction Expansion

Y(s) =1

s + 2+

2

s + 1+

1

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

=1/6

s 1 2/3

s + 2+

3/2

s + 1

y(t) =

1

6

et

2

3

e2t +3

2

et , t

0


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Impedance Method

Maps a time-domain circuit to a frequency do-

main circuit Differential equations in variable t mapped to alge-

braic equations in variable s

i(t)R

+ v(t) - + V(s) -

I(s)R

v(0)s

+ V(s) -

I(s)

1

Cs+ -

i(t)

+ v(t) -

C

si(0)

I(s) sLi(t)L

+ v(t) - + V(s) -


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Impedance Method Example

R1

v (t)in

0v (t)impedance

method

V (s)in

R1

0v (0)s

V (s)0C

-

++ +

- - +-

+

-

Cs1

V0(s) v0(0)

s

Cs +V0(s) Vin(s)

R1= 0

V0(s)Cs + 1

R1

= Vin(s)

R1+ Cs

v0(0)

s

V0(s) =1/R1

Cs + 1/R1Vin(s)

externalsource

+Cs

Cs + 1/R1

v0(0)

s

source

due to I.C.

Note equivalence to

V0(s) = G1(s)Vin(s) + G2(s)

v0(0)

s

G1: transfer function from vin to output v0 with zero

initial conditions

G2: transfer function from v0(0)/s to output v0 withzero input

V0(s) = ZSR + ZIR (superposition)


17/35


2 Transfer Functions

1. Definition

2. Stability

Condition on Transfer Function

Example: Inverted Pendulum

3. Frequency Response

Steady-State Response

Magnitude & Phase Plots

Filter Design

Effects of Poles & Zeros


18/35


Transfer Function: Definition

Differential Equation (assume m < n)

dny(t)

dtn+ an1

dn1y(t)dtn1

+ + a0 y(t) =

b0 u(t) + b1du(t)

dt+ + bm1du

m1(t)dtm1

where y(0), y(0), y(0), . . . , y(n1)(0) and

u(0), u(0), . . . , u(m1)(0) are given

Take Laplace transform of both sides(sn + an1sn1 + + a0)Y(s) + (term due to I.C.s)

= (b0 + b1s + + bmsm

)U(s) + (term due to I.C.s)

Y(s) = b0 + b1s + + bmsm

sn + an1sn1 + + a0U(s)

+I.C.(s)

sn + an1sn1 + + a0

If all initial conditions are zero, thenY(s)

U(s)=

b0 + b1s + + bmsmsn + an1sn1 + + a0

Transfer Function System Function

= H(s)

Y(s) = H(s)U(s)


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Example

R1

0v (t)

-

+

C

R2

+

-v (t)in

V0(s) =

1/Cs

R1 + 1/Cs Vin(s) V0(s)

Vin(s) =

1

R1Cs + 1 = H1(s)

R1

0v (t)

-

+

v (t)inC

R2

+

-

V0(s) =1/Cs

R2 + 1/CsVin(s) V0(s)

Vin(s)=

1

R2Cs + 1= H2(s)

Notice that H1(s) = 1H2(s)

To calculate transfer function of a circuit, the nodesfor the inputs and outputs must be identified


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Stability

A system is stable if the response y(t), y(t), . . .,y(n1)(t)

go to zero as t goes to infinity for zero inputs and arbi-trary initial conditions.

Example: 2nd Order Systems

d2y(t)

dt2+ a1

dy(t)

dt+ a0 y(t) = 0

y(0), y(0) are givenLaplace Transform:

(s2 + a1s + a0)Y(s) (sy(0) + y(0) + a1y(0)) = 0

Y(s) =

sy(0) + y(0) + a1y(0)

s2 + a1s + a0

=C(s)

(sp1)(sp2)pi distinct y(t) = K1ep1t + K2ep2t , t 0

system is stable if Re

{p1}

, Re

{p2}

< 0

General:

H(s) =b(s)

sn + an1sn1 + + a0H(s) is stable if roots of (sn + an1sn1 + + a0)are in LHP (Left-Half [of the complex] Plane)


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Stabilization: Inverted Pendulum

Non-linear model

mgl sin mlx cos = Im

x(t)

I: moment of inertia

m: mass

Small : sin , cos 1Linearized model around = 0:

I mgl = mlx

System Function:

H(s) =(s)

X(s)=

mls2Is2 mgl

poles are

mgl

I

System is unstable!


22/35


Exponential Inputs: General Case

U(s)

H(s)

Y(s)H is stable

u(t) = es0t , t 0 (s0 can be complex)Y(s) = H(s)U(s) = H(s)

1

s s0

= K1sp1 + +

Knspn +

K0s s0 , Re{pi} < 0

(assuming p1, . . . , pn, and s0 are distinct)

y(t) = K1ep1t + + Knepnt + K0es0t

K0e

s0t for large t

K0 = (s s0)H(s) 1

s s0

s=s0

= H(s0)

y(t) H(s0) es0t for large t

H(s

0) is an amplification, valuable for defining the

frequency response

Exponentials are called eigen-functions


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Exponential Inputs: Special Case

U(s)

H(s)

Y(s)

H is stable

u(t) = e+jt = (cos t +j sin t) = periodic input

U(s) =1

sj Y(s) = H(s) 1

sj

y(t) = (response due to poles of H) + H(j) ejt

H(j) ejt for large t

u(t) = cos t y(t) = Re{H(j) ejt}

H(j) is called the frequency response of H H(j) is sufficient to determine H(s)

Captures the steady-state response to sinusoidal in-puts (Q: what about general inputs?)

