Control Systems (1-135).pdf

UNIT IPART A

1. Define system.

When a number of elements or components are connected in a sequence to perform aspecific function, the group thus formed is called a system.

2. Define control system.

When the output quantity is controlled by varying the input quantity the system is calledcontrol system.

3. Define open loop system.

The control systems in which the output has no effect upon the input quantity are calledopen-loop control system.

4. Define closed loop system?Control systems in which the output has an effect upon the input quantity in such a manner

as to maintain the desired output value are called closed loop systems.

5. Define transfer function.The ratio between the laplace transform of output and laplace transform of input is called

transfer function.

6. What is negative feedback?

If the overall gain of the s/m decreases, then it is called negative feedback.7. State Newtons second law of forces.

It states that the sum of applied forces is equal to the sum of opposing forces acting on abody.8. Define block diagram.

A block diagram of a system is a pictorial representation of the functions performed byeach component and of the flow of signals.

9. What are the elements of block diagram?

www.vidyarthiplus.com

The elements of block diagram are1. Block2. Summing point3. Branch point

10. Define signal flow graph.A signal flow graph is a diagram that represents a set of simultaneous linear algebraic

equations.

11. Define node.A node is a point representing a variable or signal.

12. Define input node.The node which has only outgoing branches are called input node. This is also known as

source node.

13. Define output node.The node which has only incoming branches are called output node. This is also called as

Sink node.14. Define non-touching loops.

If the loops does not have a common node then they are said to be non-touching loops.15. Give some basic properties of signal flow graph.

1. It is applicable to linear systems only.2. A node in the signal flow graph represents the variable or signal.3. The signal flow graph of system is not unique.

16. Find the number of forward paths in the given signal flow graph.Two forward paths

P1 = G1 G2 G3 G4 G5P2 = G6 G4 G5


17. What are the components of feedback control system?

The components of feedback is are plant, feedback path elements, error detector andcontroller.

18. Why negative feedback is preferred in control system?

The negative feedback results in better stability in steady state and rejects any disturbancesignals. It also has low sensitivity to parameter variations.

19. Compare open loop and closed loop system.

Open loop Closed loopInaccurate and unreliable.Simple and economicalStable

Accurate and reliable.Complex and costlier.Great efforts are needed to design a stable system.

20. What are the basic elements used for modeling mechanical translational system?

Mass, Spring and dashpot.

21. What are the basic elements used for modeling mechanical rotational system?

1. Moment of inertia,2. Spring3. Dashpot.

23. Name the two types of electrical analogous for mechanical system.

Force Voltage analogy.Force Current analogy.


24. What is transmittance?

The transmittance is the gain acquired by the signal when it travels from one node toanother node in signal flow graph.

25. Mention the electrical analogous of simple thermal system.

The electrical analogous of simple 1st order thermal system is RC parallel circuit.

26. What is the effect of positive feedback on stability?

The positive feedback increases the error signal and drives the output to instability. Butsometimes the positive feedback is used in minor loops in control systems to amplify certaininternal signals or parameters.

27. What are the characteristics of negative feedback?

Accuracy in tracking steady state value.rejection of disturbance signals.

28. Write the force balance equation of ideal dashpot?

bdxf B fdt

29. Write the torque balance equation of an ideal rotational mass element.2

j 2

JdT Tdt

30. Write the analogous electrical elements in force-voltage analogy for the elements ofmechanical translational system.

Force, F Voltage, CMass, M inductance, LStiffness, K Capacitance, CFrictionalcoefficient, B Resistance, R


31. Write the rule when two blocks are in cascade.

Cascade rule G1-G2 = G1 x G2

32. What are the advantages and disadvantage of open loop systems?

Advantages of open loop systems

1. The open loop systems are simple and economical2. The open loop systems are easier to construct3. Generally, the open loop systems are stable

Disadvantages of open loop systems are

1. The open loop systems are in accurable and un reliable2. The changes in the output due to external disturbances are not corrected automatically

33. What are the advantages and disadvantages of closed loop systems.

The advantages of closed loop systems are

1. The closed loop systems are accurate2. The closed loop are accurate even in the presence of non- linearities3. The sensitivity of the systems are accurate even in the presence of non- linearities

The advantages of closed loop systems are

1. The closed-loop systems are complex and costlier2. The feedback in closed loop system may lead to oscillatory response3. The feed back reduces the overall gain of the system4. Stability is a major problem in closed loop system and more care is needed to design a

stable closed loop system.

G1 G2

G1 G2


34. Give some examples of control system.

The examples of control systems are

(i) Temperature control system(ii) Traffic control system(iii) Numerical control system(iv) Position control system

35. What is a linear time variant and linear time invariant system?

If the coefficients of the differential equations describing the system are constant, then themodel is linear time invariant system.

If the coefficients of differential equations are functions of time, then the model is lineartime varying system.

36. What are Analogue systems?

Systems whose differential equation are of identical form are called analogous system.

37. What is Servo Mechanism?

A Servo Mechanism is a feed back control system used to Control position (or) its derivative.

It has the following essential features.

1. It is a closed loop system2. It is used to control position, velocity (or) Acceleration3. Its characteristics include - Automatic control , Remote operation

- Fast response , High Accuracy38. Define order of a system.

The highest power of the complex variables S in the denominator of transfer function iscalled as the order of a system.

39. What is the functions of a error detector?Error detectors are used to measure the error signal in control systems. The error is the

difference between the actual magnitude of output and magnitude of the desired output.40. Give Masons gain formula.


kk K1

FTransfer functions =

Where

FK = Forward part gain of Kth forward pathK= Number of forward path, = Determinant of graph = 1-Pm1 +Pm2 - Pm3+

Pm1 = Sum of all individual loop gains taking once at a timePm2 = Sum of gain products of all possible combination of two non-touching loopsPm3= Sum of gain products of all possible combination of three non touching loopsk = Value of after eliminating the loops which touch Kth forward path.

41. Define signal flow graph.

A signal flow graph is a graphical representation of the relationship between variable of aset of linear algebraic equations.

42. What is node?

Node is a system variable which is equal to sum of all incoming signals.

43. Define Self loop.

Self loop: A path starting from one node and terminates at same node without crossing any othernode even once.

44. Define path, Non-touching loop.

Path: It is the journey from one node to any other node in the direction of branch arrow

Non-touching loop: Loops are said to be non-touching if they do not posses any common node.

45. Give the properties of signal flow graph.


1. The algebraic equations which are used to construct signal flow graph must be in the formof cost and effect relationships.

2. Signal flow graph is applicable to linear systems.3. A node in the signal flow graph represents the variable (or ) signal.4. A branch indicates the functional dependence of one signal on the other.

46. What are the basic components of an automatic control systems?

The basic components of an automatic control systems are the following.(1) Error detector(2) Amplifier and controller(3) Actuator(4) Plant(5) Sensor

47. What are sensor?

Sensors are low power transducers, which produces output signal as a measure of thecontrolled variable.

48. What are the different types of control systems?

(i) Open loop control systems(ii) Closed loop control systems(iii) Linear and Non linear control systems(iv) Time Variant and time invariant control systems

49. What is a mathematical model? What are the different types?

A mathematical model consists of a collection of equations describing the behavior of the system.

There are two types of mathematical modeling

(i) Input / output representations describing the relation between inputs and outputs of asystem

(ii) State model describing the relation between the input states and output states of asystem.

50. What is Synchro?


Synchro is an electromagnetic transducer that produces an output voltage depending uponangular displacement.

51. What are servo motors?

The motors which are used for feedback control system are called servomotor. It convertselectrical signal into angular motion.

52. What is Synchro Transmitter?

A Synchro transmitter has star connected stator winding. The rotor is a salient pole dump-bell shaped magnet with a single winding.

53. What is a Gear train? What are electrical Analogous of gears?

A gear train is a mechanical device that transmits energy from one part of the system toanother part to attain Maximum power transfer.

The electrical Analogous of gears is Transformers.

PART B


1. Write the differential equations governing the mechanical system and determine the transferfunction.

For node M1:

M12

12

d xdt + B1

1dxdt + B

ddt (x1-x) + k1x1 + K(x1-x) = 0

Taking laplace transform,

M1s2x1(s) + B1sx1(s) + Bs[x1(s)-x(s)]+ k1x1(s)+ k[(x1(s)-x(s)] =0

x1(s)[M1s2 + (B1+B)s + (k1+k)] x(s) [Bs+k]=0

x1(s)[M1s2 + (B1+B)s + (k1+k)] = x(s) [Bs+k]

x1(s) = x(s)s

21 1 1

B KM s (B B)s (k k)

For node M2 :

M22

2

d xdt + B2

dxdt + B

ddt (x-x1) + k(x-x1) + K(x-x1) = F(t)

M2s2x(s) + B2sx(s) + Bs[x(s) x1(s)] + k[x(s) x1(s)] = F(s)

x(s) [M2s2 + (B2+B)s+k] x1(s) [Bs+k] = F(s)

Substituting for x1(s),


x(s) [M2s2 + (B2+B)s+k] x(s)2

s2

1 1 1

(B k)M s (B B)s (k k)

= F(s)

22 21 1 1 2 2 s

21 1 1

Ms (B B)s (k k ) M s (B B)s k (B k)x(s)

Ms (B B)s (k k )

2

1 1 12 2 2

1 1 1 2 2 s

M s (B B)s (k k)x(s)f(s) M s (B B)s (k k) M s (B B)s k (B k)

2. Write the equations of motion in s domain. Determine the transfer function of the system.

Solution:For node (1):

2 1 1dB (x x) kx 0dt


B2S[x1(s) x(s)]+ kx1(s) = 0

X1(s) [B2s + k] B2sx(s) = 0

X1(s) =2

2

B S x(s)B S k

For Node 2

M2

2

d xdt + B1

dxdt + B2

ddt (x x1) = f(t)



Ms2 x(s) +B1Sx(s) + B2s [x(s) x1(s)] = F(s)

Ms2 + (B1+ B2)s] x(s) - B2s x1(s)] = F(s)

[Ms2 + (B1+B2)s] x(s) B2S2

2

B SB s k x(s) = F(s)

x(s)2 2

1 2 2 2

2

[Ms + (B +B )s](B s+k)-(B s)B s k = F(s)

22 2

1 2 2 2

B s kx(s)F(s) [Ms + (B +B )s](B s+k)-(B s) ]

3. Write the differential equations governing the mechanical rotational system. Obtain thetransfer function of the system?

At J1:2

11 2

dJdt

+ k (1 - ) = T2

11 2

dJdt

+ k1 - k = T


J1s21(s) + K1(s) - K(s) = T(s)

(J1 s2 + k)1(s) - K(s) = T(s)


At node J2:

J22

2

ddt+ B

ddt+ k ( - 1) = 0

J22

2

ddt+ B

ddt+ k - k1 = 0

Taking laplace,

J2S2(s) + Bs(s) K (s) - K1(s) = 0

(J2S2+ Bs + K) (s) - K 1(s) = 0

1(s) =2

2J S Bs k (s)k

Substituting for 1(s)

(J1s2+k)2

2(J s Bs x)k

(s) - K(s) = T(s)2 2 2

2 2(J s k) (J s Bs k) k (s) T(s)k

2 2 21 2

(s) kT(s) (J s k) (J s Bs k) k

4. Obtain the transfer function of the lag network.


Ej(s) = R1I(s) +1sc I(s) + I(s) R2.

