Question

Convert following the transfer function into state space representation (Marks 5) 3 +45² T($) = 54 +52 +7 Convert the following state space into a transfer function. (Marks 5) x = 11 * = x + ( u 21 y = [02]x + [2]u Evaluate the steady-state error of state-space system. (Marks 5) i [ 10] [21. *= 15 2]* +11 y = [ 02]x + [2]u Evaluate the steady-state error of state-space system. (Marks 5) -1 0x+lu x = 17 21" "11! y = [ 02]x + [2]u

Answer 1

B con vert following transfer fumetion into stale space representation - O We need from the to determine the state space equation. ginen transfer function T19) - SP+452 We have the 84+s?+ + standard state space model as ſo to ... o7 Where A = 1 А с + Ви y = cx + Du 0 0 0 - .. 1 I-am-ant-an_2-..-ai ] B = 1 oo... sy ; c= [ (by-anbo) (bon anybo) .... (bz-a,boll ; D = bo o

.. bos" + 6,5"-t... bayst bar T(S) = 99 + a,satt..tanys tam ' '; Htese (n=4] 7 (5) - Oxs'+ 1x5 + 4xs'to 1434 + Oxs® + 1x8*4 7

Now comparing with the slávdard equ bo to į b, = 1 ; 62 = 4; by=0 .a,=0 ; aq=1 ; Azzo, au 7. FOI 0 0 1 So A = 0 0 1 0 1 ooom Doon B = 10 colo0.41 ] Soi 1. [O 100 Andy u Xelo 0 1 0 Loo 01 [ X ] [ -7.0-1 o) .8.x OoOo 3. - ylo 0.41] &

© Comuat the follocoing state space, into a transfer function jc = [ 2015 In ( 7 2 1 1 y = [ 2 ] x + [2] u To obtain transfer function from state model 46) eye (81-a* + + fro from the ginen state equation A[ * ; ] ; 0;[1] L L ; Coco 2] ; D = [2] 91--15%.]-!] SI-A L I s-1 8-2 O o = SI-A 1.-7 5-2 2 [at-a)" only ) ce S-2

12 ه = "تم - 51] ال می رود (1-ي ) -1 32 و S-1 ل (2-6) (هو) ا . (1- Hs و ) و ) .ر (2- s - s ) | 3 + 2 . (وی و ). (2-s) (وسی | (34) 2 + [sr - AJ+۵ - (S) (s-2) . زد + و - ی) و +4+ 5 2 + 3s -هی

c[St-Aj& += 25+2 + 23°-65 +4. 8² - 3572 at [SI-A] + D =- 25° - 45+6 SP-35+2 CP-38 +2 state space system ② Evaluate the steady-state error of ic - 1 a °2 ]« + [ 1 ]u yolo e J si + [2] u → The laplace transform of the error f(s) = R(S) - Y(s) The output equation is .y(s) = R(S) TCS)., where t(s) is the closed loop transfer fruction. So E(S) = R(s) – RCS) 7(s) = R(S) [1 - T(s)] from Problem 2 , we know that SA T(s) = [SEA)6 +D 1718) - 25?–45.76 52-35 +2

Applying final value theorem, ess = $E(S) - It S RG) [ - (66E-a3% +)] S70 using the above values 25%-4576 S²_35+2 50 1. . - loc = SR(S) it sto $_35+2 - 25%+48-6 32-35 +2 - ess = It SRIS).-s' +-4 Pol s ?-35+2 o For unit step input, RC = cp2 sto 2 Anse [ess = -2] R(8) = 1/3 i Ramp input

@ © Some as promb lem 3, some question Evaluate the controllability and obseruability of state - space system we = ( 2 )* +(]u y = [5 3] x + [2] a → From the given stale space medel, Aaline) ; 6-13); c=[5 3) ; 0-[1] controllability To control state variables Qc = [ AB AB .... AntiB] rigelto AB = 12 •-[1] : n-124]}]-[7] ond . Tacle 32x3 – (-3)X2] = 102 laelto Thus system is controllable:

Obrenability To measuse the state variables Po = [em les cause..... (apjovc?] 14,1 = 0 Y 0.637, 0.1* :][:] TT и с. wy 19.1.0 Hence system is obseruable. Anso System is both controllable and Obseruable.