11

Control SystemsControl Systems

Somasundaran.pSomasundaran.p

M.E.(I.I.Sc)M.E.(I.I.Sc)

22

Module 1Module 1

33

IntroductionIntroduction

44

Examples Examples • ManMan

• Space-vehicle systemsSpace-vehicle systems

• Missile-guidance systemsMissile-guidance systems

• Robotic systemsRobotic systems

• Autopilot systemsAutopilot systems

• Machine toolsMachine tools

• Design of cars and trucksDesign of cars and trucks

• Control of Control of pressure,temperature,humidity,pressure,temperature,humidity,

viscocity and flow in process industries.viscocity and flow in process industries.

55

ClassificationClassification

• 1. 1. Closed-loop control systemsClosed-loop control systems

• 2. 2. Open-loop control systemsOpen-loop control systems

66

Closed-loop control systemsClosed-loop control systems

• Use of feedback controlUse of feedback control

• Measures the output & compares with Measures the output & compares with the referance input (the referance input (Required OutputRequired Output))

• Uses the difference as a means of Uses the difference as a means of controlcontrol

• Reduces the system errorReduces the system error

77

ExamplesExamples

1.The human body1.The human body

Body temperature & blood pressure areBody temperature & blood pressure are

kept constant by physiological feedbackkept constant by physiological feedback

2. Room-temperature control system2. Room-temperature control system

Measures the actual room temperature Measures the actual room temperature &&

compares with the desired room compares with the desired room temperature (Referance temperature )temperature (Referance temperature )

88

Open-loop control systemsOpen-loop control systems

• NoNo feed back of output feed back of output

• Output has Output has nono effect on control action effect on control action

• Output is Output is not not compared with the compared with the referancereferance

input.input.

Accuracy of the system depends on Accuracy of the system depends on calibration.calibration.

99

ExamplesExamples

• 1. A washing machine.1. A washing machine.

It operates on a It operates on a timetime basis. basis.

No measurement of output signal No measurement of output signal

(Cleanliness of clothes)(Cleanliness of clothes)

2. Traffic control by means of signals 2. Traffic control by means of signals operated on a operated on a timetime basis basis

1010

Advantages of closed-loop Advantages of closed-loop controlcontrol

• 1. Accuracy1. Accuracy

• 2. No effect due to external 2. No effect due to external disturbances &disturbances &

internal variations in system internal variations in system parameters.parameters.

1111

Disadvantages of closed-loop Disadvantages of closed-loop controlcontrol

• 1. Stability problem1. Stability problem

• 2. More number of components 2. More number of components increasesincreases

(a) cost(a) cost

(b) weight(b) weight

(c) size (c) size

(d) power (d) power of the systemof the system

1212

Advantages of open-loop Advantages of open-loop controlcontrol

• 1. Simple construction 1. Simple construction

• 2. Ease of maintenance2. Ease of maintenance

• 3. Less expensive3. Less expensive

• 4. No stability problem4. No stability problem

• 5. Convenient when output is hard to measure5. Convenient when output is hard to measure

oror

economically economically notnot feasible feasible

1313

Disadvantages of open-loop Disadvantages of open-loop controlcontrol

• 1. Disturbances causes errors 1. Disturbances causes errors

& &

output may be different from what is output may be different from what is desireddesired

2. To maintain the required quality in the2. To maintain the required quality in the

output, recalibration is necessary from timeoutput, recalibration is necessary from time

to time.to time.

1414

Mathematical Modeling of Mathematical Modeling of Dynamic SystemsDynamic Systems

1515

Transfer FunctionTransfer Function

• Definition:Definition:

The ratio of the Laplace transform of The ratio of the Laplace transform of the output to the Laplace transform of the output to the Laplace transform of the inputthe input

under the assumption that all initial under the assumption that all initial conditions conditions

are zero.are zero.

1616

Transfer function Transfer function G(s)=Y(s)/X(s)G(s)=Y(s)/X(s)

G(s)X(s) Y(s)

1717

Block DiagramBlock Diagram

• A pictorial representation of the A pictorial representation of the functions functions

performed by each component and performed by each component and flow of flow of

signals. signals.

1818

• The transfer functions of the The transfer functions of the components are entered in the components are entered in the corresponding blocks.corresponding blocks.

• Arrows indicate the direction of flow Arrows indicate the direction of flow of signals.of signals.

1919

2020

2121

2222

Open loop transfer functionOpen loop transfer function

• It is the Ratio of feed back signal B(s) It is the Ratio of feed back signal B(s) to the actuating error signal E(s).to the actuating error signal E(s).

• Open loop transfer function = Open loop transfer function = B(s)/E(s)B(s)/E(s)

= =

G(s)*H(s)G(s)*H(s)

2323

Feed forward transfer functionFeed forward transfer function

• It is the ratio of the output C(s) to the It is the ratio of the output C(s) to the actuating error signal E(s).actuating error signal E(s).

• Feed forward transfer function = Feed forward transfer function = C(s)/E(s)C(s)/E(s)

= G(s)= G(s)

2424

Closed loop transfer Closed loop transfer functionfunction• Closed loop transfer function = C(s)/R(s)Closed loop transfer function = C(s)/R(s)

C(s) = G(s)xE(s)C(s) = G(s)xE(s)

E(s) = R(s) - B(s)E(s) = R(s) - B(s)

= R(s) – H(s)xC(s)= R(s) – H(s)xC(s)

Hence C(s) = G(s)x[R(s) – H(s)xC(s)]Hence C(s) = G(s)x[R(s) – H(s)xC(s)]

Therefore Therefore

C(s)/R(s) = G(s)/[1+G(s)H(s)]C(s)/R(s) = G(s)/[1+G(s)H(s)]

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Some Important Block-Diagram Some Important Block-Diagram TransformationsTransformations

• 1. Combining two 1. Combining two blocks in cascade,blocks in cascade,

Original DiagramOriginal Diagram Equivalent Equivalent DiagramDiagram

I.GI.G1.G.G2=I(G=I(G1.G.G2))

G1(s) G2(s)I(s) O(s

) G1.G2

I(S) O(S)

2626

2.Moving a summing point 2.Moving a summing point behind a block, (I-B)G=(I.G)-behind a block, (I-B)G=(I.G)-(B.G)(B.G)

Original DiagramOriginal Diagram Equivalent DiagramEquivalent Diagram

G(S)

I(s) O(s)

B(s)

G

G

B(s)

O(s)I(s)

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3.Moving a summing point 3.Moving a summing point ahead of a block, (I.G) - B=[I-ahead of a block, (I.G) - B=[I-(1/G).B]G(1/G).B]GOriginal DiagramOriginal Diagram Equivalent DiagramEquivalent Diagram

G(S)I(s) O(s)

B(s)

G

1/G

I(s) O(s)

B(s)

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4.Moving a pick off point 4.Moving a pick off point behind a block.behind a block.

Original DiagramOriginal Diagram Equivalent DiagramEquivalent Diagram

G(S)I(s) O(s)

I(s)

G

1/G

I(s)O(s)

I(s)

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5.Moving a pick off point ahead 5.Moving a pick off point ahead of a block.of a block.

Original DiagramOriginal Diagram Equivalent DiagramEquivalent Diagram

G(S)I(s) O(s)

O(s)

G

G

I(s)O(s)

O(s)

3030

6a.Eliminating a feedback 6a.Eliminating a feedback looploop

Original DiagramOriginal Diagram Equivalent DiagramEquivalent Diagram

G(S)

I(s) O(s)

G/(1+GH)

O(s)I(s)

H(S)

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6b.Eliminating a feedback 6b.Eliminating a feedback looploopOriginal DiagramOriginal Diagram Equivalent DiagramEquivalent Diagram

G(s)I(s)

H(s)

O(s)

G/(1-GH)I(s) O(s)

3232

7.Special case for Rule 6 above 7.Special case for Rule 6 above (H=1)(H=1)

Original DiagramOriginal Diagram Equivalent DiagramEquivalent Diagram

G(S)

I(s) O(s)

G/(1+G)O(s)

I(s)

O(s)

3333

R C

G1 G2G3

H1

H2

3434

R C

G1 G2G3

H1

H2/G1

3535

R C

G1G2/(1-G1G2H1)G3

H2/G1

3636

R C

G1G2G3/(1-G1G2H1+G2G3H2)

3737

R C

G1G2G3/(1-G1G2H1+G2G3H2+G1G2G3)

3838

Vi(s) V0(s)

1/R1 1/C1s 1/R2 1/C2s

V1(s)

I1(s)

I2(s)

V1(s)

V0(s)

Home work

Simplify the block diagram and find V0(s) /Vi (s)

3939

Signal flow GraphSignal flow Graph

• A diagram consisting of A diagram consisting of nodesnodes which which are connected by are connected by directed branches .directed branches .

• Each node j has associated with it a Each node j has associated with it a system variable (Node signal) Xsystem variable (Node signal) Xjj

• Each directed branch Each directed branch jk jk from node from node j j to node to node kk has an associated branch has an associated branch transmittance Ttransmittance Tjkjk

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• The nodes sum up all the signals that The nodes sum up all the signals that

enter them.enter them.

• They transmit the sum signals to all They transmit the sum signals to all

out going branches.out going branches.

• The branch transmittance TThe branch transmittance Tjkjk is is

essentially the transfer function.essentially the transfer function.

• It specifies the manner in which the It specifies the manner in which the

signal at node signal at node k k depends upon that at depends upon that at

node node j.j.

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• The contribution of the signal at node The contribution of the signal at node j, Xj, to the signal at node k = the j, Xj, to the signal at node k = the product of Xj and branch product of Xj and branch transmittance Ttransmittance Tjkjk..

• Thus at the kThus at the kthth node, we have node, we have

XXkk = = ΣΣ T Tjkjk X Xjj

jj

where the summation is taken over where the summation is taken over all the branches entering the kall the branches entering the kthth node.node.

4242

Signal flow graph –An exampleSignal flow graph –An example

X1 X2 X3

X4

T12

T23

T34

T13

T42

T44

4343

The equations relating the The equations relating the system variables at the four system variables at the four nodes.nodes.• XX11=X=X11,,

• XX22=T=T1212XX11+T+T4242XX44

• XX33=T=T1313XX11+T+T2323XX22

• XX44=T=T3434XX33+T+T4444XX44

4444

Vi(s) V0(s)

1/R1 1/C1s 1/R2 1/C2s

V1(s)

I1(s)

I2(s)

V1(s)

V0(s)

Draw the signal flow graph for the block diagram shown .

4545

We can choose the five transformed We can choose the five transformed variables Vvariables Vi,I,I1,V,V1,I,I2,and Vand Vo, and draw a , and draw a five-node flow graphfive-node flow graph

• Equations are first written as follows:Equations are first written as follows:

II11(s)=V(s)=Vii(s)[1/R(s)[1/R11]- V]- V11(s)[1/R(s)[1/R11]]

VV11(s)=I(s)=I11(s)[1/sC(s)[1/sC11]-I]-I22(s)[1/sC(s)[1/sC11]]

II22(s)= V(s)= V11(s)[1/R(s)[1/R22]- V]- Voo(s)[1/R(s)[1/R22]]

VVoo(s)=I(s)=I22(s)[1/sC(s)[1/sC22] ]

4646

•In constructing the flow graph In constructing the flow graph , simply follow the cause-and-, simply follow the cause-and-effect procedure given by effect procedure given by above equations , from the above equations , from the input variable Vi(s) to the input variable Vi(s) to the output variable Vo(s).output variable Vo(s).

4747

Signal flow graphSignal flow graph

Vi

1/R1 I1

1/C1

-1/R1

V1

1/R2

I2 1/C2s

-1/R2

V0

4848

• An input variable is represented by a An input variable is represented by a source nodesource node which has only outgoing which has only outgoing branches.branches.

• An output variable is represented by a An output variable is represented by a sink sink node.node.

• A feedback loop is a closed path.A feedback loop is a closed path.

• The loop transmittance of a feedback loop The loop transmittance of a feedback loop is defined as the product of transmittances is defined as the product of transmittances of the branches forming the loop.of the branches forming the loop.

4949

• A path leading from a source node to A path leading from a source node to a sink node without passing through a sink node without passing through any node more than once is called an any node more than once is called an open path.open path.

• There may be more than There may be more than oneone open open path between a source node and a path between a source node and a sink node.sink node.

5050

Mason’s Gain formula.Mason’s Gain formula.

• The general formula for graph The general formula for graph transmittance between source node and a transmittance between source node and a sink node in a signal flow graph is as sink node in a signal flow graph is as follows:follows:

• T=T= 11 ΣΣ T Tkk ΔΔkk

ΔΔ kk

WhereWhere ΔΔ = = 1- (sum of all 1- (sum of all individual individual loop transmittance)loop transmittance) + (sum of the products of loop transmittances of + (sum of the products of loop transmittances of

all possible all possible non touching feedback loops taken non touching feedback loops taken twotwo at at a time)a time)

- (sum of the products of loop transmittances of - (sum of the products of loop transmittances of all possible all possible non touching feedback loops taken non touching feedback loops taken threethree at a time)at a time)

+ … ,+ … ,

5151

Where Where ΔΔ = 1- (sum of all individual loop transmittance) = 1- (sum of all individual loop transmittance)

+ (sum of the products of loop + (sum of the products of loop transmittances transmittances

of all possible non touching feedback of all possible non touching feedback loops taken two at a time)loops taken two at a time)

- (sum of the products of loop - (sum of the products of loop transmittances transmittances

of all possible non touching feedback of all possible non touching feedback loops taken three at a time)loops taken three at a time)

+ … ,+ … ,

ΔΔk k = value of = value of ΔΔ for that part of the graph not for that part of the graph not

touching touching

the kth open path, andthe kth open path, and

TTkk = path transmittance of the kth open path= path transmittance of the kth open path

5252

Using Mason’s Gain formula Using Mason’s Gain formula determine the graph determine the graph transmittance between node Xtransmittance between node X11 and node Xand node X44

X1

X2

X3

X4

T12T23

T42 T44

T13

T34

5353

• Two open paths:Two open paths:

(a) X(a) X11-X-X33-X-X44 TTaa = T = T1313TT3434

(b) X(b) X11-X-X22-X-X33-X-X4 4 T Tbb = T = T1212TT2323TT3434

Two feedback loops :Two feedback loops :

(A) X(A) X44-X-X44 TTAA = T = T4444

(B) X(B) X44-X-X22-X-X44 TTBB = T = T4242TT2323TT3434

No nontouching feedback loops.No nontouching feedback loops.Both feedback loops touch open paths (a) and (b).Both feedback loops touch open paths (a) and (b).

ΔΔ = 1- (T = 1- (TAA+T+TBB) = 1- T) = 1- T4444 – T – T4242TT2323TT3434

5454

• ΔΔaa = = ΔΔbb = 1 = 1

HenceHence

T= (TT= (Taa ΔΔaa + T + Tbb ΔΔbb )1/ )1/ ΔΔ

=( T=( T1313TT3434+T+T1212TT2323TT3434)/(1-T)/(1-T4444-T-T4242TT2323TT3434))

= T= T3434(T13+T12T23)/(1-T(T13+T12T23)/(1-T4444--TT4242TT2323TT3434))

5555

Using Mason’s Gain formula Using Mason’s Gain formula determine the graph determine the graph transmittance between nodes transmittance between nodes VV00 and V and Vii

T2

T4

T5

T6

T3

Vi I1 V1 I2VO

T1

T7

5656

• One open path: (a) VOne open path: (a) Vii-I-I11-V-V11-I-I22-V-Vo o

TTaa = T = T11TT22TT44TT66

Three feedback loops:Three feedback loops:

(A) I(A) I11-V-V11-I-I11 T TAA=T=T22TT33

(B) V(B) V11-I-I22-V-V11 T TBB=T=T44TT55

(C) I(C) I22-V-VOO-I-I22 T TCC=T=T66TT77 Nontouching feedback loops: Nontouching feedback loops: (A) and (C)(A) and (C)

5757

• All three feedback loops touch the All three feedback loops touch the open path.open path.

ΔΔ=1- (T=1- (TAA+T+TBB+T+TCC)+T)+TAATTCC

=1-(T=1-(T22TT33+T+T44TT55+T+T66TT77)+T)+T22TT33TT66TT77

ΔΔaa=1=1

HenceHence

T=(TT=(Taa ΔΔaa)(1/)(1/ΔΔ))

= (T= (T11TT22TT44TT66)/(1-T)/(1-T22TT33-T-T44TT55-T-T66TT77+T+T22TT33TT66TT77))

5858

For the Signal flow graph shown For the Signal flow graph shown find the Transfer Function find the Transfer Function VVOO(s)/V(s)/Vii(s)(s)

Vi

1/R1 I1

1/C1s

-1/R1

V1

1/R2

I2 1/C2s

-1/R2

V0

-1/C1s

5959

Draw a block diagram for the two-Draw a block diagram for the two-section R-C circuit shown.section R-C circuit shown.

C2

R2

C1

R1

vivo

6060

C2

R2

C1

R1

vi voi1

i2

v1

Ans: Here we will have four blocks corresponding to the four elements. In order to write the transformed voltage-current relationship for each element, we introduce three intermediate variables v1, i1, & i2 as shown.

6161

• The 4 transformed equations are :The 4 transformed equations are :

• for Rfor R11 : :

• II11(s) = [V(s) = [Vii(s) - V(s) - V11(s)](s)]1/R1/R11

• for Cfor C11 : :

• VV11(s) = [I(s) = [I11(s) – I(s) – I22(s)](s)]1/1/((sCsC11))

• for Rfor R22 : :

• II22(s) = [V(s) = [V11(s) - V(s) - Voo(s)](s)]1/R1/R22

• for Cfor C22 : :

• VV22(s) =[ I(s) =[ I22(s)] (s)] 1/1/((sCsC22))

6262

• First three equations indicate the First three equations indicate the need for summing points before the need for summing points before the blocks representing Rblocks representing R11 , C , C11 , and R , and R22 . .

• The flow of the transformed variables The flow of the transformed variables proceeds from the input variable proceeds from the input variable VVii(s)(s) through the through the intermediate variables II11(s) , V(s) , V11(s) , and I(s) , and I22(s) to the output (s) to the output variable variable VVoo(s(s).).

• It is important to note that the block It is important to note that the block diagram of a two section R-C circuit diagram of a two section R-C circuit cannot cannot be obtained by putting the be obtained by putting the block diagrams of two singe-section block diagrams of two singe-section R-C circuits in cascade.R-C circuits in cascade.

6363

Vi(s) V0(s)

1/R1 1/C1s 1/R2 1/C2s

V1(s)

I1(s)

I2(s)

V1(s)

V0(s)

Determination of the closed loop transfer function for the block diagram shown can

be done easily .

(This is the Assignment Problem No:1)

6464

C1R1

L11 2

R2

Switch S Opened att = 0

Problem : The switch across the constant current source in the circuit shown is opened at t= 0. Find the transfer function I2(s)/Ig(s) by the block diagram method.

ig

i2

6565

Ans: Here we will have four blocks corresponding to the four elements.

In order to write the transformed voltage-current relationship for each element, we introduce three intermediate variables v1, i1, & v2 as shown.

6666

C1R1

L11 2

R2

Switch S Opened att = 0

ig i1 i2

v1v2

6767

The 4 transformed equations are :The 4 transformed equations are : for Rfor R11 : :

VV11(s) =[I(s) =[Igg(s) – I(s) – I11(s)](s)] R R11 for L :for L : II11(s) = [V(s) = [V11(s) – V(s) – V22(s)](s)]1/1/((sLsL)) for C : for C : VV22(s) = [I(s) = [I11(s) – I(s) – I22(s)](s)]1/1/((sCsC))

for Rfor R22 : :

II22(s) = [V(s) = [V22(s)](s)]1/R1/Rbb

6868

Ig(s) I2(s)

R1 1/sL 1/sC 1/R2

I1(s)

V1(s)

V2(s)

I1(s)

I2(s)

Determination of the closed loop transfer function for the block diagram shown can

be done easily .

(This is the Assignment Problem No:4)

V2(s)

6969

Module 2Module 2

Stability AnalysisStability Analysis

7070

General stability requirements General stability requirements for closed loop control for closed loop control systems.systems.• An unstable feedback system can not An unstable feedback system can not

perform any useful control function.perform any useful control function.

• It is therefore of utmost importance It is therefore of utmost importance to be able to determine , from the to be able to determine , from the characteristics of its forward and characteristics of its forward and feedbackfeedback

paths, whether a system with paths, whether a system with feedback is feedback is stablestable..

7171

Here we shall examine the Here we shall examine the fundamental requirements for fundamental requirements for system system stabilitystability• For the system shown, the ratio of the For the system shown, the ratio of the

transformed output C(s) to the transformed output C(s) to the transformed input R(s) is given by the well transformed input R(s) is given by the well known equation, C(s)/R(s)= known equation, C(s)/R(s)= G(s)/[1+G(s)H(s)]G(s)/[1+G(s)H(s)]

G(s)

H(s)

R(s)

C(s)

7272

In general, both G(s) and H(s) will In general, both G(s) and H(s) will be ratios of polynomials in s.be ratios of polynomials in s.

• Hence C(s)/R(s) will have the Hence C(s)/R(s) will have the following general formfollowing general form

C(s)/R(s)=aC(s)/R(s)=ammssmm+a+am-1m-1ssm-1m-1+….+a+….+a11s+as+a00

________________________________________________________________

bbnnssnn+b+bn-1n-1ssn-1n-1+….+b+….+b11s+bs+b00

Where the polynomials in both the numerator and the denominator Where the polynomials in both the numerator and the denominator

have been written.have been written.

7373

• Let the input function r(t) is a unit step Let the input function r(t) is a unit step function U(t), although the subsequent function U(t), although the subsequent discussion will apply to a general input discussion will apply to a general input function.function.

• We have R(s) =£ [U(t)]=1/s andWe have R(s) =£ [U(t)]=1/s and• aammssmm+a+am-1m-1ssm-1m-1+….+a+….+a11s+as+a00

________________________________________________________________

s(bs(bnnssnn+b+bn-1n-1ssn-1n-1+….+b+….+b11s+bs+b00))

= 1 K= 1 K00 K K11 K K22 K Knn __ __ __ ___ __ __ __ ___ bbnn s s-s s s-s11 s-s s-s22 s-s s-snn

+ + +…

C(s) =

+

7474

• Where the constants KWhere the constants K00,K,K11,K,K22… K… Knn can be can be found by Heaviside’s expansion theorem.found by Heaviside’s expansion theorem.

• The output or response function is then The output or response function is then

• c(t) = £-1 [C(s)]c(t) = £-1 [C(s)]

• = 1 = 1 KK00++KK11℮℮ss11tt++KK22eess22tt+… ++… +KKnneessnntt] U (t)] U (t)

bbnn

7575

• In the last equation the first term is In the last equation the first term is step function and represents the step function and represents the steady state response.steady state response.

• The exponential terms are transient The exponential terms are transient responses for which the constants Kresponses for which the constants K11, , KK22,…, K,…, Knn depend upon the amplitude depend upon the amplitude and the form of the input function.and the form of the input function.

7676

• The co-efficient of t in the exponents, sThe co-efficient of t in the exponents, s11, s, s22, …, s, …, snn are the roots of the polynomial in the are the roots of the polynomial in the denominator of the closed loop transfer function denominator of the closed loop transfer function G(s) G(s)

• 1+G(s)H(s)1+G(s)H(s)

• Now consider the typical root, SNow consider the typical root, Skk in general it in general it can be a complex number:can be a complex number:

• SSkk = = σσkk + j + jωωkk

The term in the output function corresponding The term in the output function corresponding to the root sto the root skk is is

KKkke e ((σσkk + j + jωωkk)t)t = K = Kkkeeσσkktt e e jjωωkktt

bbnn b bnn

7777

• The amplitude of above term will increase The amplitude of above term will increase indefinitelyindefinitely with time when with time when σσkk (the real (the real part of spart of skk) is ) is positivepositive. .

• On the other hand, when On the other hand, when σσkk is is negativenegative the amplitude will the amplitude will decreasedecrease exponentially exponentially with time and the transient will with time and the transient will subsidesubside..

• We therefore cannot tolerate a We therefore cannot tolerate a positivepositive σσkk

which will result in a runaway or which will result in a runaway or an an unstableunstable situation situation

7878

• If If σσkk = 0, we have a transitional case = 0, we have a transitional case which will give rise to sustained which will give rise to sustained oscillations of constant amplitude.oscillations of constant amplitude.

This situation will also be considered This situation will also be considered unstableunstable..

The roots of a polynomial are called The roots of a polynomial are called the the zeroszeros of the polynomial for of the polynomial for obvious reasons.obvious reasons.

7979

• ss11,s,s22,…,s,…,skk,…, s,…, snn are are zeroszeros of the of the denominatordenominator of the closed loop transfer of the closed loop transfer function C(s)/R(s).function C(s)/R(s).

Hence they are also the Hence they are also the zeroszeros of the of the polynomial polynomial 1+G(s)H(s)1+G(s)H(s) , ,

since C(s)/R(s) =G(s)/[1+G(s)H(s)]since C(s)/R(s) =G(s)/[1+G(s)H(s)]

Thus we conclude that the Thus we conclude that the necessary & necessary & sufficientsufficient condition that a feedback condition that a feedback system be system be stablestable is that all the is that all the zeroszeros of of 1+G(s)H(s)1+G(s)H(s) have have negative real partsnegative real parts..

8080

• The The zeroszeros of the polynomial in the of the polynomial in the denominatordenominator of a rational function make of a rational function make that function go to that function go to infinityinfinity; they are ; they are called the called the polespoles of the function. of the function.

• SS11, S, S22, …, S, …, Skk, …, S, …, Snn are poles of the are poles of the closed loop transfer function C(s)/R(s).closed loop transfer function C(s)/R(s).

8181

• Hence the stability condition stated Hence the stability condition stated above can be restated as follows:above can be restated as follows:

The The necessary & sufficientnecessary & sufficient condition that a feed back system be condition that a feed back system be stablestable is that all the is that all the polespoles of its of its closed loop transfer function have closed loop transfer function have negativenegative real partsreal parts..

8282

• The task of solving for The task of solving for all the rootsall the roots of of equation 1+G(s)H(s)=0 is frequently a equation 1+G(s)H(s)=0 is frequently a tedious one, especially when it is an tedious one, especially when it is an equation of equation of high degreehigh degree in s. in s.

• Hence the method of determining the Hence the method of determining the stability without stability without actually finding all the actually finding all the rootsroots of the equation 1+G(s)H(s)=0 will of the equation 1+G(s)H(s)=0 will be studied next.be studied next.

8383

• The Routh-Hurwitz Stability CriterionThe Routh-Hurwitz Stability Criterion

Consider the following equation of the nConsider the following equation of the nthth degree in sdegree in s

aannssnn+a+an-1n-1ssn-1n-1+…+a+…+a11s+as+a00=0, =0, (1)(1)

where the where the coefficientscoefficients a ann, a, an-1n-1, …, a, …, a11, a, a00 are are all of the same signall of the same sign and and none is zeronone is zero. .

Some roots will have Some roots will have +ve real+ve real partsparts and and the system will be the system will be unstable unstable if all the if all the coefficients of its characteristic equation coefficients of its characteristic equation do not have the same signdo not have the same sign or if or if all the all the terms are not present.terms are not present.

8484

aannssnn+a+an-1n-1ssn-1n-1+…+a+…+a11s+as+a00=0, =0, (1)(1)

• Step 1Step 1. Arrange the coefficients of the . Arrange the coefficients of the given equation in two rows in the given equation in two rows in the following fashion:following fashion:

Row 1: aRow 1: ann aan-2n-2 aan-4n-4 … …

Row 2: aRow 2: an-1n-1 aan-3n-3 aan-5n-5 … …

Where the Where the arrow arrow direction indicate the direction indicate the order of arrangement.order of arrangement.

8585

Step 2Step 2. Form a third row from the first . Form a third row from the first and second rows as follows :and second rows as follows :

Row 1: aRow 1: ann aan-2n-2 aan-4n-4 … …

Row 2: aRow 2: an-1n-1 aan-3n-3 aan-5n-5 … …

Row 3Row 3: : bbn-1n-1 bbn-3n-3 bbn-5n-5 … where … where

bbn-1n-1= = -1-1 aann aan-2n-2

aan-1n-1 aan-1n-1 aan-3 n-3 , ,

bbn-3n-3= -1= -1 aann aan-4n-4

aan-1n-1 aan-1n-1 aan-5n-5 , ,

8686

Step 3Step 3. Form a fourth row from the second . Form a fourth row from the second and third rows:and third rows:

Row 2: aRow 2: an-1n-1 aan-3n-3 aan-5n-5 … …

Row 3: bRow 3: bn-1n-1 bbn-3n-3 bbn-5n-5 … …

Row 4Row 4: : ccn-1n-1 ccn-3n-3 ccn-5n-5 … …

Where ,Where , ccn-1n-1 = -1 a = -1 an-1n-1 a an-3n-3

bbn-1n-1 b bn-1n-1 b bn-3n-3

CCn-3n-3 = -1 a = -1 an-1n-1 a an-5n-5

bbn-1n-1 b bn-1n-1 b bn-5n-5 ……

8787

Step 4Step 4. continue this procedure of . continue this procedure of forming a forming a

new row from the two preceding rows new row from the two preceding rows until only zeros are obtained.until only zeros are obtained.

In general an array of In general an array of n+1n+1 rows will rows will result with the last two rows each result with the last two rows each containing a single element.containing a single element.

8888

StatementStatement : : Routh-HurwitzRouth-Hurwitz criterioncriterion states that the system whose states that the system whose characteristic equation is eqn (1), is characteristic equation is eqn (1), is stablestable if and only if and only if all the elements in if all the elements in the first columnthe first column of the array formed by of the array formed by the coefficients in the above manner the coefficients in the above manner have the have the same algebraic signsame algebraic sign. .