E(s) =1sc I(s) + R2 I(s).

E(s) = I(s)2

21 SCR1 R I(s)

sc sc

I (s) =0

2

SCE (s)1 SCR

Substituting in (1),

Ei(s) =0 1 2

2

SCE (s) R sc 1 SCR1 SCR sc

= E0 (s)1 2

2

1 SC[R R ]1 SCR

0 2

i 1 2

E (s) 1 SCRE (s) 1 SC[R R ]

This circuit is also called lag compensator.

5. Obtain the transfer function of the lead network shown.


Z =1 1

11

1R x Rsc1 1 SR C.R sc

Ei(s) = ZI(s) + I(s) R2 ----(1)

E0(s) = I(S) R2

I(s) =0

2

E (s)R

Sub in (1).

Ei(s) = I(s)[Z + R2]

=0

22

E (S) [Z R ]R

0 2

i 2

E (S) RE (S) Z R

Sub Z, Transfer Function =2

12

1

RR R

1 SR C

2 1

1 2 1

R (1 SR C)R R (1 SR C)

1 2 2

1 1 2 2

SR R C RR SR R C R

=SS

where1 2

1 1 2

R R1 ; =R c R R C


6. Reduce the given block diagram to its canonical form d hence obtain the equivalent transferfunction.C(S)R(S)

Solution :


1 2 3 4

1 2 1

1 2 3 42

1 2 1

G G (G G )1 G G HTransfer Function = G G (G G )1 .H1 G G H

1 2 3 1 2 4

1 2 1 1 2 3 2 1 2 4 2

G G G G G G )1 G G H G G G H G G G H

7. Obtain C(S)|R(S) using block diagram reduction rules.

1 2 3 4

1 2 1 1 2 3 4 2

G G (G G )1 G G H G G (G G )H


8. Find by using Masons gain formula.

P1 = G1 G2 G3 G4

P2 = G5 G4

1 = 1- 0 = 1

2 = 1 [- G2H1] = 1 + G2H1.

= 1 [ -G2H1 G1G2G3G4H2 G5 G4 H2] + [G2 H1G5 G4 H2]

= 1 + G2H1 +G1G2G3G4H2 + G5 G4 H2 +G2 G5 G4 H1 H2

1 2 3 4 4 5 2 1

2 1 1 2 3 4 2 5 4 2 2 5 4 1 2

G G G G G G (1 G H )C(S)R(S) 1 G H G G G G H G G H G G G H H

9. Draw the signal flow graph and find.

C(S)R(S)

C(S)R(S)


P1 = G1 G2

1 = 1- [0] = 1

= 1 [ -G1H1 G1G2 G2H2] + [G1H1G2H2]

= 1 + G1H1 +G1G2 +G2H2 + G1 G2 H1H2

TF =1 1P

1 2

1 1 1 2 2 2 1 2 1 2

G G1 G H G G G H G G H H

10. Write the differential equations governing the mechanical system. Draw the force-voltageand force current electrical analogous circuits and verify by writing mesh and node equations.

For M1,

M12

1 11 12 1 22

d x dx dB B (x x )dt dt dt

+k1(x1-x2) = f(t)

For M2,

M22

2 22 2 2 12 2 12

d x dx dB K x B (x x )dt dt dt

+k1(x2-x1) = 0On replacing the displacements by velocity,


i.e.2

2

d x dv dx; v;x vdtdt dt dt

M11

1 1 12 1 2 1 1 2dv B v B (v v ) k (v v )dt f(t)dt

M22

2 2 2 2 12 2 1 1 2 1dv B v k v dt B (v v ) k (v v )dt 0.dt

Force Voltage Analogous circuit.

The Mesh equations are,

11 1 1 12 1 2 2

1

22 2 2 2 12 2 1 2

2 1

1( ) ( , ) ( )

1 1( ) ( , ) 0

diL R i R i i i i dt e tdt cdiL R i i dt R i i i i dtdt C c

Force current Analogous circuit

The node equations are


11 1 1 2 1 2

1 12

21 2 2 2 1 2 1

2 2 1

1 1 ( ) ( ) ( )

1 1 1 1( ) ( )12

dVC V V V V V dt i tdt R R

dVC V V dt V V v v dtdt R L R L

11. Write the differential equations governing the mechanical rotational system shown. Drawthe torque-voltage and torque current electrical analogous circuits and verify by writing meshand node equations.

For J12

1 11 1 1 1 22

22

2 22 2 2 2 1 2 12

( ) ,

( ) 0

d dJ B k Tdt dt

for JdJ B K Kdt dt

on replacing the angular displacements,2

22

11 1 1 1 1 2

22 2 2 2 2 1 2 1

d dt we get

( )

, ( ) 0

d dw dB w wdt dt dt

dJ B K w w dt TdtdJ B K w dt k w w dtdt

Torque voltage Analogous Circuit:


The mesh equations are,

L1

1 21

2 2 12 1

1 11 , ( , ) ( )

2 1 12 2 ( 2 ( )

diL R i i i e tdt CdiL R i i dt i i dtdt C C

Torque current Analogous circuit

Writing the node equations,

11 1 1 2

1 1

21 2 2 2 1

2 2 1

1 1 ( , ) ( )

1 1 1( ) 0

dVC V V V dt i tdt R L

dVC V V dt V V dtdt R L L

12. Derive the transfer function of armature controlled dc motor.

The speed of DC motor is directly proportional to armature voltage and inverselyproportional to flux in field winding. In armature controlled DC motor the desired speed isobtained by varying the armature voltage. This speed control system is an electro mechanicalcontrol system. The electrical system consist of the armature and the field circuit. The mechanicalsystem consist of the rotating part of the motor and load connected to the shaft of the motor.


Let Ra = Armature resistance, La = Armature inductance Hia = Armature current, AVa = Armature Voltageeb = Back emf, VKt Torque constant NM/AT = Torque developed by motor, Nm = Angular displacement, radJ = Movement of inertial of motor and load, kg- m2/radB = Frictional coeff of motor and load, NMKb = Back emf constant VBy KVL,

iaRa+Lqia

b ad

e Vdt

Torque of DC motor is proportional to the product of flux and current. Since flux is constant inthis system, the torque is proportional to ia aloneT iaT = kt ia

The mechanical system of motor is shown below

The differential equation governing the mech, system of motor is given by,

J2

2d dB Tdt dt


The back emf of DC machine is proportional to speed of shaft,,b

de

dt

Back emf eb = b bd

e Kdt on taking laplace transform,

Ia(S)Ra +LasIa(S)+Eb(S)= Va(S) (1)T(S) = KtIa(S)JS2(S) +BSQ(S) = T(S)Eb(S) = Kbs(S)On equating, we get,KtIa(s) = JS2+BS) (S)Ia(S) = JS2 +BS) / Kt (S)Eqn (1) Can be written as,(Ra+sLa) Ia (S)+Eb(S) =Va(S)Substituting for Eb(s) and Ia(S),(Ra+sLa) JS2+BS Q(S)+ KbSQ(s) = Va(S)

-------------Kt

(Ra+sLa) JS2+BS +Kb KtS Q(s) = Va(S)-------------

Kt

2

2 3 2

2

2

( )( ) ( )( )

( )(

( [ ) ( ]

[

a a a b t

t

a a a b t

t

a a a a

t

a a b l

S KtV S R SL JS Bs K k s

KR JS R BS LaJs L BS K k s

Ks JL s JR BL S BR kBX

KJR BL BRa K k

s s SJla Jla

2

2 3 2

2

2

( )( ) ( )( )

( )(

( [ ) ( ]

[

a a a b t

t

a a a b t

t

a a a a

t

a a b l

S KtV S R SL JS Bs K k s

KR JS R BS LaJs L BS K k s

Ks JL s JR BL S BR kBX

KJR BL BRa K k

s s SJla Jla

13. Derive the transfer function of field controlled DC motor.

The speed of a DC motor is directly proportional to armature voltage and inverselyproportional to flux. In field controlled DC motor the armature voltage is kept constant and thespeed is varied by varying the flux of the machine. Since flux is directly proportional to fieldcurrent, the flux is varied by varying field current. The speed control system is anelectromechanical control system.The electrical system consists of armature and field circuit and the mech, system consists of


rotating part of the motor and load

Rf = Field resistance ,Lf = Field inductance, Hif = Field current, AVf = Field voltage VT = Torque developed by motor, NmKif = Torque constant Nm/AJ = Moment of inertia of motorB = Frictional coefficient of motorBy KVL,

Rf if +Lf +if

fVt

The torque of DC motor is proportional to product of flux and armature current. Sincearmature current is constant in this system, the torque is proportional to flux alone, but flux isproportional to field current.Tif

T = k if

The differential equation governing the mechanical system is ,2

2d dJ B Tdt dt

Taking laplace transform

RfIf(S) +LfsIf(S)=Vf(S) -(1)

T(S) = Ktf If(s) - (2)

Js2(S) +BS(S) =T(S)-(3)Equating (2) and (3)


Ktf If(S) = Js2(S) +BS(S)

If (S) = S(Js+B) / ktf (s)

The eqn (1) can be written as

(Rf+SLf) If (s) =Vf(S)

on sub, for If (s),

(Rf+SLf)S( ) ( )f

tf

Js Bs V s

K

( )( ) ( ) )

1 1

(1 )(1 )

tf

f f f

tf

ff

f

m

f m

ksV s S R sl B SJ

kSL SJSR BR B

KS ST ST

where motor gain constant, Km=tf

f

kR B

Field time constant, Tf = Lf/ Rf

Mechanical time constant, Tm = J/B.

14. Use Masons gain formula for determining the overall T.F. of the system shown.

Solution:

Maisons Gain formulakp

=kT


k = No. of forward path

Pk Forward path gain of kth forward path

- 1 - sum of individual loop gain +sum of possible combination

+of two Nontouching loop gain

Sum of possible combination of three Non touching loop gain .....

d. Forward Path Gain:

P1 = G1 G3 G4 G6 G7

P2 = G1 G3 G4 G6 G7

P3 = G1 G2 G5 G6 G7

P4 = G1 G3 G5 G6 G7 k = 4

b) Individual loop Gain:


L1 = G6 H1 L2 = G7H2

L3 = G2 G4 G6 G7 H3

L4 = G3 G4 G6 G7 H3

L5 = G3 G5 G6 G7 H3

L6 = G2 G4 G6 G7 G5 H3


C. Two Non touching Loop Gain:

L12 = G6 H1 G7 H2

d)

1

2

3

4

= 1 - 0 = 1 = 1 - 0 = 1

1 - 0 = 1 = 1 - 0 = 1

1 1 2 2 3 3 4 4P =

P P PT

1 2 3 4 5 6 12 = 1 - L L L L L L L

6 1 7 2 2 4 6 7 3 3 4 6 7 3

3 5 6 7 3 2 4 6 7 5 3

6 1 7 2

G = 1 -

.....

H G H G G G G H G G G G HG G G G H G G G G G H

G H G H

Over all gain of system is

1 2 4 6 7 1 3 4 6 7

1 3 5 6 7

6 1 7 2 2 4 6 7 3 3 4 6 7 3

3 5 6 7 3 2 4 6 7 5 3 6 7 1

G 1 11

=

1 G G

G G G G G G G G GG G G G G

TFG A G H G G G G H G G G G H

G G G H G G G G H G G H

15. Obtain the Transfer function of the mechanical system shown.

Step: 1


Draw the freebody diagram based on Mass element (M1)

According to Newtons Law

Fk1+fB1+fm1+fB+fk = 0 ____ (1)

Let

Fk1 Opposing force offered by spring K1

FB1 Opposing force offered by Dashpot B1

FM1 Opposing force offered by Mass M1

FB Opposing force offered by Dashpot, B

Fk Opposing force offered by spring, k.

fk1 = K1X1 ;

211

1 1 B2

11 1 k 1

; f

; f

m

B

Bd x xd xf Mdfdf

dxf B k x xdf

Substitute above valve in equ (1)

K1X1+M1 2 11 11 12 x xd x dxBd B k x xdt dfdf

Taking LT on both sides

K1X1(s)+M1S2X1(s)+BS(X1(s)-X(s)-X(s)+B1SX1(s)+ K(X1(s)-X(s) = 0

21 1 1 1( ) - BS+KM S BS B S K K X S X S 1 21 1 1

s

MsB K XX s

S B B S K K


Step: 2

Draw the free body diagram based on Mass M

f (t) = fm2 + fB2 + fB + fk

2 12 2 12Bd x xd x dxf t M B k x x

d df df

Taking LT on both sides

22 2 1

1

22 2 2

1 1 1

( )

s

F s M S x s B Sx s BS x s x s

k x s X s

BS k Bs k x sF s M S B S BS K X s

M S B B K K

2

1 1 122 2 2

21 1 1

( )( )

sM S B B k kM S B S BS k

BS KF s x s

M S B B S K K

T.F.

Draw Analog electrical circuitsForce Voltage Analog y

21 1 1

22 22 2 1 1 1

X s M S B B S K KF s M S B B S K M S B B S K K BS K


1 1 1 1 1 1

2 2

1

L B R K F V L B R K C X i

X i

M CM

Force current Analog y

1 1 1 1 1 12 2 2 2 1 1

2

K L B R F i =iM C k L B R x V B R x V

M C

16. The network shown figure modifies the error signal. It voltage Vi of a Servomechanism to a. Find the T.F. Vo/Vi neglecting any load on the output terminals. Evaluate the function (a) p ra sine signal voltage of 1.0 v at angular frequency w = 20 red/sec and b) for a step function inputsignal voltage of 1.oV.

Solution:


By voltage divider theorem 2 1

21

21 1

o

i

V s RRV s R RRR cs

S = j w

2 1

21 1

o

i

V jw RRV jw R jR cw

w = 20 rad/sec c = 0.2 uFR1 = 200k R2=20 k

W C R1 = 20 x 0.2 x 10-6 x 200 x 103

R1/(1jR1CW) = 200/(1+j1)

0

0

2020020

1

0.128 39.805

j

i

VV

jiV vV

(b)0 2

12

1

( )( )

1i

V s RRV s R

R cs

0 ( ) 25( ) 275i

V s SV s S

Vi (s) = (S+25)/S(S+275)

= 2220

1 1 10111 275

Inverse LT.V ( ) 0.1 0.9 t

s STaking

t e

17. The solenoid shown in Fig. Produces a magnetic force proportional to the current I in theV0(t)=0.1 [1+9e-222t]


coil = kii. The coil has resistance (R) and inductance (L). Write the differential equation.Determine the transfer function 2

x sE s

.

Figure:Solution:

The solenoid connected in the left end of the system develops a force proportional to thecurrent passing through it i.e., = ki i. The solenoid has a mass M1 and has a frictional coefficientB1. The coil also has inductance L and resistance R.

When an emf e(t) is applied to the solenoid then a current i(t) results.

1

di te t L Ri t

dtf t k i t

Total anticlockwise moment = 1f t lTotal anticlockwise moment

22 2

2 1 12 2 2 2 2 1 1 12 2

dx dx d xd xM k x B l B M ldt dt dt dt

we know that for a balanced torque system total anticlockwise moment is equal to total clock wisemoment.


2 22 2 1 1

1 2 2 2 2 2 1 1 12 2

2 22 2 2 1 1

2 2 2 2 1 12 21

21 1 2 2

1 1 2 2 221

d x dx dx d xf t l M k x B l B M ldt dt dt dt

l d x dx dx d xf t M B k x B Ml dt dt dt dt

d x dx l dxf t M B B k xdt dt l dt

To find the transfer function 2X sE s

Taking Laplace transform on both sides of Eq. and we have

i

i

E sE s R sL I s I s

R sLF s k I s

k E sR sL

Taking Laplace transform on both sides of Eq. 2 221 1 1 2 2 2 2

2

lF s M s B s X s M s B s k X sl

we can find from the geometry of the Fig. that

1 1 2 2 1 2

11 2

2

l / x l / x for small value of x & xlX s X sl

Substituting Eq. and Eq. in Eq. we have

i 2 21 21 1 2 2 2 2 2

2 1

2

2 21 22 1 2 2 2

2 1

k E s l lM s B s X s M s B s k X sR sL l l

X (s) kE s l lR sL M s B s M S B s k

l l


18. For the spring, damper and mass system shown in figure find the differential equationsgoverning the system.

Figure:Solution:The forces acting on mass M1 are shown in figure.

Where fk1=k1(x1-x2)b1 1 1 2

21

m1 1 2

df B (x -x )dtd xf Mdt

The force balance equation in mass M1 is

m1 b1 k12

11 1 1 2 1 1 22

f f f 0

d x dM B (x -x )+k (x -x )=0dt dt

The forces acting on mass M2 are shown in figure. The force balance equation is

22 2

2 2 2 2 1 2 12

1 2 1

d x dxM k x B k x xdt dtdB x x f tdt

fb1

M1

fml1

fk1

Tx1


Figure:Exercise 1.1

Obtain the differential equations of the following mechanical systems

(i) Figure:

(ii) Figure:

19. Using the block diagram reduction technique, find C/R.

Solution: I Reduction: moving takeoff point from point A to B. i.e., along the direction of flow.So divide the takeoff point path by block gain G2.

M1

f(t)

fB1

fb2fm2fk1

fk1


II Reduction:

1. Eliminating two blocks in parallel by a single block.2. Cascading two blocks.

III Reduction:

1. Two blocks in cascade.2. Move summing point ahead of block H3 along direction of flow.

IV Reduction: Eliminating summing point by multiplying signs.


V Reduction: Reduce feedback loop.

G=G1G2; H = H1 H3. Type of feedback is negative.

VI Reduction: Cascade two blocks and reduce feedback loop.

51 24 3 2 3

1 2 1 3 2

GG GG G G ; H =H H1 G G H H G

Type of feedback is positive.

Feedback path reduction= G1 GH

1 2 3 4 1 4 5

1 2 1 3 1 2 3 4 2 3 1 4 5 2 3

1 2 3 4 1 4 5

1 2 1 3 1 2 3 4 2 3 1 4 5 2 3

G G G G G G GR S C S1 G G H H G G G G H H G G G H H

C S G G G G G G GR S 1 G G H H G G G G H H G G G H H

20. Using Masons gain formula, find C/R of the signal flow graph shown in figure.

Figure:Solution:

By Masons gain formula, Transfer functionk

k k1

F


Forward path Fk:

Figure:

F1 = 1 2 3 4G G G G ; [R-X1-X2-X3-X5-X6-C]F1 = 1 5 8 4G G G G ; [R-X1-X2-X4-X5-X6-C]

Determinant of graph: m1 m2 m3.1 P P P Pm1=Individual loop gains

Figure:

P11=G1 G2H1; [X1-X2-X3-X1]P21=G3 G4H2; [X3-X5-X6-X3]P31=G1 G5 G8 G4H1H2; [X1-X2-X4-X5-X6-X3- X1]P41=G5 G6; [X2-X4-X2]P51=G7; [X4-X4] Self loop.

pm2=Sum of product of all possible combinations of two non-touching loops.

P12=P11* P51= G1 G2 G7 H1


P22=P21* P41= G3 G4 G5 G6 H2P32=P21* P51= G3 G4 G7 H2

pm3 = Sum of product of all possible combination of three non-touching loops. pm3 = 0

11 21 31 41 51 12 22 32

1 2 1 3 4 2 1 5 4 8 1 2 5 6 7

1 2 7 1 3 4 5 6 2 3 4 7 2

1 P P P P P P P P

1 G G H G G H G G G G H H G G G +G G G H +G G G G H +G G G H

Calculation of k:

1 = 1-G7; as all loops touch forward path F1 except self loop.

2 = 1; as all loops touch forward path F2

1 1 2 2

1 2 3 4 7 1 4 5 8

1 2 1 3 4 2 1 4 5 8 1 2 5 6 7

1 2 7 1 3 4 5 6 2 3 4 7 2

F FTransfer function =

G G G G 1 G G G G G 1T.F

1 G G H G G H G G G G H H G G G +G G G H +G G G G H +G G G H

21. Write the equations of motion in s-domain for the system shown in figure. Determine thetransfer function of the system.

FigureSOLUTION

Let, Laplace transform of x(t) = X(s)Laplace transform of f(t) = F(s)

Let x1 be the displacement at the meeting point of spring and dashpot. Laplace transformof x1 is X1(s).


The system has two nodes and they are mass M and the meeting point of spring anddashpot. The differential equations governing the system are the force balance equations at thesenodes. The equations of motion in the s-domain are obtained by taking Laplace transform of thedifferential equations.

The free body diagram of mass M is shown in figure. The opposing forces are marked asfm, fb1 and fb2.

21 12

2 2 1

;

( )m b

b

d x dxf M f Bdt dtdf B x xdt

Figure:

By Newtons second law the force balance equation is1 22

1 2 12

( )( ) ( )

m b bf f f f td x dx dM B B x x f tdt dt dt

On taking Laplace transform2

1 2 12

1 2 2 1

( ) ( ) [ ( ) ( )] ( )[ ( ) ] ( ) ( ) ( )Ms X s B sX s B s X s X s F sMs B B s X s B sX s F s

The free body diagram at the meeting point of spring and dashpot is shown in figure. Theopposing forces are marked as fk and fb2.

