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Control Systems 10ES43

Dept. of ECE/SJBIT

Question paper solution

UNIT-1

1. Compare linear and nonlinear control system. ( 4 marks, Dec 2012)

Linearcontrol system: obey super position theorem, stability depends only on root

location, do not exhibit limit cycle, can be analyzed by Laplace, Fourier and Z transforms

Nonlinear control system: do not obey super position theorem, stability depends on root

location and initial condition of input, exhibit limit cycle, cannot be analyzed by Laplace,

Fourier and Z transforms.

2. Distinguish between open loop and closed loop control system. Describe two

examples for each.

(10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010

Control systems are classified into two general categories based upon the control action

which is responsible to activate the system to produce the output viz.

1) Open loop control system in which the control action is independent of the out put.

2) Closed loop control system in which the control action is some how dependent upon the

output and are generally called as feedback control systems.

Open Loop System is a system in which control action is independent of output. To each

reference input there is a corresponding output which depends upon the system and its operating

conditions. The accuracy of the system depends on the calibration of the system. In the presence

of noise or disturbances open loop control will not perform satisfactorily.

EXAMPLE - 1 Rotational Generator

The input to rotational generator is the speed of the prime mover ( e.g steam turbine) in r.p.m.

Assuming the generator is on no load the output may be induced voltage at the output terminals.

Actuating signal

nput

output input

Controller

System


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Speed of the

Prime mover

Induced Voltage

Output

Inputs

Time Cleanliness of clothes

Rotational Generator

Fig 1-2 Rotational Generator

EXAMPLE – 2 Washing machine

Most ( but not all ) washing machines are operated in the following manner. After the clothes to

be washed have been put into the machine, the soap or detergent, bleach and water are entered in

proper amounts as specified by the manufacturer. The washing time is then set on a timer and the

washer is energized. When the cycle is completed, the machine shuts itself off. In this example

washing time forms input and cleanliness of the clothes is identified as output.

Washing Machine

Fig 1-3 Washing Machine

EXAMPLE – 3 WATER TANK LEVEL CONTROL

To understand the concept further it is useful to consider an example let it be desired to maintain

the actual water level 'c ' in the tank as close as possible to a desired level ' r '. The desired level

will be called the system input, and the actual level the controlled variable or system output.

Water flows from the tank via a valve Vo ,and enters the tank from a supply via a control valve

Vc. The control valve is adjustable manually.

Fig 1-4 b) Open loop control

Fig –1.4 a) Water level control

A closed loop control system is one in which the control action depends on the output. In

closed loop control system the actuating error signal, which is the difference between the input

signal and the feedback signal (output signal or its function) is fed to the controller.

WATER

TANK

Desired Water

level r

Actual Water

level c

Valve VC

Valve VO

Waterin

Water

out

C


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Fig –1.5: Closed loop control system

EXAMPLE – 1 – THERMAL SYSTEM

To illustrate the concept of closed loop control system, consider the thermal system shown in fig-

6 here human being acts as a controller. He wants to maintain the temperature of the hot water at

a given value ro C. the thermometer installed in the hot water outlet measures the actual

temperature C0C. This temperature is the output of the system. If the operator watches the

thermometer and finds that the temperature is higher than the desired value, then he reduce the

amount of steam supply in order to lower the temperature. It is quite possible that that if the

temperature becomes lower than the desired value it becomes necessary to increase the amount

of steam supply. This control action is based on closed loop operation which involves human

being, hand muscle, eyes, thermometer such a system may be called manual feedback system.

Fig 1-6 a) Manual feedback thermal system b) Block diagram

EXAMPLE –2 HOME HEATING SYSTEM

The thermostatic temperature control in hour homes and public buildings is a familiar example.

An electronic thermostat or temperature sensor is placed in a central location usually on inside

wall about 5 feet from the floor. A person selects and adjusts the desired room temperature ( r )

Control

elements

System /

Plant

Feed back elements

controller

Forward path

Reference

input

Actuating/

error

signal

Error

detector

Feed back signal

Controlled

output

Desired hot

water.temproc

Brain of

operator

(r-c)

Muscles

and Valve

Actual

Water temp

Co C +

C

Thermometer

+

Human operator

Thermometer

Hot water

Drain

Cold water

Steam Steam


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say 250

C and adjusts the temperature setting on the thermostat. A bimetallic coil in the

thermostat is affected by the actual room temperature ( c ). If the room temperature is lower than

the desired temperature the coil strip alters the shape and causes a mercury switch to operate a

relay, which in turn activates the furnace fire when the temperature in the furnace air duct system

reaches reference level ' r ' a blower fan is activated by another relay to force the warm air

throughout the building. When the room temperature ' C ' reaches the desired temperature ' r '

the shape of the coil strip in the thermostat alters so that Mercury switch opens. This deactivates

the relay and in turn turns off furnace fire, which in turn the blower.

Fig 1-7 Block diagram of Home Heating system.

A change in out door temperature is a disturbance to the home heating system. If the out side

temperature falls, the room temperature will likewise tend to decrease.

3. For the two port network shown, obtain the transfer functions, i) V2 (s) / V1 (s) and

ii) V1(s) / I1(s) ( 8 marks , Dec 2012)

Ans. Taking laplace transform of the given network

Desired temp.roc Relay

switch

Actual

Temp.

Co C Furnace Blower House +

Outdoor temp

change

(disturbance)


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Applying KVL


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4. For the system shown in fig. i) draw the mechanical network ii) write the differential

equations iii) obtain torque to voltage analogy. ( 8 Marks, Dec 2012)


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5. Write the differential equations for the mechanical system shown in fig. 1(b) and

obtain F-V and F-I analogous electrical circuits. ( 10 Marks), Jan 2009

6. Mention merits and demerits of open loop and closed loop control systems and give

an example for each. ( 6 Marks), July 2009

An advantage of the closed loop control system is the fact that the use of feedback makes

the system response relatively insensitive to external disturbances and internal variations

in systems parameters. It is thus possible to use relatively inaccurate and inexpensive

components to obtain the accurate control of the given plant, whereas doing so is

impossible in the open-loop case.

From the point of view of stability, the open loop control system is easier to build

because system stability is not a major problem. On the other hand, stability is a major

problem in the closed loop control system, which may tend to overcorrect errors that can

cause oscillations of constant or changing amplitude.

It should be emphasized that for systems in which the inputs are known ahead of time and

in which there are no disturbances it is advisable to use open-loop control. closed loop

control systems have advantages only when unpredictable disturbances it is advisable to

use open-loop control. Closed loop control systems have advantages only when

unpredictable disturbances and / or unpredictable variations in system components used

in a closed –loop control system is more than that for a corresponding open – loop control

system. Thus the closed loop control system is generally higher in cost.

7. Obtain the transfer function of an armature controlled DC servomotor.

(6 Marks), July 2009, Dec 2010

A DC motor is used in a control system where an appreciable amount of shaft power is

required. The DCmotors are either armature-controlled with fixed field, or field-

controlled with fixed armature current.DC motors used in instrument employ a fixed

permanent-magnet field, and the control signal is appliedto the armature terminals.


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Fig. 1. (a) Schematic diagram of an armature-controlled DC servo motor, (b) Block

diagram

In order to model the DC servo motor shown in Fig. 1, we define parameters and

variables as follows.

Ra = armature-winding resistance, ohms

La = armature-winding inductance, henrys

ia = armature-winding current, amperes

if = field current, amperes

ea = applied armature voltage, volts

eb = back emf, volts

θ = angular displacement of the motor shaft, radians

T = torque delivered by the motor, lb-ft

J = moment of inertia of the motor and load referred to the motor shaft, slug-ft2

.f = viscous-friction coefficient of the motor and load referred to the motor shaft, lb-

ft/rad/sec

The torque T is delivered by the motor is proportional to the product of the armature

current iaandthe air gap flux Ψ, which is in turn is proportional to the field current

Ψ = Kf if


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UNIT-2

1. Illustrate how to perform the following in connection with block diagram

reduction rules:

i) Shifting a take - off point after a summing point.

ii) Shifting a take - off point before a summing point. ( 4 marks, Dec 2012)

Shifting a take - off point before a summing point

2. The performance equations of a controlled system are given by the following set

of linear algebraic equations:

i) Draw the block diagram.

ii) Find the overall transfer function C(s)/ R(s) using block diagram

reduction technique.