If the elements in the first column are If the elements in the first column are not all of the same sign, not all of the same sign, the number of the number of sign changessign changes in the in the first columnfirst column = = the the number of roots with +ve real parts.number of roots with +ve real parts.

8989

• Problem 1Problem 1• Apply Routh-Hurwitz criterion to the following Apply Routh-Hurwitz criterion to the following

equation and find the stability of the system.equation and find the stability of the system.SS44+2s+2s33+3s+3s22+4s+5=0+4s+5=0

Solution:Solution:Routh arrayRouth array11stst row: row: 11 33 5522ndnd row: row: 22 44 0033rdrd row: row: 11 5544thth row: row: -6-655thth row: row: 55Since the number of sign changes in the first Since the number of sign changes in the first

column are two, the given eqn has two roots column are two, the given eqn has two roots with +ve real parts. Hence the system is with +ve real parts. Hence the system is unstableunstable..

9090

• Problem 2Problem 2• Apply Routh-Hurwitz criterion to the following Apply Routh-Hurwitz criterion to the following

equation and find the stability of the system.equation and find the stability of the system.SS44+7s+7s33+17s+17s22+17s+6=0+17s+6=0

Solution:Solution:Routh arrayRouth array11stst row: row: 11 1717 6622ndnd row: row: 77 171733rdrd row: row: 14.5814.58 6644thth row: row: 14.1214.1255thth row: row: 66

Since all the coefficients in the first column Since all the coefficients in the first column are of the same sign (+ve) the given eqn has are of the same sign (+ve) the given eqn has no roots with +ve real parts. Hence the no roots with +ve real parts. Hence the system is system is stablestable

9191

• Problem 3Problem 3

• Apply the Routh-Hurwitz criterion to the Apply the Routh-Hurwitz criterion to the following equation and find the stability of the following equation and find the stability of the system:system:

SS55+2s+2s44+2s+2s33+4s+4s22+11s+10=0+11s+10=0

• SolutionSolution

Here all the coefficients are +ve and none is Here all the coefficients are +ve and none is zero. Hence we proceed to write the Routh zero. Hence we proceed to write the Routh array as follows:array as follows:

11stst row: row: 11 22 1111

22ndnd row: row: 22 44 1010

33rdrd row: row: 00 66

• We cannot proceed further because the first We cannot proceed further because the first element in the 3element in the 3rdrd row is zero. row is zero.

9292

• The situation can be remedied by The situation can be remedied by multiplying the original eqn by a factor multiplying the original eqn by a factor (s+a),(s+a), where where a>0a>0, and then applying , and then applying the same procedure.the same procedure.

• The reasoning is that multiplication by The reasoning is that multiplication by (s+a) will change the vanishing (s+a) will change the vanishing element without changing the number element without changing the number of roots with +ve real parts.of roots with +ve real parts.

• Any +ve real number can be used for a; Any +ve real number can be used for a; the simplest choice is the simplest choice is a=1a=1..

• Multiplying the given eqn by (s+1), we Multiplying the given eqn by (s+1), we have the new eqn have the new eqn

• SS66+3s+3s55+4s+4s44+6s+6s33+15s+15s22+21s+10=0+21s+10=0

9393

• SS66+3s+3s55+4s+4s44+6s+6s33+15s+15s22+21s+10=0+21s+10=0

The Routh arrayThe Routh array now becomes : now becomes :

• 11 44 1515 1010

• 11 22 77 (after dividing by 3)(after dividing by 3)

• 11 44 55 (after dividing by 2)(after dividing by 2)

• -1-1 11 (after dividing by 2)(after dividing by 2)

• 11 11 (after dividing by 5)(after dividing by 5)

• 22

• 11

9494

• There are There are two changes of signtwo changes of sign in the in the first column of the Routh array, which first column of the Routh array, which indicates that there are indicates that there are two rootstwo roots having +ve real parts in the given eqn. having +ve real parts in the given eqn.

• Hence the system is Hence the system is unstable.unstable.

9595

• Problem Problem • Apply the Routh-Hurwitz criterion to the Apply the Routh-Hurwitz criterion to the

following equation and find the stability of the following equation and find the stability of the system:system:SS55+s+s44+4s+4s33+24s+24s22+3s+63=0+3s+63=0

• SolutionSolutionHere all the coefficients are +ve and none is Here all the coefficients are +ve and none is zero. zero. Hence we proceed to write the Routh Hence we proceed to write the Routh array as follows:array as follows:11stst row: row: 11 44 3322ndnd row: row: 11 2424 636333rdrd row: row: -1-1 -3-3 (after dividing by 20) (after dividing by 20)44thth row: row: 11 33 (after dividing by 21) (after dividing by 21)55thth row: row: 00

??. .

9696

We cannot proceed further because of We cannot proceed further because of the the zerozero in the fifth row. in the fifth row.

The earlier technique of multiplying The earlier technique of multiplying the original eqn by the original eqn by a factor (s+1) a factor (s+1) failsfails here because the here because the whole fifth row whole fifth row ((which happens to have only one which happens to have only one element) element) vanishes.vanishes.

9797

• Examination reveals that this situation Examination reveals that this situation will occur whenever the array has will occur whenever the array has two two consecutiveconsecutive rowsrows in which in which the ratio of the ratio of the corresponding elements is constant.the corresponding elements is constant.

• When this happens, When this happens, all the elements in all the elements in the following row will vanish. the following row will vanish.

• This is an indication that the given eqn This is an indication that the given eqn has at least has at least one pair of rootsone pair of roots which lie which lie radiallyradially opposite to each otheropposite to each other and and equidistant from the origin.equidistant from the origin.

9898

• The row can be completed by forming The row can be completed by forming an an auxiliary polynomialauxiliary polynomial in s in s22, using the , using the elements in the elements in the last non vanishing rowlast non vanishing row as its as its coefficientscoefficients; the coefficients of ; the coefficients of the derivative of this aux.ploynomial the derivative of this aux.ploynomial has to be taken for the elements in the has to be taken for the elements in the following row.following row.

• In the problem, the elements in the last In the problem, the elements in the last non vanishing row are 1 and 3; hence non vanishing row are 1 and 3; hence the aux. polynomial in sthe aux. polynomial in s22 is is

• ss22 +3 +3

9999

• The derivative of this polynomial is The derivative of this polynomial is

2s2s, whose coefficient , whose coefficient 22 is to be taken as the is to be taken as the element in the row following the last non element in the row following the last non vanishing row.vanishing row.

• The Routh array now becomes The Routh array now becomes

• 11 44 33

• 11 2424 6363

• -1-1 -3-3

• 11 33

• 22

• 33

100100

• There are two changes of sign in the There are two changes of sign in the first column and hence there two first column and hence there two roots with +ve real parts. Hence the roots with +ve real parts. Hence the system is unstable. system is unstable.

• The rootsThe roots of the eqn formed by the of the eqn formed by the aux polynomial aux polynomial

SS22+3=0 +3=0 are also the roots of the are also the roots of the original eqn.original eqn.

• In this problem they are In this problem they are s= s= ++j√3j√3

which have zero real partswhich have zero real parts

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ROOT-LOCUS ANALYSISROOT-LOCUS ANALYSIS

• The The locuslocus ofof rootsroots of characteristic of characteristic equation of the closed loop system equation of the closed loop system as the as the

gaingain is varied from is varied from zero to infinityzero to infinity is is the the root-locusroot-locus..

102102

For the system shown , the closed loop For the system shown , the closed loop transfer function is C(s) = G(s)transfer function is C(s) = G(s)

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

The characteristic equation of the closed The characteristic equation of the closed loop system isloop system is

1+G(s)H(s)=01+G(s)H(s)=0

G(s)

H(s)

R(s) C(s)

103103

• i.e., G(s)H(s) = -1 i.e., G(s)H(s) = -1 (1)(1)

• Here we assume that G(s)H(s) is a ratio of Here we assume that G(s)H(s) is a ratio of

• polynomials in s.polynomials in s.

• Since s is a complex quantity, G(s)H(s) is Since s is a complex quantity, G(s)H(s) is alsoalso

• a complex quantity.a complex quantity.

• Since G(s)H(s) is a complex quantity, Since G(s)H(s) is a complex quantity, Equation Equation (1)(1) can be split in to two equations can be split in to two equations

by equating the angles and magnitudes of by equating the angles and magnitudes of both sides to obtain the following:both sides to obtain the following:

Angle condition : G(s)H(s)= Angle condition : G(s)H(s)= + 180+ 18000 (2k+1) (2k+1)

• Magnitude condition : G(s)H(s) = 1Magnitude condition : G(s)H(s) = 1

104104

• The The values of svalues of s that fulfill that fulfill both the both the angle and magnitude conditionsangle and magnitude conditions are are the roots of the characteristic equation.the roots of the characteristic equation.

• A plot of the points in the complex A plot of the points in the complex plane satisfying the plane satisfying the angle condition angle condition alonealone is the root locus. is the root locus.

105105

General Rules for constructing General Rules for constructing ROOT LOCIROOT LOCI

• 11. . Locate the poles and zeros of Locate the poles and zeros of G(s)H(s)G(s)H(s)

• on the s plane.on the s plane.

• The root locus branches start from The root locus branches start from open-loop polesopen-loop poles and terminate at and terminate at zeroszeros

• (finite zeros (finite zeros oror zeros at infinity) zeros at infinity)

• The open-loop zeros are the zeros of The open-loop zeros are the zeros of G(s)H(s)G(s)H(s)

106106

• Number of branchesNumber of branches of root-locus of root-locus equals the equals the number of open-loop polesnumber of open-loop poles ((nn) since it is , generally, greater than ) since it is , generally, greater than the number of open-loop zeros (the number of open-loop zeros (mm).).

• The number of individual root-locus The number of individual root-locus branches terminating at branches terminating at finitefinite open- open-loop zeros is equal to the number (loop zeros is equal to the number (mm) ) of the of the

• open-loop zeros.open-loop zeros.

• The remaining The remaining n-m n-m branches branches terminate atterminate at

• infinity along asymptotes.infinity along asymptotes.

107107

• If we include poles and zeros at If we include poles and zeros at infinityinfinity, , the number of open-loop poles equals the number of open-loop poles equals that of open-loop zeros.that of open-loop zeros.

• Hence it is always true that , the root Hence it is always true that , the root loci loci startstart at the at the poles poles of G(s)H(s) and of G(s)H(s) and endend at the at the

zeroszeros of G(s)H(s), as K varies from zero of G(s)H(s), as K varies from zero to infinity , where the poles & zeros to infinity , where the poles & zeros include both those at include both those at finitefinite s plane and s plane and those at those at infinityinfinity..

108108

• 2. Determine the root loci on the real 2. Determine the root loci on the real axisaxis..

• Root loci on the real axis are determined Root loci on the real axis are determined by by polespoles and and zeroszeros lying lying on iton it..

• The complex-conjugate poles & zeros of The complex-conjugate poles & zeros of G(s)H(s) have G(s)H(s) have no effectno effect on the on the root loci root loci on the real axis.on the real axis.

• For constructing root loci on the real axis, For constructing root loci on the real axis, choose a test pointchoose a test point on it. on it.

• If the total number of real poles & real If the total number of real poles & real zeros to the right of this zeros to the right of this test pointtest point is is oddodd , ,

• then then this pointthis point lies on a root locus. lies on a root locus.

109109

• 3. Determine the asymptotes of root loci3. Determine the asymptotes of root loci• If the test point s is located far from the If the test point s is located far from the

origin, then the angle of each complex origin, then the angle of each complex quantity may be considered the same.quantity may be considered the same.

• Therefore, the root loci for very large Therefore, the root loci for very large values of s must be values of s must be asymptotic to straight asymptotic to straight lineslines whose angles are given by: whose angles are given by:

• Angles of asymptotesAngles of asymptotes = = +180+18000(2k+1)(2k+1)• n-mn-m• where k=0,1,2,…where k=0,1,2,…• n = number of finite poles of n = number of finite poles of

G(s)H(s) G(s)H(s) • m=m= number of finite zeros of number of finite zeros of

G(s)H(s) G(s)H(s)

110110

• The The numbernumber of distinct of distinct asymptotesasymptotes = = n-m.n-m.

• k=0 corresponds to the asymptotes with k=0 corresponds to the asymptotes with the smallest angle with the real axis.the smallest angle with the real axis.

• Although k assumes an infinite number of Although k assumes an infinite number of values, as k is increased the angle values, as k is increased the angle repeatsrepeats itself.itself.

• All the asymptotes intersect on the real All the asymptotes intersect on the real axis.axis.

• The The pointpoint at which they do so is obtained at which they do so is obtained as follows.as follows.

• s= s= (sum of poles)-(sum of zeros)(sum of poles)-(sum of zeros)• n-mn-m

111111

4.Find the breakaway and break-in points.4.Find the breakaway and break-in points.Because of the conjugate symmetry of rootBecause of the conjugate symmetry of rootloci, the breakaway and break-in points loci, the breakaway and break-in points either lie on the real axis or occur in complex-either lie on the real axis or occur in complex-

conjugate pairs.conjugate pairs.

If root locus lies between If root locus lies between two open-loop polestwo open-loop polesthen there exists at least then there exists at least one breakaway point one breakaway point between two poles. between two poles.

Similarly if root locus lies between Similarly if root locus lies between two open-loop two open-loop zeroszeros

then there exists at least then there exists at least one break-in point one break-in point between two zeros.between two zeros.

112112

If root locus lies between anIf root locus lies between an open-loop open-loop polepole

and a zero (and a zero (finite or infinitefinite or infinite) on the real ) on the real axis,axis,

then there may exist then there may exist no no breakaway or breakaway or break-inbreak-in

pointspoints or or there may exist there may exist both both breakaway or break-in points.breakaway or break-in points.

113113

Suppose that the characteristic Suppose that the characteristic equation is given by equation is given by

B(s) + K A(s) =0B(s) + K A(s) =0

• Breakaway and break-in pointsBreakaway and break-in points correspondcorrespond

• to multiple roots of the characteristic to multiple roots of the characteristic

• equation.equation.

• Hence they can be determined from Hence they can be determined from the roots of the roots of

• dK =0 dK =0 (1)(1)

• dsds

114114

• ItIt is important to note thatis important to note that Breakaway Breakaway andand break-in pointsbreak-in points must must be roots of be roots of Equation(1), but Equation(1), but not allnot all roots of roots of Equation(1) are breakaway andEquation(1) are breakaway and break-in break-in pointspoints . .

• If a If a real rootreal root of Equation(1) of Equation(1) lieslies on on the the rootroot locus portion of the real axislocus portion of the real axis, then it , then it isis an actual breakaway and an actual breakaway and break-in break-in point.point.

• If a If a real rootreal root of Equation(1) is of Equation(1) is notnot on the on the root locus portion of the real axis, then root locus portion of the real axis, then this root is this root is neitherneither a breakaway a breakaway nor nor aa break-in point.break-in point.

115115

• If two roots of Equation (1) are If two roots of Equation (1) are a a complex-conjugatecomplex-conjugate pairpair and if it is not and if it is not certain whether they are on root loci, certain whether they are on root loci, then it is necessary to check the then it is necessary to check the corresponding corresponding K valueK value..

• If the value of If the value of KK is is positivepositive, then that , then that point point isis an actual a breakaway an actual a breakaway oror aa break-in point.break-in point.

• If the value of If the value of KK is is negativenegative, then that , then that point is point is neitherneither a breakaway a breakaway nor nor aa break-in point.break-in point.

116116

• 5. Determine the angle of departure 5. Determine the angle of departure (angle of arrival) of the root locus from (angle of arrival) of the root locus from a complex pole (at a complex zero). a complex pole (at a complex zero).

• To sketch the root loci with accuracy, To sketch the root loci with accuracy, we must find the we must find the direction of root locidirection of root loci near the complex poles & zeros.near the complex poles & zeros.

• Angle of departureAngle of departure from from a complex pole a complex pole == 1801800 0 - (sum of the angles of vectors - (sum of the angles of vectors to the to the complex polecomplex pole in question from in question from other poles)+ (sum of the angles of other poles)+ (sum of the angles of vectors to thevectors to the complex complex polepole in question in question from zeros)from zeros)

117117

• Angle of arrivalAngle of arrival at a at a complex zerocomplex zero == 1801800 0

- (sum of the angles of vectors to the - (sum of the angles of vectors to the complexcomplex zerozero in question from other in question from other zeros)+ (sum of the angles of vectors to zeros)+ (sum of the angles of vectors to the the complex zerocomplex zero in question from in question from poles).poles).

118118

• 6. Find the points where the root loci 6. Find the points where the root loci may cross the imaginary axis may cross the imaginary axis

• The points where the root loci may The points where the root loci may cross the jcross the jωω axis can be found by axis can be found by (a) (a) Use of Routh’s stability criterion Use of Routh’s stability criterion oror(b) Letting s= j(b) Letting s= jωω in characteristic equation, in characteristic equation,

equating both the real part and imaginary equating both the real part and imaginary part to zero, and solving for part to zero, and solving for ωω & K. & K.

• The values of The values of ωω thus found give the thus found give the frequencies at which root loci cross the frequencies at which root loci cross the imaginary axis.imaginary axis.

• The K value gives the gain at the The K value gives the gain at the crossing point.crossing point.

119119

• 7. Taking a series of test points in the 7. Taking a series of test points in the broad neighborhood of the origin of the s broad neighborhood of the origin of the s plane, sketch the root loci.plane, sketch the root loci.

• Determine root loci in the broad Determine root loci in the broad neighborhood of the jneighborhood of the jωω axis & the origin. axis & the origin.

• The most important part of the root loci The most important part of the root loci is on neither the real axis nor the is on neither the real axis nor the asymptotes but the part in the broad asymptotes but the part in the broad neighborhood of the jneighborhood of the jωω axis & the origin. axis & the origin.

• The shape of the root loci in this The shape of the root loci in this important region in the s plane must be important region in the s plane must be obtained with sufficient accuracy.obtained with sufficient accuracy.

120120

1.Consider the system shown 1.Consider the system shown in fig. Draw the root locus & in fig. Draw the root locus & determine the range of K for determine the range of K for stability.stability.

K

S(s+1)(s+2)

R(s) C(s)

121121

• For the given system, G(s)= For the given system, G(s)=

• H(s)=1H(s)=1

• The angle condition becomes The angle condition becomes

• G(s)H(s) = G(s)H(s) =

= 0- s- s+1 - s+2 = 180= 0- s- s+1 - s+2 = 18000(2k+1), (2k+1), (k=0,1,2….)(k=0,1,2….)

• The magnitude condition is The magnitude condition is

• G(s) = = 1G(s) = = 1

KS(s+1)(s+2)

KS(s+1)(s+2)

+

_

KS(s+1)(s+2)

122122

• Step 1Step 1. . Locate the poles & zeros of Locate the poles & zeros of G(s)H(s) on the s planeG(s)H(s) on the s plane. .

• The poles are s=0, -1, -2.The poles are s=0, -1, -2.

• There are no zeros.There are no zeros.

σ

jω

0 1 2-1-2-3-4

j1

-j2

-j1

j2

j3

-j3

xx x

123123

• The location of poles are indicated byThe location of poles are indicated by crossescrosses..

• ((The location of zeros,if any, are indicated The location of zeros,if any, are indicated byby smallsmall circlescircles))

• The number of branches of root loci = n-The number of branches of root loci = n-mm

• =3-0=3-0

• ==3 3

• Where,Where,

• nn = number of finite poles of G(s)H(s) = = number of finite poles of G(s)H(s) =33

• mm== number of finite zeros of G(s)H(s) =number of finite zeros of G(s)H(s) =00

124124

• Step2. Step2. Determine the root loci on the Determine the root loci on the real axisreal axis..

• To determine the root loci on the real To determine the root loci on the real axis, we select test point s on the –ve axis, we select test point s on the –ve real axis between 0 and -1real axis between 0 and -1..

• The number of poles & zeros to the The number of poles & zeros to the right of the test point is 1(odd right of the test point is 1(odd number). number).

• Hence this test point lies on the root Hence this test point lies on the root loci.loci.

• Thus, root loci exists on the –ve real Thus, root loci exists on the –ve real axis between 0 & -1axis between 0 & -1

125125

• If we select a test point between -1 & If we select a test point between -1 & -2, the number of poles to the right -2, the number of poles to the right of the test point is an even number of the test point is an even number (2).(2).

• Hence this test point is not on the Hence this test point is not on the root loci.root loci.

• Thus, root loci does not exists on the Thus, root loci does not exists on the –ve real axis between 0 & -1.–ve real axis between 0 & -1.

• If we select a test point to the right If we select a test point to the right of -2, the number of poles to the of -2, the number of poles to the right of the test point is an odd right of the test point is an odd number (3).number (3).

126126

• Hence this test point is on the root Hence this test point is on the root loci.loci.

• Thus, root loci does exists on the –ve Thus, root loci does exists on the –ve real axis between -2 & -∞.real axis between -2 & -∞.

• Step 3Step 3. . Determine the asymptotes of Determine the asymptotes of the root loci.the root loci.

• Angles of asymptotesAngles of asymptotes = = 18018000(2k+1)(2k+1)

• n-mn-m

• where k=0,1,2,…where k=0,1,2,…

• n = 3 n = 3

• m=m= 00

+_

127127

• Angles of asymptotesAngles of asymptotes = = 18018000(2k+1)(2k+1)

• 33

• Number of asymptotes = n-m = 3Number of asymptotes = n-m = 3

• When k=0, Angles of asymptotesWhen k=0, Angles of asymptotes=+60=+6000, -60, -600.0.

• When k=1, Angles of asymptotesWhen k=1, Angles of asymptotes=+180=+18000, -, -18018000

• Point of intersection of Point of intersection of asymptotes = asymptotes =

• (sum of poles)-(sum of zeros)(sum of poles)-(sum of zeros)

• n-mn-m

• = (0+-1+-2) – 0 = -1= (0+-1+-2) – 0 = -1

• 3 3

+_

128128

• The asymptotes are drawn as shownThe asymptotes are drawn as shown

• Step 4Step 4: : Determine the break away point Determine the break away point • The break away points can be determined The break away points can be determined

from the roots of dK =0from the roots of dK =0• ds ds • The characteristic equation isThe characteristic equation is• 1+G(s)H(s)=01+G(s)H(s)=0• i.e., 1+ K =0i.e., 1+ K =0

• or, K=-(sor, K=-(s33+3s+3s22+2s)+2s)• By setting dK =0 , we obtainBy setting dK =0 , we obtain• dsds• dK = -(3sdK = -(3s22+6s+2)=0+6s+2)=0• ds ds

• or or s= -0.4226, s= -1.5774s= -0.4226, s= -1.5774

S(s+1)(s+2)

129129

• Since the break away point must lie on Since the break away point must lie on a root locus between 0 & -1, it is clear a root locus between 0 & -1, it is clear that s= -0.4226 corresponds to the that s= -0.4226 corresponds to the actual break away point.actual break away point.

• Point s= -1.5774 is Point s= -1.5774 is notnot on the root on the root locus. locus.

• Hence it is not an actual break away Hence it is not an actual break away point.point.

• In fact evaluation of values of K In fact evaluation of values of K corresponding to s= -0.4226, s= -corresponding to s= -0.4226, s= -1.5774 yields1.5774 yields

• K=0.3849 K=0.3849 (+ve)(+ve) for s= -0.4226 for s= -0.4226

• K=- 0.3849 K=- 0.3849 (-ve)(-ve) for s= -1.5774 for s= -1.5774

130130

• Step : 5. Determine the points where Step : 5. Determine the points where the root loci cross the imaginary axis the root loci cross the imaginary axis

• Method 1Method 1

• By use of Routh’s stability criterion By use of Routh’s stability criterion

• The characteristic eqn is The characteristic eqn is

• i.e. si.e. s33+3s+3s22+2s+K=0+2s+K=0

• The Routh’s array becomesThe Routh’s array becomes

1+ K = 0 s(s+1)(s+2)

131131

• The value of K that makes the sThe value of K that makes the s11 term in the first column = 0 is K=6term in the first column = 0 is K=6

• The crossing points can be found by The crossing points can be found by solving the aux. eqn obtained from solving the aux. eqn obtained from the sthe s22 row; i.e. row; i.e.

• 3s3s22+K=3s+K=3s22+6=0 which yields +6=0 which yields

S3 1 2

S2 3 K

S1 6-K 3

S0 K

132132

• S= j √2S= j √2

• The frequencies at the crossing The frequencies at the crossing points on the imaginary axis are thus points on the imaginary axis are thus ωω= √2= √2

• The gain value corresponding to the The gain value corresponding to the crossing points is K=6crossing points is K=6

• Method 2Method 2

• Let s= jLet s= jωω in the characteristic eqn, in the characteristic eqn, equate both the real and imaginary equate both the real and imaginary part to zero, and solve for part to zero, and solve for ωω & K. & K.

• Now, the characteristic eqn is Now, the characteristic eqn is

• ss33+3s+3s22+2s+K=0+2s+K=0

+_

+_

133133

• Let s= jLet s= jωω,,

• (j(jωω))33+3(j+3(jωω))22+2(j+2(jωω)+K=0)+K=0

• Or (K-3Or (K-3ωω22)) +j(2+j(2ωω- - ωω33)=0)=0

• equating both the real and imaginary equating both the real and imaginary part to zero, we obtain part to zero, we obtain

• K-3K-3ωω22 = 0, 2 = 0, 2ωω- - ωω3 3 =0=0

• From which From which ωω= √2= √2

• K = 6K = 6

+_

134134

• The frequencies at the crossing points on The frequencies at the crossing points on the imaginary axis are thus the imaginary axis are thus ωω= √2= √2

• The gain value corresponding to the The gain value corresponding to the crossing points is K=6crossing points is K=6

• Step No 6: Step No 6: Choose a test point in the Choose a test point in the broad neighborhood of the jbroad neighborhood of the jωω axis and the axis and the origin and apply the angle condition.origin and apply the angle condition.

• If a test point is on the root loci, then the If a test point is on the root loci, then the

• sum of three angles, sum of three angles, θθ11+ + θθ22+ + θθ3, 3, must be must be 1801800 0

• If the test point does not satisfy the angle If the test point does not satisfy the angle condition, select another test point until it condition, select another test point until it satisfy the angle condition.satisfy the angle condition.

+_

135135

• Continue this process and locate Continue this process and locate sufficient number of points satisfying sufficient number of points satisfying the angle condition.the angle condition.

S+2

θ1

S+1S

θ2θ30-1-2

j1

σ

jω

136136

σ

jω

0 1 2-1-2-3-4

j1

-j2

-j1

j2

j3

-j3

xx x600

asymptotes

K ∞

K ∞

∞ K

K=6

K=6

137137

• Problem 2: Sketch the root locus plot Problem 2: Sketch the root locus plot of system having G(s)=K(s+2) , of system having G(s)=K(s+2) , H(s)=1H(s)=1

• ss22+2s+3+2s+3

138138

• Step 1Step 1. . Locate the poles & zeros of Locate the poles & zeros of G(s)H(s) on the s planeG(s)H(s) on the s plane. .

• The poles are s=-1+j√2, -1-j√2 The poles are s=-1+j√2, -1-j√2

• The zero is s= -2.The zero is s= -2.

σ

jω

0 1-1-2

-j1

j1

j2

o

j√2j√2

-j√2-j√2

x

x

139139

• Step2. Step2. Determine the root loci on the Determine the root loci on the real axisreal axis..

• To determine the root loci on the real To determine the root loci on the real axis, we select test point s on the –ve axis, we select test point s on the –ve real axis between 0 and -2real axis between 0 and -2..

• The number of poles & zeros to the The number of poles & zeros to the right of the test point is nil. right of the test point is nil.

• Hence this test point Hence this test point does notdoes not lies on lies on the root loci.the root loci.

• Thus, root loci Thus, root loci does notdoes not exist on the – exist on the –ve real axis between 0 & -2ve real axis between 0 & -2

140140

• Taking a test point to the left of s= -Taking a test point to the left of s= -2, we find that the 2, we find that the number of real number of real poles & zerospoles & zeros to the right of test to the right of test point is odd (=1).point is odd (=1).

• Hence the root loci exist on the –ve Hence the root loci exist on the –ve real axis from s=-2 to -∞ as shownreal axis from s=-2 to -∞ as shown

141141

σ

jω

0 1-1-2

-j1

j1

j2

o

j√2j√2

-j√2-j√2

x

x

142142

• Step 3Step 3. . Determine the asymptotes of Determine the asymptotes of the root loci.the root loci.

• Angles of asymptotesAngles of asymptotes = = 18018000(2k+1)(2k+1)

• n-mn-m

• where k=0,1,2,…where k=0,1,2,…

• n = number of poles= 2 n = number of poles= 2

• m= number of zeros=m= number of zeros=11

• Angles of asymptotesAngles of asymptotes = = 18018000(2k+1)(2k+1)

• 11

• Number of asymptotes = n-m = 1Number of asymptotes = n-m = 1

• When k=0, Angles of When k=0, Angles of asymptotes=asymptotes=18018000

+_

+_

143143

• Step 4Step 4: Determine the angle of : Determine the angle of departure from the complex conjugate departure from the complex conjugate poles.poles.

• The The presence of a pair of complex presence of a pair of complex conjugate polesconjugate poles demands determination demands determination of the angle of departure from these of the angle of departure from these poles.poles.

• Knowledge of this angle is important Knowledge of this angle is important since the root locus near a complex since the root locus near a complex pole yields information as to whether pole yields information as to whether the locus the locus originating from the complex originating from the complex polespoles migrates towards real axis or migrates towards real axis or extends toward the asymptotes.extends toward the asymptotes.