2 2 1 k 1

b2

2 1 1

( ); f newton's second law, f 0

( ) 0

b

k

df B x x KxdtBy f

dB x x Kxdt


Figure:On taking Laplace transform

2 1 12 1 2

21

2

[ ( ) ( )] ( ) 0( ) ( ) ( ) 0

( ) ( )

B s X s X s KX sB s K X s B sX s

B sX s x sB s K

Substituting for X1 (s) from equation2 2

1 2 22

2 21 2 2 2

22

2 21 2 2 2

[ ( ) ] ( ) ( ) ( )[ ( ) ]( ) ( )( ) ( )

( )( ) ( ) ( ) ( )

B sMs B B s X s B s X s F sB s KMs B B s B s K B sX s F sB s K

X s B s KF s Ms B B s B s K B s

RESULT

The differential equations governing the system are

1.2

1 2 12 ( ) ( )d x dx dM B B x x f tdt dt dt 2. 2 1 1( ) 0dB x x Kxdt

The equations of motion in s-domain are

1. 2 1 2 2 1[ ( ) ] ( ) ( ) ( )Ms B B s X s B sX s F s 2. 2 1 2( ) ( ) ( ) 0B s K X s B sX s

The transfer function of the system is

22 2

1 2 2 2

( )( ) [ ( ) ]( ) ( )

X s B s KF s Ms B B s B s K B s


UNIT IIPART A

1. What do you mean by time response of the system?It is the output of closed loop system as a function of time. It is denoted by c(t).2. What are standard test signals?The standard test signals are1. Step input2. Ramp input3. Parabolic input4. Impulse input5. Sinusoidal signals

3. Define ramp signal.The ramp signal is a signal whose value increase linearly with time from an initial value of zero att=0.4. Define impulse signal.A signal which is available for very short duration is called impulse signal.5. Define order of a system.The order of the system is given by the order of the differential equations governing the system6. Define damping ratio.The damping ratio is defined as the ratio of the actual damping to the critical damping.7. How the system is classified based on depending ratio?The system is classified into 4 types1. Undamped system2. Underdamped system3. Critically damped system4. Over damped system


8. What are time domain specifications?The time domain specifications areDelay timeRise timePeak timeMaximum overshootSetting time, ts

9. Define delay time.

It is the time taken for response to reach 50% of the final value, for the very first time

10. Define risetime.

It is the time taken for the response to raise from 0 to 100% for the very first time.

11. Define Peaktime.

It is the time taken for the response to reach the peak value for the very first time.

12. Define peak overshoot.

It is defined as the ratio of the maximum peak value measured from final value to the final value

13. Define settling time.

It is defined as the time taken by the response to reach and stay within a specified error.

14. Define steady state error.The steady state error is the value of error signal e(t), when t tends to infinity.15. What is the drawback of static error coefficient?The drawback in static error coefficients is that it does not show not show the variation of errorwith time and input should be a standard input.16. What is transient and steady state response?

The transient response is the response of the system when the input changes from one stateto another. The response of the system as t is called steady state response.17. What is the importance of test signals?


The test signals can be easily generated in laboratories and are used to predetermine theperformance of the system.

18. Define step signal.

The step signal is a signal whose value changes from o to A and remains constant at A for t>0.

19. What will be the nature of response of a second order system with different types ofdamping?

(1) Undamped system - Oscillatory2) Underdamped system - damped oscillatory3) Critically damped - exponentially rising4) For overdamped - exponentially rising

20. What is type number of a system? What its significance?

The type number is given by number of poles of loop transfer function at the origin. The typenumber of the system decides the steady state error.

21. What are static error constants?

The kp, kv and ka are called static error constants. These constants are associated with steady stateerror in a particular type of a system and for a standard input.

22. Define velocity error constant.

The velocity error constant kv = l+ S G(S)H (s) The steady state error in type-1 system for unitramp input is given by 1 /kv

23. Mention two advantages of generalized error constants over static error constants.

1. Generalized error series gives error signal as a function of time.2. Generalized error constant is used to determine the steady state error for any type of input.

24. What is the effect on system performance when a proportional controller is introduced in a


system?

1. It improves the steady state tracking accuracy relative stability and disturbance signalrejection

2. Increases loop gain of the system.

25. What is the disadvantage in proportional controller?

The disadvantage is that it produces a constant steady state error.

26. What is the effect of PI controller on the system performance?

1. The PI controller increases the order of the system by one, which results in reducing thesteady state error.

2. The system becomes less stable

27. What is the effect of PD controller on the system performance?

The effect of PD controller is to increase the damping ratio of the system and so the peakovershoot is reduced.

28. Why derivative controller is not used in control system?

The derivative controller produces a control action based on rate of change of error signal and itdoes not produce corrective measures for any constant error.

29. A second order system bas a dumping ratio of 0.6 and natural frequency of oscillation is 10rad/sec. determine the damped freq of oscillation

2

2

1

10 1 (0.6)8 / sec.

d n Z

rad

30. For the system given, find the type and order of the system.2( ) ( ) ( 1)( 6 8)

kG s H sS S S S

Type = 1Order = 431. The damping ratio of a system is 0.75 and the natural frequency of oscillation is 12 rad/sec.


determine peak time2

2

p

1

12 1 (0.75)7.94 rad/sec

t7.94

0.396 sec

n

d

d Z

32. A unity feedback system has a open 100p transfer function of10( ) ( 1)( 2)G S S S Determine

the steady state error for unit step input.0 0

10( ) ( 1)( 2)5

11

16

s sp

p

p

K L G S LS S

k

lssk

33. Give the expression for rise time (Apr 96).

2-1

0

1 = tan

r

d

t

zwhere

z

34. Give the expression for peak overshoot (Apr -96).

2% 100

1p

zM e xZ

Where z = damping ratio.

35. What are the types of controllers that are used in closed loop system? Apr 98).

Proportional Integral controller (pI)Proportional +derivative controller (PD)Proportional+ Integral+ derivative controller (PID)


What is the difference between type and order? (April -98)

Type order

i) Specified for loop TF i) Specified for any transfer functionii) Given by number of poles ii) Given by number of lying at the origin

poles of transfer function

36. Define positional error coefficient (Apr 97).

Kp = L+ G(s)H(s)s0

37. What is the effect of PD controller on the system performance?

1. Increases damping ratio2. Reduces peak overshoot3. Reduces settling time

38. What are the demands of good time response?

1. Less Settling time2. Less overshoot3. Less rise time4. Smallest steady State error.

39. What is rate feedback controller?

This is achieved by feeding back the derivative of output signal internally using a tachogeneratorand comparing with signal proportional to error.

40. What is step signal ?

The step signal is a signal whose value changes from zero to A at t=0 and remains constantat A for t>0. The mathematical representation of step signal is

r(t) =Au(t)Where u(t) =1 for t 0

U(t) = 0 for t< 041. What is a parabolic signal?


The parabolic signal is a signal whose value varies as a square of time from an initial valueof Zero at t=0. This parabolic signal represents constant acceleration input to the signal. Themathematical representation of parabolic signal in

2Atr(t) ;t 02

0;t 0

42. What is positional Error Coefficient?

Steady state Error for a step input isP

11 K where KP is the positional error coefficient. The

positional Error coefficient is giver by P s 0K limG(s)H(s)

43. Enumerate the advantages of generalized error coefficient.

The advantages of generalized Error coefficients are

1. It gives error signal as a function of time.2. It can be used to determine the steady state error for any type of input.

44. A system has a transfer function 2C(s) 4R(s) S 1.6S 4

For the unit step response, the settling time for 2% tolerance band is

s n nn

s

4t ;Given w 2; 2 w 1.6w

4t 5sec and =0.40.4 2

45. What are the roles of controller in Control systems?

The main role of controller is to modify the error and to achieve better control action.

46. What is a proportional Controller?

The controller which produces the output signal in proportional to the error signal is calleda proportional controller. The transfer function of proportional controller in KP.

47. What is a PI Controller?


PI Controller is a proportional plus integral controller which produces on output signalconsisting of two terms one proportional to error signal and other proportional to the integral ofthe error signal.

48. What is PD controller?

PD Controller is a proportional plus derivative controller which produces an output signalconsisting of two terms, one proportional to error signal and other proportional to the derivativeof the error signals.49. For the following differential equation 2 2d y dy2 4 8y 0dtdt Determine the damping ratio.

2

2

d y dy2 4 8y 0dtdt

on taking Laplace Transform,

2S2Y11(s)+4sy1(s)+8y(s)=0

S2Y11(s)+2Sy1(s)+4Y(s)=0Compare the above equation with the standard form of second order characteristic equation.S2+2Wns+Wn2=0Wn2=4Wn=2rad/sec.2wn=2=

n

1 0.5w

50. What in the type and order of the given system is whose open loop transfer function?

The number of poles lie at origin of S-plane gives the types of system.The highest power of the complex variables S in the denominator of transfer function is called asthe order of a system.Type 1Order 2

51. A first order system is shown in fig below. What is its time response to a unit step input.


Solution:C(s) 1R(s) 1 ST

52. Give the relation ship between static and dynamic error coefficients.

op

1v

2a

1C1 K

1CK1C

K

53. Give the steady state error for step and velocity input .

PART B1. Obtain the response of first order system for unit step input

ssP

ssv

1e for step input =1+K

1e for velocity input=K


The closed loop transfer function of first order system

( ) 1 ----(1)( ) 1

c s

R s s

If the input is unit step input, r(t) =1 and R(S) = 1/s

Response in S-domain, c(s) = 1 / 1+zs1 1

,

11

(1/ )1/

( 1/ )

s zs

sz z s

z

S S z

By Partial fraction,1( )

1/( 1/ )

1 ( 1/ )

s= 0

1 (1/ )

1

A BC s

z S S zS S z

A s z Bsz

Put

A zz

A

S= -1/z

1/z =B(-1/Z)B=-1

( ) 1 11/

Put

C Ss s z

Taking inverse laplace transform,

1 1

/

1 1( )1/

( ) 1 t zC t L L

S S ZC t e

2. Derive the response of undamped second order system for unit step input.


The standard form of closed loop transfer function of second order system is,2

2 2

2

2 2

( )( ) 2

undamped system, Z = 0C(S)R(S)

s

C S WnR S S zwn wnFor

WnS Wn

when the input is unit step input, r(t) = 1 of R(s) = 1/s2

2 2

2

2 2

( ) ( ),

1

wnC s R Ss wn

wn

s s wn

By partial fraction,2

2 2 2 2( ) ( )wn A Bs cC s

s s wn s wn s

wn2 = A(s2+wn2) + Bs+C)sput S = 0

w2 = A(wn2)A = 1

Equating coefficient of s20 = A+BB= -1

Equating coefficient of S

O=C2 2

1 (1)( )n

sC Ss s w

Taking inverse laplace transform,


c(t) = L-1 (1/s) L-1 2 nns

s w

c(t) = 1 cos wnt

3. Obtain the response of second order underdamped system with unit step input.

The standard form of second order system is2

2 2( )( ) 2

n

n n

wC sR s S zw s w

we know r(t) = 1R(S) = 1/sWe know r(t) =1R(S) = 1/s

2

2 21( )

