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MAS 239 U as = 5 -4 = ' ✗ 4) Poles Clt) Problem b) If - - Gist (s) ✗ = 40 ✗ (t ) + = 2s 2 K, + 041 2 V61 + duct + 2 V41 dt e- real poles "t e- yz + 10£ 2 1s at 63,2% . T constant C final = step Amplitude Cs 40 different Ko = (25+2) UH + dxltl V41 dt 2 : = ✗ (s ) 14 + step Time 14s + dt a) - -1 = 26+1 ) = (5+145+40) ✗ G) d Zeroes -10 , (5+414+10) (5) (s) ✗ ZV Ole Andre 15.02.21 (5+4) (5+10) Poles ✗ C) . 2 ( Stl ) G) (5) by assignment I Problem ✗ Mandatory - = K Ts , 1,02 300 = 2 = + = 1,02s = 150 0,025 -11 - l 150 , = steady 0,02 state Seconds 300 rad /s Tuntland Problem 3 There as ( is 961 ( - 2) ( . 5- y input -4+-52 Which ) Stl -12) Is stable is this also ? Wn k 5+25 = s at poles system = step Unit , complex (5+1) s 0,5 = under damped a = G (5) final two are This b) ( -1 correct ( final " Wns . K= U complex poles Wn ? 0,5 = I = 0,5 10 C. (5) Wn C) 5+85+20 = 22+-1412 = 4 = Under Problem It The seeing The over = 4,47 0,89 = 2hr5 dg 2T5 swn g5 = damped , There are two at -4+-52 4 would be derivative the Integral time pace needed in control of control case as decreases overshoot and settling time by change removes steady state error by sunning up error