MAHATMA GANDHI INSTITUTE OF TECHNICAL EDUCATION & RESEARCH CENTER

ELECTRONICS & COMMUNICATIONELECTRONICS & COMMUNICATIONSUB:CONTROL SYSTEM ENGINEERING

2nd year (4th sem)

ENROLLMENT NO. : NAME:

140330111001 AMETA KRUSHNAKANT G.

140330111009 POTTBATNI SANJAY A.

140330111012 YAGNIK GAGAN R.

Guided by: Prof. Shruti Dholariya

Transfer FunctionsTransfer Functions

STATE VARIABLE MODELS

We consider physical sytems described by nth-order ordinary differential equation. Utilizing a set of variables, known as state variables, we can obtain a set of first-order differential equations. We group these first-order equations using a compact matrix notation in a model known as the state variable model.

The time-domain state variable model lends itself readily to computer solution and analysis. The Laplace transform is utilized to transform the differential equations representing the system to an algebraic equation expressed in terms of the complex variable s. Utilizing this algebraic equation, we are able to obtain a transfer function representation of the input-output relationship.

With the ready availability of digital computers, it is convenient to consider the time-domain formulation of the equations representing control system. The time domain techniques can be utilized for nonlinear, time varying, and multivariable systems.

dt

)t(dy)t(x

)t(y)t(x

2

1

=

=

yycy)t(uW,yk2

1E,ym

2

1E 2

22

1 δ−δ=δ==

Kinetic and Potential energies, virtual work.

Therefore we will define a set of variables as [x1 x2], where

Lagrange’s equation ( ) ( )y

2121 Qy

EE

y

EE

dt

d =∂−∂−

∂−∂

21 EEL −=Lagrangian of the system is expressed as Generalized Force

)t(uxkxcdt

dxm

)t(ukxdt

dyc

dt

ydm

122

2

2

=++

=++

Equation of motion in terms of state variables.

We can write the equations that describe the behavior of the spring-mass-damper system as the set of two first-order differential equations.

)t(um

1x

m

kx

m

c

dt

dx

xdt

dx

122

21

+−−=

= This set of difefrential equations describes the behavior of the state of the system in terms of the rate of change of each state variables.

As another example of the state variable characterization of a system, consider the RLC circuit shown in Figure 3.

u(t)

Current source

L

C

RVc

Vo

iL

ic

( ) 2c

2

c22L1 vC

2

1dti

C2

1E,iL

2

1E === ∫

The state of this system can be described in terms of a set of variables [x1 x2], where x1 is the capacitor voltage vc(t) and x2 is equal to the inductor current iL(t). This choice of state variables is intuitively satisfactory because the stored energy of the network can be described in terms of these variables.

Figure 3

Therefore x1(t0) and x2(t0) represent the total initial energy of the network and thus the state of the system at t=t0.

Utilizing Kirchhoff’s current low at the junction, we obtain a first order differential equation by describing the rate of change of capacitor voltage

Lc

c i)t(udt

dvCi −==

Kirchhoff’s voltage low for the right-hand loop provides the equation describing the rate of change of inducator current as

cLL viRdt

diL +−=

The output of the system is represented by the linear algebraic equation

)t(iRv L0 =

We can write the equations as a set of two first order differential equations in terms of the state variables x1 [vC(t)] and x2 [iL(t)] as follows:

212

21

xL

Rx

L

1

dt

dx

)t(uC

1x

C

1

dt

dx

−=

+−=Lc i)t(udt

dvC −=

cLL viRdt

diL +−=

The output signal is then 201 xR)t(v)t(y ==

Utilizing the first-order differential equations and the initial conditions of the network represented by [x1(t0) x2(t0)], we can determine the system’s future and its output.

The state variables that describe a system are not a unique set, and several alternative sets of state variables can be chosen. For the RLC circuit, we might choose the set of state variables as the two voltages, vC(t) and vL(t).

In an actual system, there are several choices of a set of state variables that specify the energy stored in a system and therefore adequately describe the dynamics of the system.

The state variables of a system characterize the dynamic behavior of a system. The engineer’s interest is primarily in physical, where the variables are voltages, currents, velocities, positions, pressures, temperatures, and similar physical variables.

The State Differential Equation:

The state of a system is described by the set of first-order differential equations written in terms of the state variables [x1 x2 ... xn]. These first-order differential equations can be written in general form as

mnm11nnnn22n11nn

mm2121nn22221212

mm1111nn12121111

ububxaxaxax

ububxaxaxax

ububxaxaxax

++++=

++++=++++=

Thus, this set of simultaneous differential equations can be written in matrix form as follows:

+

=

m

1

nm1n

m111

n

2

1

nn2n1n

n22221

n11211

n

2

1

u

u

bb

bb

x

x

x

aaa

aaa

aaa

x

x

x

dt

d

n: number of state variables, m: number of inputs.

