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UNIT - 8Introduction to State variable analysisQ1) Obtain the state space representation for the given electircal system

A1) The state model is given as

Fig 1 State ModelThe 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 asY(t)=CX(t)+DU(t)V0=Vc= x1(t) ….(a)Hence output equation becomesV0= 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=

Q2) Obtatin the state space representation for the given electircal system

A2) 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).

Fig 2 State ModelHere 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=

Q3) Write the solution for non-homogenous state equation?A3) Let the scalar state equation be

=ax+bu

-ax=bu

Multiplying both sides by e-at

x(t) = eat x(0) +d

Then for non-homogeneous state equation

=Ax+Bu

The solution x(t) can be given as

x(t) = eAt x(0) +d

The above equation is the required solution. Q4) Write solution for homogenous state equation?A4) The first order differential equation is given as

=ax x(0)=x0

x(t)=x0=

Consider state equation

=Ax(t); x(0)=x0

The solution will be of the form

x(t)=a0+a1t+a2t2+a3t3+….+aiti

substituting value in above equation

a1+2a2t2+3a3t3+……….=A[a0+a1t+a2t2+a3t3+….+aiti]

a1=Aa0

a2=Aa1=A2a0

ai=Aia0

Solution for x(t) will be

x(t)=[I+At+A2t2+………+Aiti]x0

The matrix exponential form can be written as

The solution x(t) will be x(t)=x0

=Ax(t)+Bu(t) and x(0)=x0

- Ax(t)=Bu(t)

Multiplying both sides by

[ - Ax(t)]==Bu(t)

Integrating both sides w.r.t t we get

= d

Multiplying both sides by

x(t)=x(0)+d

x(0) is called as Homogeneous solution

dis the Forced solution

At t=t0

x(t)=x(t0)+d

The above equation is the required solution

Q5) Derive the state equation and explain the state variables?A5) 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.