( 8 marks, Dec 2012)


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3. Draw the corresponding signal flow graph for the given block diagram is shown

in fig. and obtain the overall transfer function by Mason’s gain formula.


; ;

- G1 G2 H2; L2 = - G2 G3 H2; L3 = - G2 H1;


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4. Obtain the C/R ratio for the block diagram shown using block diagram

reduction technique ( 06 Marks, July 2009)

5. Find C(S)/ R(S) by Mason’s gain formula ( 06 Marks, July 2009)

6. Explain briefly the following termsi) Forward path ii) path gain iii) loop gain

iv) canonical form (08 Marks, July 2009)

Path: A path is a traversal of connected branches in the direction of branch arrow.

Loop: A loop is a closed path.

Self loop: It is a feedback loop consisting of single branch.

Loop gain: The loop gain is the product of branch transmittances of the loop.

Nontouching loops: Loops that do not posses a common node.


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Forward path: A path from source to sink without traversing an node more than

once.

Feedback path: A path which originates and terminates at the same node.

Forward path gain: Product of branch transmittances of a forward path.

7. Define the transfer function. Explain Mason’s gain formula for determining the

transfer function from signal flow graphs. ( 06 marks, Dec 2010)

The relationship between an input variable and an output variable of a signal flow graph

is given by the net gain between input and output nodes and is known as overall gain of

the system. Masons gain formula is used to obtain the over all gain (transfer function) of

signal flow graphs.

Masons Gain Formula

Gain P is given by

kkkPP

1

Where, Pk is gain of kth forward path,

∆ is determinant of graph

∆=1-(sum of all individual loop gains)+(sum of gain products of all possible

combinations of two nontouching loops – sum of gain products of all possible

combination of three nontouching loops) + ∙∙∙

∆k is cofactor of kth

forward path determinant of graph with loops touching kth

forward

path. It is obtained from ∆ by removing the loops touching the path Pk.

8. For the block diagram shown in fig.Q2 (b), determine the transfer function

Q2(s)/Q(s) using block diagram reduction algebra. (8 Marks, Dec 2010)


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8. For the system described by the signal flow graph shown in fig. Q2(c), obtain the

closed loop transfer function C(s) / R(s) , using Mason’s gain formula.

( 6 Marks, Dec 2010)

There are two forward paths.

The gain of the forward path are: P1=abcd, P2=ac

There are five loops with loop gains:

L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befg

There is one combination of Loops L1 and L2 which are nontouching with loop gain

product L1L2=ehcg

∆ = 1+eh+cg+bei+efd+befg+ehcg

Forward path 1 touches all the five loops. Therefore ∆1= 1.

Forward path 2 does not touch loop L1.Hence, ∆2= 1+ eh

The transfer function T = ehcgbefgefdbeicgeh

ehacabefPP

X

X

1

12211

1

3

9. Obtain the closed loop transfer function C(s)/ R(s) for the signal flow graph of a

system shown in fig. Q2(b) using Mason’s gain formula. ( 10 Marks, June 2012)

a e b f

-h

-g

-i

X1 X5

X4 X3

X2

c

d

X3 1


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There are three forward paths.

The gain of the forward path are: P1=G1G2G3G4G5

P2=G1G6G4G5

P3=G1G2G7

There are four loops with loop gains:

L1=-G4H1, L2=-G2G7H2, L3= -G6G4G5H2 , L4=-G2G3G4G5H2

There is one combination of Loops L1 and L2 which are nontouching with loop gain

product L1L2=G2G7H2G4H1

∆ = 1+G4H1+G2G7H2+G6G4G5H2+G2G3G4G5H2+ G2G7H2G4H1

Forward path 1 and 2 touch all the four loops. Therefore ∆1= 1, ∆2= 1.

Forward path 3 is not in touch with loop1. Hence, ∆3= 1+G4H1.

