Control Systems Angle of asymptotes for complimentary root locus is (2q)*180/P (2q+1)*180/P (2q+1)*180/(P-Z) (2q)*180/(P-Z) (2q)*180/P (2q+1)*180/P (2q+1)*180/(P-Z) (2q)*180/(P-Z) ANSWER DOWNLOAD EXAMIANS APP
Control Systems The characteristic equation of a control system is given by What are the angles of asymptotes of the root loci for k ≥ 0? 0°, 120°, 240° 0°, 180°, 300° 120°, 180°, 240° 60°, 180°, 300° 0°, 120°, 240° 0°, 180°, 300° 120°, 180°, 240° 60°, 180°, 300° ANSWER DOWNLOAD EXAMIANS APP
Control Systems For a transfer function H (s) = P(s) / Q(s), where P (s) and Q (s) are polynomials in s. Then the degree of P (s) and Q (s) are same. the degree of P (s) is always greater than the degree of Q (s). degree of P (s) is independent of degree of Q (s). maximum degree of P (s) and Q (s) differ at most by one. the degree of P (s) and Q (s) are same. the degree of P (s) is always greater than the degree of Q (s). degree of P (s) is independent of degree of Q (s). maximum degree of P (s) and Q (s) differ at most by one. ANSWER DOWNLOAD EXAMIANS APP
Control Systems Settling time for 5% tolerance band is ------------- 5T. 4T. 2T. 3T. 5T. 4T. 2T. 3T. ANSWER DOWNLOAD EXAMIANS APP
Control Systems The impulse response of a system is c(t) = -te-t + 2 e-t (t>0). Its closed loop transfer function is (2s+1)/s² (2s+1)/(s+1)² (2s+2)/(s+1)² (2s+2)/s² (2s+1)/s² (2s+1)/(s+1)² (2s+2)/(s+1)² (2s+2)/s² ANSWER DOWNLOAD EXAMIANS APP
Control Systems The initial slope of Bode plot for a transfer function having no poles at the origin is 0 dB/decade -10dB/decade. -20dB/decade. +10dB/decade. 0 dB/decade -10dB/decade. -20dB/decade. +10dB/decade. ANSWER DOWNLOAD EXAMIANS APP
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