Question

Question 5 Following differential equations defines input-output relationships of a system with y as output and r as inputs. d’yı + dy 2 + y, + 5 y, = 10 r, dt ? dt. dy 2 + 1 + 7y, = 8r2 dt dt at a) Define suitable state variables and find the state equation and output equation. [8marks] b) Find system matrix (A), input matrix (B) and output matrix (C). [5marks] c) Draw the state space diagram and find systems transfer function matrix. [7marks]

Answer 1

Question (a)

heyat ayat + y2 + 5y2 = 101

dyz - dy1 + 7yı = 8r dt dt*

From the second equation, we get

dyz - dyı – 7y, +8r2 dtdt

Substituting this equation in the first one, we get

dy 1-7y. + 8r2 + y2 + 5y = 10r1 dt?

ay-% -6y2 + 5y2 = 10ri — Brz

Let us take the state variables as

x1 = y1

X2dt

x3 = y2

Let the inputs be

H=In

U2 = r2

From the above definition, we get one state space equation as

So

x1 = x2

Now substituting the state variables in the differential equation we get

ay-% -6y2 + 5y2 = 10ri — Brz

x2 – x2 - 6x1 + 5x3 = 10u1 - 8uz

x2 = 6x1 + x2 - 5x3 + 10u1 - 8uz

Substituting the state variables in , we get

= - - 7y1+ 8r -

x'z = -x, – 7x1 + 8uz

So the state equations are

x1 = x2

x2 = 6x1 + x2 - 5x3 + 10u1 - 8uz

x'z = -x, – 7x1 + 8uz

The output equations are

yi = y1 = x1

y2 = y2 = x3

Question (b)

The state space representation in matrix form is

93-C E630

[ ៩ 31 -[:] : 1= []

So the matrices are

A = 0 6 1-7 1 1 -1 0 1 -5 0 1

B = O 01 10 -8 lo 8]

1 0 9=3 Lo 0 ]

Question (c)

The state space diagram is as shown below

11 13

Transfer function is given by

Y(s) UC =C(SI - A]-1B+D

[1 0 [s1 – A] = SO 1 Lo 0 0 0 1] - 0 6 -7 1 1 -1 0 1 -5 0]

Is [sl - A] = -6 17 -1 5-1 1 01 5 s

IsI - A = S(S2-5-5)+1(-65 – 35)

|si - Al = 53 – 52 – 5s - 6s - 35

|si - Al = 53 – 52 - 115 - 35

Co factor Matrix of [sl - A] = s(S-1) -(-s) -5 -(-65 – 35) s2 (5) -6-7(5-1)] -(s + 7) s(s -1)-6]

[s- s 6s +351 - 7s Co factor Matrix of [s1 – A] = s 52 -s-7 1-5 -55 52 - 5 - 61

s2-5 Adjoint Matrix of [sl – A] = 6s +35 11-7s S 52 -S-7 -5 -5 -5s 52-5-61

So the inverse

[sl – A]-? = a 2 Adjoint Matrix of [sl – A]

[sl - Aj” = 3 – 52 - 115 - 35 S2- SS 6s +35 52 11-75 -5 -7 -5 -55 52-5-61

So the transfer function will be

Y(s) UC =C(SI - A]-1B+D

Y(S) 1 s2-5 1153 – 52 - 115 - 35 265 + 35 P11-7s s -5 10 01 52 -5s 10 -8 -S-7 SP - $-61108 ] U(S) Lo

Y(S) PO ==-5 -115– 356 91 (S) 0 10s -8s – 40 -8s2 - 40s 'L-10s - 70 8s +56 +8s2 - 85 - 48] 1052 U(s) 53 – 52 - 1

Y(s) U(s) 105 1052 "1-10s - 70 -85 – 40 -882 – 405 852 + 8 53 – 52 – 118

Y(s) 10s US = 53 – 52 - 115 -351-10s - 70 -8s - 407 852 +8]

Y(s) U(S) 10s 33 - 52 - 115 - 35 -105 – 70 153 – 52 - 115 - 35 -8s - 40 53 - 52 - 115 - 35 852 + 8 53 - 52 - 115 - 35

So the transfer function matrix is

105 53 - 52 - 115 - 35 -10s -70 153 - 52 - 115 - 35 -8s - 40 53 - 52 - 115 - 35 8 + 14 53 - 52 - 115 - 35