2 n nwnC s

s s zw s w

Applying partial fraction,

2 2

2 2 2

2 2

( )2

( 2 ) ( )0( )

1

s

n n

n n n

n n

A Bs CC ss s zw w

w A s zw s w Bs C SPutsw A w

A

Equating coefficients of s20 = A(1) + BB = -1

Equating coefficient of S,0 = 27 wn A+CC = -27wn

2 221( )

2n

n n

s zwC ss s zw s w

Adding and subtracting 22 nz w to the denominator of second term,


t td

212 2 2 2 2 22

212 2 2 2 2( 2 ) ( )

212 2 2( 2 ) (1 )

( ) [1 coswd - sinw ]zwnt zwntnd

S zwns S zw s w z w z wn n n n

S zwns S zw s zw w z wn n n n

S zwns S zw w zn n

zwc t e e

w

2 2

2 2

2n

2 2 2 2

21( 2 ) (1 )

21( ) , wd = w 1-z

21( ) ( )

n

n n

n

n d

n n

n d n d

S zws S zw w z

S zws S zw w

whereS zw zw

s S zw w s zw w

Taking L-1zw

t n( ) [1 cosw - sind wd

1 [cos sin ]21

1 [cos sin ]21

21 [ 1 cos sin ]21

21 [sin cos 1 ]21

1 (sin cos cos21

zwnt zwnt tC t e e wd

zwnzwnt t te w wd d

wn z

zzwnt t te w wd d

zzwnt

e t tz w z wd d

zzwnt

e t tw z w zd d

zzwnt

e twd

z

sin ](Re note)w tz fered

1 sin( )21

21( = tan + )

zwnte t

wdz

zwhere

z

Note:


1-z2 SinQ = 1-z21 cos = Z

tan = 1-z2/ z

z

4. Desire the response of second order critically damped system with unit step input.

The standard form of closed loop transfer function is2

2 2( )( ) 2 n n

C S wnR s s zw s w

for critical damping z=12

2 2( )( ) 2 n n

C S wnR s s zw s w

=2

2 2( )n

n

w

s w

for unit step input, R(s) = 1/S2

2

2

2

1( ) ( )

( ) ( )

n

n

n

n

wC Ss S w

wC SS S w

Applying partial fraction,2

2 2( ) ( )n

n n n

w A B CS s w S S w S w

wn2 = A (S+wn)2+B (S) (S+wn) + CSPut S = 0,Wn2 = A[wn)2A =1Put S = - wnW2n = C (-wn)C = -wn


Put S = 1Wn2 = A(1+wn)2 +B(1+wn)+C(1)Wn2 = 1+wn2+2wn+B(1+wn) +-wnWn2 = 1+wn2+wn+B(1+wn)B(1+wn) =- wn2 +1+wn2+wnB(1+wn) = = 1+wn

B= 1

c(s) = 21 1

( )n

n n

w

s s w s w

Taking L-1C(t) = 1-e-wnt (1+wnt)

5. Derive the expression for rise time Response of second order system for underdamped caseis,

2

2

2

( ) 1 ( )1

t = tr, c(t) = c(tr) =1

( ) 1 ( ) 11

sin( ) 01

0,sin( ) 0

zwnt

d

zwntr

r d r

zwntr

d r

zwntrd r

eC t Sin w tz

Ate

c t Sin w tz

ew t

z

Since e w t

sin = 0, =0, , 2wdtr + = wdtr = -

2-1 1

=tan

d

trw

zwhere

2d

21

2

w 1

1tan

1

n

r

n

and w z

z

ztw z


6. Drive the expression for peak time.

Differentiate C(t) with respect to t and equate to 0

i.e 2

( ) 0

( ) 1 ( 0)1

t tp

n pd

dc t

dte zw t

C t Sin w tz

differentiating w, r, t t,

22

2

2d n

2

2

2

2

2

( ) ( sin( )1

cos( )1

w = w 1-z

( ) ( )sin( )1

[ sin( ) 1 7 cos( )1

1cos( )

1

[ s1

zwnt

n

zwnt

d d

zwnt

n d

zwntz

n d d

zwntd

zwnt

n

d ec t zw w t

dt ze

w t wz

Put

d ec t zw w t

dt ze

w e z w w tz

wn ze w t

z

ew e z

z

in( ) 1 7 cos( )z

d dw w t

2

2

2

[cos sin( ) cos( )]1

[sin( ) cos cos ( )sin ]1

[sin( ) ]1

t

t

tnd d

e zwn

nd d

zwnnd

we zwn w t sin w t

z

ww t w t

z

we w t

z


=2

2

2

sin1

d t = tp, ( ) 0

dt

sin 01

0 sin 0sin 0, 0, , 2 .....

1

zwntnd

zwntnd p

zwntpd p

d p

d

p

n

we w t

z

at c t

we w t

z

e w t

w t

tpw

tw z

7. Derive the relationship between the maximum overhead and damping ratio % peakovershoot,

% Mp( ) ( ) 100( )

c tp cx

C where c(tp) = peak response at t = tp c() = Final steady state

value

The unit step response of the second order s/m is given( ) 1 sin( )

21-

-e t= , c(t) = c( ) = 1 sin( )21

1 0

( ) 1-zw tn p1-e

t = tp c(t) = c(t ) = ( )p 21-z-zw tn p1-e

t / c(t ) ( / )p p 21-z-zwn1-e sin( )

21=

1-

zwnte

c t wdtz

At wdtz

c

At Sin w t pd

Put wd Sin w wd d

w zn

21 100

2z

2-z / 11-e

sin ( sin ( + )=-sin0121-z

z1+e- 2 21-zc(t ) 1p 21-z

2c(t ) 1+e -z / 1p

( ) ( )% 100

( )

121 1 100

1

%

z

z

z

c t cpM xPc

e

zx

Mp e


( ) 1 sin( )21

-

-e t= , c(t) = c( ) = 1 sin( )21

1 0

( ) 1-zw tn p1-e

t = tp c(t) = c(t ) = ( )p 21-z-zw tn p1-e

t / c(t ) ( / )p p 21-z-zwn1-e sin( )

21=

1-

zwnte

c t wdtz

At wdtz

c

At Sin w t pd

Put wd Sin w wd d

w zn

21 100

2z

2-z / 11-e

sin ( sin ( + )=-sin0121-z

z1+e- 2 21-zc(t ) 1p 21-z

2c(t ) 1+e -z / 1p

( ) ( )% 100

( )

121 1 100

1

%

z

z

z

c t cpM xPc

e

zx

Mp e

by88

8. Derive the equation for setting time (ts).

The response of second order system has two components, they are,

1. Decaying exponential component, 21nzw te

z

2. Sinusoidal component, sin(wdt+)

In this decaying exponential term dampens (or) reduces the oscillations produced bysinusoidal component. Hence the setting time is decided by the exponential component. Thesettling time can be found out by equating exponential component to percentage tolerance errors.For 2% tolerance, at t = ts;

20.02

1

n szw te

z

for least values of z,e-zwnts = 0.02-zwnts = ln 0.02-zwnts = -4ts = 4 / Zwn= 4T,


where T=1 /zwn = Time constant of the system

For 5% error,e-zwnts = 0.05-zwnts = ln(0.05)-Zwnts = -3ts = 3 / zwn=3T,where T = 1 / zwn

9. A unit step input is applied to the unity feedback system for which open loop transfer

function16( ) ( 8)G s S S . Find

1. Its closed loop transfer function2. Natural frequency of oscillation3. Damping ratio z4. Damped frequency of oscillations, wd

Solution

(i)

2

( ) ( )) ( ) 1 ( ) ( )16( 8)

161 ( 8)168 16

c s G siR s G s H s

s s

s s

s s

ii) The standard equation is,S2+2zwns+wn2Comparing S2+8S +16 with standard equation,Wn2 = 16 2zwn =8Wn =16 z= 8 / 8= 4 rad / sec z = 1iv) wd = wn 1-z2= 41-1wd = 0 rad/sec


10. A second order system is given by 2( ) 25( ) 6 25

C SR S S S

Find its risetime, peak overshoot andsettling time if subjected to unit step input. Also calculate expression for its output response.

Solution:

Comparing transfer function with2

2 22n

n n

w

S zw s w Wn2 = 25 2zwn = 6wn = 5

21

21

6 0.610

1tan

1 0.6tan

0.6

z

z

z

=0.9272 radians.wd = wn1-z2

=51-0.62= 4 rad/sec

0.92724

0.5535sec

rt

wd

p 2

40.785 sec

%M 1001

pd

tw

ze x

z

= 9.48/


n-zw t

2

3

2

3

4 1.33 sec

1-eC(t) = sin( )1-z

1sin(4 0.9272)

1 0.61 1.5625 sin(4 0.9272)

s

n

d

t

t

tzw

w t

et

e t

11. Obtain the response of unity feedback system whose open loop transfer function is4( ) ( 5)G s andS S when the input is unit step.

Solution:

2

( ) ( )( ) 1 ( )

4( 5)

41 ( 5)45 4

( ) 4( ) ( 4)( 1)1 4

( 4)( 1)

C S G SR S G S

S S

S S

S SC sR s S S

S S S

By partial fraction,( )

4 14 ( 4)( 1) ( )( 1) ( )( 4)

S = 0,4= A(4) (1)

A=1

A B CC sS S S

A S S B S S C S SPut

Put S = -44= B(-4) (-4+1)= B (-4) (-3)4 = 12B


B = 1/3Put S = -14= C(-1)(-1+4)=C(-1)(3)4=-3C

C =-4/3

1

4

1 1/ 3 4 / 3( )4 1

1( ) 1 4 / 33

t t

C Ss S s

Taking L

C t e e

12. For servomechanisms with open loop transfer function given below explain what type ofinput signal give rise to a constant steady state error and calculate their values.

