PracticeProblemsMidterm_solu

Hence we start writing down the equation for y as

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Unformatted text preview: ction with input r /R Hence, we start block diagramdown the equation for= 0). follows: writing to be zero (w = v Y as By definition of e, we see that E/R = 1 Y /R. Hence, we start writing down the equation for Y as follows: Y = GU = GD(R Y = GU = GD(R H Y ). H Y ). 7 7 Solving for the ratio Y /R, we have Y (s) GD = . R (s) 1 + GDH The transfer function from r to e is Q(s) := GD s(s2 + (11 b)s + 10 a) =3 1 + GDH s + 11s2 + 10(b + 1)s + 10a E (s) =1 R (s) 28. Find the condition on the parameters a and b such that the closed-loop system is stable. Solution: Routh table is given by 1 11 c 10a 10(b + 1) 10a where c = 110(b + 1) 10a. The system is stable iff c > 0 and 10a > 0 hold, or equivalently, 0 < a < 11(b + 1) 29. What is the system type for tracking? Assume stability and a 6= 0. Solution: When a 6= 10, the transfer function Q(s) has one zero at the origin, and hence it is Type 1 When a = 10 and b 6= 11, there are two zeros at the origin and Q(s) is Type 2 If a = 10 and b = 11, then there are three zeros at the origin...
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