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 (0.5)e-2tu(t). e-tu(t). 2e-tu(t). e-2tu(t). (0.5)e-2tu(t). e-tu(t). 2e-tu(t). e-2tu(t). ANSWER DOWNLOAD EXAMIANS APP
Control Systems Poles and zeros are arranged alternatively on negative real axis, then type of network is/are RL network. Both 2 and 3. LC network. RC network. RL network. Both 2 and 3. LC network. RC network. ANSWER DOWNLOAD EXAMIANS APP
Control Systems A function y (t) satisfies the following differential equation : Where, δ (t) is the delta function. Assuming zero initial condition and denoting the unit step function by u(t), y(t) can be formed as etu(t). et. e-tu(t). e-t. etu(t). et. e-tu(t). e-t. ANSWER DOWNLOAD EXAMIANS APP
Control Systems Lowest critical frequency is due to pole and it may be present at the origin or nearer to the origin, then the type of network is RL. LC. Any of the above. RC. RL. LC. Any of the above. RC. ANSWER DOWNLOAD EXAMIANS APP
Control Systems Which of the following represents the value of resonant frequency (ωr)? 2*(ωn) √(1-ζ²) (ωn) √(1-ζ²) (ωn) √(1-ζ²/2) (ωn) √(1-2ζ²) 2*(ωn) √(1-ζ²) (ωn) √(1-ζ²) (ωn) √(1-ζ²/2) (ωn) √(1-2ζ²) ANSWER DOWNLOAD EXAMIANS APP
Control Systems At which frequency does the Bode magnitude plots for the functions K/S2 have gain crossover frequency ω = K2 rad/sec. ω = 0 rad/sec. ω = K rad/sec. ω = √K rad/sec. ω = K2 rad/sec. ω = 0 rad/sec. ω = K rad/sec. ω = √K rad/sec. ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS