Control Systems The open loop transfer function for unity feedback system is given by 5(1+0.1s) / (s(1+5s)(1+20s)) Find the steady state error for a step input of magnitude 10 is equal to infinite. 5. 2. zero. infinite. 5. 2. zero. ANSWER DOWNLOAD EXAMIANS APP
Control Systems For a lead compensator, the poles lie: on origin. on RHS. on LHS. on LHS before zero. on origin. on RHS. on LHS. on LHS before zero. ANSWER DOWNLOAD EXAMIANS APP
Control Systems Given the Laplace transform of f(t) = F(s), the Laplace transform of (f(t)e-at) is equal to eas*F(s) e-as*F(s) F(s+a) F(s)/(s+a) eas*F(s) e-as*F(s) F(s+a) F(s)/(s+a) ANSWER DOWNLOAD EXAMIANS APP
Control Systems Consider the following statements for phase lead compensation: The maximum phase lag occurs at the arithmetic mean of the two corner frequencies. both 1 and 2 Phase lead compensation shifts the gain crossover frequency to the right nether 1 nor 2 The maximum phase lag occurs at the arithmetic mean of the two corner frequencies. both 1 and 2 Phase lead compensation shifts the gain crossover frequency to the right nether 1 nor 2 ANSWER DOWNLOAD EXAMIANS APP
Control Systems The magnitude condition for root locus is .................. |G(s)H(s)| = 1 |G(s)H(s)| = ∞ |G(s)H(s)| = 2 |G(s)H(s)| = 0 |G(s)H(s)| = 1 |G(s)H(s)| = ∞ |G(s)H(s)| = 2 |G(s)H(s)| = 0 ANSWER DOWNLOAD EXAMIANS APP
Control Systems The impulse response of an initally relaxed linear system is e-2tu(t). To produce a response of te-2tu(t), the input must be equal to e-2tu(t). 2e-tu(t). (0.5)e-2tu(t). e-tu(t). e-2tu(t). 2e-tu(t). (0.5)e-2tu(t). e-tu(t). ANSWER DOWNLOAD EXAMIANS APP
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