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The block diagram of a closed-loop control system is shown in the figure. R(s) , Y(s) , and ?(?) are the Laplace transforms of the time-domain signals ?(?), ?(?), and ?(?), respectively. Let the error signal be defined as e(t) = r(t) - y(t) . Assuming the reference input r(t) = 0 for all ?, the steady-state error e(∞) ,due to a unit step disturbance ?(?), is _________ (round off to two decimal places).
  • a)
    -0.09
  • b)
    -0.11
Correct answer is between '-0.09,-0.11'. Can you explain this answer?
Most Upvoted Answer
The block diagram of a closed-loop control system is shown in the fig...
Method 1
Given block diagram of a closed loop control system is shown below,
Given : r(t) = 0 and d(t) = u(t)
Error is given by
e(t) = r(t) - y(t)
Taking Laplace transform of above equation
E(s) = R(s) - Y(s)
According to the equation
Reference input r(t) = 0
i.e. R(s) = 0
Substituting the value of R(s) = 0, in the above equation of E(s)
E(s) = R(s) - Y(s)
R(s) = 0
E(s) = -Y(s) … (i)
From the block diagram
Substituting the value of Y(s) from equation (ii) in equation (i),
E(s) = -Y(s)
Given d(t) = u(t)
Hence, D(s) = 1/s
Substituting the value of D(s) = 1/s in equation (iii)
Steady state error is given by,
Hence, the steady state error e(∞) due to unit step disturbance d(t) is -0.1.
Method 2
Given block diagram of a closed loop control system is shown below,
Here, R(s) = 0, so above block diagrams can reduced as,
Thus, transfer function of above block diagram is,
According to question,
Error signal, e(t) = r(t) - y(t)
e(t) = - y(t) [ ∵ r(t) = 0]
E(s) = -Y(s)
So, equation (i) becomes as,
Steady-state error e(∞) under unit step disturbance is,
Hence, the steady state error e(∞) due to unit step disturbance d(t) is -0.1.
Free Test
Community Answer
The block diagram of a closed-loop control system is shown in the fig...
Method 1
Given block diagram of a closed loop control system is shown below,
Given : r(t) = 0 and d(t) = u(t)
Error is given by
e(t) = r(t) - y(t)
Taking Laplace transform of above equation
E(s) = R(s) - Y(s)
According to the equation
Reference input r(t) = 0
i.e. R(s) = 0
Substituting the value of R(s) = 0, in the above equation of E(s)
E(s) = R(s) - Y(s)
R(s) = 0
E(s) = -Y(s) … (i)
From the block diagram
Substituting the value of Y(s) from equation (ii) in equation (i),
E(s) = -Y(s)
Given d(t) = u(t)
Hence, D(s) = 1/s
Substituting the value of D(s) = 1/s in equation (iii)
Steady state error is given by,
Hence, the steady state error e(∞) due to unit step disturbance d(t) is -0.1.
Method 2
Given block diagram of a closed loop control system is shown below,
Here, R(s) = 0, so above block diagrams can reduced as,
Thus, transfer function of above block diagram is,
According to question,
Error signal, e(t) = r(t) - y(t)
e(t) = - y(t) [ ∵ r(t) = 0]
E(s) = -Y(s)
So, equation (i) becomes as,
Steady-state error e(∞) under unit step disturbance is,
Hence, the steady state error e(∞) due to unit step disturbance d(t) is -0.1.
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The block diagram of a closed-loop control system is shown in the figure. R(s) , Y(s) , and ?(?) are the Laplace transforms of the time-domain signals ?(?), ?(?), and ?(?), respectively. Let the error signal be defined as e(t) = r(t) - y(t) . Assuming the reference input r(t) = 0 for all ?, the steady-state error e(∞) ,due to a unit step disturbance ?(?), is _________ (round off to two decimal places).a)-0.09b)-0.11Correct answer is between '-0.09,-0.11'. Can you explain this answer?
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