UNIT - 4

Introduction to State Space Analysis

The state of a system is a minimal set of variables known as state variables. The knowledge of these variables at any instance of time together with the knowledge of the inputs for the same instance of time, determines the complete behaviour of the system. The fewer drawbacks in the transfer function method for representing any system led to the use of state variables in analysis of system. Few advantages are listed below:

- The state space can be used for linear or nonlinear, time-variant or time-invariant systems.
- It is easier to apply where Laplace transform cannot be applied.
- The nth order differential equation can be expressed as 'n' equation of first order.
- It is a time domain method.
- As this is time domain method, therefore this method is suitable for digital computer computation.
- On the basis of the given performance index, this system can be designed for an optimal condition.

Representation of state space: The system shown below has ‘m’ inputs, ‘p, outputs and ‘n’ number of state variables. The state equation gives us the relation between the state variables and the inputs.

So, the above system shown can be described through equations as

=f1 (x1, x2,…….xn , u1, u2……… ,um)

= f2 (x1, x2,…….xn , u1, u2……… ,um)

=f1 (x1, x2,…….xn , u1, u2……… ,um) (1)

The above set of equations can be represented as

=f(x(t),u(t)) (2)

As we are concerned for time invariant system, for which the term in (1) is linear combination of state variables and input. So,

(t)=2tx1+x2+u1+u2 (3)

=a11x1(t)+a12x2(t)+…a1nxn(t)+b11u1(t)+b12u2(t)+…….+b1mum(t)

=a21x1(t)+a22x2(t)+…a2nxn(t)+b21u1(t)+b22u2(t)+…….+b2mum(t)

=an1x1(t)+an2x2(t)+…annxn(t)+bn1u1(t)+bn2u2(t)+…….+bnmum(t)

The above equation can be represented in matrix form as given below

(4)

The above coefficients aij and bji in equation (4) can be written in vector matrix form as

=Ax(t)+Bu(t) (5)

The output of the system can be represented by linear combination of state variables and inputs.

y1=c11x1(t)+c12x2(t)+…….+c1nxn+d11u1+………..+d1mum

y2=c21x1(t)+c22x2(t)+…….+c2nxn+d21u1+………..+d2mum

yr=cr1x1(t)+cr2x2(t)+…….+crnxn+dr1u1+………..+drmum (6)

The equation (6) can be represented in matrix form as (7)

Where coefficients cij and dji are constants. The output equation is given as

Y=CX+DU (8)

Calculation of transfer function from state model:

Fig: State model

The above figure shows the state model of a system with two inputs u1(t) and u2(t), and having outputs y1(t) and y2(t). As discussed in above section 4.1, we know that the output equation is given as

Y(t)=Cx(t)+Du(t) (9)

(10)

Taking L.T of equation (9)

Y(s)=CX(s)+DU(s) (11)

Taking L.T of equation (10)

SX(s)=AX(s)+BU(s) (12)

X(s)=[SI-A]-1BU(s) (13)

Y(s)=CX(s)+DU(s)

Y(s)=C{[SI-A]-1BU(s)} + DU(s)

= C{[SI-A]-1B} + D (14)

[SI-A]-1=

The denominator of equation (14) is the characteristic equation

[SI-A]=0

Physical variable form: The state variables selected here are the physical quantities of the system, which can be measured. The selection of these variables can be directly related to the physical system because the solution of state equation is related to the time variation of the system variables.

The number of energy storing elements in any system is equal to the number of state variables. Below shown are few electrical circuits, just to brush up the concept of energy storing elements and state variable relation.

Now calculating the state equation and output equation for state variable analysis.

State Equation and Output equation:

Number of outputs=Number of output equation

Fig State Model

Considering multiple input and multiple output system, with two inputs u1(t)and u2(t), and two outputs y1(t) and y2(t) respectively.

y1(t)=c11x1(t)+c12x2(t)+d11u1(t)+d12u2(t)

y2(t)=c21x1(t)+c22x2(t)+d21u1(t)+d22u2(t)

=+

The output equation is given as

Y(t)=CX(t)+DU(t)

Y(t)= X(t)=

C= D=

U(t)=

# Now finding State Equation

Number of energy storing elements= Number of state variables ==a11x1(t)+a12x2(t)+b11u1(t)+b12u2(t)

==a21x1(t)+a22x2(t)+b21u1(t)+b22u2(t)

=+

The state equation is then given as

=Ax(t)+Bu(t)

Question: Obtatin the state space representation for the given electircal system

Solution: The state model is given as

The state model shows that there are two energy storing elements L, C. As we already know that Number of state variables is equal to the number of energy storing elements.Hence we have two state variables[x1(t) and x2(t)]. We have one output V0(taken across capacitor) and input u(t).

