
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=VILRVC == = …………….(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= A 
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 IIL=IC = = = =x2(t)+ ……..(a) =[0 1/C]+ [1/C 0] …….(b) == Applying KVL in the given electrical circuit we get VC=VL+ILR VCILR=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= 
= 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 
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= 
=+ 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)=+ =+ 
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= 
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 
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= 
T(s)= = P1=2/s P2=8/s2 L1=10/s L2=100/s2 
=x2 =10x2100x1+u =+ 
==x2 ==x3 = The above differential equation than becomes +6+11 x2+6 x1=u (t) =u6(t)11 x26 x1 Hence the state equation will be = + [u] 
=+ =x1 =6x15x2 From Equation (14) = C{[SIA]1B} + D =C{}1B+D =C B+D =[1 0] + 0 =+0 = 
Y(t)=Cx(t)+Du(t) (1) (2) Taking L.T of equation (1) Y(s)=CX(s)+DU(s) (3) Taking L.T of equation (2) SX(s)=AX(s)+BU(s) (4) X(s)=[SIA]1BU(s) (5) Y(s)=CX(s)+DU(s) Y(s)=C{[SIA]1BU(s)} + DU(s) = C{[SIA]1B} + D (6) [SIA]1= The denominator of equation (6) is the characteristic equation [SIA]=0 