The column matrix consisting of the state variables is called the state vector and is written as

=

n

2

1

x

x

x

x

THE TRANSFER FUNCTION FROM THE STATE EQUATION

The transfer function of a single input-single output (SISO) system can be obtained from the state variable equations.

uBxAx +=xCy =

where y is the single output and u is the single input. The Laplace transform of the equations

)s(CX)s(Y

)s(UB)s(AX)s(sX

=+=

where B is an nx1 matrix, since u is a single input. We do not include initial conditions, since we seek the transfer function. Reordering the equation

[ ])s(BU)s(C)s(Y

)s(BU)s()s(BUAsI)s(X

)s(UB)s(X]AsI[1

φ=φ=−=

=−−

Therefore, the transfer function G(s)=Y(s)/U(s) is

B)s(C)s(G φ=Example:

Determine the transfer function G(s)=Y(s)/U(s) for the RLC circuit as described by the state differential function

[ ] xR0y,u0C

1x

L

R

L

1C

10

x =

+

−

−=

[ ]

+−=−

L

Rs

L

1C

1s

AsI

[ ]

LC

1s

L

Rs)s(

sL

1C

1

L

Rs

)s(

1AsI)s(

2

1

++=∆

−+

∆=−=φ −

Then the transfer function is

[ ]

LC1

sLR

s

LC/R

)s(

LC/R)s(G

0C

1

)s(

s

)s(L

1)s(C

1

)s(LR

s

R0)s(G

2 ++=

∆=

∆∆

∆−

∆

+

=

2R(s)

1/s

-8

4x1 1/s 1 1/sx3 Y(s)

1

-4

-1.5

2R(s)

-8

1/sx2 1/s 0.75

1

1

Block diagram with x1 defined as the leftmost state variable.

[ ] [ ]0Dand75.011C

0

0

2

B,

010

004

5.148

A

==

=

−−−=

We can use the function expm to compute the transition matrix for a given time. The expm(A) function computes the matrix exponential. By contrast the exp(A) function calculates ea

ij for each of the elements aijϵA.

∫ ττ+= τ−t

0

)t(AAt d)(uBe)0(xe)t(x

∫ τττ−φ+φ=t

0

d)(uB)t()0(x)t()t(x

For the RLC network, the state-space representation is given as:

[ ] [ ]0Dand01C,0

2B,

31

20A ==

=

−−

=

The initial conditions are x1(0)=x2(0)=1 and the input u(t)=0. At t=0.2, the state transition matrix is calculated as

>>A=[0 -2;1 -3], dt=0.2; Phi=expm(A*dt)

Phi =

0.9671 -0.2968 0.1484 0.5219

The state at t=0.2 is predicted by the state transition method to be

=

−=

== 6703.0

6703.0

x

x

5219.01484.0

2968.09671.0

x

x

0t2

1

2.0t2

1

The time response of a system can also be obtained by using lsim function. The lsim function can accept as input nonzero initial conditions as well as an input function. Using lsim function, we can calculate the response for the RLC network as shown below.

t

u(t)

DuCxy

BuAxx

+=+=

System

Arbitrary Input Output

t

y(t)

y(t)=output response at t

T: time vector

X(t)=state response at t

t=times at which response is computed

Initial conditions

(optional)

u=input

[y,T,x]=lsim(sys,u,t,x0)

THE DESIGN OF STATE VARIABLE FEEDBACK SYSTEMS

The time-domain method, expressed in terms of state variables, can also be utilized to design a suitable compensation scheme for a control system. Typically, we are interested in controlling the system with a control signal, u(t), which is a function of several measurable state variables. Then we develop a state variable controller that operates on the information available in measured form.

State variable design is typically comprised of three steps. In the first step, we assume that all the state variables are measurable and utilize them in a full-state feedback control law. Full-state feedback is not usually practical because it is not possible (in general) to measure all the states. In paractice, only certain states (or linear combinations thereof) are measured and provided as system outputs. The second step in state varaible design is to construct an observer to estimate the states that are not directly sensed and available as outputs. Observers can either be full-state observers or reduced-order observers. Reduced-order observers account for the fact that certain states are already available as system outputs; hence they do not need to be estimated. The final step in the design process is to appropriately connect the observer to the full-state feedback conrol low. It is common to refer to the state-varaible controller as a compensator. Additionally, it is possible to consider reference inputs to the state variable compensator to complete the design.

note that the initial condition is included in the transformationsF(s) - f(0+)=Ltf t( )d

d

⋅

s0

∞

tf t( ) es t⋅( )−⋅

⌠⌡

d⋅-f(0+) +=

0

∞

tf t( ) s− es t⋅( )−⋅ ⋅

⌠⌡

d−f t( ) es t⋅( )−⋅=

0

∞vu

⌠⌡

d

we obtain

v f t( )anddu s− es t⋅( )−⋅ dt⋅

and, from which

dv df t( )u es t⋅( )−where

u v⋅ uv⌠⌡

d−=vu⌠⌡

dby the use of

Ltf t( )d

d

0

∞

ttf t( ) e

s t⋅( )−⋅d

d

⌠⌡

d

Evaluate the laplace transform of the derivative of a function

The Laplace Transform

Practical Example - Consider the circuit.