The output equation is then given as

Y(t)=CX(t)+DU(t)

V0=Vc= x1(t) ….(a)

Hence output equation becomes

V0= x1(t)

y(t)=[1 0]+[0]u(t)

So, C=[1 0] D=[0]

Now writing the state equation

=Ax(t)+Bu(t)

For that applying KVL in the above circuit

V=ILR+L+Vc

State Equation is =Ax(t)+Bu(t)

=

x1(t)=Vc

=

IL=C

=

==

=(1/C) x2(t) ……….(b)

==

VL=L

==VL/L

From KVL

L=VL=V-ILR-VC

==

= …………….(c)

From equation (b) and (c)

= [0 1/c]+ [0 0]u(t)

=[-1/L -R/L] + [1/L 0]

Now writing the state equation

=Ax(t)+Bu(t)

=+

Hence A= B=

Question: Obtatin the state space representation for the given electircal system

Solution: The state model shows that there are two energy storing elements L, C. As we already know that Number of state variables is equal to the number of energy storing elements.Hence we have two state variables[x1(t) and x2(t)]. We have one output V0(taken across capacitor) and input u(t).

Here output is V0. But from above electrical circuit V0=ILR

V0=x2(t)R

y(t)= V0= [0 R] + [0] u(t)

The output equation is given as

Y(t)=CX(t)+DU(t)

C=[0 R] D=[0]

Now finding state equation,we apply KCL in the given electrical circuit

I=IC+IL

=

But I-IL=IC

=

=

=

=x2(t)+ ……..(a)

=[0 1/C]+ [1/C 0] …….(b)

==

Applying KVL in the given electrical circuit we get

VC=VL+ILR

VC-ILR=VL=L

=

=[1/L -R/L] + [0] u(t) ………….(c)

From equation (b) and (c) we have

Now writing the state equation

=Ax(t)+Bu(t)

=+

Hence A= B=

Note: We should always take voltage across the inductor L, and current through capacitor C.

Mechanical form: In this type of systems the potential and kinetic energy of a mass are function of position and velocity of the mass respectively. Hence, the velocity and position are selected as state variables.

Question: Write the state equation for the mechanical system shown below?

The free body diagram is

The position and velocity are state variables hence,

x1(t)=x(t)

x2(t)=

==x2(t)

=

The system equation will be

=

=

=

The state equations are

= x2(t)

=

=+f(t)

The phase variables are defined as the state variables obtained from any one of the system variables and its derivatives. The nth order linear differential equation for given input y(t) and output u(t) is given as

y(n)+a1y(n-1)+………..+an-1+any=bou(m)+……+bm-1+ bm u. ….. (15)

The initial conditions are expressed as y(0),,…..,y(n-1)(0).

The transfer function assuming all initial conditions zero is

T(s)==[b0sm+b1sm-1+….+bm-1s+bm/[sn+a1sn-1+……..+an-1s+an] ……(16)

Considering simple case when there are no zeros in the system, the transfer function becomes

T(s)==[b]/ [sn+a1sn-1+……..+an-1s+an] …….(17)

For above transfer function the differential equation is given as

y(n)+a1y(n-1)+………..+an-1+any=b u …….(18)

Let x1=y

x2=

…….

xn=y(n-1)

From (18) the differential equation is given as

……….

=-anx1-an-1x2-…….-a1xn+bu

=Ax+Bu ……….(19)

Note: Matrix A is known as Bush form or companion form. Matrix B has all elements zero except the last element.

The output equation is given as y=x1

y=Cx ……….(20)

Where C =[1 0 0………..0]

The phase variable formation for the transfer function can be obtained by the below mentioned two canonical forms.

Calculation of state model from Transfer function

This can be implemented by three ways as listed below:

a) By Block diagram method (Jordan’s Canonical form)

b) By Signal flow graph

c) By differential equation

By Block diagram method (Jordan’s Canonical form)

Q1) The closed loop transfer function is given as T(s)= . Calculate the state model?

