Question

2. Consider the pole-zero plot of a transfer function H(s) given in Figure P14.2. (a) If the dc gain is -10, find HG). (b) Compute the impulse response. (c) Compute the step response. CHECK: Your answer to (b) should be the derivative of your answer to (c), since the delta function is the derivative of the step function. (d) If the input is 10 ), find the pos- itive number a such that the response does not have a term of the form ker(s). Find the zero-state response under this condition. *4 Figure P14.2

Answer 1

→ Solution (a) Poles are located at -2 43 and -us zeros are located at 4, and -4. Trans for function is written in the form below. H(s) K(s+4) (s-4) K (5-16) (S+2) ( 5+41) (5-43) (5+2) ($ +16) where K is DC gain = -10 H(s) = 10 (516) (5+2) ($+16) (6) impulse response is obtained by applying inverse Laplace transform on 4(s). -10(5-16) (5+2) (546) KE) kit) = { [ -10 (8-16) - S+2) (5+16)) Apply partial tractions to get the terms. ). A S+2 -lo (5-16) (BSC) 5+16 (5+2) ($+16) A($+15) +(85+6) (+2) (5+2) ($7167 A = (5+2) HOS) IS-2 5-015-16) = 6 Equaling & terms, ) B: -1o-A = -16. -10 = A + B Equating terms, O 2B+C - C -2B → CE 32

HIS) = 16 (5-2) Song - St103 som lever e] aos het) = (582t - (16 cusut + 2 sinut) ] urt) hie) = (62t - 16 Gb&t 2 Sin 4t.), ult-) step response is obhained as. Y(3) - H(s)u(s) , where urs) - 5 for step input. Y(5) -10 (S-16) -10 (5-16) S (5+2) (8+16) s SCS+2) ($716) Y(5) "ht be + Cesta A: YS).s Is.o. -10.6-16), S. 2.16 B = Y() (suz) | sona e ro (4-16)- .-3 Equating & terms, A (5+2) ($16) + AS ($+6) + (CS+D) s(s+2) A37165+28+32) + B 5+16BS + es + $D+20+20S. A+B+C c -A-B -5+3 = -2 =0 - (0:-2 Equating š terms, -10 = 2A + D + 20 D -10-2A-20 = -10 - 10 + 4 = TD=-16] -16

1637. - $22 - 403) - G enerate Apply inverse Laplace transform. y(t) = (5 - 3é2t - 26% 4- Sin44) 1/) (d) input term = 10e at ut). In the function to nullify the term (5+2). acol because 10 ett ult) = 2 502) 10 The term (5+2) becomes (5²4) so, the exponential term is vanished. zero state response is 10 Y() -lo (SP-16) Y(s) = (5+2) ( Š+16) (5-2) -100 (5-16) (5-4) (8+16)