The transfer function

T=

2174225432254627214

14721654154321332211

1

1

HHGGGHGGGGHGGGHGGHG

HGGGGGGGGGGGGGPPP

sR

sC

10. Derive an expression for the closed loop transfer function of a negative feedback

system. ( 4 Marks, Dec 2011)

B(s)

R(s) C(s) C(s)

B(s)

Fig 2.4

G(s)

H(s)

+ -


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The ratio of the feed back signal B(s) to the actuating error signal E(s) is called the open

loop transfer function.

open loop transfer function = B(s)/E(s) = G(s)H(s)

The ratio of the output C(s) to the actuating error signal E(s) is called the feed forward

transfer function .

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

If the feed back transfer function is unity, then the open loop and feed forward transfer

function are the same. For the system shown in Fig1.4, the output C(s) and input R(s) are

related as follows.

C(s) = G(s) E(s)

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

= R(s) - H(s)C(s) but B(s) = H(s)C(s)

Eliminating E(s) from these equations

C(s) = G(s)[R(s) - H(s)C(s)]

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

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

C(s) G(s)

=

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

C(s)/R(s) is called the closed loop transfer function.


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UNIT-3

1. Derive the expression for peak time. ( 4 marks, Dec 2012)

sec

2. The loop transfer function of transfer function is given by

i) Determine the static error coefficients

ii) Determine steady state error coefficients for the input r(t) = 2t2 + 5t +10



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3.


Comparing with standard 2nd order system equation


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Ts= 2 Sec

4. Derive expressions for peak response time tpand maximum overshoot Mp of an

under damped second order control system subjected to step input

( 6 Marks, June 2012, Dec 2010, Dec 2011)

W. K.T. for the second order system,

T.F. = C(S) = Wn2 .------------------------------(1)

R(S) S2+2 WnS+ Wn

2

Assuming the system is to be underdamped ( < 1)

Rise time(tr):

W. K.T. C(tr) = 1- e- Wnt

Cos Wdtr + .sinwdtr

1-2

Let C(tr) = 1, i.e., substituting trfor t in the above Eq:

Then, C(tr) = 1 = 1- e- Wntr

Coswdtr + .sinwdtr

1-2

Cos wdtr + sinwdtr= tan wdtr= - 1-2 = wd .

1-2


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jW

Thus, the rise time tr is , jWd

tr= 1 . tan-1

- w d = - secsWn 1-2

Wn

Wd wd

When must be in radians. -

Wn

Peak time :- (tp)

Peak time can be obtained by differentiating C(t) W.r.t. t and equating that

derivative to zero.

dc = O = Sin Wdtp Wn . e- Wntp

dt t = t p 1-2

Since the peak time corresponds to the 1st peak over shoot.

Wdtp = = tp = .

Wd

The peak time tp corresponds to one half cycle of the frequency of damped oscillation.

Maximum overshoot :- (MP)

The max over shoot occurs at the peak time.

i.e. At t = tp = .

Wd

Mp = e –( / Wd)

or e –( / 1- 2)

Settling time :- (ts)

An approximate value of ts can be obtained for the system O < <1 by using the

envelope of the damped sinusoidal waveform.

Time constant of a system = T = 1 .

Wn

Setting time ts= 4x Time constant.

= 4x 1 .for a tolerance band of +/- 2% steady state.

Wn

S- Plane


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5. For a unity feedback control system with G(s) = 10(S+2) / S2 (S+1). Find

i) The static error coefficients

ii) Steady state error when the input transform is

( 6 Marks, June 2012)

i) G(s) H(s) = 10(S+2) / S2 (S+1) ------ it is a type II system

ii) Kp = G(s) H(s)

Kp = 0

Kv = S G(s) H(s)

Kv = 0

Ka = S2 G(s) H(s)

Ka = 20

iii) Ess =


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UNIT-4

1. Explain Routh-Hurwitz’s criterion for determining the stability of a system and

mention any three limitations of R-H criterion.

