EC2255 - CONTROL SYSTEMS

UNIT – CONTROL SYSTEM MODELING

CONTROL SYSTEMS:

A number of elements or components are connected in a sequence to perform a specific

function, the group thus form is called a SYSTEM.

In a system, when the output quantity is controlled by varying the input quantity, the system

is called CONTROL SYSTEM.

Open loop system:

Any physical system which does not automatically correct the variation in its output is called

an open loop system.

Closed loop system:

Control systems in which the output has an effect upon the input quantity in order to maintain

the desired output value are called closed loop system.

Advantages of open loop system:

It is much simpler and less expensive

No sensors are needed to measure the variables to provide feed back

The open loop systems are easier to construct

Generally the open loop system are stsble

Disadvantages of open loop system:

This systems are inaccurate and unreliable

The changes in the output due to external disturbance are not corected automatically

There is no compensation for any disturbances entering into the system

Its performance is highly depends on the properties of the system

Advantages of closed loop system:

The closed loop system are accurate

It is less affected by noise

It compensates for disturbances

Less sensitive to system parameter variation

Disadvantages of closed loop system:

These systems are more complex and more expensive

The feedback in closed loop system may lead to oscillatory response

The feedback reduces the overall gain of the system

If the closed loop system is not properly designed, the feedback system may lead to

undesirable response

Mathematical models of control systems

Relate the input and output of system with mathematical relation called differential equation.

A system or the mathematical model of a system is linear if it satisfies ar obeys the principle

of superposition and homogenity.

Transfer function

The ratio between laplace transform output to laplace transform input with zero initial

condition called transfer function

Modeling of electrical system

Basic elements

Resister :

Inductor :

Capacitor:

Mathematical Models of Electrical Elements

PROBLEM 1:

Find the transfer function relating the capture voltage Vc(S) and the input voltage V(S) as shown in

the figure.

To find:

apply KVL,

R i(t) + L + = V(t)

(Voltage drop across capacitor is the output voltage Vc(t) )

Note: OUTPUT

Vc(t) Vc(S)

V(t) V(S)

substitute i(t) = C

R C + L + = V(t)

R C + LC + Vc(t) = V(t)

Taking laplace transform,

R C S Vc(S) + L C S2 Vc(S) + Vc(S) = V(S)

Vc(S) [L C S2 + R C S + 1] = V(S)

T.F = =

PROBLEM 2:

Obtain the transfer function of electrical network as shown in the figure.

To find:

Node voltage at V1

+ C1 + =

Take L.T.

+ C1S V1(S) + - =

V1(S) = (1)

Node voltage at V2

=0

Taking L.T.

- + C2 S V2(S) =

V1(S) = V2(S) [ 1 + R2 C2 S] (2)

Substitute equ. (2) in (1)

V2(S) [ 1+ R2 C2 S] =

V2(S) =

V2(S) = E(S)

Mechanical Translation and Rotational System:

Modeling of Mechanical Systems:

Translation System:

Input [Force] Output [Linear Displacement]

Rotational System

Input [Torque] Output [Angular displacement]

Elements of Translation and Rotational systems:

Basic Components of Mechanical Translation Systems:

Mass (M):

Dash Part (B):

Friction between Two Moving Surfaces

Opposing Force is given by,

Taking Laplace transform

Spring (K):

Spring between two moving points

Taking Laplace transform

Basic Components of Mechanical Rotational Systems:

Moment of Inertia:

Taking Laplace Transform

Friction:

Spring:

Analogy Systems:

Differential equation for transfer function mechanical and differential equation of electrical

equation is equal called analogous system.

Force – Voltage Analogy:

Force – Current Analogy:

PROBLEM 3:

Draw the equivalent mechanical system of the given system. Hence write the set of equilibrium

equations for it and obtain electrical analogous circuits using,

(i) F – V Analogy and

(ii) F – I Analogy

Equivalent Circuit

Block Diagram of Closed Loop Systems:

Block Diagram Reduction Techniques:

A block diagram with several summers and pick off points can be reduces to a single block,

by using block diagram algebra, consisting of the following rules.

PROBLEM 4:

Find the overall transfer function of the system, using block diagram reduction technique.

Solution:

Step1: Using rule 6, the feedback loop with G2 and H2 is replaced by a single block

Step2: The two blocks in the forward path are in cascade and can be replaced using rule 1.

Step3: Finally, the loop is replaced by a single block using rule 6

The transfer function can be simplified as,

PROBLEM 5:

Reduce the block diagram and obtain

Solution:

Step1: Move pick off point 2 to left of G2 and combine G2 G3 in cascade. Further G2 G3 and G4

have same inputs and the outputs are added at summer III. Hence they can be added and

represented by a single block.

Step 2: Moving the pick off point 1 to right of block (G2 G3 + G4),

Step 3: Absorbing loop with (G2 G3 + G4) and H2

The transfer function of the System is

Signal Flow Graph Representation of Control System:

Mason’s gain formula is given by,

Where Mk is the kth forward path gain, Δ is the determinant of the graph, is given by,

Where Pmr is the product of the gains of mth possible combination of r non touching loops.

Or

Δ = 1 – (sum of gains of individual loops) + (sum of gain products of possible

combinations of 2 non touching loops) – (sum of gain products of all possible

combinations of 3 non touching loops) + …….

and Δk is the value of Δ for that part of the graph which is non touching with k th forward path.

PROBLEM 6:

Find the transfer function for the following signal flow graph

Solution:

No. of forward paths K = 2

Block Diagram to Signal Flow Graph:

PROBLEM 7:

Obtain the overall transfer function of the system shown by block diagram, using signal flow graph

techniques.

Solution

Name the various summing and take off points to draw the signal flow graph,

The corresponding signal flow graph is,

System Equation to Signal Flow Graph:

PROBLEM 8:

Construct the signal flow graph for the following set of simultaneous equations.

Solution

The value of the variable is the algebraic sum of all the signals entering at the node,

representing that variable. The variables are X1, X2, X3, X4 while the gains are A21, A23, A31, A32, A33,

A42 and A43

So selecting nodes representing variables and simulating differential equations

Therefore the resultant signal flow graph is

Previous Year University Question Paper:

April/May – 2010

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May/June – 2011

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