Unit-1

Introduction to Control Systems

1.1 Basic Components of a control system

A control system is a set of mechanical or electronic devices that control the other devices. In the control system behaviours of the system are desired by differential equations.

Fig 1 Block diagram for the closed-loop control system

The above figure shows a feedback system with a control system. The components of the control system are:

i)                   The actuator takes the signal transforms it accordingly and causes some action.

ii)                 Sensors are used to measure continuous and discrete process variables.

iii)              Controllers are those elements that adjust the actuators in response to the measurement.

Key takeaway:

1)     The feedback control system is having a sensor.

2)     These sensors help to improve accuracy by generating an error signal.

1.2 Feedback and its effect

Open-loop systems:

This is the loop of the control system without any feedback. In this, the control action is not dependent on the desired output.

Fig. 2 open-loop control system

Examples of open-loop systems are the traffic signals, Automatic washing machine, and in fields control d.c. motor.

G(s) = c(s)/R(s)          -------(1)

Closed-loop systems:

This is a type of control system with feedback. In this type of system, the control action is dependent on the desired output.

Fig 3 Closed Loop control system

The error signal is again fed to the controller to the error and get the desired output.

G(s) – forward amplification

H(s) _ Reverse amplification

The above system is with negative feedback.

Positive feedback is used only in oscillators and another user is not known as discussing only negative feedback.

[ R(s) – c(s) H (S) ] G(s) = c(s)

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

C(s)/R(S) = G(S)/1+G(s) H(s)

S0, the transfer function of closed Loop system is

 C(s)/ R(S) = G(s)/ 1+G(S) H(S)      --------(2)

     1+ G(S) H(s) = 0 [ characteristic Equation]

The transfer function of a linear time-invariant system is the ratio of Laplace Transform

Comparisons between Open Loop and Closed Loop system

Key takeaway:

1)     The open-loop systems are without feedback.

2)     The closed-loop systems use feedback. In a closed Loop, system stability depends on system components

1.3 Types of feedback control systems

There are two types of feedbacks

i)                   Positive Feedback

Fig 3 Closed Loop control system

In positive feedback systems, the feedback signal is added to the input signal. The positive feedback is used only in oscillators. The transfer function is given as

C(s)/ R(S) = G(s)/ 1-G(S) H(S)

ii)                Negative Feedback

In negative feedback, the error signal is inverted at the input signal. The discussion in this unit is based on this negative feedback system only. The transfer function is given by

C(s)/ R(S) = G(s)/ 1+G(S) H(S)   

Key takeaway

1.4 Block diagrams Reduction and signal flow graphs

1.4.1 Block Diagram Reduction

Advantages of Block diagram reduction technique:

1. Very simple to Construct the Block diagram of complicated electrical & mechanical systems.

2. The function of the individual elements can be visualized form a block diagram

3. Individual, as well as overall performance of the system, can be studied by the figure shown in Block diag.

4. Overall CLTF can be easily calculated by Block diagram reduction rules.

Disadvantages of Block diagram reduction technique:

It does not include any information above the physical construct of the system (a completely mathematical approach).

a) Source of energy is generally not shown in the block diagram so w.gdiff. block diagram can be drawn for the same function

CLTf:-ve feedback

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

CLTF:-+ve feedback

C(S)/R(S) = G(S)/1-G(s)H(S)

Rules of Block diagram Algebra:

Block in cascade

Moving summing point after a block

Moving summing point ahead of the block

Moving to take-off point after a block

Moving to take-off point ahead of a block

Eliminating a feedback Loop

Q. Reduce given B.D to canonical (simple form) and hence obtain the equivalent Tf = c(s)/ R(S)?

Sol:-

C(S)/R(S) = (G1G2) (G3+G4)/1+G1G2H1)/1-G1,G2(G3+G4) H2/1+G1G2H1

= G1G2(G3+G4)/1+G1G2H1-G1G2H2(G3+G4)

=G1G2(G3+G4)/1+(H1-H2)(G1G2) (G3+G4)

C(s)/R(S) = G1G2(G3+G4)/1+(H1-H2(G3+G4)) G1 G2

C(s)/R(s)=  G1(G3+G2)/(1-G1G3X1) (1-G2X2) H1

= G(G3+G2)/(1-G3G1H1) (1-G2H2) + G1H1(G3+G2)

= G1(G3+G2)/1-G3G1H1-G2H2+G1H1(G3+G2H1

=G1(G3+G2)/1-G3H2+G1G2H1(1+G3H2)

1.4.2 Signal flow graph

Block Diagram

SFG

Q:-

Ra+cb =c

c/R= a/1-b

Diagram:-

RULES:-

1. The signal travels along a branch in the direction of an arrow.

2. The input signal is multiplied by the transmittance to obtain the o/p.

3. I/p signal at a node is the sum of all the signals entering at that node.

4. A node transmits a signal at all branches leaving that node.

Q. The SFG shown has forward path and singles isolated loop determine overall transmittance relating X3 and X1­?

Sol:-

X1- I/p node

X2-Intenmediale node

X3- o/p node

ab- forward path (p)

bc- 1 loop (L)

At node XQ:

X2 = x1a + x3c [Add i/p signals at  node]

At node x3:

x2b =x3

 (x1a+x3c) b = x3

X1ab = x3 (1-bc)

X1 = x3 (1-bc)/ab

Ab/(1-bc) = x3/x1

T= p/1-L

X1:- I/p node    x2, x3,x4,x5,Qnlexmedili node

X0:- o/p node  abdeg:- forward path

  bc, ef :- Loop [isolated] 

x2 = ax1+c x3

x3= bx2

x4 = d x3+f x5

x5 = e x4

x6= g x5

x6 = g(e x4) = ge [dx3+ e f x5]

xb = ge [d (bx2) + f (e x4)]

xb = ge [ db (ax1+cx3) + fe (dx3+ fx5)]

xb = ge [db (ax1+cb (ax1+x3) +fe[cdbx2]+

 f( e [db (ax1+ cx3)

x2 = ax1 + cb (x2)  x4 = d bx2 + f exq

x2 = ax1 + cbx2          = db (d4) + fe/1-cb

x2 = ax1/(1-cb)   xy = db x2 + f x6/g

xy = db [ax1]/1-cb + f xb/g

x5 = c db ( ax1)/1-cb + efxb/g

xb = gx5

= gedb (ax1)/1-cb + g efxb/g

Xb = gx5

gedb (ax1)/1-cb + g efxb/g

(1-gef/g) xb = gedb ax1/1-ab

Xb/x1 = gedb a/ (1- ef – bc + beef

Xb/x1 = p/ 1- (L1+L2) + L1 L2 for isolated loops

Key takeaway

1.5 Modeling of Physical systems: electrical networks, mechanical systems elements, free body diagram, analogous Systems

1.5.1 Electrical Networks

Applying Kirchoff’s voltage Law

V= Ri +Ldi/dt +1/c

V= Rdq/dt + L d2q/dt2+q/c

Now ByKirchoff's current low

I= V/R+ 1/L + cdv/dt

But V= dø/dt

I= 1/R dø/dt + 1/L ø+cd2ø/dt2

But v= dø/dt

I= 1/R dø/dt+ 1/L. ø+c d2ø/dt2

Q2) Find V0(s)/Vi(s)?

Sol: Let     Z1   = R2 11 1/c2

=R2*1/c2s/R2+1/c2s

Z1= R2/1+R2c2s

 Let     Z2 = R1+1/c1s

Z2= 1+R1c1s/c1s

V0(s)/vi(s) = z2/z1+z2

      = 1+R1c1s/c1s/R2/1+R2c2s+1+R1c1s/c1s

V0(s)/v0(s) =(c1+R1c1s) (1+R2c2s)/R2c1s+1+sR1c1s2R1R2c1c2

1.5.2 Mechanical Systems and equations of mechanical systems

1.5.2.1 Translational mechanical system:

The motion that takes place along a straight line is called translational motion. The forces that resist motion are - 

Inertia: A body with mass ‘N’ acceleration ‘a’ will produce inertia fm(t)= malt N’ acceleration ‘a’ will produce inertia

fm(t)= M a(t)

In terms of velocity

Fm(t) = M dv(t)/dt

In terms of displacement 

Fm(t)= Md2/dt2µt

Damping force: The damping force is proportional to velocity for vis case function

 F0(t) = B(t) = B dx(t) /dt

 B Damping coefficient N/M sec

Fig 4 Damping Force

Spring force: A spring stores potential energy. The releasing force of a spring is proportional to the displacement.

Fk(t) x(t)

Fk(t) = k x (t)

Fk(t) = k

 K = string constant N/m.

Fig 5 Spring Constant

1.5.2.2 Rotational mechanical system:

The motion of a body about a fixed axis is called Rotational motion. The types of torques which can resist the motion are

Inertial Torque: - The inertial torque is the product of the moment of inertia I and angular acceleration.

TI(t) = J(t)

TI(t) = Jd/dt w(t)

TI(t) = Jd2ø(t)/dt2(N-m)

W (t) – Angular velocity

Ø(t) = Angular displacement 

Damping Torque: - it is a product of damping efficient B and angular velocity w

T0 (t) = B w(t)

T0(t) = B d/dt ø(t)

Spring Torque: - It is the product of torsional stiffness and angular displacement

Tø(t) = k ø(t)

K = N.m/rad

                              Fig 6 Spring Torque

D’Alembert’s Principle

For anybody, the algebraic sum of externally applied forces and the forces resisting motion in any given direction is zero.

Ex.

Fig 7 Mechanical System

In the above figure applying D’Alembert’s principle to write the equation of motion

 As force f(t) actsdownwards all other forces (of k,B,x)actsOpposite to itconsidering f(t) as the and all other forces negative.

F(t)+fm (t) + fD(t) +fx(t) =0

f(t)- M d2x(t)/dt2+B dx(t)/dt +x(t)

Q.1 Draw the free body diagram and write the differential equation for the system below.