( 10 Marks, June 2012)

Routh’s Stability Criterion:

It is also called Routh’s array method or Routh-Hurwitz’s methodRouth suggested a method

of tabulating the coefficients of characteristic equationin a particular way. Tabulation of

coefficients gives an array called Routh’s array.

Consider the general characteristic equation as,

F(s) = a0 sn + a1 s

n-1 + a2 s

n-2 + ….. + an = 0.

Method of forming an array :

Sn a0 a2 a4 a6 ……….

Sn-1

a1a3 a5 a7

Sn-2

b1 b2 b3

Sn-3

c1 c2 c3

- - - -

- - - -

S0

an

Coefficients of first two rows are written directly from characteristics equation.


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From these two rows next rows can be obtained as follows.

a1 a2 – a0 a3 a1 a4 – a0 a5 a1 a6 – a0 a7

b1 = , b2 = , b3 =

a1 a1a1

From 2nd

& 3rd

row , 4th row can be obtained as

b1 a3 – a1 b2 b1 a5 – a1 b3

C1 = , C2 =

b1 b1

This process is to be continued till the coefficient for s0 is obtained which will be an. From this

array stability of system can be predicted.

Routh’scriterion :

The necessary & sufficient condition for system to be stable is “ All the terms in the first column

of Routh’s array must have same sign. There should not be any sign change in first column of

Routh’s array”.

If there are sign changes existing then,

1. System is unstable.

2. The number of sign changes equals the number of roots lying in the right half of the

s-plane.

Limitation of Routh’scriterion :

1. It is valid only for real coefficients of the characteristic equation.

2. It does not provide exact locations of the closed loop poles in left or right half of s-plane.

3. It does not suggest methods of stabilizing an unstable system.

4. Applicable only to linear system.


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2. A unity feedback control system is characterized by the open loop transfer function

3. Define: i) Marginally stable systems; ii) absolutely stable system; iii) conditionally

stable systems. ( 06 Marks, June 2012)

Definition of stable system:

A linear time invariant system is said to be stable if following conditions are satisfied.

1. When system is excited by a bounded input, output is also bounded & controllable.

2. In the absence of input, output must tend to zero irrespective of the initial conditions.

Unstable system:

A linear time invariant system is said to be unstable if,

1. for a bounded input it produces unbounded output.

2. In the absence of input, output may not be returning to zero. It shows certain output

without input.

Besides these two cases, if one or more pairs simple non repeated roots are located on the

imaginary axis of the s-plane, but there are no roots in the right half of s-plane, the output

response will be undamped sinusoidal oscillations of constant frequency & amplitude. Such

systems are said to be critically or marginally stable systems.

Critically or Marginally stable systems:

A linear time invariant system is said to be critically or marginally stable if for a bounded

input its output oscillates with constant frequency & Amplitude. Such oscillation of output

are called Undamped or Sustained oscillations.


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For such system one or more pairs of non repeated roots are located on the imaginary axis as

shown in fig6(a). Output response of such systems is as shown in fig6(b).

C(t)

Constant Amplitude &frequency oscillations

X J 2 ------------------------------------

----------------------------- steady state output

X J 1 ------------------------------------

t

X - J 1

X - J 2

Fig 6(a)

non repeated poles on J axis.

If there are repeated poles located purely on imaginary axis system is said to be unstable.

C(t)

J 1 x x

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

J 1 x x steady state output

s-plane t

Conditionally Stable: A linear time invariant system is said to be conditionally stable, if for a

certain condition if a particular parameter of the system, its output is bounded one. Otherwise if

that condition is violated output becomes unbounded system becomes unstable. i.e. Stability of

the system depends the on condition of the parameter of the system. Such system is called

conditionally stable system.

S-plane can be divided into three zones from stability point of view.


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J axis

Left half of s-plane Right half of s-plane

Stable UnstableReal

S-Plane

(repeated)

unstable (non repeated roots)

Marginally stable

4. Define the term stability applied to control system sand what is the difference

between absolute stability and relative stability. ( 4 Marks, Dec 2012)

A linear time invariant system is stable if following conditions are satisfied.

For bounded input, the response of the system should be bounded and

controllable.

In absence of input the output should be zero.