20( 2) 101) ( ) 2) G(S) =( 1)( 3) (S+2)(S+3)SG S

S S s

103) ( ) 2( 1) 2)

20( 2)) ( )( 1)( 3)

( ) H(s)S 0

20(S+2)=L+S, (1)

S(S+1)(S+3)20(2) 40

3 31 3

40

0.075

10( ) ( )( 2)( 3)

G(S)+(S)s-->0

10=L+

(S+2)(s+3)0

10 5 =

6 31 1

1 1

G SS S S

si G s

s s s

K L SG Sv

ess Kv

ii G SS S

K Lp

s

ess kp

5/ 3


103) ( ) 2( 1) 2)

20( 2)) ( )( 1)( 3)

( ) H(s)S 0

20(S+2)=L+S, (1)

S(S+1)(S+3)20(2) 40

3 31 3

40

0.075

10( ) ( )( 2)( 3)

G(S)+(S)s-->0

10=L+

(S+2)(s+3)0

10 5 =

6 31 1

1 1

G SS S S

si G s

s s s

K L SG Sv

ess Kv

ii G SS S

K Lp

s

ess kp

5/ 3

10; ( ) 12( 1)( 2)

2 ( ) ( )

S-->0

=Lt S2G(S) s-->0

102=L+S 2( 1)( 2)

0

102

5

38

0.375

) ( )

1 15

0.2a

H SS S S

K L S G S H Sa

S S S

s

ess

ess

iii G s

k

13. For a unity feedback control system the open loop transfer function , 210( 2)( ) .( 1)

SG SS S

findthe position, velocity and acceleration error constants 01

Solution:

Position error constant, kp = L+ G(S) H(S)S 0


= L+ 10(S+2)----------

S 0 S2 (s+1)=

Velocity error constant, Kr L+SG(S)H(S)

S02

2

0

10 S ( 1)( 2)

102

51 1

50.2

S

a

LS S S

essk

14. For a unity feedback control system the open loop transfer function,

G(S)= 210( 2)

( 1).s

s S . Find the error constants.

Solution:

Position error constant, Kp = L+G(S) H(S)S0

210( 2)

( 1)0

SLS s

S

Velocity error constant Kr = L+S G(S) H(S)s02

10( 2)( 1)0

SLS s

S


Acceleration error constant, ka L+S2 G(S) H(S)S0

210( 2)

( 1)0

20 //

SLS s

S

For a unity feedback control system the open loop transfer function G(S) = 210( 2)

( 1)S

S s . Find the

steady state error when the input is R(S) where R(S) = 2 33 2 1( )

3R S

s S S

Solution:

e(t) = r(t) Co + r(t) C1 +r(t)01

.... ( )21 1

n nCC r tn

r(t) = L-1[R(s)]

213 2

3 212

3 262( ) 2 2 / 36

( ) 1 / 31( )

1 ( ) ( )1

1 ( )

tt

tt

tr t t

r t

F SG S H S

G S

21 ( 1)210( 2) ( 1) 10( 2)1 2 ( 1)

3 2

3 2 10 203 2

( )0 3 2 10 20 s-->0 S-->0

=0

S Ss S S S

S S

S S

S S s

S SC L F s L

S S s


dC1=L+ ( )

ds0

3 2

3 2 10 20

0

3 2 2 3 2 2( 10 20)(3 2 ) ( )(3 2 103 2 2( 10 20)

0

01

2 ( ) ( ( ))2 20 s->0

3d 20=L+

ds

F s

s

d S SL

ds S S S

s

S s s s s s s s sLt

S s s

S

C

d d dC L F s L F s

ds ds ds

s

s

270 403 2 2( 10 20)

s s

s s s

3 2 2 2 3 2

3 2 2

3 2 4

2

4

0 1

2

S 10 20) (60 140 40) 20 70 40 )2( 10 20)(3 2 10)

10 20)

020 40 1

20 102( ) ( ) ( ) ( )

211 2 (0) ( 2 )(0) 1/3 1/10 1/23 6 3

1/60

sS s s s s ss s s s sL

s s s

s

x

Ce t r t c cr t r t

t tt x x

ess

1( )

60 t-> t->

1= 60

L e t L

15. Measurement conduct on a servomechanism show the system response to be c(t) = 1+0.2 e-


60t -1.2e -10t when subject to a unit step input. Obtain an expression for closed loop transferfunction. Determine the undamped natural frequency and damping ratio.

Solution:-

C(t) = 1+0.2e-bot 1.2e-10t

2 2 2

2

1 1 1( ) 0.2 1.260 10

1 1 1( ) 0.2 1.260 10

( 60)( 10) 0.23( 10) 1.2 ( 60( 60)( 10)

70 600 0.25 2 1.2 72( 60)( 10)

1 600( 60)( 10)

600( ) ( ) ( 60)( 10)( ) 600( ) 70

C ss s s

C SS s s

s S S s SS S S

s s s s s

S S S

s s s

C s R ss s

C sR s s s

600

Comparing with standard eqn,2

2 2

2

n

2600

60024.49 rad /sec

2zw 70

n

s

n n

n

n

w

s zw w

w

w

702

70224.49

1.43

n

zw

Z


16. Consider a Unity feed back system having TF 2cs a

R s S Ks a Determine the OLTF and

steady state error coefficients.Solution:

2cs a

R s S Ks a

If G(s) is the forward transfer friction for unity feed back system,

2

2

2

2

( )1 ( )

( )1 ( )

( )1 ( )

1 1 1( )1( )( ) ( )

cs G sR s G s

G s aG s S Ks a

G s S Ks aG s a

S KsG s a

S KsG s a

aG sS s k

Positional error coefficient

1

1

0 ( ) ( )

= 0 1

P

P

K S G s H saS

S S K

K

velocity error coefficient

1

1

0 . ( ) ( )

= 0 1

V

V

K S S G s H saS S

S S K

aKK

Acceleration co efficient]


12

12

0 ( ) ( )

= 0 1

0

a

a

K S S G s H saS S

S S K

K

For a unity feed back system,

360.72

G sS s

Determine the characteristic equation and hence, calculate damping ratio, peak time, settling timePeak overshoot and number of cycles completed before output settles for unit step input.

Characteristic equation of system

S(s+0.72)+36=0S2+0.72s+36=0 ----------(1)

Comparing equ(1) with std. equation

S2+2hWns+Wn2=0--------(2)2nn

n

n

2d n

2

W 362GW 0.72W 6rad/ sec0.72G

2 W0.72G 0.062 6

W W 1 G

= 6 1-(0.06) 5.989rad/ sec

360.72

( ) 136

.

1 ( ) 0.72 36

G sS s

H sG s

T FG s H s S s


Pd

sn

P 2

Peak time t 0.5245secW 5.989

4 4Settling time t 11.11 sec.GW 0.06 6

GPeak overshoot % M e 100%1 G

P 2

0.06%M e 100%1 (0.06)

=82.79%

Number of cycles completed before settling=Settling time /no. of cycles per second

s

d

tW2

211.11 10.595.989

The closed loop T.F. of fourth order system is

4 3 2

1S 2S 10s 35s 50s 24

Determine the

response of the system when a step input of 10 is applied to the input. Hence, calculate the steadystate output of the system.

R(t)=10

10R(s)s

2s 1T(s)

2 s 1 s 2 s 3 s 4

t 2 3t 4t

2s 1C(s)R(s) 2 s 1 s 2 s 3 s 4

5 5 15 25 35c(s)24s 6(s 1) 4(s 2) 6(s 3) 24(s 4)

Taking inverse LT5 5 15 25 35c(t)= e e e e24 6 4 6 24


Obtain the Unit impulse response and unit step response of a unity feed back system whose oftenloop T.F. is

2

2s 1G(s)s

i. Let R(s) = 1/s unit step input

2

2

2

2

2

C(s) G(s)R(s) 1 G(s)H(s)

12sC(s) s2s 1R(s)

1 (1)s

C(s) 2s 1R(s) s 2s 1

2s 1C(s) R(s)s 2s 1

2s 11 =s s 2s 1

2

(2s 1)C(s)S(s 2s 1)

By partial fraction

2 2

2s 1 A Bs CS(s 2s 1) s S 2s 1

2

2s 1s (s 2s 1)

2

2

A(s 2s 1) Bs Cs(s 2s 1)

22s 1 =A(s 2s 1) Bs C s


2comparing S coefficientPut s=0

0 A B1=A

B 1

Comparing S coefficientZ=2A+CC=2-2AC=2-2(1)=0

2 2

22

1 12

-12

-12 2

s 02s 1 1s 2s 1 s s 2s 1

2s 1 1 ssS s 2s 1 s 1

1 SC(t) L C(s) Ls s 1

1 S =Ls s 1

S 11 1 =Ls s 1 S 1

-12

-t t

t

1 1 1 =Ls s 1 S 1

=1-e te

C(t) 1 e t 1

(ii) Input = Unit impulseR(s) =1

2

2 2

C(s) G(s)R(s) 1 G(s)H(s)

2s 1C(s) R(s)S 2s 1

2s 1 2s 1 =1S 2s 1 S 2s 1

C=0


1 12

-1 -12 2 2

-1 t t2

-t

2S 1C(t) L C(s) LS 2s 1

S S 1 S 1 1 =L =Ls 1 s 1 S 1

1 1 =L e tes 1 S 1

C(t)=e 1 t

A unit ramp input is applied to a unity feed back system whose transfer function is2

100s 5s 100 Find the time response and steady state error.

2

2

2 2

C(s) 100R(s) S 5s 100unit Ramp input

1R(s) =S

100C(s)S S 5s 100

By partial fractions we get

2 2

2 22 2 2

s 31 1 2o 4C(s)

2os S S 5s 100

5 5s s1 1 1 2 2C(s)2os S 2os 5 75 5 75s s

2 2 2 2

Taking Inverse LT We get


2.5t 2.5t

2.5t 2.5t

1 1 75 75C(t) t e cos t 4.04e sin t2o 2o 2 2

R(t) te(t) R(t) C(t)

1 1 75 75 = e cos t 4.04e sin t20 20 2 2

Steady state error

2.5t 2.5tss

t

ss

1 1 75 75C e cos t 4.04e sin t2o 2o 2 2

1C2o

H

17. A second order system has 40%r peak overshoot and settling time of 2 sc for unit step input.Find resonant peak gain and reasonant freq.

Solution:

Peak Overshoot =40%G

p 2

G

2

G

2

G

2

e%M 100%1 G

e40 100%1 G

e0.4 100%1 G

ln(0.4)1 G

G 0.2799


sn

n

n

r 2

2r n

4tGW

42GW

2W 7.1428rad/ sec.0.2799

1Reasonant Peak M2 0.2799 1 0.2799

=1.86072

Reasonant frequency W W 1 2G

=7.142 1-

22 0.2799

=6.5586rad/sec.

18. For a unity feed back second order system, the open loop transfer function is

2n

2n

G ss s 2

Calculate the generalized error coefficients and find error series.

Solution:

2n

n

2n

2 2n

2n

o 2 2s 0 s 0n n

G ss s 2

H s 1

s 2 s1F s1 G s H s s 2 s

s 2 sC lim F s lim 0s 2 s

1 s 0

2 2 2n n n n n

2s 0n n

3n

4n n

dC lim F sds

s 2 s 2s 2 s 2 s 2s 2lim

s 2 s s

2 2


2 2 22n n n n

2 22s 0 s 0 2 2n n

22 2n n n

2 2n n n2

n 4s 0 2 2n n

4 2n n n n2

n 8n

4

n

2s 2 s 2 s s 2 sd F s dC lim limds ds s 2 s

s 2 s 2 s

s 2 s 2s 22 lim

s 2 s

2 22

1 42

2

o 2

2

2n n

e t C r t C r t ....

