Electronics and Communication Engineering - Automatic Control Systems - Discussion
Discussion Forum : Automatic Control Systems - Section 4 (Q.No. 24)
24.
Given figure shows the magnitude plot for a system. The steady state error for unit step input is
Discussion:
3 comments Page 1 of 1.
Amod kumar jha said:
8 years ago
But here low-frequency line is not interesting at 40 db, it is interesting somewhere else above 40 db.
So, K can be calculated by another method,
20log(k) = 40
K = Kp = 100,
Ess = 1/(1+100) = 1/101
The Answer should be option A.
So, K can be calculated by another method,
20log(k) = 40
K = Kp = 100,
Ess = 1/(1+100) = 1/101
The Answer should be option A.
Kusiroll said:
8 years ago
Steady state error of a bode plot constant Kp(step input) are located at the intersection of the low frequency asymptote with the w =1 line.
Therefore steady state error of a step input = 1/1+ Kp where 1/1+40 = 1/41.
Therefore steady state error of a step input = 1/1+ Kp where 1/1+40 = 1/41.
Rakesh said:
7 years ago
Yes, the answer is 1/101.
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