144144

• Determination of angle of departure.Determination of angle of departure.

σ

jω

0 1-1-2

-j1

j1

j2

o

j√2j√2

-j√2-j√2

x

x900

550

145145

• Angle of departureAngle of departure from from a complex pole a complex pole == 1801800 0 - (sum of the angles of vectors to - (sum of the angles of vectors to the the complex polecomplex pole in question from other in question from other poles)+ (sum of the angles of vectors to poles)+ (sum of the angles of vectors to thethe complex complex polepole in question from zeros) in question from zeros)

• Angle of departureAngle of departure from from the complex the complex pole at pole at s=-1+j√2 =s=-1+j√2 = 1801800 0 ––(90(9000)+55)+5500=145=14500

• (Since the root locus is symmetric about (Since the root locus is symmetric about the real axis),the real axis),

Angle of departureAngle of departure from from the complex the complex pole at pole at s=-1-j√2 =s=-1-j√2 = -145-14500

146146

• Step:4. Step:4. Determine the break-in pointDetermine the break-in point • The break-in point can be found fromThe break-in point can be found from• the roots of dK=0the roots of dK=0• dsds• Since the characteristic equation is given Since the characteristic equation is given

by by • 1+G(s)H(s)=0, we have1+G(s)H(s)=0, we have• 1+ K (s+2) 1+ K (s+2) = 0= 0• ss22+2s+3+2s+3• • Hence, K= -(sHence, K= -(s22+2s+3)+2s+3)• (s+2)(s+2)• we have, dK = we have, dK = --(2s+2)(s+2)-(s(2s+2)(s+2)-(s22+2s+3)+2s+3)• ds (s+2)ds (s+2)2 = 02 = 0

• which gives swhich gives s22+4s+1=0+4s+1=0•

147147

• i.e., s= -3.7320 or s= -0.2680i.e., s= -3.7320 or s= -0.2680

• Note that point s= -3.7320 is on the Note that point s= -3.7320 is on the root locus.root locus.

• Hence this point is an actual break-in Hence this point is an actual break-in point.point.

• Since s= -0.2680 is Since s= -0.2680 is notnot on the root on the root locus it is locus it is notnot an actual break-in point. an actual break-in point.

148148

• Step 5. Step 5. Sketch root-locus plot based on Sketch root-locus plot based on the information obtained in the the information obtained in the foregoing steps. foregoing steps.

• To determine root loci accurately, To determine root loci accurately, several points must be found using several points must be found using angle condition.angle condition.

•

149149

• Root locus plotRoot locus plot

σ

jω

0 1-1-2

-j1

j1

j2

o

j√2j√2

-j√2-j√2

x

x

1450

-1450

-3.73

Break-in point

150150

• It is seen that root locus in the complex s It is seen that root locus in the complex s plane is a part of a circle.plane is a part of a circle.

• We can prove that root locus is circular in We can prove that root locus is circular in shape by deriving the eqn for the root shape by deriving the eqn for the root locus, using the angle condition.locus, using the angle condition.

• For the present system, the angle For the present system, the angle condition is condition is

• S+2 – s+1- jS+2 – s+1- j√2 – s+1+ √2 – s+1+ jj√2 = 180√2 = 18000(2k+1)(2k+1)

• If s= If s= σσ+j+jωω is substituted in to this last is substituted in to this last eqn, we obtain eqn, we obtain

• σσ+2+j+2+jωω - - σσ+1+j+1+jωω - j- j√2 – √2 – σσ+1+j+1+jωω + + jj√2 = √2 =

• 18018000(2k+1)(2k+1)

+ _

+ _

151151

• Which can be written as Which can be written as

• TanTan-1-1( ) - Tan( ) - Tan-1-1( ) - Tan( ) - Tan-1-1( ) = ( ) = 18018000(2k+1)(2k+1)

ωσ+ 2

ω - √2 σ + 1

ω + √2 σ + 1

+ _

152152

• Sketch the root locus plot for the Sketch the root locus plot for the system having open loop transfer system having open loop transfer function given by G(s)H(s) = function given by G(s)H(s) = k k

• s(s+4)(ss(s+4)(s22+4s +13) +4s +13)

153153

• Step 1Step 1. . Locate the poles & zeros of Locate the poles & zeros of G(s)H(s) on the s planeG(s)H(s) on the s plane..

• The open-loop poles are,The open-loop poles are,

• s =s = 0 0, s= , s= - 4- 4, s = , s = -2 + j 3-2 + j 3, s = , s = --2 - j 32 - j 3

• The number of open loop poles =The number of open loop poles = 4 4

• The number of open loop zeros = The number of open loop zeros = 00

• Hence number of branches of root Hence number of branches of root locuslocus

• = 4 – 0 = 4= 4 – 0 = 4

154154

x

- 1 0-2-3-4x

j1

j2

j3

j4

-j1

-j2

-j3

x

x

155155

• Step 2. Step 2. Determine the root loci on Determine the root loci on the real axisthe real axis..

• Root loci on the Root loci on the real axisreal axis is is determined by poles and zeros lying determined by poles and zeros lying on on itit..

• If we select any test point between 0 If we select any test point between 0 & -4, we find that number of poles to & -4, we find that number of poles to the the rightright of test point is = 1 ( of test point is = 1 (oddodd))

• Hence the segment of real axis Hence the segment of real axis between between

0 & -4 is on the root loci.0 & -4 is on the root loci.

156156

x

- 1 0-2-3-4x

j1

j2

j3

j4

-j1

-j2

-j3

x

x

157157

• Step 3Step 3. . Determine the asymptotes of Determine the asymptotes of the root loci.the root loci.

• Angles of asymptotesAngles of asymptotes = = 18018000(2k+1)(2k+1)

• n-mn-m

• where k=0,1,2,…where k=0,1,2,…

• n = 4 n = 4

• m=m= 00

• number of asymptotes = n-m = 4number of asymptotes = n-m = 4

• Angles of asymptotes =Angles of asymptotes =

• + 45+ 4500, - 45, - 450 0 ,+135,+1350 0 , -135, -13500

158158

• Point of intersection of asymptotes Point of intersection of asymptotes (on the real axis) (on the real axis)

• or the centroid = or the centroid =

• (sum of poles)-(sum of (sum of poles)-(sum of zeros)zeros)

• n-mn-m

• = (0+-4+-2+j3+ -2-j3) – 0 = (0+-4+-2+j3+ -2-j3) – 0 = -2 = -2

• 44

159159

• Step 4Step 4: : Determine the break away point Determine the break away point • The break away points can be The break away points can be

determined from the roots of dK =0determined from the roots of dK =0• ds ds • The characteristic equation isThe characteristic equation is• 1+G(s)H(s)=01+G(s)H(s)=0• i.e., 1+ k = 0i.e., 1+ k = 0• s(s+4)(ss(s+4)(s22+4s +13) +4s +13)

• i.e., s(s+4)(si.e., s(s+4)(s22+4s +13) + k = 0+4s +13) + k = 0

•

160160

• K = - ( sK = - ( s44 + 8s + 8s33 + 29 s + 29 s22 + 52s) + 52s)

• d K = -(4s3 + 24 s2 + 58 s +52)d K = -(4s3 + 24 s2 + 58 s +52)

dsds

put d K = 0put d K = 0

dsds

Thus (4s3 + 24 s2 + 58 s +52) = 0Thus (4s3 + 24 s2 + 58 s +52) = 0

The break away points are given The break away points are given by roots of the above equation. One by roots of the above equation. One of the roots is s = -2 .of the roots is s = -2 .

The other roots are found as follows:The other roots are found as follows:

161161

• (s+2) is a factor ,(s+2) is a factor ,

• Hence we have (4s3 + 24 s2 + 58 s Hence we have (4s3 + 24 s2 + 58 s +52)+52)

• (s+2) (s+2)

• = (4s2 + 16s + 26)= (4s2 + 16s + 26)

• and thus (s+2) (4s2 + 16s + and thus (s+2) (4s2 + 16s + 26) =026) =0

• The The two complex break away pointstwo complex break away points are given by roots of (4s2 + 16s + are given by roots of (4s2 + 16s + 26) =026) =0

• i.e. , s= i.e. , s= -2 + j 1.58-2 + j 1.58 , , -2 – j 1.58-2 – j 1.58

162162

• Step 5. Step 5. Determination of angle of departure.Determination of angle of departure.

• Presence of a pair of complex conjugate Presence of a pair of complex conjugate poles demand determination of poles demand determination of angle of angle of departure from these poles.departure from these poles.

• Angle of departure from pole at -2+j3Angle of departure from pole at -2+j3

• = 180= 18000-(-(φφ11+ + φφ22+ + φφ33) )

• where where φφ1 1 = = 18018000- tan - tan -1-1 (3/2) (3/2)

• φφ2 2 = = tan tan -1-1 (3/2) (3/2)

• φφ33 = 90 = 9000

•

163163

x

- 1 0-2-3-4x

j1 φ1

j2

j3

j4

-j1

-j2

-j3

x

x

164164

• Hence angle of departure from pole at Hence angle of departure from pole at -2+j3 -2+j3

• = 180= 18000-(-(φφ11+ + φφ22+ + φφ33))

• = = 18018000--(180(18000- tan - tan -1-1 (3/2)+ (3/2)+tan tan -1-1 (3/2)+ (3/2)+ 909000) )

• = - = - 909000

• And angle of departure from pole at -And angle of departure from pole at -2-j32-j3

• = + = + 909000

165165

• Step :6. Determine the points where Step :6. Determine the points where the root loci cross the imaginary axisthe root loci cross the imaginary axis

• By use of Routh’s stability criterionBy use of Routh’s stability criterion• The characteristic equation isThe characteristic equation is• 1+G(s)H(s)=01+G(s)H(s)=0

• i.e., 1+ k = 0i.e., 1+ k = 0• s(s+4)(ss(s+4)(s22+4s +13) +4s +13)

i.e., s(s+4)(si.e., s(s+4)(s22+4s +13) + k = 0+4s +13) + k = 0

• i.e., si.e., s44 + 8s + 8s33 + 29 s + 29 s22 + 52s+k = 0 + 52s+k = 0

166166

• Routh’s array :Routh’s array :• ss44 1 29 K 1 29 K• ss33 8 52 8 52• ss22 22.5 K 22.5 K• ss11 (52 – 0.35K)(52 – 0.35K) • ss00 K K • For stability , For stability , (52 – 0.35K)(52 – 0.35K) ≥ 0 ≥ 0 • or K ≤ 148.6or K ≤ 148.6• when K = when K = 148.6148.6 , root loci intersects , root loci intersects

the the jjωω axis . axis .

167167

• Corresponding value of s is found fromCorresponding value of s is found from

• auxiliary equation for sauxiliary equation for s22 row. row.

• i.e., 22.5 si.e., 22.5 s22 + K = 0 + K = 0

• or 22.5 sor 22.5 s22 + 146.8 = 0 + 146.8 = 0

• Or s = Or s = +j 2.56+j 2.56, , - j 2.56- j 2.56

168168

x

x

xx0

-4

j3

-j3

-2

450

-450

1350

-1350

K=∞K=∞

K=∞K=∞

169169

• Home WorkHome Work

• Draw the root locus plot for the Draw the root locus plot for the system having open loop transfer system having open loop transfer functionfunction

G(s)H(s) = KG(s)H(s) = K

(s2 + 2s +2) (s2 + 2s + (s2 + 2s +2) (s2 + 2s + 5)5)

170170

Frequency Response AnalysisFrequency Response Analysis

• Frequency Response means the steady Frequency Response means the steady state response of a system to a state response of a system to a sinusoidalsinusoidal

Input.Input.

• In Frequency Response Analysis, we vary In Frequency Response Analysis, we vary

the frequency of the input signal over a the frequency of the input signal over a certain range and study the resulting certain range and study the resulting response.response.

171171

Bode PlotBode Plot

• A sinusoidal transfer function may be A sinusoidal transfer function may be represented by two separate plots, one represented by two separate plots, one giving the magnitude versus frequency,giving the magnitude versus frequency,

• And other the phase angle (in degrees) And other the phase angle (in degrees) versus frequency.versus frequency.

• A Bode diagram consists of two graphs:A Bode diagram consists of two graphs:

• One is a plot of the logarithm of the One is a plot of the logarithm of the magnitude of a sinusoidal transfer magnitude of a sinusoidal transfer function; other is a plot of the phase function; other is a plot of the phase angle; angle;

172172

• both are plotted against frequency in both are plotted against frequency in logarithmic scale .logarithmic scale .

• The standard representation of The standard representation of logarithmic magnitude of sinusoidal logarithmic magnitude of sinusoidal transfer function transfer function

• G( jG( jωω) is 20 log ) is 20 log G( jG( jωω) , where the ) , where the base of logarithm is 10.base of logarithm is 10.

• The unit used in this The unit used in this representation of representation of magnitude is the decibel, (dB).magnitude is the decibel, (dB).

• The curves are drawn in semi log paper, The curves are drawn in semi log paper, using the log scale for frequency and using the log scale for frequency and the linear scale for both magnitude (in the linear scale for both magnitude (in decibel)decibel)

• And phase angle (in degrees).And phase angle (in degrees).

173173

• The open loop transfer function of a closed The open loop transfer function of a closed loop control system may assume following loop control system may assume following general form:general form:

• G(s)H(s) = K [(1+ sTG(s)H(s) = K [(1+ sT11) (1+ sT) (1+ sT22) …] ) …] ωωnn22

• ssN N [(1+sT[(1+sTaa)(1+sT)(1+sTbb)…])…](s(s22+2+2ζωζωnns+s+ωωnn22))

• The sinusoidal transfer function is obtained The sinusoidal transfer function is obtained by substituting s= jby substituting s= jωω, thus, thus

• G(G(jjωω)H()H(jjωω) = K [(1+ ) = K [(1+ jjωωTT11) (1+ ) (1+ jjωωTT22) …] ) …] ωωnn22

• (j(jωω))NN[(1+[(1+jjωωTTaa)(1+ )(1+ jjωωTTbb)…])…]((ωωnn22-- ωω2 2 ++

j2j2ζωζωnnωω))

• N = Type of the systemN = Type of the system

174174

• Initial SlopeInitial Slope of Bode plot : of Bode plot :

Type of Type of system Nsystem N

Initial Slope Initial Slope

-20N -20N dB/decadedB/decade

Intersection Intersection with 0 dB with 0 dB axis ataxis at

00 0 dB/decade0 dB/decade Parallel to Parallel to 0dB axis0dB axis

11 -20dB/decade-20dB/decade = K= K

22 -40dB/decade-40dB/decade = K = K 1/21/2

33 -60dB/decade-60dB/decade = K = K 1/31/3

…… …… … …

NN -20NdB/decade-20NdB/decade = K = K 1/N 1/N

175175

• Each first order term , i.e. , (1+jEach first order term , i.e. , (1+jωωT ) T ) termterm

• In the denominator will contribute -In the denominator will contribute -20 dB/decade .20 dB/decade .

