1. (1pt) Consider a second order underdamped system with one finite zero whose transfer function is Win (s + 2) H(s) = 2 ($2 + 208 + W? ) The unit-step response of the system v(t) and the impulse response of the system h(t) can be expressed as v(t) = 1+ Wn - 202+ 22 e-ot cos (wat - 0) ) u(t) zV1 - 2 and h ( t ) = Vwn - 202 + 22 Whe ot sin (wat - 0 + B) \\ u(t) zV1 - $2 where sin B = ( and cos B = V1 - 2. Assume wn = 1,o = 0.5 and z is the last digit of your student ID divided by 10. If the number is 0, use z = 1. Compute the peak time to > 0 to reach the first peak in seconds and the overshoot Mp in percentage of the step-response.\f. Peak time (tp ) to = n to = A = 3 . 63 seconds 0.866 to = 36 3 seconds . Ourshoot ( Mp): The percentage Durshoot Mp is Mp : X100 ). - 0'5 1 Mp + e V1 - 015 2 X100 % = 4 e 0 8 66 * 100%. MD = 0.163 x 100% Mp = 16:30 Answer: Peak time to = 3.63 seconds Percentage Quershoot (Mp'/ ) = 16.31%Answer: the given transfer function describes a second ouder underdamped System with One finite zero. including natural frequency ( wn), ing HA -0: 0.5, Zero, 2= 0.4 Given Pecansfer function. HCS ) = won2 (s+z ) 2 (s2 + 2 05 + won?) . Calculate damping ratio E: 0 = 015 0:5 wn 1 4 = 0.5 . Damped frequency wd : 1 = WJ1- E2 = 1XJ1- 0.52 = 10.75 us = 0'B 66 elad/ sec. . Phase angle ( Q) : Q = tan-1/ wn - EZ 1-Z JI- EZ 0 : tan1 1 - 10. 5x0.4) = 0 = tan1 / 1 - 0.2 - 0.411-0.50 (0.4x0.866, Q = tan ( 2 . 31) 8 = -66. 587 Q = - 66 . 587 X A - -1. 16 radians 180 9 = - 1.16 gladians