The location of roots of the characteristic equation for a stable system is in

left half of an s-plane.

Absolute stability: same above 3 conditions.

Relative stability: The response decays to zero much faster as compare to the poles

close to jw – axis.

When poles are located far away from jw-axis or imaginary axis in LHS of S-plane.

4. Using Routh’s stability criterion determine the stability of the following systems:

i) Its open loop transfer function has poles at s = 0, s= -1, s = -3 and s = -

5. Gain K = 10.


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ii) It’s a type 1 system with an error constant of 10/sec and poles at s = -3

and s = -6 ( 8 Marks, Dec 2012)

i) Poles at s = 0, s= -1, s = -3 and s = -5. Gain K = 10.

+ +13s+50 = 0

No sign change in first column therefore it is a stable system.

ii) It is a type – 1 system

There is a pole at s =0

poles at s = -3 and s = -6 and Kv = 10/sec

Type - 1

K = 180

Thus the characteristic equation


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S3+9S

2+18S+180 = 0

There are two sign changes so two roots are located in RHS of s-plane. Hence the system

is unstable.

5. Using R-H Criterion determine the stability of the system having the characteristic

equation, S4 + 10S

3 + 36S

2 + 70S + 75 = 0 has roots more negative than S = -2.



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UNIT – 5

1. Sketch the root locus for a unity feedback control system with open loop transfer

function: ( 12 Marks, june 2012)

25)6ss(s

KG(s)

1) Root locus is symmetrical about real axis.

2) There are no open loop zeros (m=0). There are three open loop poles at s=-0,

-3 j4(n=3). Three branches of root loci start from the open loop poles when K= 0 and all

the three branches go asymptotically when K .

3) Entire negative real axis lies on the root locus as there is a single pole at s=0 on the real

axis.

4) The asymptote angle is A = .2,1,01,1,0,)12(180

mnqmn

q

Angle of asymptote are A = 60 , 180 , 300 .

5) Centroid of the asymptote is mn

zeros)of(sumpoles)of(sumσA

0.23

)33(

6) The brake points are given by dK/ds =0.

j18.0434K

j2.0817and2s

02512s3sds

dK

25s)6s(s25)6ss(sK

1,2

1,2

2

232

For a point to be break point, the corresponding value of K is a real number greater than

or equal to zero. Hence, S1,2are not break points.

7) Angle of departure from the open loop pole at s=0 is 180 . Angle of departure from

complex pole s= -3+j4 is

zeros) from inquestion polecomplex a to vectorsof angles theof (sum

poles)other fromquestion in polecomplex a to vectorsof angles theof (sum

180p


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87.36)903

4tan180(180 1

p

Similarly, Angle of departure from complex pole s= -3-j4 is

36.87or323.13)270(233.13180φp

8) The root locus does cross the imaginary axis. The cross over point and the gain at the

cross over can be obtained by

Rouths criterion

The characteristic equation is 0K25s6ss 23 . The Routh’s array is

For the system to be stable K < 150. At K=150 the auxillary equation is 6s2+150=0.

s = ±j5.or

substitute s= j in the characteristic equation. Equate real and imaginary parts to zero. Solve

for and K.

The plot of root locus is shown in Fig.5.

6s6

K150s

K6s

251s

0

1

2

3

1500,Kj50,ω

025ωjωK)6ω(

0Kjω25jω6jω

0K25s6ss

22

23

23


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2. The open loop transfer function of a feedback control system in

Check whether the following points are on the root locus. If so, find the value of K at

these points, i) S = -1.5 ii) S = -0.5 + j2 ( 6 Marks, Dec 2012)


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3. Sketch the root locus plot for a negative feedback control system characterized by

an open loop transfer function,

Comment on stability. (14 Marks, Dec 2012)

4. Define brake away / in point on a root locus. Explain any one method of

determining the same.

