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If r = 1 in the G(s) = 
then the compensator can give the minimum phase at a frequency of
then the compensator can give the minimum phase at a frequency of- a)
- b)0.777 rad / s
- c)1 rad / s
- d)
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
If r = 1 in theG(s) = then the compensator can give the minimum phase ...
Concept:
Lead and Lag compensators:
Gc(s) = (1 + aTs) / (1 + Ts) where a and T > 0, a > 1 (lead) & a < 1 (lag)
∠Gc(s) = ϕ = tan-1 ωaT – tan-1 ωT
ωm is the geometric mean of the two corner frequencies 1/T and 1/aT
ωm = 1/T√α
Lead and Lag compensators:
Gc(s) = (1 + aTs) / (1 + Ts) where a and T > 0, a > 1 (lead) & a < 1 (lag)
∠Gc(s) = ϕ = tan-1 ωaT – tan-1 ωT
ωm is the geometric mean of the two corner frequencies 1/T and 1/aT
ωm = 1/T√α
Calculations:
Given Gc(s) =
r = 1
By substituting r value we get Gc(s) =
Here T = 1 and a = 0.1
Given Gc(s) =
r = 1
By substituting r value we get Gc(s) =
Here T = 1 and a = 0.1
Most Upvoted Answer
If r = 1 in theG(s) = then the compensator can give the minimum phase ...
Concept:
Lead and Lag compensators:
Gc(s) = (1 + aTs) / (1 + Ts) where a and T > 0, a > 1 (lead) & a < 1 (lag)
∠Gc(s) = ϕ = tan-1 ωaT – tan-1 ωT
ωm is the geometric mean of the two corner frequencies 1/T and 1/aT
ωm = 1/T√α
Lead and Lag compensators:
Gc(s) = (1 + aTs) / (1 + Ts) where a and T > 0, a > 1 (lead) & a < 1 (lag)
∠Gc(s) = ϕ = tan-1 ωaT – tan-1 ωT
ωm is the geometric mean of the two corner frequencies 1/T and 1/aT
ωm = 1/T√α
Calculations:
Given Gc(s) =
r = 1
By substituting r value we get Gc(s) =
Here T = 1 and a = 0.1
Given Gc(s) =
r = 1
By substituting r value we get Gc(s) =
Here T = 1 and a = 0.1
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