Question

onsider a DC electric motor. Assume the armature inductance is negligible. Consider the harmature voltage ea as the input and the angular displacement @t) as the ulput. By measuring an input and the system's response as functions of time, the transfer function is then found to be 10s(s +1) What are the motor time-constant and the motor gain? a. b. iIt the input is 2 u(l), find the output i.e. the angular displacement et) c. Find the angular speed o(t)

Answer 1

Solution:

Transfer function H(S) = 10/S(S + 1) (a) H(S) = 10/S(S + 1) = 0[1/S - 1/S + 1] taking inverse laplace both side we have L^-1 [H(S)] = L^-1 {10 [1/s - 1/s + 1]} or, h(t) = 10[1 - e^-t] u(t) hence h(t) = 10(1 - e^-t/T) u(t) time constant T = 1 (b) input X(S) = L{2u(t)} = 2/s H(S) = 10/S(S + 1) or, Y(S)/X(S) = 10/S(S + 1) or, Y(S) = 10/S(S + 1) X(S) = 10/S(S + 1) 1/S = 10/S^2(S + 1) = 10[1/S^2 - 1/S(S + 1)] = 10[1/S^2 - 1/S - 1/S + 1] taking inverse laplace transform both side we have L^-1 [Y(s)] = L^-1 {10[1/s^2 - 1/s + 1/s + 1]} or, y(t) = 10[t - 1 - e^-t] u(t) hence theta(t) = 10[t - 1 + e^-t] u(t) (c) w(t) = d/dt theta(t) = d/dt 10[t - 1 + e^-t]^u(t) = 10(1 - e^-t) u(t)