Question

4. Consider the transfer function, Y(s)_ 3 F(s) + s(s2 + 2s + 4) (a) Qualitatively, what is the time response y(t) if f(t) represents a unit-step input? What is the value of y(t) when time is sufficiently large? What is the time constant that we may use to evaluate the "speed" of response? (b) Repeat step (a) if f(t) represents an impulse input. What is y(t) when time is sufficiently large?

Answer 1

4) (5) S (5 +25+4) a) Given ye) Some 10 Ple) is a step input Stepoput means f(t) = 1 and F(S) ='s By Substituting f(s) 's in een ☺ 465) . ? S(5'+25+) Y() = 3.6%) S (5 +25+4) ( 403) = 5+25+4) fraction methad this fs can be By using Partial written as By taking a.com As(s+25+4) + B (1.125+4) +(3+0) ses+25+4) 3 s' (5+257 AS3+2 As++63 +265+46 +68+05 3 A+c=0-10 SA+87020 4A+200- 48-3 B = 3/4

4 / 3 =g by substituting 8=3/4 in 4A +26 =D een @ 0 40=-710) by Substituting A = -3/8 in egn® Atc=0 C- 2A+B+0 -0 H + +D=0 Y(3) -ştat S19574 by Substituting n, o,c, o values it becomes Y = + + st25t4 LSY LU? =445) for getting yet) apyly inverse Capace on both Sides tek syit}} - i'$465) }

Yet) = it's y533 Yet)=zę * 5+ in the pasan? : 378%} + 4 5 4 3 tansso? 10 + (1) + 515 sy} heve sºn25+4 can be written as (8+004(0) +*+ * 75 6+) - 7 Southers) y(t) = - + + + resist - yit) at fufficiently Serge time means in stority (oo) 's' apparathes Pero'. 5:15467} sin vzt. t'approaches YO) = TSHsty el sin vzco). yet) = -34%60) + = CS560) - :-39+200) + 0-0 the 4 (t) = os Time constant qu. (5 +2.5+4) given - it is in the form of

2844+3 = 46) for getting ylt apply inverse laplace on both sides 400) = '{4033 443 243 +24 +273_} 5'+25+y 5+25+4 can be written as ($+DP+(B)2 YCH= ,43 +7° 72, (-173 +2° +*2 {+up+673) S+1) Lt - el estostogos = - v3t +1-3,247! / B3 PBB Y (5+1)+83) yet) = $ - * 8t + 1-um] et sinet yit) at sufficiently large time means "t'approaches infinity then et vet e m e cortev) - Osing po) = 34-0-0 4 Lt) = 24. t-10

quash +2875+1 hese qual SO T- b) f(s) = 1. flt) is an impiese input mean then y) = 3 s(s+25+4) foi finding yet) we use partial fraction method. sc8+23+4) A EP+25+y) +8+075 = 3 As'+265+ 4A +B5Y+es = 3 A+B=0 0 2A +c= D 4A = 3 -> A = 3/y Substitute in een 0 then A+B=0 BE-A -3 A) = 3/4 Substitute in een then 2A+C=0 11949 +6=0 (=-3 B By substituting A,B,C values 465 becomes Y() = § + GS + Stasty štasty.