• Each first order term , i.e. , (1+jEach first order term , i.e. , (1+jωωT ) T ) termterm

• In the numerator will contribute +20 In the numerator will contribute +20 dB/decade.dB/decade.

• Each second order term iEach second order term in the n the denominator will contribute -40 denominator will contribute -40 dB/decade dB/decade

176176

• Corner frequency:Corner frequency:

Determination of Determination of corner frequencycorner frequency is is vital, since the slope of magnitude plot vital, since the slope of magnitude plot will change at each will change at each corner frequencycorner frequency . .

• Each first order term , i.e. , (1+jEach first order term , i.e. , (1+jωωTT ) ) term has a term has a corner frequencycorner frequency of of ωω = = 1/T1/T

• Each second order term , i.e. ,Each second order term , i.e. ,

• ωωnn22

• (s(s22+2+2ζωζωnns+s+ωωnn22) term has ) term has a a

corner frequencycorner frequency of of ωωnn . .

•

177177

• Draw the Bode plot for the following Draw the Bode plot for the following

• Open-loop transfer function Open-loop transfer function

• G(s)H(s) = 4 G(s)H(s) = 4

• s(1+0.5s)(1+0.08s) s(1+0.5s)(1+0.08s)

• And determine (a) the gain marginAnd determine (a) the gain margin

• (b) the phase margin(b) the phase margin

• (c) the closed loop (c) the closed loop stability stability

178178

• G(s)H(s) = 4 G(s)H(s) = 4 • s(1+0.5s)(1+0.08s) s(1+0.5s)(1+0.08s) • Step1Step1: Check whether given transfer : Check whether given transfer

function is in normalized form. function is in normalized form. • Here, G(s)H(s) is in normalized formHere, G(s)H(s) is in normalized form

• Then put s= jThen put s= jωω • G(jG(jωω)H(j)H(jωω) = 4 ) = 4 • jjωω(1+0.5 j(1+0.5 jωω)(1+0.08 j)(1+0.08 jωω ) )

• Step 2Step 2:: the corner frequencies are the corner frequencies are• ωω = 1/0.5 = 2 = 1/0.5 = 2• and and ωω = 1/0.08 = 12.5 = 1/0.08 = 12.5•

179179

• The starting frequency of the Bode The starting frequency of the Bode plot is taken as lower than the lowest plot is taken as lower than the lowest corner frequency.corner frequency.

• Step 3Step 3. Determine the initial slope . Determine the initial slope and point of intersection with and point of intersection with frequency (frequency (ωω))axis:axis:

• Since system is type 1, the initial Since system is type 1, the initial slope of the Bode plot is -20 db/ slope of the Bode plot is -20 db/ decade.decade.

• point of intersection with frequency point of intersection with frequency ((ωω))axis axis

occurs at occurs at ωω = K = 4 = K = 4

180180

• Step 4Step 4: Determine change in slope at : Determine change in slope at each corner frequency and plot the each corner frequency and plot the magnitude plot.magnitude plot.

• The initial slope of -20 db/ decade The initial slope of -20 db/ decade continues till the lowest corner continues till the lowest corner frequency (frequency (ωω = 2) = 2) is reached.is reached.

• The corner frequency (The corner frequency (ωω = 2) = 2) is due to is due to the first order term in the the first order term in the denominator, (1+j0.5denominator, (1+j0.5ωω) which will ) which will contribute contribute -20 db/ decade.-20 db/ decade.

181181

• Hence total slope becomes (-20 + -20) Hence total slope becomes (-20 + -20) db/ decade = -40 db/ decade db/ decade = -40 db/ decade

• ( Note that a ( Note that a decadedecade is a frequency is a frequency band from band from ωω11 to to 1010ωω11, where , where ωω1 is 1 is any frequency)any frequency)

• ( Also note that an ( Also note that an octaveoctave is a is a frequency frequency band from band from ωω11 to to 22ωω11, where , where ωω1 is any frequency)1 is any frequency)

• ( Octave = ( Octave = 1/0.3011/0.301decade)decade)• Hence 20 db/ Hence 20 db/ decade = 20 x 0.3 db/ decade = 20 x 0.3 db/

octaveoctave• = 6 = 6 db/ octavedb/ octave•

182182

• The slope of -40 db/decade will The slope of -40 db/decade will continue till the next corner continue till the next corner frequency (frequency (ωω = 12.5) is reached. = 12.5) is reached.

• The corner frequency (The corner frequency (ωω = 12.5) = 12.5) is is due to the first order term in the due to the first order term in the denominator, (1+j0.08denominator, (1+j0.08ωω) which will ) which will contribute contribute -20 db/ decade.-20 db/ decade.

• Hence total slope becomes (-40 + -Hence total slope becomes (-40 + -20) db/ decade = -60 db/ decade.20) db/ decade = -60 db/ decade.

• The slope of -60 db/decade will The slope of -60 db/decade will continue for frequencies greater than continue for frequencies greater than ωω = 12.5. = 12.5.

183183

• Step 5: calculate G(jStep 5: calculate G(jωω) H(j) H(jωω) for ) for frequencies between frequencies between ωω = 1 to 100: = 1 to 100:

• G(jG(jωω) H(j) H(jωω)= )=

• -90-9000 – tan – tan-1-1(0.5(0.5ωω)- )- tantan-1-1(0.08(0.08ωω) )

ωω

rad/secrad/sec11 22 8 8 1010 2020 5050

G(jG(jωω) ) H(jH(jωω

degreedegree

--121121

--144144

--202202

--207207

--234234

--252252

184184

1 10 100 4

10

20db

0

-10

-20

ω

-20db/decade

-40db/decade

-20db/decade

52.7

-1800

-1500

-1200

-2100

-2400

-2700

185185

• The gain margin is defined as negative The gain margin is defined as negative of gain in db at phase cross over of gain in db at phase cross over frequency.frequency.

• phase cross over frequency is 5 rad/secphase cross over frequency is 5 rad/sec

• And corresponding gain value = -13 dbAnd corresponding gain value = -13 db

• Hence gain margin = 13 dbHence gain margin = 13 db

• The phase margin is given byThe phase margin is given by

• 18018000 + G(j + G(jωω)H(j)H(jωω) at gain ) at gain cross cross over frequency .over frequency .

• The gain cross over frequency is 2.7 The gain cross over frequency is 2.7 rad/sec rad/sec

• And corresponding phase angle = -150And corresponding phase angle = -15000

186186

• Hence Hence phase marginphase margin = 180 = 18000 + (- + (-15015000))

• = = 30300 0

• The The phase marginphase margin & gain margin are & gain margin are both positive.both positive.

• Hence system is stable.Hence system is stable.

• If phase cross over frequency is If phase cross over frequency is greater than gain cross over greater than gain cross over frequency, system will be stable.frequency, system will be stable.

187187

• Draw the Bode plot for the following Draw the Bode plot for the following

• Open-loop transfer function Open-loop transfer function

• G(s)H(s) = 30 G(s)H(s) = 30

• s(1+0.5s)(1+0.08s) s(1+0.5s)(1+0.08s)

• And determine (a) the gain marginAnd determine (a) the gain margin

• (b) the phase margin(b) the phase margin

• (c) the closed loop (c) the closed loop stabilitystability

188188

• Draw the Bode plot for the following Draw the Bode plot for the following

• Open-loop transfer function Open-loop transfer function

• G(s)H(s) = 48(s+10) G(s)H(s) = 48(s+10)

• s(s+20)(ss(s+20)(s22+2.4s+16) +2.4s+16)

• And determine (a) the gain marginAnd determine (a) the gain margin

• (b) the phase margin(b) the phase margin

• (c) the closed loop (c) the closed loop stability stability

189189

• G(s)H(s) = 48(s+10) G(s)H(s) = 48(s+10)

• s(s+20) (ss(s+20) (s22+2.4s+16) +2.4s+16)

• Step1Step1: Check whether given transfer : Check whether given transfer function is in normalized form. function is in normalized form.

• Here, G(s)H(s) is not in normalized form.Here, G(s)H(s) is not in normalized form.

• G(s)H(s) = 48 x 10 [s/10 +1] G(s)H(s) = 48 x 10 [s/10 +1] 1616

• 20x 20x 1616x s[s/20 +1] x s[s/20 +1] (s(s22+2.4s+16)+2.4s+16)

• = 1.5(0.1s +1) x 16 = 1.5(0.1s +1) x 16

s( 0.05s + 1) (ss( 0.05s + 1) (s22+2.4s+16)+2.4s+16)

190190

Then put s= jThen put s= jωω • G(jG(jωω)H(j)H(jωω) =1.5(0.1j) =1.5(0.1jωω +1) x 16 +1) x 16 jjωω ( 0.05 j ( 0.05 jωω + 1) [ j2.4 + 1) [ j2.4ωω + + (16- (16- ωω22))]]

• Step 2Step 2:: the corner frequencies are the corner frequencies are• ωω = 1/0.1 = 10, = 1/0.1 = 10,• ωω = 1/0.05 = 20, and = 1/0.05 = 20, and • ωω = = ωωn n =√ 16 = 4 since =√ 16 = 4 since

ωωnn2 2 = 16 = 16

• in the quadratic term.in the quadratic term.• •

191191

• The starting frequency of the Bode The starting frequency of the Bode plot is taken as lower than the lowest plot is taken as lower than the lowest corner frequency.corner frequency.

• Step 3Step 3. Determine the initial slope . Determine the initial slope and point of intersection with and point of intersection with frequency (frequency (ωω))axis:axis:

• Since system is type 1, the initial Since system is type 1, the initial slope of the Bode plot is -20 db/ slope of the Bode plot is -20 db/ decade.decade.

• point of intersection with frequency point of intersection with frequency ((ωω))axis axis

occurs at occurs at ωω = K = 1.5 = K = 1.5

192192

• Step 4Step 4: Determine change in slope at : Determine change in slope at each corner frequency and plot the each corner frequency and plot the magnitude plot.magnitude plot.

• The initial slope of -20 db/ decade The initial slope of -20 db/ decade continues till the lowest corner continues till the lowest corner frequency (frequency (ωω = 4) = 4) is reached.is reached.

• The corner frequency (The corner frequency (ωω = 4) = 4) is due to is due to the second order term (quadratic the second order term (quadratic term) in the denominator, 16 term) in the denominator, 16

• [ j2.4[ j2.4ωω + + (16- (16- ωω22))] ]

• which will contribute which will contribute -40 db/ decade.-40 db/ decade.

193193

• Hence total slope becomes (-20 + -40) db/ Hence total slope becomes (-20 + -40) db/ decade = -60 db/ decade decade = -60 db/ decade

• ( Note that a ( Note that a decadedecade is a frequency band is a frequency band from from ωω11 to to 1010ωω11, where , where ωω1 is any 1 is any frequency)frequency)

• ( Also note that an ( Also note that an octaveoctave is a frequency is a frequency band from band from ωω11 to to 22ωω11, where , where ωω1 is any 1 is any frequency)frequency)

• ( Octave = ( Octave = 1/0.3011/0.301decade)decade)

• Hence -60 db/ Hence -60 db/ decade = -60 x 0.3 db/ octavedecade = -60 x 0.3 db/ octave

• = -18 = -18 db/ octavedb/ octave

•

194194

• The slope of -60 db/decade will The slope of -60 db/decade will continue till the next corner continue till the next corner frequency (frequency (ωω = 10) is reached. = 10) is reached.

• The corner frequency (The corner frequency (ωω = 10) = 10) is due is due to the first order term in the to the first order term in the numerator, (1+j0.1numerator, (1+j0.1ωω) which will ) which will contribute contribute +20 db/ decade.+20 db/ decade.

• Hence total slope becomes (-60 + Hence total slope becomes (-60 + +20) db/ decade = -40 db/ decade.+20) db/ decade = -40 db/ decade.

• The slope of -40 db/decade will The slope of -40 db/decade will continue till the next corner continue till the next corner frequency (frequency (ωω = 20) is reached. = 20) is reached.

195195

• The corner frequency (The corner frequency (ωω = 20) = 20) is due is due to the first order term in the to the first order term in the denominator, (1+j0.05denominator, (1+j0.05ωω) which will ) which will contribute -20contribute -20db/ decade.db/ decade.

• Hence total slope becomes (-40 + -20) Hence total slope becomes (-40 + -20) db/ decade = -60 db/ decade.db/ decade = -60 db/ decade.

• The slope of -60 db/decade will The slope of -60 db/decade will continue for frequencies greater than continue for frequencies greater than ωω =20 =20

196196

• Step 5: calculate G(jStep 5: calculate G(jωω) H(j) H(jωω) for ) for frequencies between frequencies between ωω = 1 to 100: = 1 to 100:

• G(jG(jωω) H(j) H(jωω)= tan)= tan-1-1(0.1(0.1ωω)) -90 -9000 – –

• tantan-1-1(0.05(0.05ωω)- )- tantan-1-1 2.4 2.4ωω/(16- /(16- ωω2 2

)) ωω

rad/secrad/sec11 44 5 5 1010 2020 5050

G(jG(jωω)H(j)H(jωω

degreedegree

- 96- 96 --169169

--204204

--235235

--244244

--256256

197197

The Nyquist Criterion:The Nyquist Criterion:

198198

Mathematical background:Mathematical background:

• Consider the denominator of overall Consider the denominator of overall transfer function:transfer function:

• Do(s) = 1+G(s)H(s)Do(s) = 1+G(s)H(s)• = K(s-s= K(s-s11)(s-s)(s-s22)(s-s)(s-s33)…(s-s)…(s-smm) ) • (s-s(s-saa)(s-s)(s-sbb)(s-s)(s-scc)…(s-s)…(s-snn) )

Equation(1)Equation(1)

• Where s1,s2,…,sm are zeros of Do(s) Where s1,s2,…,sm are zeros of Do(s) and sa, sb, … sn are poles of Do(s) and and sa, sb, … sn are poles of Do(s) and s is a complex variable : s = s is a complex variable : s = σσ + j + jωω

199199

• The function Do(s) apparently will also be The function Do(s) apparently will also be complex.complex.

• Let us write Do(s) = u(s) + j v(s)Let us write Do(s) = u(s) + j v(s)• When s takes on different values, both When s takes on different values, both

u(s) and v(s) will vary.u(s) and v(s) will vary.• It is manifestly impossible to represent It is manifestly impossible to represent

the function Do(s) graphically in a single the function Do(s) graphically in a single diagram for all complex values of s.diagram for all complex values of s.

• However, since the zeros, s1,s2,…,sm and However, since the zeros, s1,s2,…,sm and the poles, sa, sb,…, sn, the poles, sa, sb,…, sn, completely completely characterizecharacterize the function Do(s), a simple the function Do(s), a simple plot in the s-plane showing the plot in the s-plane showing the location of location of zeros and poleszeros and poles, provides a neat way of , provides a neat way of specifying properties of Do(s) graphically. specifying properties of Do(s) graphically.