The KVL equation is

4 i t( )⋅ 2ti t( )d

d⋅+ 0 assume i(0+) = 5 A

Applying the Laplace Transform, we have

0

∞

t4 i t( )⋅ 2ti t( )d

d⋅+

e

s t⋅( )−⋅⌠⌡

d 0 40

∞

ti t( ) es t⋅( )−⋅

⌠⌡

d⋅ 2

0

∞

tti t( ) e

s t⋅( )−⋅d

d

⌠⌡

d⋅+ 0

4 I s( )⋅ 2 s I s( )⋅ i 0( )−( )⋅+ 0 4 I s( )⋅ 2 s⋅ I s( )⋅+ 10− 0

transforming back to the time domain, with our present knowledge of Laplace transform, we may say thatI s( )

5

s 2+:=

0 1 20

2

4

6

i t( )

t

t 0 0.01, 2..( )≡

i t( ) 5 e2 t⋅( )−⋅≡

The Partial-Fraction Expansion (or Heaviside expansion theorem)

Suppose that

The partial fraction expansion indicates that F(s) consists of

a sum of terms, each of which is a factor of the denominator.

The values of K1 and K2 are determined by combining the

individual fractions by means of the lowest common

denominator and comparing the resultant numerator

coefficients with those of the coefficients of the numerator

before separation in different terms.

F s( )s z1+

s p1+( ) s p2+( )⋅

or

F s( )K1

s p1+

K2

s p2++

Evaluation of Ki in the manner just described requires the simultaneous solution of n equations.

An alternative method is to multiply both sides of the equation by (s + pi) then setting s= - pi, the

right-hand side is zero except for Ki so that

Kis pi+( ) s z1+( )⋅

s p1+( ) s p2+( )+s = - pi

s -> 0t -> infinite

Lim s F s( )⋅( )Lim f t( )( )7. Final-value Theorem

s -> infinitet -> 0

Lim s F s( )⋅( )Lim f t( )( ) f 0( )6. Initial-value Theorem

0

∞sF s( )

⌠⌡

df t( )

t5. Frequency Integration

F s a+( )f t( ) ea t⋅( )−⋅4. Frequency shifting

sF s( )d

d−t f t( )⋅3. Frequency differentiation

f at( )2. Time scaling

1

aFs

a

⋅

f t T−( ) u t T−( )⋅1. Time delaye

s T⋅( )−F s( )⋅

Property Time Domain Frequency Domain

The Transfer Function of Linear Systems

V1 s( ) R1

Cs+

I s( )⋅ Z1 s( ) R

Z2 s( )1

CsV2 s( )1

Cs

I s( )⋅

V2 s( )

V1 s( )

1

Cs

R1

Cs+

Z2 s( )

Z1 s( ) Z2 s( )+

The Transfer Function of Linear Systems

The Transfer Function of Linear Systems

The Transfer Function of Linear Systems

Block Diagram Models

Block Diagram Models

Original Diagram Equivalent Diagram

Original Diagram Equivalent Diagram

Block Diagram Models

Original Diagram Equivalent Diagram

Original Diagram Equivalent Diagram

Block Diagram Models Example 2.7

Signal-Flow Graph Models

For complex systems, the block diagram method can become difficult to complete. By using the signal-flow graph model, the reduction procedure (used in the block diagram method) is not necessary to determine the relationship between system variables.

Y1 s( ) G11 s( ) R1 s( )⋅ G12 s( ) R2 s( )⋅+

Y2 s( ) G21 s( ) R1 s( )⋅ G22 s( ) R2 s( )⋅+

a11 x1⋅ a12 x2⋅+ r1+ x1

a21 x1⋅ a22 x2⋅+ r2+ x2

Example 2.8

Y s( )

R s( )

G 1 G 2⋅ G 3⋅ G 4⋅ 1 L 3− L 4−( )⋅ G 5 G 6⋅ G 7⋅ G 8⋅ 1 L 1− L 2−( )⋅ +

1 L 1− L 2− L 3− L 4− L 1 L 3⋅+ L 1 L 4⋅+ L 2 L 3⋅+ L 2 L 4⋅+

Example 2.10

Y s( )

R s( )

G 1 G 2⋅ G 3⋅ G 4⋅

1 G 2 G 3⋅ H 2⋅+ G 3 G 4⋅ H 1⋅− G 1 G 2⋅ G 3⋅ G 4⋅ H 3⋅+

Y s( )

R s( )

P1 P2 ∆2⋅+ P3+

∆

P1 G1 G2⋅ G3⋅ G4⋅ G5⋅ G6⋅ P2 G1 G2⋅ G7⋅ G6⋅ P3 G1 G2⋅ G3⋅ G4⋅ G8⋅

∆ 1 L1 L2+ L3+ L4+ L5+ L6+ L7+ L8+( )− L5 L7⋅ L5 L4⋅+ L3 L4⋅+( )+

∆1 ∆3 1 ∆2 1 L5− 1 G4 H4⋅+