Sol:- The free body diagram for M1 will be

F(t) = M1 d2/dt2x1+ B1 d/dt (x1-x2)+ k1(x1-x2)

Similarly for M2we have

K1(x1-x2) + B1d/dt(x1-x2) = k2x2+M2d2/dt2+B2dx2/dt

1.6 Sensors and encoders in control systems

1.6.1 Sensors

A sensor is a device that converts the physical input into a readable output. The basic sensor which we all have known is the keyboard switch sensors.

The four most common sensors are temperature sensors, pressure sensors, level sensors, and flow sensors (flowmeters). Other sensors that may be used in process applications are:

1.6.1 Encoders

Optical Encoders

They are used to convert linear or rotary displacement into digital codes.  They are of two types

i)                   Absolute Encoder: The output is a digitally coded signal with distinct digital code.

ii)                 Incremental Encoder: The output is pulse for each increment of resolution.

Incremental Encoder

These encoders have four parts LED source, a rotary disc a stationary mask, and a sensor. The disc has alternate opaque and transparent segments. As the disc rotates during half of the incremental cycle the transparent sector of the rotating and stationary disc comes in contact. Allowing the light from LED to reach the sensor by generating an electrical pulse.

The output waveform of the sensor of an encoder is triangular or sinusoidal depending upon the resolution. The resolution of an incremental encoder is given by

Resolution = 360o/N

N= number of sectors of disc.

Fig 8 Incremental Encoder

Fig 9 Incremental Encoder Disc track pattern

In a dual-channel encoder, two optoelectronic channels are employed. They are installed in the same rotating disc and are 90o to each other such that their output waves also have 90o displacements. A circuit that senses the relative time phase of the output of the two channels determines the direction of rotation of the disc.

Absolute Encoders

It is a digital transducer as its output does not require any conditioning and it can read the information directly. It has binary code etched on the rotating disc which has many tracks. Each particular least significant increment of resolution has a unique code meaning transparency as 1 and opaqueness as 0.

Key takeaway

The optical encoders have simple construction, low cost, ease of application, and are the most popular encoders.

The main disadvantage of absolute encoders is that the output count is volatile when power is off.

1.7 Modeling of armature controlled and field controlled DC servomotor.

Armature Controlled dc motor:

In armature-controlled motors the time constant is small and hence response is fast. The efficiency is better than field control. The transfer function is calculated below.

Fig.10 Armature controlled d.c servomotor.

Applying kVL in the armature circuit

V = Raia+ Ldia/dt +E-----(3)

Where

V= applied voltage

Ra = armature resistance

La = Armature inductance

Ia= Armature current

If= field current

E= Induced emf in armature

T= Torque developed by the motor

But   E= Kb w

E= kbdø/dt  --------(4)

W =angular velocity

Kb= back emf

T= k ia-----(5)

The equation for torque will be

 T= Jd2ø/dt2+Bdø/dt --------(6)

Taking Laplace of equation (3), (4), (5), (6)

 V(S) – E(S) = I1(s) (Ra+sLa)

 E(s) = kb s(s)

T(s) = k Ia (s)

T(S)= (s2J+SB) (S)

T(S) = (SJ+B) s(S)

(S)/v(s)= k/(Ra+sla) (Js +B) s+ kkbs ------(a)

The Block diagram representation for the above equation (a) is shown below.

Fig. 11 Block diagram for armatures controlled d.c. motor

Field controlled D.C servomotor:

For field-controlled D.C motor, the ratioL/R is large which means the timing controller for the field circuit is large

Fig.12 circuit for field controlled D.C motor

Applying KVL in field circuit

Vf= Rf If +Lfd If/dt------(7)

 T = k øIa

ø If

ø= kfIf 

:. T= kkfIa If

Let k, = KIa

T= k’ Kf If    --------(8)

Dynamic equation for torque  is

T= Jd2/dt2 +Bd/dt   ------- (9)

Taking Laplace of above equations

Vf(s) = Rf If (c)+SLf If(s) [Rf+SLf]

If= Vf(S)/Rf+ SLf--------(10)

T(s) =(S) [s2J+SB] --------(11)

T(S) = kkfVf(s)/Rf+SLf------ (12)

From equation (11) and (12) we get

(s)/ vf(s) = kkf/s (sJ+B) (Rf+sLf)

The block diagram representation of above equation (12) is shown in figure below

Fig. 13 Block diagram for field-controlled d.c. servomotor

Key takeaway:

1. In armature-controlled motors the time constant is small and hence response is fast. The efficiency is better than field control.

2. For field-controlled D.C motor, the ratio L/R is large which means the timing controller for the field circuit is large

References:

1. (Schaums Outlines Series) Joseph J. Distefano III, Allen R. Stubberud, Ivan J. Williams,

“Control Systems”, 3rd Edition, TMH, Special Indian Edition, 2010.

2. A. Anand Kumar, “Control Systems”, Second Edition, PHI Learning private limited, 2014.

3. William A. Wolovich, “Automatic Control Systems”, Oxford University Press, 2011.