(6 Marks, Dec 2010)

Breakaway and breakin points.The breakaway and breakin points either lie on the real

axis or occur in complex conjugate pairs. On real axis, breakaway points exist between

two adjacent poles and breakin in points exist between two adjacent zeros. To calculate

these polynomial 0ds

dK must be solved. The resulting roots are the breakaway / breakin

points. The characteristic equation given by Eqn.(7), can be rearranged as

)z(s)z)(szK(sA(s)

and )p(s)p)(sp(s B(s) where,

(14)0KA(s)B(s)

m21

n21


Dept. of ECE/SJBIT

The breakaway and breakin points are given by

)15(0Bds

dABA

ds

d

ds

dK

Note that the breakaway points and breakin points must be the roots of Eqn.(15), but

not all roots of Eqn.(15) are breakaway or breakin points. If the root is not on the root

locus portion of the real axis, then this root neither corresponds to breakaway or breakin

point. If the roots of Eqn.(15) are complex conjugate pair, to ascertain that they lie on

root loci, check the corresponding K value. If K is positive, then root is a breakaway or

breakin point.

5. For the following characteristic polynomial 20)4s)(s4s(s

KG(s)

2 draw the

root locus for 0 K . (8 Marks, Dec 2010)

1) Root locus is symmetrical about real axis.

2) There are no open loop zeros (m=0). There are three open loop poles (n=3) at s = -0, -4, -

2 j4. Three branches of root loci start from the three open loop poles when K= 0 and to

infinity asymptotically when K .

3) Root locus lies on negative real axis between s = 0 to s = -4.0 as there is one pole to the

right of any point s on the real axis in this range.

4) The asymptote angle is A = 3,2,1,01,)12(180

mnqmn

q

Angle of asymptote are A = 45 , 135 , 225 , 315 .

5) Centroid of the asymptote is mn

zeros)of(sumpoles)of(sumσA

0.24

)0.40.20.2(

6) The root locus does branch out, which are given by dK/ds =0.

100Kj2.45,2.0s

64;K2.0,s

40)16s2)(4s(s

08040s32s16s8s4s

08072s24s4s

0ds

dKbygivenispointBreak

80s)36s8s(s

20)4s4)(ss(sK

2

1

2

223

23

234

2


Dept. of ECE/SJBIT

The root loci brakeout at the open loop poles at s = -2.0, when K = 64 and breakin and

breakout at s=-2+j2.45, when K=100

7) The angle of departure at real pole at s=0 is 180 and at s=-4 is 0 . The angle of

departure at the complex pole at s = -2 + j4 is -90 .

zeros) from inquestion polecomplex a to vectorsof angles theof (sum

poles)other fromquestion in polecomplex a to vectorsof angles theof (sum

180p

The angle of departure at the complex pole at s = -2 – j4 is 90

8) The root locus does cross the imaginary axis, The cross over point and gain at cross over

is obtained by either Routh’s array or substitute s= j in the characteristic equation and

solve for and gain K by equating the real and imaginary parts to zero.

Routh’s array

The characteristic equation is 0K80s36s8ss 234

Ks26

8K2080s

K26s

808s

K361s

isarrayRouthsThe

0

1

2

3

4

90)90.463(116.6-180

900

8tanθ,4.36

2

4tanθ

116.6)atan2(4,-2θ

116.6or63.42-

4tanθ

p

1

3

1

2

1

1

1

90270

)270296.6(243.4-180p


Dept. of ECE/SJBIT

For the system to be stable K > 0 and 2080-8K > 0. The imaginary crossover is given by

2080-8K=0 or K = 260.

At K = 260, the auxiliary equation is 26s2+260 = 0. The imaginary cross over occurs at s=

j 10.

or

The root locus plot is shown below

260K0K36ωω

10js;10j0,ω080ω8ω

zerotopartsimaginaryandrealEquate

080ω8ωjK36ωω

0Kjω80jω36jω8jω

jωsput

0K80s36s8ss

24

3

324

234

234


Dept. of ECE/SJBIT

UNIT – 6

1. State the advantages and limitations of frequency domain approach.


Advantages:

2. Determine the transfer function, of a system whose asyptotic gain plot is shown in fig.

( 10 Marks, Dec 2012)

Starting slope at 0dB = no pole or zero at the origin.

Wc1 = 1 factor is 1 / 1+ T1 S = 1/ 1+S

Shift at W = 1 is 0dB.