2 1 4e t r t r t ....

19. An unit feed back system has G(s) = 1

s 1 2s . The input to the system is described by r(t)=2+4t+6t2+2t3. Determine the generalized error coefficients and express the steady state erroras a function of time.

Solution:

2 3

2

2 2

2

0 2s 0 s 0

1 s 0 s 0

1Given G s H s 1; r t 2 4t 6t 2ts 1 2s

1 1 F ss+G s H s 1 G s

s 1 2s1 2s s =1 2s s 1 2s s 11+

s 1 2s

2s s C lim F s lim 02s s 1

d d C lim F s limds ds

2

2

2s s2s s 1


2 2

1 2s 0 2

2s 0 2

2 2

2 2 2 2s 0 s 0 2

22 2

4s 0 2

0 1 2

2s s 1 4s 1 2s s 4s 1C lim

2s s 1

4s 1 =lim 1

2s s 1

d d 4s 1C lim F s limds ds 2s s 1

2s s 1 4 2 4s 1 2s s 1 4s 1 =lim

2s s 1

=2C 0; C 1; C 2

Error signal

n n0 1 2r t r t Ce t r t C r t C C ...2! n! Given r(t) = 2+4t+6t2r(t) =4+12tr(t) =12

e(t) =r(t) C0 + r(t) C1 + 2r(t)C2!

=(2+4t+6t2)0+(4+12t)+ 12 22!

=12t +16s's t

e lime(t)

20. The open loop transfer function of a unity feedback system is given by KG(s)s(Ts 1)

where K and T are positive constants. By what factor should the amplifier gain be reduced sothat the peak overshoot of unit step response of the system is reduced from 75% to 25%.[April 1996, Madras University]

Solution:Given: KG(s)

s(Ts 1)

Characteristic equation is given by K1 0s(Ts 1)

C.E. = Ts2+s+K=0


C.E.=s2 s K 0T T

Case(i):%Peak overshoot Mp=75% Mp =0.752

1 2 2

1

ln 0.75ln 0.75

0.0912

Case (ii):% Peak overshoot Mp = 25(or) Mp = 0.25

2

1 2 2

1

ln 0.25ln 0.25

0.0437

Comparing 2 2 2n ns Ks 0 with s 2 s 0.T T

n

n

KT12T

K 12T T

Squaring on both sides, we get

22

2 2

2

1 22 2

K 14T T1 TK

T 41K

4T1 1K ,K

4T 4T

The factor by which K1has to be increased2 21

2 2 12

1 22

2

1

2 1

K 4T 0.0912K 0.40374T

K0.0510

K

K 0.0510K


21. A servo mechanism is used to control the angular position 0 of a mass through a commandsignal i. The moment of inertia of load is 200 kg-m2 and the motor torque at load is 6.88 104N/m/rad of error. The damping torque coefficient is 5 104 N-m/rad/sec. Find the timeresponse for a step input of 1 radian.

Solution:

Figure:

Given

2 4

3

2

J 200kg m ; kT=6.88 10 N/m/ radf 5 10 N m/ rad/ sec

kTG s and H s 1Js fs

The closed loop transfer function T(s) G s

1 G s H s

That is,

T

20 T2

Ti T2

T

2 T

ks kJs fsT s ks Js fs k1

Js fsk / J = kfs sJ J

Substituting the values kT, f, J in the above equation, we have

4

3 4 22

6.88 10344200T s

5 10 6.88 10 s 25s 344s s200 200

Comparing the equation with the standard form of second order system as shown in equation wehave


2n n

n

344 344 18.547252 25 0.6739

2 18.547

We know for < 1, the unit step response is given by equation

nt 20 n22

-12.5t 0 -1

et 1 sin 1 t1

1 =1-1.353e sin 13.7t 47.63 =tan

22. A system shown in figure is initially at equilibrium with r =1 and d = 0. a step functiondisturbance d(t) = u(t) is then initiated at t = 0. Determine the response c(t) for t > 0.(AU April 04).

Figure:Solution:The given system is in equilibrium with r=1.

c(t) = 1 at t = 0.But we want the response from t=0. So, Let us take this value (c (t) =1) as new reference. (ie)

r(t) =c(t) = 0 at t=0.

C s can be found by assuming r t 0.D s

Figure:


2

2 2

2

2

2 2C s 2s s 42s 4 s 4 .3 2 2s s 4 6D s s 4 2s 8s 61 .

2s s 4 2s s 44s 2s =

2s 8s 6 s 4s 3For step disturbance

1 1 2s D s C s .s s s 4s 3

2s 2 C ss 4s 3 s 1 s 3

A B =s 1 s 3

t 3t

1 1 =s 1 s 3

C t e e u t

23. A second order position control system has OLTF KG(s)s(1 0.2s)

where K is the gain. Findthe value of K so that steady state error shall not exceed 0.5o when input shaft rotated as 5 r.p.m.

Solution:

GivenAngular velocity = 5 r.p.m

2

oss

5 2rad/ sec

60

rad/ sec or r(t) = t rad6 6

R(s) =6s

e 0.5 radian360

We know

ss s 0

s 0

sR se lim

1 G s H s

sR sGp lim360 1 G s H s


We know

2

s 0 s 0

s6s 6lim lim

K K360 1 s s1 0.2s 1 0.2s

=6K

360 6KK 60

24. A unity feedback control system has an amplifier with gain KA = 10 and gain ratio, G(s) =1/s(s+2) in the feed forward path. A derivative feedback, H(s) = sK0 is introduced as a minorloop around G(s). Determine the derivative feedback constant, K0 so that the system dampingfactor is 0.6.

SOLUTION

The given system can be represented by the block diagram shown in figure.

FigureHere, KA = 10; G(s) = 1( 2)s s and H(s) = sK0The closed loop transfer function of the system can be obtained by block diagram reduction

techniques.

Step 1: Reducing the inner feedback loop.

Figure


20 00

20

1( ) 1 1( 2)

11 ( ) ( ) ( 2) 21 .( 2)1 = s (2 )

G s s sG s H s s s sK s s sKsKs s

K s

Figure.Step 2. Combining blocks in cascade.

FigureStep 3. Reducing the unity feed back path.

Figure

The closed loop transfer function, 20

( ) 10( ) (2 ) 10

C sR s s K s

The given system is a second order system. The value of K0 can be determined bycomparing the system transfer function with standard form of second order transfer function.

22 2 2

0

( ) 10( ) 2 (2 ) 10

nn n

C sR s s s s K s

On comparing we get

2 1010 3.162 rad/sec.

n

n

00

2 22 2

= 2 0.6 3.162-2 = 1.7944

nn

KK


RESULT

The value of constant, K0 = 1.7944

25. A unity feedback control system has an loop transfer function, G(s) = 10/s(s+2). Find the risetime, percentage overshoot, peak time and settling time for a step input of 12 units.

SOLUTION

(Note: The formula for rise time, percentage overshoot and peak time remains same for unitstep and step input).

The unity feedback system is shown in figure

Figure: Unity feedback system

The closed loop transfer function, ( ) ( )( ) 1 ( )C s G sR s G s

Given that, G(s) = 10/s (s+2)

2

10( ) 10 10( 2)

10( ) ( 2) 10 2 101 ( 2)C s s sR s s s s s

s s

The values of damping ratio and natural frequency of oscillation n are obtained bycomparing the system transfer function with standard form of second order transfer function.

22 2 2

( ) 10( ) 2 2 10

nn n

C sR s s w s s s

On comparing we get,2 10

10 3.162 rad/sec.n

n


2 22 1 0.3162 3.162

n

n

2 21 1

2 2d

1 1 0.316tan tan 1.249 rad0.3161 3.162 1 0.316 3 rad/secn

Rise time, 1.249 0.63 sec.3r dt

Percentage overshoot, 2 20.316

1 1 0.316% 100 100 = 0.3512 100=35.12%

PM e e

Peak overshoot = 35.12 12 units = 4.2144 units100 Peak time, 1.047 sec3P dt

.

Time constant, T = 1 1 1 sec0.316 3.162n For 5% error, Settling time, tS = 3T = 3 secFor 2% error, Settling time, tS = 4T = 4 sec.

RESULTRise time tr = 0.63 secPercentage overshoot, %MP = 35.12%Peak overshoot = 4.2144 units, (for a input of 12 units.)Peak time, tP = 1.047 secSettling time, tS = 3 sec for 5% error

= 4 sec for 2% error

26. For a unity feedback control system the open loop transfer function G(s) = 10(s+2)/s2 (s+1).Find (a) the position, velocity and acceleration error constants, (b) the steady state error whenthe input is R(s) where R(s) 2 33 2 13s s s

SOLUTION(a) To find the static error constants

For a unity feed back system, H(s) = 1Position error Constant , KP = 0 0( ) ( ) ( )s sLt G s H s Lt G s


= 2010( 2)( 1)ssLt s s

Velocity error constant, KV = 0 0. ( ) ( ) . ( )s sLt sG s H s Lt sG s

= 2010( 2)( 1)ssLt s s s

Acceleration error constant, Ka = 2 20 0( ) ( ) ( )s sLt s G s H s Lt s G s = 2 20

10( 2) 10 2 20( 1) 1ssLt s s s

(b) To find steady state error Ist method

The steady state error for non standard input is obtained using generalized error series,given below.

The error signal, e(t) = r(t) 20 1( ) ( ) ....... ( ) .......2! !nCCC r t C r t r t n

Given that, R(s) = 2 33 2 13s s s Input signal in time domain, r(t) = 1 1 2 33 2 1[ ( )] 3L R s L s s s

=

2 21 1 12 23 3 2! 3 6t tt t

2233

1( ) ( ) 2 2 26 31 r( ) ( ) ( ) 3

r( ) ( ) r( ) 0

d tr t r t tdtd dt r t r tdt dtd dt r t tdt dt

The derivatives of r(t) is zero after second derivative. Hence we have to evaluate only threeconstants C0, C1 and C2 .

The generalized error constants are given by2

0 1 2 20 0 0( ); C ( ) ; C ( )s s sd dC Lt F s Lt F s Lt F sds ds

F(s) =

22

2

11 1 110( 2)1 ( ) ( ) 1 ( ) 1 10 21 1

s ssG s H s G s s s s

s s

=3 2

3 2 10 20s s

s s s


3 20 3 20 0( ) 010 20s s

s sC Lt F s Lt s s

3 2

1 3 20 0

3 2 2 3 2 2

3 2 20

3 23 2 20

( ) 10 2010 20 3 2 3 2 10 = ( 10 20)

20 70 40 = 0( 10 20)

s s

s

s

d d s sC Lt F s Ltds ds s s ss s s s s s s s sLt s s ss s sLt s s s

2

2 20 0 C ( ) ( )s sd d dLt F s Lt F sds ds ds

3 2

3 2 2020 70 40 = ( 10 20)s

d s s sLt ds s s s 3 2 2 2

3 2 3 2 23 2 40

( 10 20) (60 140 40)(20 70 40 )2 ( 10 20)(3 2 10) = ( 10 20)s

s s s s ss s s s s s s sLt s s s

= 420 40 120 10

Error signal,

. .. 20

2

( ) ( ) ( ) ( ) 2!1 1 1 1 = 2 0 2 03 6 3 3 10 2!1 = 60

lCe t r t C r t C r t

t tt

Steady state error, 1 1( ) 60 60ss t te Lt e t Lt

IInd method

The error signal in s-domain, E(s) = ( )1 ( ) ( )R sG s H s

Given that R(s) = 2 33 2 13s s s

210( 2)( ) and H(s) = 1( 1)

sG s s s


2 3 2 32

2 2

3 2 1 3 2 13 3( ) 10( 2) ( 1) 10( 2)1 ( 1) ( 1)

s s s s s sE s s s s ss s s s

2 2 22 2 2 3 2

3 ( 1) 2 ( 1) 1 ( 1)( 1) 10( 2) ( 1) 10( 2) 3 ( 1) 10( 2)

s s s s s ss s s s s s s s s s s s

The steady state error ess can be obtained from final value theorem.

Steady state error , ( ) . ( )ss t se Lt e t Lt s E s 2 2 2

2 2 2 3 23 ( 1) 2 ( 1) 1 ( 1)

( 1) 10( 2) ( 1) 10( 2) 3 ( 1) 10( 2)ss ss s s s s se Lt s s s s s s s s s s s s s

2 2

2 2 23 ( 1) 2 ( 1) ( 1)

( 1) 10( 2) ( 1) 10( 2) 3 ( 1) 10( 2)1 10 0 60 60

ss s s s sLt s s s s s s s s s s

160sse

III rd method

Error signal in s domain, E(s) = ( )1 ( ) ( )R sG s H s

( ) 1 ,( ) 1 ( ) ( )E sR s G s H s Given that 2

10( 2)( ) ( ) 1( 1)sG s andH ss s

22

2

( ) 1 ( 1)10( 2)( ) ( 1) 10( 2)1 ( 1)

E s s ssR s s s s

s s

3 2 2 3 2 33 2 2 3 ....10 20 20 10 20 40

s s s s s ss s s s s s


2 3

2 32 3

2 3 3 4 522 3

.. ..

2 3

20 40sE(s)=R(s) ...20 4020 101 1( ) ( ) ... 2 20 2020 40

taking inverse laplace transform,1 1e(t) = ( ) ( ) ...20 40

3 2 1 that R(s)= s 3

s ss

s ss s s s s sss R s s R s

Onr t r t

Given s s

3 4 5

3 4 5 6

64 5

2 20 20

2 4 40 40-3 310 40 40

s s s

s s s s

ss s

2 21 1 2 33 2 1 1 1 1( ) ( ) 2 23 3 3 2! 3 6t tr t L R s L t ts s s

.

.. .23... ..3

1( ) ( ) 2 2 26 31( ) ( ) ( ) 3

( ) ( ) ( ) 0

d tr t r t tdtd dr t r t r tdt dtd dr t r t r tdt dt

Error signal in time domain, e(t) = ..1 1 1 1( )20 20 3 60d r tdt

Steady state error, 1 1( ) 60 60ss t te Lt e t Lt

RESULT(a) Position error constant, KP =

Velocity error constant, Kv = Acceleration error constant Ka = 20

(b)2 3

state error 13 2 1when R(s)= 60s 3

ss

Steadye

s s

27. For servomechanisms with open loop transfer function given below explain what type ofinput signal give rise to a constant steady state error and calculate their values.

(i) G(s) = 20( 2) ;( 1)( 3)s

s s s

(ii) G(s) = 10 ;2 3s s (iii) G(s) = 2

10( 1)( 2)s s s


SOLUTION

(i) G(s) = 20( 2)( 1)( 3)s

s s s

Let us assume unity feed back system, H(s) = 1

The open loop system has a pole at origin. Hence it is a type-1 system. In systems withtype number-1, the velocity (ramp) input will give a constant steady state error.

The steady state error with unit velocity input, 1/ss ve K

Velocity error constant, 0 0( ). ( ) ( )v s sK Lt sG s H s Lt sG s = 0

20(s+2) 20 2 40 s ( 1) 3 1 3 3sLt s s s

Steady state error, 1 3 0.07540sse K (ii) G(s) =

102 3s s

Let us assume unity feed back system H(s) = 1

The open loop system has no pole at origin. Hence it is a type-0 system. In systems withtype number-0, the step input will give a constant steady state error.

The steady state error with unit step input = 11ss Pe K Position error constant, = 0 0( ). ( ) ( )P s sK Lt G s H s Lt G s

= 010 10 52 2 2 3 3sLt s s

(iii) G(s) = 2 10( 1)( 2)s s s Let us assume unity feed back system H(s) = 1The open loop system has two poles at origin. Hence it is a type-2 system. In systems with

type number-2, the acceleration (parabolic) input will give a constant steady state error.The steady state error with unit acceleration input, 1/ss ae KAcceleration error constant , 2 20 0( ) ( ) ( )a s sK Lt s G s H s Lt s G s

2

2010 10 51 2 1 2a sK Lt s s s s


Steady state error, 1 1 0.25ss ae K RESULT

1. Steady state error in system (i) with unit velocity input = 0.0752. Steady state error in system (ii) with unit step input = 0.3753. Steady state error in system (iii) with unit acceleration = 0.2

28. Consider a unity feed back system with a closed loop transfer function 2( )( )C s Ks bR s s as b

.Determine the open loop transfer function G(s). Show that the steady state error with unitramp is given by a Kb

.

SOLUTION

For unity feedback system, H(s) = 1.The closed loop transfer function, ( ) ( ) ( )( ) 1 ( ) ( ) 1 ( )

C s G s G sR s G s H s G s

Let M(s) = ( ) ( )( ) 1 ( )C s G sR s G s

( )1 ( )G sG s = M(s)

On multiplication we get,

G(s) = M(s) [1+G(s)]=M(s)+M(s).G(s)G(s) M(s).G(s) = M(s)G(s) [1-M(s)] = M(s)G(s) = ( )1 ( )

M sM s

Given than M(s) = 2Ks bs as b

Open loop transfer function

G(s) = 2 22

( )1

Ks bKs bs as bKs b s as b Ks b

s as b


= 2 2 [ ]Ks b Ks b Ks b

s as b Ks b s a K s s s a K

Velocity error constant, 0 0( ) ( ) ( )V s sK Lt sG s H s Lt sG s = 0s

Ks b bLt s s s s K a K

Steady state error with velocity input, 1ssV

a Ke K b

RESULTOpen loop transfer function, G(s) = [ ]

Ks bs s a K

Steady state error with velocity input, sse = a Kb

UNIT IIIPART A

1. Define frequency response.

The frequency response is the steady state response of a system when the input to the system is asinusoidal signal.

2. Write any three advantages of frequency response analysis.

1. The absolute and relative stability of the closed loop system can be estimated from theknowledge of their open loop response.

2. The practical testing of systems can be easily carried with available sinusoidal signalgenerators and precise measurement equipments.

3. The transfer function of complicated system can be determined experimentally by frequencyresponse plots.

3. What are frequency domain specifications?

1. Resonant peak, Mr.2. Resonant frequency, or3. Bandwidth4. Cut-off rate


5. Gain margin6. Phase margin

4. Define resonant peak.

The maximum value of the magnitude of closed loop transfer function is called the resonant peak.

5. Define resonant frequency.

The frequency at which the resonant peak occurs is called resonant frequency.

6. Define bandwidth.

The bandwidth is the range of frequency for which the system gain is more than -3dbr.

7. Define cut-off frequency.

The frequency at which the gain is -3db is called cut-off frequency.

8. Define cut off rate.

The slope of the log-magnitude curve near the cut off frequency is called cut off rate.

9. Define gain margin.

The gain margin is defined as the reciprocal of the magnitude of open loop transfer function atphase cross over frequency.

10. Define phase margin.

The phase margin is the amount of additional phase lag at the gain cross over frequency requiredto bring the system to the verge of instability.

11. What are the graphical techniques available for frequency response analysis?

Bode plotPolar plotNichols plotM and N circlesNichols chart.


12. What are the two graphs of bode plot?

A plot of the magnitude of a sinusoidal transfer function Vs log A plot of the phase angle of a sinusoidal transfer function Vs log

13. What are the advantages of bode plot?

Multiplication of magnitudes can be converted into addition.An approximate log-mag curve can be sketched.

14. What do you mean by polar plot?

The polar plot of a sinusoidal transfer function G(jw) is a plot of the magnitude of G(jw) is a plotof the magnitude of G(jw) Vs the phase angle of G(jw) on polar coordinates as is varied fromzero to infinity.

15. Draw the bode plot for G(s)=1

(1 4 )(1 7 )s s

16. Define phase cross over frequency.

It is the frequency at which the phase of G)j) is -1800.

17. Define gain cross over frequency.

It is the frequency at which the magnitude of G(j) is unity.

18. Determine the resonant frequency of a 2nd order system whose 2( ) 64( ) 10 64

C SR S S S

2 2N r64 27w 10 w 1 27N nw w

2108 / sec 7= 0.625 =8 1-2(0.625)2 8 =3.741

nw rod


19. Write the expression for resonant peak and resonant frequency.

Resonant peak, Mr= 21

2 1

20. Define corner frequency.

The magnitude plot can be approximated by asymptotic straight lines. The frequenciescorresponding to the meeting point of asymptotes are called corner frequency.

21. The damping ratio of a second order system is 0.5 calculate resonant peak.

Mr=2 2

1 12 1 2 0.5 1 (0.5)

. 1.154.Mr

22. What is approximate bode plot?

In approximate bode plot, the magnitude plot of a first and second order factors areapproximated by two straight lines, which are asymptotes to exact plot.

23. What is the value of error in approximate magnitude plot of a first order factor at the cornerfrequency?

The error is 3mdb, mmultiplicity factor. Positive error for numerator and negative error fordenominator factor.

24. What is minimum phase system?

The minimum phase systems are systems with minimum phase transfer functions. In minimumphase transfer functions, all poles and zeros will lie on the left half of S plane.

25. What is all pass systems?

The all pass systems are systems with all pass transfer functions. In all pass transfer functions, themagnitude is unity at all frequencies and the transfer function will have anti-symmetric pole zeropattern.

26. What is non minimum phase transfer functions?


A transfer function which has one or more zeros in the right half S-plane is known as non-minimum phase transfer function.

27. Draw the polar plot of G(s)=1

1 ST

28. Sketch the polar plot of G(S)= 2 1 2 31

(1 )(1 )(1 )S ST ST ST

29. What is Nichols chart?

The Nichols chart consists of M ad N contours superimposed on ord

Date post:	10-Jan-2016
Category:	Documents
Upload:	rajugeeth
View:	255 times
Download:	0 times

Control Systems (1-135).pdf

Documents