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The Transfer Function of lead and lag compensators have _____ and ______ phase angles respectively which the system ______ and ______ respectively.
- a)-ve, -ve, slower, faster
- b)+ve, +ve, slower, faster
- c)+ve, -ve, faster, slower
- d)-ve, +ve, faster, slower
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The Transfer Function of lead and lag compensators have _____ and ____...
Concept:
We know:
Hc(s) = s+as+b;
when α = zero/pole = α/b is greater than 1 i.e., α > 1 then it is lag compensator.
else when α < 1, then it is lead compensator.
Calculation:
For lead compensator: Hc(s) = s+a/s+b
ϕ = tan−1(ω/1) − tan−1(ω/2)
i.e., phase angle will be positive for ω > 0
similarly, for lag compensator: Hc(s) = s+2/s+1
We know:
Hc(s) = s+as+b;
when α = zero/pole = α/b is greater than 1 i.e., α > 1 then it is lag compensator.
else when α < 1, then it is lead compensator.
Calculation:
For lead compensator: Hc(s) = s+a/s+b
ϕ = tan−1(ω/1) − tan−1(ω/2)
i.e., phase angle will be positive for ω > 0
similarly, for lag compensator: Hc(s) = s+2/s+1
ϕ = tan−1(ω/2) −tan−1(ω)
i.e. phase angle is negative for ω > 0.
Now,
For lead compensator, gain crossover frequency (ωgc) increase which causes an increase in B.W. and hence the speed of the system improves.
For lag compensator; Gain crossover frequency (ωpc) decreases which causes decreases in B.W. and hence speed of the system decreases.
i.e. phase angle is negative for ω > 0.
Now,
For lead compensator, gain crossover frequency (ωgc) increase which causes an increase in B.W. and hence the speed of the system improves.
For lag compensator; Gain crossover frequency (ωpc) decreases which causes decreases in B.W. and hence speed of the system decreases.
Most Upvoted Answer
The Transfer Function of lead and lag compensators have _____ and ____...
Concept:
We know:
Hc(s) = s+as+b;
when α = zero/pole = α/b is greater than 1 i.e., α > 1 then it is lag compensator.
else when α < 1, then it is lead compensator.
Calculation:
For lead compensator: Hc(s) = s+a/s+b
ϕ = tan−1(ω/1) − tan−1(ω/2)
i.e., phase angle will be positive for ω > 0
similarly, for lag compensator: Hc(s) = s+2/s+1
We know:
Hc(s) = s+as+b;
when α = zero/pole = α/b is greater than 1 i.e., α > 1 then it is lag compensator.
else when α < 1, then it is lead compensator.
Calculation:
For lead compensator: Hc(s) = s+a/s+b
ϕ = tan−1(ω/1) − tan−1(ω/2)
i.e., phase angle will be positive for ω > 0
similarly, for lag compensator: Hc(s) = s+2/s+1
ϕ = tan−1(ω/2) −tan−1(ω)
i.e. phase angle is negative for ω > 0.
Now,
For lead compensator, gain crossover frequency (ωgc) increase which causes an increase in B.W. and hence the speed of the system improves.
For lag compensator; Gain crossover frequency (ωpc) decreases which causes decreases in B.W. and hence speed of the system decreases.
i.e. phase angle is negative for ω > 0.
Now,
For lead compensator, gain crossover frequency (ωgc) increase which causes an increase in B.W. and hence the speed of the system improves.
For lag compensator; Gain crossover frequency (ωpc) decreases which causes decreases in B.W. and hence speed of the system decreases.
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Question Description
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