Question

6.13 Consider the circuit of Figure P6.13. (a) Determine the transfer function G(s) = 1 (s)/V(s). (b) Determine the time constant for the system. (c) Determine the response of the system when v(t) = 10u(1) V. 5022 100 12 + iz -0.2 UF (1) FIG. P6.13

Answer 1

502 1002 u حاد + 0-24 F vity) By transforming the above netwolk into s-domain. 1002 sor L ->(s) + m ->215) Kloopt y tics)-izes) Dloot loop-2 ves) CS By writting KVL, in input Loop=1 - VS)+ j(s). 50 + s. 1ics) – izcs) : Ls) [50+^] - ines) = ves) - -Ves) + PLCs). 50 + izes). 100 =0 50 i(s) + 100 126) = v(s) h(s)+ 2i2cs) = vis) 100 1265)+ + zes) – ics)) = 0 izcs) [100++] = Tics) 2 50 CS

12cs) [it 10065 CS izcs) = rics) 1 +10025 2 from Equation vis) 50 ics) + 1 (s) 1 + 100 CS rics +10015+! VCS) hus [ J- 50 1+100 CS 50x (2 +100C5) 1 + 1000s .: Trasferr forresten ves) 31(5) (1+100 cs ) Ills) Gisl= vis) 50 Ⓡ (2+100 CS ). where e=0.2HF (ь finding the time constant :- To find the time constant, replace all sources their internal resistance. by 1002 502 M um Req 501) 100 A Rear 50% 100 50+100 =_50x100 1503 ..T= Rear C =100 100+ 0.2x10 6 100 3 T = 20 HSBC

(6) finding the response of the system, for the Input vit) = loult) 15 Х s nt 2 S 5 1000 VO = 10 s Gis) = Ills) VIS) => 1,16) = GS). Ves) Ils)= 10 (1+lools 2+10015 1000 (s+ Tood) 1000 (s+ Ills) =/S+ 1/1000 S+ 2/100C By applying partial fractions B St 2/1000 55 A t - 1 5 Al A- 2 St 111000 S+ 2/1000 160 6--0 1 S+ 11000 1s 2 دام 2 BE S 100C I, + ist 10 St 2 1000 applying inverse By 2.t Till Ils) = to Laplace transform, IL[I1(s)] = Ilt) = * 66t) + ē tooc ult) It) = 0.1[1+ 8 lot ).uit) 10