Question

Name: Problem 2 (30 points) (a) Based on the lecture notes of Week 5, obtain the state transition matrix 0(t) = At for A= 1; 0 11 I. L3 21: (b) Compute the zero-input state response and zero-input output response at time t1 = 2 for the causal LTI state-space system ż= Az + Bx, y = Cz, where A= 3.). B=(Q), c = [1 0], 20) = [-]

Answer 1

Problem 2 I- A The state transition matrix &(t)= eat for a State matrix A is given by, eAt = ['((SI-A)-) where I is the identity matris Now, (SI-A) - [5 07 To 17 Los/ [32] (SI-A) = 15 - 7 (-38-2 |

Neu, ($1-A)' - - *(க-2)-3 UN Aleu 8( -2 -3 = 29-3 =(8-3) (3+1) Thus, (SI-A) = 8-2 Tearow (S-3)(s+I) (S-3)(s+1) மே

s-> -1 S-2 = A + B & By Partial fraction ? (8-3)(8+1) (6-3) (8+1) A= Lt (6-3) x(4-2) = { By residue ? 3 (4-3) (6+1) 4 method B= Lt (+1)x (4-2) 2 = 3 (8-3) (441) (1,3) (34) 63) * A = Lt (4-3) x = 12 6-7 B = Lt (H) x I = -1 -yol (4-3) (1) 4 -3) (4+1) 4

هه ر (9) (9) A = It (4-3) x 3 (1+) (3-و 3% A - 3 A 4 3- - - x3 (ا+ه) ا -3 4 (الا) (3-و) ا-دلا + B A - (2 + 1) (3) (او) (دو) A= Lt (6-3) x & (1+) (3-و) 93 : A

B= It (4+1) X & dal (8-3) (3+1) B = - - - 1 14.) = 1 / 2 Therepote (ST-A)"Acest ten thoa tron) 2 4 19 3 – 3 4(9-3) 4(+1) 3 + 1 +(-3) 4(x+1) Now, L'" (63.1–)") = +2+3e-t) +(ate-t) (met) (e*et)

which is the required state transition materise $(t). Note; L (e-at) = 1 I leat) - I . S-a b) We have the zero-input state response z(t) = eAt zlo) where e At is the State transition Matrix. since here the State matrix A is same as the part a, so eat will be same as parta

: 2(t) = (1 4 Bet) - letet) 7.7 ( # (etet) #fortet) - = 1 + 6 - ) - * (* -t) (este-t)- (se#get) 1

li ent -e-t 0.135 ! -0.135 Y(t)= cz(t) = [O] [ Y(t)= e-t Y(6) 9 = 0.135 Thankyou.. ..