Question

Q-3) ine transfer function of a system nas a zero at s = 1, and poles at s =-2 and 5 = -3. The steady state value of the response to a unit step is -2. a) Find the impulse response of the system. b) Find the response to a unit ramp. c) Find the response to sint. @ Restart to install the latest Wine updates We haven't been able to update device. Leave it on and plugged we'll try to restart outside of your hours. Or, pick a time that works Pick a time

Answer 1

Giveni 3) The Loansfer function of a zero at sol system has and pales SS - 2 and Y(s) Then the at SE-3 k (3-1) Transfer Function will be (Ste) (543 is & Y(s) is output. Let X(s) input Gain. = H(S) X(s) where к response * Given steady state value of the unit step is Then unit step responce , Y(s) = X(s). k(sal) (5+2)(s+8) Laplace transform of unit step Lluct) ] = Ś = x(s) = Ś unit step responce X(s) k(sal) s(s+2) (st) value of y(s) 11 * steady state LE s. Y(s) so -2 (Given) It & K(S-1) (5+2)(542) - 2 » sto k=12 k (0-1) (012) (0+3) -2 > 6 Transfer function, H(S) = 12 (5-1) (5+2) (5+3)

a) Impulse response of the system: The output Cyct) the input is impulse function is Impulse response. ie when &(t) =S (H) 9(K)= Impulse when known as Response IX(s). Laplace transform of Impulse function, L[sct)] Impulse Response Y(s) = 12 (3-1) (5+2) (5+3) By partial Fractions Method, Yos) + 12 (S-1) (s+2) (s+3) - A (5+2) (5+3) where 12 (S-1) 12 (-2-1) (-2+3) A = -36 at s=-2 at s=-2 B = (5+3) Y(s) 12(S-1) (5+2) 12 (23-1) (-3+2) B=48 at S=-3 at sa-3 - 36 Y(s) = 48 ste st3 By Applying Inverse Laplace transform, 48 I [Y(s) L S+3 St2 [a] kie eat ult) sta :Impulse Responce УСЕ) - -3t alt) + 48 e -36 e ult)

6) Response to unit xamp Here input is unit ramp. I Laplace transform unit ramp L[NCE) = x(s) S2 For the input X(t) = *(6) the output yes) 12(3-1) sei (5+2) (5+3) By using patial fractions Method A Ys) = В + + 12 (5-1) (st2) (st3) alu ola 52 (5+2) (5+3) 12(5-1) = A s(ste) (st3) + B (s+2)(s+3) +€ 5°(5+3)+osť(+2) in equation ③ * Put Sco both sides 12 6 BE2 12 (0-1) = A CO) + B (ote) (0+3)+c (o) to (o) - BE both sites on * Put in equation si - 2 36 C=-9 4 12 (-2-1) = A (o) + B(0) + € (-2)262+3) +D() > C= * Put s=-3 in in equation o both sides Dat 16 = - 48 12 (-3-1) = A (0) + B(0) + C (0) + D(-3)2 -3+2)+0= 9 * By comparing s3 Co-efficients on both sites in equation a A t cts AS-C-0--G9)-16 .-(-9)- 16 DA= 3 2 16/3 .. Y(s) + - 9 (312) S2 (543) Inverse Laplace transform A + By Applying E'[*] = ucts K; E' [ 1 ] = kikulk) : E' [k Je kentuck) : output y(t) = YCH = { -] uct) -et f et qe 16 e 3t 3

c) response to the input &CE) sint By using sinusoidal steady state Analysis we know, 3(-) : (6)= A.sin(wt+0) A.lHewl sin(wf+0+0) (H (w) | Le Inpat output Angular By comparing inputs Act (Amplitude) & Wel (Frequency ) Transfer function H(s) - 12 (S-1) (ste) (543) To get transfer function in terms of Frequency (w) o put saja H( w) = 12 (jw-1) (Jwtz) (jwt3) (justz) = swetu (tan! (2) (wal) Situz L 180 - tan! (w) ; (Jwtz) = Jweta (tan' (wa) ко LAILOI 1812&2 Icht ΤΑΙ .((*1- % -%, Bllel 12. It w2 > H(w) 180 - kan' (w) a tan' (wo) a tan' (ww.) Sutwa Satwa WEI 12. Hz H(w) / [180-tau (il-fran (2) - tan" () = 24/900 w디 54+12 Satie (H(w), 2.4 $690° NOW outputs YCH = A. 1 t (w), sin (wt tot +8) IH (wo) O coº; $890 A = 1 = 2.4 2.4 cos os (L) .! loutput YCEE 2.4 sin(t +90)