Motivates frequency plots!


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Steady-State Response: Example

H(s)PilotsCommand

PitchAngle

H(s) =(s + 10)(s 6)

s2 + 26s + 6

Q: How can you reach a steady-state angle 0?

If input command is u(t) = U0, then

limt(t) = H(0)U0 =

606

U0 = 10 U0 = 0U0 = 0.10

Unit Step Response

0 5 10 15 20 25 30 35 40 45 50

10

5

0

t (sec)

amplitude

Unit Step Response: U0 = 1 0 = 10


25/35


Circuit Analysis: Sinusoidal Input

-

+

V(s) = 1 + RCs

R= H(s)

V(s)

I(s)R C

I(s)

If i(t) = cos t, then i(t) = Re[ejt]

Steady-state response to ejt is v(t) = R

1+RCjejt

Steady-state response to cos t is Re

R1+RCje

jt

More precisely:H(j) = |H(j)| ej() , () = H(j)

v(t) = |H(j)| Amplification

cos(t + () Phase Shift

)

I(j )

+ V(j ) -

Impedance Method - "Phasors"

+ v -

i I(j )

+ V(j ) -

R R

L i

+ v - + V(j ) -

j LI(j )

1

i

+ v -

C j C


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Frequency Plots

inv (t)j t= e

j C1

j C1 + R 1 + j RCH(j ) =

1

=

Ex:

+-

v(t)C

R

+

-

|V(j)| = V(j) V(j) =

11+jRC 11jRC

12

=

11+2R2C2

12

MagnitudePlot

H(j )

V(j) =1

1 +jRC 1jRC

1jRC =1jRC

1 + w2R2C2

V(j) = tan1(

RC)

PlotPhase

H(j )


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Poles & Zeros

H(s) = K

(s

z1)(s

z2)

(s

zm)

(sp1)(sp2) (spn) , m nj

p1

z1

p

ppz

z

2

34

3

2

|H(j)| = |K(j z1)(j z2) (j zm)||(j p1)(j p2) (j pn)|

= |K|m

i=1 |j zi|

ni=1

|j

pi

|H(j) = K+ mi=1

(j zi)n

i=1(j pi)


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Bode Plots

H(s) = K

mi=1(s

zi)

ni=1(spi)

20log |H(j)| = 20log |K| + mi=1

20 log |j zi| n

i=120 log |j pi|

H(j) =

K+

m

i=1

(j

zi)

n

i=1

(j

pi)

(converts multiplication to addition)

1st-order example: (sT + 1)

Magnitude: |jT + 1| = (2T2 + 1)1220 log(2T2 + 1)

1

2 = 10 log(2T2 + 1)

Phase: tan1 T1

1TT

0.1 10T

Approx. Figure

0

deg

1T

10T

0

20 db/dec20

db

4545 deg/dec

90


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Example

H(s) = K

s10

+ 1

(s + 1)( s100 + 1)

101

100

101

102

103

30

60

90

0

Frequency (rad/sec)

Phasedeg

101

100

101

102

103

60

40

20

0

Frequency (rad/sec)

GaindB

Bode Plot for Example (K=1)


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Band-Pass Filters

2

1

H (j )2

H (j )1

Q: How good is H = H1 H2 as a band-pass filter?


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Band-Pass Filters

H1

low-pass filter with cutofflH2 high-pass filter with cutoffh

H = H1 H2 is a band-pass filter withbandwidth (h l), l h

H (j )1

H (j )

dB

dB

2

h

dB

h

H(j )

(log scale)

(log scale)

(log scale)


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Effects of Poles & Zeros on Frequency Response

R L C

1

Ls

1

R

Cs+ +

1

LCs 2 LR

s+ + 1

Ls

Z(s) =

=Z(s)

For w0 =1LC

, Q = R

CL Z(s) = R

1Q

s

0

s0

2+ 1

Q

s

0

+ 1

Study the case when Q is large:

poles at

s0

2+ 1Q

s

0

+ 1 = 0

p1, p2 = 02Q

j02

1 4

Q2

0

2Q+j0

Note: As Q increases, magnitude of Re{pi} decreases(poles p1 and p2 move closer to imaginary axis)

p1

p2

j

(j - p )1

(j - p )

2

(j - 0)

jIm

Re

Complex Plane


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H(j) =R

0

Q

j

j + 02Q j0

j + 02Q +j0

=jR

0Q

02Q

2+j

w0Q

+ (2 20)

=

H(j)

R for

0, Q large

MagnitudePlot

H(j )

H(j )

PlotPhase


35/35



H(j) =R

0

Q

j

j + 02Q j0

j + 02Q +j0

p1, p2 = 02Q j02

1 4

Q2

0

2Q+j0

Universal Resonance Curves

0 1 2 3 4 5 60

5

10Magnitude & Phase of H(jw) vs w; Effects of Varying w0, Q

Magnitude w0=2 w0=3 w0=4Q=10

0 1 2 3 4 5 6100

0

100

Phase w0=2

w0=3

w0=4Q=10

0 1 2 3 4 5 60

5

10

Magnitude

Q=10

Q=5Q=2

w0=3

0 1 2 3 4 5 6100

0

100

Phas

e

Q=10Q=5 Q=2

w0=3

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