200200

• For each value , For each value , ssoo , of s, there corresponds , of s, there corresponds a value of Do(a value of Do(ssoo) = u() = u(ssoo) + j v() + j v(ssoo) which can ) which can be represented by a point in Do(s) plane.be represented by a point in Do(s) plane.

• If we now allow s to take on successive If we now allow s to take on successive values along a closed contour C in the values along a closed contour C in the

• s-plane , a corresponding closed contours-plane , a corresponding closed contour гг, , as shown in the next slide, will be traced.as shown in the next slide, will be traced.

• We say that the contour C in the s-plane is We say that the contour C in the s-plane is mapped mapped ontoonto the Do(s) plane as the the Do(s) plane as the contourcontour гг by the functional relationship by the functional relationship given in given in

• Equation(1)Equation(1)

201201

• Mapping of a contour C in S-plane onto a contour Mapping of a contour C in S-plane onto a contour гг in Do(s) plane in Do(s) plane

• contour Ccontour C contour contour гг

• (a) (a) s-planes-plane (b) (b) Do(s) planeDo(s) plane

jω

σ

jv

u

202202

• We know that the exact shape of contour We know that the exact shape of contour гг depends critically on that of contour depends critically on that of contour CC ; ;

• but why have we drawn contour but why have we drawn contour гг clockwise for a clockwise contour clockwise for a clockwise contour CC as as shown?shown?

• In order to answer this question we have In order to answer this question we have to pursue further by examining to pursue further by examining fourfour different cases listed below: different cases listed below:

• ((11) contour ) contour C C encircles neither zeros nor encircles neither zeros nor poles of Do(s), (poles of Do(s), (22) contour ) contour C C encircles encircles zeros contour zeros contour C C encircles poles,(encircles poles,(33) contour ) contour C C encircles poles but not zeros, (encircles poles but not zeros, (44) ) contour contour C C encircles both poles and zeros.encircles both poles and zeros.

203203

• (1) (1) Contour C encircles neither zeros nor polesContour C encircles neither zeros nor poles::

• It is clear that a closed contour in the s-plane It is clear that a closed contour in the s-plane maps onto a closed contour in the Do(s) plane.maps onto a closed contour in the Do(s) plane.

• Contours C and Contours C and гг essentially divide the s & essentially divide the s & Do(s) plane into two regions, one inside and Do(s) plane into two regions, one inside and one outside the contours.one outside the contours.

• If the contour C in the s-plane does not encircle If the contour C in the s-plane does not encircle any poles or zeros, then the contour any poles or zeros, then the contour гг in the in the Do(s)-plane cannot encircle either the origin or Do(s)-plane cannot encircle either the origin or the point at infinity. the point at infinity.

• As we trace the contour C in the clockwise As we trace the contour C in the clockwise direction (in the direction of arrow), the interior direction (in the direction of arrow), the interior

204204

• the interior of the contour which the interior of the contour which contains neither zeros nor poles is contains neither zeros nor poles is always on the right of the contour.always on the right of the contour.

• Correspondingly, in the Do(s) plane the Correspondingly, in the Do(s) plane the region which contains neither the origin region which contains neither the origin nor the point at infinity should also be on nor the point at infinity should also be on the right of the contour as the right of the contour as гг is traced.is traced.

• In other words, the interior of the In other words, the interior of the contour C in the s-plane is mapped onto contour C in the s-plane is mapped onto the Do(s) plane as the interior of the the Do(s) plane as the interior of the contour contour гг, , and if C is described in and if C is described in clockwise direction, so is clockwise direction, so is гг . .

205205

Mapping of a contour C which encircles Mapping of a contour C which encircles a zero of Do(s)a zero of Do(s)

jw

σ

s1 C

jv

u

г

(a) S plane(b) Do(s) plane

206206

Mapping of a contour C which encircles Mapping of a contour C which encircles two zeros of Do(s)two zeros of Do(s)

jw

σ

s1 C

jv

u

г

(a) S plane(b) Do(s) plane

s2

207207

• (2) Contour C encircles zeros but not (2) Contour C encircles zeros but not poles:poles:

• As we trace the contour C in the direction As we trace the contour C in the direction of arrow, the interior of the contour, of arrow, the interior of the contour, which is always on the right of the which is always on the right of the contour and which contains a zero, s1, contour and which contains a zero, s1, maps onto Do(s)-plane as the contour maps onto Do(s)-plane as the contour гг..

• гг is correspondingly traced with its is correspondingly traced with its interior, which contains the origin, interior, which contains the origin, always on its right.always on its right.

• Hence if C is described in a clockwise Hence if C is described in a clockwise direction, so is direction, so is гг . .

208208

Mapping of a contour C which encircles Mapping of a contour C which encircles a pole of Do(s)a pole of Do(s)

jw

σ

Sa

C

jv

u

г

(a) S plane(b) Do(s) plane

X

209209

• The contour The contour гг encircles the origin of the encircles the origin of the Do(s) plane Do(s) plane onceonce in clockwise direction in clockwise direction because the contour C encircles because the contour C encircles one zeroone zero in the s-plane in the clockwise direction.in the s-plane in the clockwise direction.

• In general, if the contour C encircles Z In general, if the contour C encircles Z zeros in the clockwise direction, the zeros in the clockwise direction, the corresponding corresponding

• contour contour гг encircles encircles the originthe origin of the of the Do(s) plane Z times in clockwise directionDo(s) plane Z times in clockwise direction ..

• A zero of multiplicity A zero of multiplicity kk counts as counts as kk zeros. zeros.

210210

• (3) (3) Contour C encircles poles but not Contour C encircles poles but not zeros:zeros:

• Here, as we trace the contour C in the Here, as we trace the contour C in the direction of arrow, the region on the direction of arrow, the region on the right which contains a pole, Sa, maps right which contains a pole, Sa, maps on to the on to the

• Do(s)-plane as the region also on the Do(s)-plane as the region also on the right of the contour right of the contour гг, , which contains which contains the point at infinity as the point at infinity as гг is described. is described.

• Hence interior of contour C maps onto Hence interior of contour C maps onto the exterior of the contour the exterior of the contour гг . .

• Since Since

211211

Mapping of a contour C which encircles Mapping of a contour C which encircles a pole of Do(s)a pole of Do(s)

jw

σ

Sa

C

jv

u

г

(a) S plane(b) Do(s) plane

X

212212

Contour C which encloses the entire Contour C which encloses the entire right half of the complex s- planeright half of the complex s- plane

jw

σ

(a) S plane

S ∞

S = +j∞

S = -j∞

S = 0

213213

Module 3Module 3

Analysis of Discrete SystemsAnalysis of Discrete Systems

214214

•Stability Analysis in z-planeStability Analysis in z-plane::• Juri's Stability Test:Juri's Stability Test:

• Consider the characteristic polynomialConsider the characteristic polynomial

• F(z) = aF(z) = annzznn+ a+ an-1n-1 z zn-1n-1 + … + a + … + a00 =0 =0, , aan n

>0>0

• Like Routh-Hurwitz criterion , Juri's Like Routh-Hurwitz criterion , Juri's Stability Test consists of two parts:Stability Test consists of two parts:

• 1. A simple test for necessary conditions 1. A simple test for necessary conditions &&

• 2. A test for sufficient conditions2. A test for sufficient conditions..

• The necessary conditions for stability areThe necessary conditions for stability are

• F(1) > 0F(1) > 0 , , (-1)(-1)nnF(-1)>0F(-1)>0

215215

• The sufficient condition for stability The sufficient condition for stability can be established by the following can be established by the following method.method.

• MethodMethod: :

• Prepare a table of coefficients of the Prepare a table of coefficients of the characteristic polynomial as shown:characteristic polynomial as shown:

•

216216

RoRoww

zz00

zz11 zz22 zz33 …… zzn-kn-k …… zzn-2n-2 zzn-2n-2 zznn

11 aa00 aa11 aa22 aa33 …… aan-kn-k …… aan-2n-2 aan-1n-1 aann

22 aann aan-1n-1 aan-2n-2 aan-3n-3 …… aakk …… aa22 aa11 aa00

33 bb00 bb11 bb22 bb33 …… bbn-kn-k …… bbn-2n-2 bbn-1n-1

44 bbn-1n-1 bbn-2n-2 bbn-3n-3 bbn-4n-4 …… bbk-1k-1 …… bb11 bb00

55 cc00 cc11 cc22 cc33 …… ccn-kn-k …… ccn-2n-2

66 ccn-2n-2 ccn-3n-3 ccn-4n-4 ccn-5n-5 …… cck-2k-2 …… cc00

......

2n-2n-55

……

ss00

……

ss11 ……

ss22

……

ss33

…… ……

2n-2n-44

ss33 ss22 ss11 ss00

2n-2n-33

rr00 rr11 rr22

217217

• where where • bbkk = a = a00 a an-kn-k

• aann a ak k

• cckk = b = b00 b bn-1-kn-1-k

• bbn-1n-1 b bkk

• ddkk = c = c00 c cn-2-kn-2-k

• ccn-2n-2 c ck k •

• … … … … … …•

218218

• The The sufficientsufficient conditions for stability conditions for stability are :are :

• aa0 0 < a< ann

• bb00 > b > bn-1n-1

• cc0 0 > c > cn-2n-2 n-1n-1

constraintsconstraints

• ……

• ……

• rr00 > r > r22

219219

• ProblemProblem::

• Determine stability of the system Determine stability of the system described by the characteristic described by the characteristic equation:equation:

• 2z2z44 + 7z + 7z33 + 10 z + 10 z22 +4z +1 =0 +4z +1 =0

220220

• Necessary conditionsNecessary conditions::

• F(1) = 2+7+10+4+1=24 > 0 F(1) = 2+7+10+4+1=24 > 0 satisfiedsatisfied

• (-1)(-1)44F(-1)= 2-7+10 – 4+1 > 0 F(-1)= 2-7+10 – 4+1 > 0 satisfiedsatisfied

•

RowRow zz00 zz11 zz22 zz33 zz44

11

22

33

44

55

11

22

-3-3

-1-1

88

44

77

-10-10

-10-10

2020

1010

1010

-10-10

-10-10

2020

77

44

-1-1

-3-3

22

11

221221

• Applying the stability constraints:Applying the stability constraints:

• 1 < 21 < 2 satisfied satisfied

•

• -3 > -1-3 > -1 satisfied satisfied

• 8 > 208 > 20 not not satisfiedsatisfied

• Hence system is Hence system is unstableunstable

222222

HOME WORKHOME WORK

Determine stability of the system Determine stability of the system described by the characteristic described by the characteristic equationequation::

zz44 – 1.7 z – 1.7 z33 + 1.04 z + 1.04 z22 – 0.268 z + – 0.268 z + 0.024=00.024=0

223223

DIFFERENCE EQUATIONSDIFFERENCE EQUATIONS

• Before attempting to solve difference Before attempting to solve difference equations, knowledge of the following equations, knowledge of the following theorem is vital.theorem is vital.

• The Shifting TheoremThe Shifting Theorem::

• If If ZZ[f(t)] = [f(t)] = FF(z) , then(z) , then

• ZZ[f(t+T)] = z[[f(t+T)] = z[FF(z) – f(0)](z) – f(0)]

224224

• COROLLARY 1COROLLARY 1::

• If If ZZ[f(t)] = [f(t)] = FF(z) , then(z) , then

• ZZ[f(t+2T)] = z[f(t+2T)] = z22[[FF(z) – f(0)]- zf(T)(z) – f(0)]- zf(T)

• COROLLARY 2COROLLARY 2::

• If If ZZ[f(t)] = [f(t)] = FF(z) , then(z) , then

• ZZ[f(t- nT) U(t-nT] = z[f(t- nT) U(t-nT] = z-n-n[[FF(z)] (z)]

•

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• ProblemProblem::

• Solve the difference equationSolve the difference equation

• x(k+2) - 3x(k+1) + 2x(k) = 4x(k+2) - 3x(k+1) + 2x(k) = 4k k ;;

• x(0)=0, x(1)=1x(0)=0, x(1)=1

• Solution:Solution:

• Taking the z-transform of the Taking the z-transform of the difference equation,difference equation,

[z[z22 XX(z)- z(z)- z22xx(0) – z(0) – zxx(1)]-3[z(1)]-3[zXX(z)-z(z)-zxx(0)]+ (0)]+ 22XX(z) (z)

= z/(z-4)= z/(z-4)

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(z(z22-3z+2) -3z+2) XX(z) = z(z) = z22xx(0)+z[(0)+z[xx(1)-3(1)-3xx(0)]+z/(z-4) (0)]+z/(z-4)

XX(z) = (z) = {{zz22xx(0)+z[(0)+z[xx(1)-3(1)-3xx(0)](0)]}}// (z(z22-3z+2) -3z+2)

++ zz//{{(z-4) (z(z-4) (z22-3z+2)-3z+2)}}

Substituting the initial conditions,Substituting the initial conditions,

XX(z) = z(z) = z//{{(z-1)(z-2)(z-1)(z-2)} } + + zz//{{(z-1)(z-2)(z-(z-1)(z-2)(z-4)4)}}

= [= [-z-z//(z-1)(z-1)++zz//(z-2)(z-2)] ] ++

[[z/z/3(z-1)3(z-1) --z/z/2(z-2) 2(z-2) ++ z/ z/6(z-4)6(z-4)]]

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• Taking inverse z-transform:Taking inverse z-transform:

• x(k) = [ -1 + (2)x(k) = [ -1 + (2)kk ] ] ++ [ 1/3 – (1/2) (2) [ 1/3 – (1/2) (2)kk

++

• (1/6) (4)(1/6) (4)kk ]]

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• Home WorkHome Work::

• Solve the difference equation:Solve the difference equation:

• c(k+2) + 3c(k+1) +2c(k) = u(k); c(0) c(k+2) + 3c(k+1) +2c(k) = u(k); c(0) = 1,= 1,

• c(k) = 0 for k<0 c(k) = 0 for k<0

• [[HintHint: c(1) needed in the solution can be : c(1) needed in the solution can be found by letting k=-1, thus:found by letting k=-1, thus:

• c(1) + 3 c(0) +2c(-1) = u(-1)c(1) + 3 c(0) +2c(-1) = u(-1)

• hence c(1) =hence c(1) =33 ]]

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Module 4Module 4

State Space AnalysisState Space Analysis

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• StateState: :

• The state of a dynamic system is the The state of a dynamic system is the smallestsmallest set of variables ( called set of variables ( called state variables) such that the state variables) such that the knowledge of these knowledge of these variables variables at t = at t = tt00, together with the knowledge of , together with the knowledge of the the inputinput for t for t ≥ t≥ t00, , completely completely determine the behavior determine the behavior of the of the systemsystem for any time t for any time t ≥ t≥ t00..

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• State variablesState variables::

• The state variables of a dynamic system The state variables of a dynamic system are the variables making up the are the variables making up the smallest smallest setset of of variablesvariables that determine the that determine the statestate of the dynamic system.of the dynamic system.

