The block diagram of a closed-loop control system is shown in the figure. R(s), Y(s), and D(s) are the Laplace transforms of the time-domain signals r(t), y(t), and d(t), respectively. Let the error signal be defined as e(t) = r(t) − y(t). Assuming the reference input r(t) = 0 for all t, the steady-state error e(∞), due to a unit step disturbance d(t), is _________ (rounded off to two decimal places).
Correct Answer
Solution & Step-by-step Explanation
In the range: -0.1 - -0.1
Calculation:
Given
r(t) = 0
e(t) = r(t) - y(t)
e(t) = - y(t)
E(s) = - Y(s) → (1)
Circuit can be redraw as
Y(s) = [ 10 E(s) + D(s) ] × 1s(s+10)\ \
From (1)
E(s) = [ 10 E(s) + D(s) ] × 1s(s+10)\ \
For unit step disturbance d(t)
D(s) = 1/ s
E(s) = - [ 10 E(s) + 1s \ \ ] × 1s(s+10)\ \
E(s) = -110s+s^2(s+10)\ \
steady-state error is given as
Calculation:
Given
r(t) = 0
e(t) = r(t) - y(t)
e(t) = - y(t)
E(s) = - Y(s) → (1)
Circuit can be redraw as
Y(s) = [ 10 E(s) + D(s) ] × 1s(s+10)\ \
From (1)
E(s) = [ 10 E(s) + D(s) ] × 1s(s+10)\ \
For unit step disturbance d(t)
D(s) = 1/ s
E(s) = - [ 10 E(s) + 1s \ \ ] × 1s(s+10)\ \
E(s) = -110s+s^2(s+10)\ \
steady-state error is given as