20 log K = 0dB

Therefore K = 1

Next change in slope occur at W2 = Wc2 .


Dept. of ECE/SJBIT

3. List the effects of lead compensation. ( 4 Marks, Dec 2012)

The lead compensation adds a dominant zero and a pole, this increases the

damping of the closed loop system.

Increasing damping means less overshoot, less rise time , less settling

time.

It improves phase margin.

Since it improves phase and gain margin result in improving the relative

stability.

4. Explain Nyquist’s stability criterion. ( 4 Marks, Dec 2010 , June 2010)

It states that the number of unstable closed-loop poles is equal to the number of unstable

open-loop poles plus the number of encirclements of the origin of the Nyquist plot of the

complex function D(S) .

This can be easily justified by applying Cauchy’s principle of argument to the function D(S)with

the -plane contour. Note that Z and P represent the numbers of zeros and poles, respectively, of

D(S) in the unstable part of the complex plane. At the same time, the zeros of D(S) are the

closed-loop system poles, and the poles of D(S) arethe open-loop system poles (closed-loop

zeros).


Dept. of ECE/SJBIT

The above criterion can be slightly simplified if instead of plotting the function ,

D(s) = 1+ G(S) H(S)

we plot only the function G(S) H(S) and count encirclement of the Nyquist plot of G(S) H(S)

around the point ( -1, j0) so that the modified Nyquist criterion has the following form. The

number of unstable closed-loop poles (Z) is equal to the number of unstable open-loop poles (P)

plus the number of encirclements (N) of the point ( -1, j0) of the Nyquist plot of G(S) H(S) ,

that is

Z = P + N


Dept. of ECE/SJBIT

UNIT – 7

1.

(Dec 2012)

a)

GH(j ) GH(j )

0 1.5626 0

0.5 1.5 -24.73

1 1.34 -47.72

5 0.307 -151.54

10 0.01 -198.22

100 0.0001 -261.18

Im


Dept. of ECE/SJBIT

c)


Dept. of ECE/SJBIT

UNIT – 8

1.

June 2012

state-transition matrix is a matrix whose product with the state vector at an initial

time gives at a later time . The state-transition matrix can be used to obtain the

general solution of linear dynamical systems. It is also known as the matrix exponential.

Properties of state transition matrix

Let A ∈ Rn×n. Then the following hold:

(a) eAt1

eAt2

= eA(t1+t2)

for all t1, t2 R;

(b) AeAt

= eAt

A for all t R;

(c) (eAt

)−1

= e−At

for all t R;

(d) det eAt

= e(trA)t

for all t R;

(e) If AB = BA, then eAt

eBt

= e(A+B)t

for all t R.

Dec 2012

State: A state of a system is the current value of internal elements of the system, that

change separately (but not completely unrelated) to the output of the system. In essence,

the state of a system is an explicit account of the values of the internal system

components.


Dept. of ECE/SJBIT

State Variables

The state variables represent values from inside the system, that can change over time. In

an electric circuit, for instance, the node voltages or the mesh currents can be state

variables. In a mechanical system, the forces applied by springs, gravity, and dashpots

can be state variables.

We denote the input variables with u, the output variables with y, and the state variables

with x. In essence, we have the following relationship:

Where f(x, u) is our system. Also, the state variables can change with respect to the

current state and the system input:

Where x' is the rate of change of the state variables. We will define f(u, x) and g(u, x) in

the next chapter.

State space: In a state space system, the internal state of the system is explicitly

accounted for by an equation known as the state equation. The system output is given in

terms of a combination of the current system state, and the current system input, through

the output equation. These two equations form a system of equations known collectively

as state-space equations. The state-space is the vector space that consists of all the

possible internal states of the system. Because the state-space must be finite, a system can

only be described by state-space equations if the system is lumped.

b) advantages:

1. it can be applicable for for nonlinear and time varying system.

2. It can be applied to MIMO

3. Any type of of input can be considered for designing the system

4. Method involves matrix algeebra can be made conviniently adapted for

the digital computers.


Dept. of ECE/SJBIT

c)

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