Sol: The transfer function can be simplified using partial fraction method as

=

S+2=As+10A+Bs+5B

Equating coefficients of s from both sides

A+B=1

Equating coefficients of s0 from both sides

10A+5B=2

Solving above equations we get

A=-3/5

B=8/5

The transfer function will be

T(s)=

=-+

Number of poles= Number of energy storing elements

The output equation is given as

y(t)=k1x1(t)+k2x2(t)

y(t)= [k1 k2] + [0] [u(t)]

y(t)= [-3/5 8/5] + [0] [u(t)]

C=[-3/5 8/5]

D=0

From the above block diagram

=u(t)-5 x1

=-5x1+u(t)

=-10x2+u

Therefore, the state equation is given as

=+

A=

B=

Q2) The closed loop transfer function is given as T(s)= . Calculate the state model?

Sol: The transfer function can be simplified using partial fraction method as

=+

s+7=A[s2+7s+12] +B[s2+6s+8] +C[s2+5s+6]

Equating coefficients of s2 from both sides

A+B+C=0

Equating coefficients of s from both sides

7A+6B+5C=1

Equating coefficients of s0 from both sides

12A+8B+6C=7

Solving above equations and finding values of A, B and C

A=5/2

B=-4

C=3/2

The transfer function will be

T(s)=+

=+

The output equation will be

y=k1x1+k2x2+k3x3

y=[k1 k2 k3][5/2 -4 3/2]

C=[5/2 -4 3/2]

D=[0]

The state equation is given as

=u(t)-2x1

=-2x1+u(t)

=-3x2+u

= -4x3+u

Therefore

= + [u]

A=

B=

Inferences from the above matrices (Jordan’s canonical form)

a) Matrix [A] is diagonal matrix and the diagonals are closed loop poles.

b) Matrix [B] is always of the form

c) Matrix [C] contains residue of poles

d) Matrix [D] is always zero When number of poles ≠ number of zeros

e) When number of poles = number of zeros only change comes in matrix [D]. It is the ratio of coefficient of highest power of s in closed loop transfer function.

Q3) The closed loop transfer function is T(s)=. Find the state equation?

Sol: Here the number of poles = number of zeros

T(s)=

=

=

=2+

=2+

Solving above by partial fraction method

88= As+2A+Bs+3B

A=-B

2A+3B=88

A=-88

B=88

The transfer function becomes

T(s)= 2-

The output equation will be

y(t)=2U+k1x1+k2x2

y=[-88 88]+[2]u

The output equation will be

=u(t)-3 x1

=-3x1+u(t)

=-2x2+u(t)

Therefore, the state equation is given as

=+

A=

B=

b) By signal flow graph

The transfer function id given as

T=

Limitations:

1) =1 for all k

2) All loops should be touching

Number of terms in numerator= Number of forward paths

Number of terms in denominator= Number of loops

Q1) The CLTF T(s)=. Find the state equation?

Sol: T(s)=

=

P1=2/s

P2=-8/s2

L1=-10/s

L2=-100/s2

The state equation will be

=x2

=-10x2-100x1+u

=+

C) By using differential equation:

Q1) Find the state equation from the given differential equation

Sol: Let y=x1

==x2

==x3

=

The above differential equation than becomes

+6+11 x2+6 x1=u

(t) =u-6(t)-11 x2-6 x1

Hence the state equation will be

= + [u]

Calculation of Transfer function from State model:

Refer Section 4.1

Q1) For the given data below compute the transfer function of the system

A= B= C=[1 0] and D=[0]

Sol: The state equation will be

=+

=x1

=-6x1-5x2

Equation (14) in section 4.1

= C{[SI-A]-1B} + D

=C{-}-1B+D

=C B+D

=[1 0] + 0

=+0

=

In armature-controlled motors the time constant is small and hence response is fast the efficiency is better than field cannot. The transfer function is calculated below.

Fig Armature controlled d.c servomotor.

Applying kVL in armature circuit

V = Raia+ Ldia/dt +E-----(21)

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 motor

But Eb= Kb w

Eb= kbdø/dt --------(22)

W =angular velocity

Kb= back emf

T= k ia-----(23)

The equation for torque will be

T= Jd2ø/dt2+Bdø/dt --------(24)

- In mechanical systems the energy is stored in mass as the function of position and velocity. Therefore, the state variables are . The current ia is another state variable.

Let x1=

x2= and x3=ia

=x2

Eb= kbdø/dt

T= Jd2ø/dt2+Bdø/dt

T= k ia

Jd2ø/dt2+Bdø/dt= k ia

V(t)-Eb=Raia+ Ldia/dt

V(t)- kbdø/dt = Raia+ Ldia/dt

From above equations we can write

J+B x2=k x3

V(t)- kb x2=Ra x3+L

Simplifying above two equations we get

=

=

The matrix is given as

= + v(t)