• If at least n variables xIf at least n variables x11,x,x22,…,x,…,xnn are are needed to needed to completely describecompletely describe the the behavior of a dynamic system (so that behavior of a dynamic system (so that once the input is given for t once the input is given for t ≥ t0 and the ≥ t0 and the initial state at t = t0 is specified, the initial state at t = t0 is specified, the future state of the system is completely future state of the system is completely determined), then such n determined), then such n variables are a variables are a set of state variables.set of state variables.

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• State vectorState vector::• If n state variables are needed to If n state variables are needed to

completely describe the behavior of a completely describe the behavior of a system, then these n state variables system, then these n state variables can be considered the n components can be considered the n components of a vector of a vector XX . .

• Such a vector is called state vector.Such a vector is called state vector.• A state vector is thus a vector that A state vector is thus a vector that

determines uniquely the system state determines uniquely the system state

X X (t)(t) for any time t for any time t ≥ t≥ t0, 0, once the state once the state at at

t = tt = t0 0 is given and input u(t) for t is given and input u(t) for t ≥ t≥ t0 0

isis specified. specified.

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• State spaceState space: :

• The n-diamensional space whose The n-diamensional space whose coordinate axes consists of the xcoordinate axes consists of the x11 axis, axis,

xx2 2 axis, xaxis, x33 axis, … , x axis, … , xnn axis is called a axis is called a state space.state space.

• Any state can be represented by a Any state can be represented by a point in point in

state space.state space.

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• State-space equationsState-space equations: : • In state-space analysis we are concerned In state-space analysis we are concerned

with three types of variables : with three types of variables : input input variablesvariables, , output variablesoutput variables, and , and state state variablesvariables..

• For any time-invariant system , the For any time-invariant system , the state state equationequation has the following general form: has the following general form:

• x(t) = A x(t) + B u(t)x(t) = A x(t) + B u(t)• and the and the output equationoutput equation is : is :• y(t) = C x(t) + D u(t)y(t) = C x(t) + D u(t)• where A is called state matrix, B is input where A is called state matrix, B is input

matrix, C is output matrix & D is direct matrix, C is output matrix & D is direct transmission matrix.transmission matrix.

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• State-transition Matrix:State-transition Matrix:• The solution of homogeneous state The solution of homogeneous state

equation equation • x = Ax can be written as x = Ax can be written as x(t) = x(t) = ΦΦ(t) (t) x(0)x(0) wherewhere ΦΦ(t) (t) is an n x n matrix and is is an n x n matrix and is

the unique solution of the unique solution of ΦΦ(t)(t) = = AA ΦΦ(t) , (t) , ΦΦ(0) =I(0) =I• • •

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• To verify this, note thatTo verify this, note that

• x(0) = x(0) = ΦΦ(0) x(0) = x(0)(0) x(0) = x(0)

• and and x(t)x(t) = = ΦΦ(t) (t) x(0) x(0)

• = = A A ΦΦ(t)(t) x(0)x(0)

• = = AA x(t) x(t)

• Hence the proof.Hence the proof.

• Also , Also , ΦΦ(t) = e (t) = e At At = ₤ = ₤ -1 -1 [ (sI-A) [ (sI-A) -1-1 ] ]

• Note that Note that ΦΦ-1-1(t) = e (t) = e -At -At = = ΦΦ(-t) (-t)

•

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• Properties of state-transition matrix:Properties of state-transition matrix:• For the time-invariant systemFor the time-invariant system• x = Axx = Ax • for which for which ΦΦ(t) = e (t) = e At At we have we have • 1. 1. ΦΦ(0) = e (0) = e A0A0 = I = I• 2.2. ΦΦ(t) = e (t) = e At At = (= (e e –At–At ) ) -1 -1 =[ =[ΦΦ(-t)(-t)] ] -1-1 • or or ΦΦ -1-1(t)(t) = = ΦΦ(-t)(-t)

• 33. . ΦΦ(t1+t2) = e (t1+t2) = e A(t1+t2)A(t1+t2) = e = e At1 At1 e e At2At2 • ==ΦΦ(t1) (t1) ΦΦ(t2) (t2) •

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• 4. [4. [ΦΦ(t)(t)]] n n = = ΦΦ((nnt) t)

• 55. . ΦΦ(t2-t1) (t2-t1) ΦΦ(t1-t0) = (t1-t0) = ΦΦ(t2-t0) (t2-t0)

• = = ΦΦ(t1-t0) (t1-t0) ΦΦ(t2-t1)(t2-t1)

•

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• Problem:Problem:

• Obtain the state-transition matrix Obtain the state-transition matrix ΦΦ(t) (t) of the following system :of the following system :

• x1 0 1 x1x1 0 1 x1

• x2 = -2 -3 x2 x2 = -2 -3 x2

• Find also the inverse of state Find also the inverse of state transition matrix.transition matrix.

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• For this system,For this system,

• A = 0 1A = 0 1

• -2 -3 -2 -3

• The state-transition matrix The state-transition matrix ΦΦ(t) (t) is is given bygiven by ΦΦ(t) = e (t) = e At At = ₤ = ₤ -1 -1 [ (sI-A) [ (sI-A) -1-1 ] ]

• since sI-A = s 0 0 1since sI-A = s 0 0 1

• 0 s - -2 -3 0 s - -2 -3

• = s -1= s -1

• 2 s+3 2 s+3

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• The inverse of (s I-A) is given by The inverse of (s I-A) is given by

• (sI-A) (sI-A) -1-1 = 1 s+3 1 = 1 s+3 1

• (s+1)(s+2) -2 s (s+1)(s+2) -2 s

• s+3 1 s+3 1

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

• = -2 s= -2 s

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

•

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• Hence Hence ΦΦ(t) = e (t) = e At At = ₤ = ₤ -1 -1 [ (sI-A) [ (sI-A) -1-1 ] ]

• = 2e = 2e –t–t – e – e-2t-2t e e-t-t – e – e-2t-2t

•

• -2e -2e –t–t +2e +2e -2t-2t -e -e-t-t +2e +2e-2t -2t

• since, since, ΦΦ -1-1(t)(t) = = ΦΦ(-t) , we obtain the (-t) , we obtain the inverse of state transition matrix as inverse of state transition matrix as follows : follows :

• ΦΦ -1-1(t)(t)==ΦΦ(-t)= (-t)= 2e 2e tt – e – e2t2t e ett – e – e2t2t

• -2e-2ett +2e +2e 2t2t -e -ett +2e +2e2t 2t

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• ProblemProblem::

• Obtain the state-transition matrix Obtain the state-transition matrix ΦΦ(t) (t) of the following system :of the following system :

• x1 -1 -0.5 x1x1 -1 -0.5 x1

• x2 = 1 0 x2 x2 = 1 0 x2

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• For this system ,For this system ,

• A = -1 -0.5A = -1 -0.5

• 1 0 1 0

• The state-transition matrix The state-transition matrix ΦΦ(t) (t) is is given bygiven by ΦΦ(t) = e (t) = e At At = ₤ = ₤ -1 -1 [ (sI-A) [ (sI-A) -1-1 ] ]

• since sI-A = since sI-A = s+1 0.5 s+1 0.5

• -1 s -1 s

• (sI-A) (sI-A) -1-1 = s+1 0.5 = s+1 0.5 -1-1

• -1 s -1 s

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• (sI-A) (sI-A) -1 -1 = = 1 s -0.5 1 s -0.5

• ss2 2 + s+0.5 1 s+1 + s+0.5 1 s+1

• = s+0.5-0.5 -0.5= s+0.5-0.5 -0.5

• ss2 2 + s+0.5 s+ s+0.5 s2 2 + s+0.5 + s+0.5

• 1 s+0.5+0.51 s+0.5+0.5

• ss2 2 + s+0.5 s+ s+0.5 s2 2 + s+0.5 + s+0.5

•

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• we have we have ΦΦ(t) = e (t) = e At At = ₤ = ₤ -1 -1 [ (sI-A) [ (sI-A) -1-1 ] ]

•

• = = ee-0.5t-0.5t(cos 0.5t – sin 0.5t)(cos 0.5t – sin 0.5t) -e-e-0.5t -0.5t sin sin 0.5t0.5t

• 2e2e-0.5t -0.5t sin 0.5tsin 0.5t ee-0.5t-0.5t(cos 0.5t – sin (cos 0.5t – sin 0.5t)0.5t)

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• For the electrical For the electrical network shown network shown determine the state determine the state model. Consider imodel. Consider i11, i, i22 and vand vcc as state as state variables. The output variables. The output variables are ivariables are i11 and i and i22..

• Current supplied by Current supplied by ee11 is i is i11 & that by e & that by e2 2 is is ii22..

i1i1

i1

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• The differential equations are The differential equations are written below: written below:

• RR11ii11 + L + L11 d dii11+ + vvcc = e= e11

• dt dt

• RR22ii22 + L + L22 d dii22+ + vvcc = e= e22

• dt dt

• And c dAnd c dvvcc = = ii11 + + ii22• dtdt

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• On rearranging above equations:On rearranging above equations:

• ddii11 = - R = - R11 i i11 – 1 – 1 vvcc + 1 e + 1 e11

• dt Ldt L11 L L11 L L11

• ddii22 = - R = - R22 – 1 – 1 vvcc + 1 e + 1 e22

• dt Ldt L22 L L22 L L22

• And dAnd dvvcc = 1 = 1 ii11 + 1 + 1 ii22• dt C C dt C C

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• The output equation is :The output equation is :

• yy11 = = ii11

• and yand y22 = = ii22

• Put xPut x11 = = ii11, x, x22 = = ii22 and xand x33 = = vvcc..

•

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• xx11 -R -R11/L/L11 0 -1/L 0 -1/L11 xx11

xx22 = 0 -R = 0 -R22/L/L22 -1/L -1/L22 xx22 ++

• xx33 1/C 1/C 0 1/C 1/C 0 xx33

• 1/L1/L11 0 0 ee11

• 0 1/L0 1/L2 2 ee22

• 0 0 00 0 0

• and and yy = = yy11 1 0 1 0 xx11

• yy22 = 0 1 = 0 1 xx22

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Home workHome work

• ProblemProblem::

• Obtain state space Obtain state space representation for representation for the electrical the electrical network shown.network shown.

253253

• Select v1(t), v2(t) and i(t) as state Select v1(t), v2(t) and i(t) as state variables.variables.

• The equations describing dynamic The equations describing dynamic behavior of the system are:behavior of the system are:

• v1- e + C1 dv1 + i = 0, at node 1v1- e + C1 dv1 + i = 0, at node 1

• R1 dtR1 dt

• C2dv2 + v2 – i = 0 , at node 2C2dv2 + v2 – i = 0 , at node 2

• dt R2 dt R2

• Ldi + v2 – v1 = 0 Ldi + v2 – v1 = 0

• dtdt

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• Rearranging the above equations Rearranging the above equations yield :yield :

• dv1 = - v1 – i + edv1 = - v1 – i + e

• dt R1C1 C1 R1C1dt R1C1 C1 R1C1

• dv2 = - v2 + Idv2 = - v2 + I

• dt R2C2 C2dt R2C2 C2

•

• di = v1 - v2di = v1 - v2

• dt L Ldt L L

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• Since v1, v2, i are selected as state Since v1, v2, i are selected as state variables, therefore, variables, therefore,

• x1 = v1x1 = v1• x2 = v2 x2 = v2 • and x3 = i and x3 = i • Hence state space representation Hence state space representation

can be obtained as :can be obtained as :• x1 = - 1 x1 – 1 x3 + 1 e x1 = - 1 x1 – 1 x3 + 1 e • R1C1 C1 R1C1R1C1 C1 R1C1• x2 = - 1 x2 + 1 x3 x2 = - 1 x2 + 1 x3 • R2C2 C2 R2C2 C2

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• x3 = 1 x1 – 1 x2x3 = 1 x1 – 1 x2

• L LL L

• In matrix form, state space In matrix form, state space representation:representation:

• x1 -1/R1C1 0 1/C1 x1x1 -1/R1C1 0 1/C1 x1

• x2 = 0 -1/R2C2 1/C2 x2x2 = 0 -1/R2C2 1/C2 x2

• x3 1/L - 1/L 0 x3x3 1/L - 1/L 0 x3

• ++ 1/R1C1 1/R1C1

• 0 e0 e

• 0 0

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• Module 4 (contd…)Module 4 (contd…)

•Solution of State Equation:Solution of State Equation:

•The solution of nonhomogeneous The solution of nonhomogeneous state equation can be obtained asstate equation can be obtained as

• X(t) = eX(t) = eAtAt X(0) + X(0) + 00∫∫t t eeA(t-A(t-гг))Bu(Bu(гг)d)dгг,,

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• x1 0 1 x1 0x1 0 1 x1 0

• x2 = -2 -3 x2 + 1 x2 = -2 -3 x2 + 1 uu

• Obtain the time response of the Obtain the time response of the above system, where u(t) is the unit above system, where u(t) is the unit step function occurring at t=0, or u(t) step function occurring at t=0, or u(t) =1(t)=1(t)

259259

• For the given system:For the given system:

• A = 0 1 B = 0A = 0 1 B = 0

• -2 -3 1 -2 -3 1

• The state transition matrix The state transition matrix ΦΦ(t) = e (t) = e At At

• = ₤ = ₤ -1 -1 [ (sI-A) [ (sI-A) -1-1 ] ] isis

• = 2e = 2e –t–t – e – e-2t-2t e e-t-t – e – e-2t-2t

•

• -2e -2e –t–t +2e +2e -2t-2t -e -e-t-t +2e +2e-2t -2t

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• The response to the unit-step input is The response to the unit-step input is then obtained as:then obtained as:

• X(t) = eX(t) = eAtAt X(0) + X(0) + 00∫∫t t eeA(t-A(t-гг))Bu(Bu(гг)d)dгг,,

• = = eeAtAt X(0) + X(0) +

• 00∫∫t t 2e 2e –(t- –(t- гг)) – e – e-2(t- -2(t- гг)) e e-(t--(t-гг)) – e – e-2(t--2(t-гг)) 0 0

•

• -2e -2e –(t- –(t- гг)) +2e +2e-2(t- -2(t- гг) ) -e -e-(t- -(t- гг)) +2e +2e-2(t- -2(t- гг)) 1 1 1 1

ddгг

•

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• x1(t) 2ex1(t) 2e-t-t – e – e-2t-2t e e-t-t – e – e-2t -2t x1(0)x1(0)

• x2(t) = -2ex2(t) = -2e-t-t +2e +2e-2t-2t -e -e-t-t +2e +2e-2t -2t x2(0) x2(0)

• + + ½ - e½ - e-t-t + ½ e + ½ e-2t-2t

• ee-t -t - e - e-2t-2t

• If the initial state is zero If the initial state is zero oror XX(0) =0(0) =0

• oror x1(0) = 0 x1(0) = 0

• x2(0) = 0 x2(0) = 0

• Then Then XX(t) can be simplified , thus:(t) can be simplified , thus:

•

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• x1(t) ½ - ex1(t) ½ - e-t-t + ½ e + ½ e-2t-2t

• x2(t) = ex2(t) = e-t -t - e - e-2t-2t

263263

Thank you Thank you