Control systems miscellaneous
- The system with transfer function
K s2(1 + sT)
is operated in closed-loop with unity feedback. The closed-loop system is—
-
View Hint View Answer Discuss in Forum
C.E. = 1 + G (s) H (s) = 0
= 1 + K 1 = s2 + s3T + K = 0 s2(1 + sT)
There is missing term of s1. Hence the system is unstable.Correct Option: B
C.E. = 1 + G (s) H (s) = 0
= 1 + K 1 = s2 + s3T + K = 0 s2(1 + sT)
There is missing term of s1. Hence the system is unstable.
- Given that the transfer function G (s) is
State the type and order of the system—K s2(1 + sT)
-
View Hint View Answer Discuss in Forum
Type = 2, Order = 3
Correct Option: A
Type = 2, Order = 3
- The transfer function G (s) is
. This open-loop system will be—K s2(1 + sT)
-
View Hint View Answer Discuss in Forum
C.E. of the given O.L.T.F. ⇒ 1 + G (s) = 0
⇒ 1 + K = 0 s2(1 + sT)
⇒ s2 + s3T + K = 0
⇒ s3 T + s2 + K = 0
system is unstable since there is missing terms of s1.Correct Option: A
C.E. of the given O.L.T.F. ⇒ 1 + G (s) = 0
⇒ 1 + K = 0 s2(1 + sT)
⇒ s2 + s3T + K = 0
⇒ s3 T + s2 + K = 0
system is unstable since there is missing terms of s1.
- A system has the transfer function (1 – s)/ (1 + s). It is a—
-
View Hint View Answer Discuss in Forum
A system with
T.F = 1 - s T + s
is a non-minimum phase system because a non-minimum phase system have at least one pole and/or zero in the R.H. side plane for nonminimum phase system as ω → ∞, phase angle φ ≠ – 90° (n – m). In this case, phase is – tan–1 ωT – tan–1 ωT
= – 2 tan–1 ωT = – 180° as ω → ∞.Correct Option: A
A system with
T.F = 1 - s T + s
is a non-minimum phase system because a non-minimum phase system have at least one pole and/or zero in the R.H. side plane for nonminimum phase system as ω → ∞, phase angle φ ≠ – 90° (n – m). In this case, phase is – tan–1 ωT – tan–1 ωT
= – 2 tan–1 ωT = – 180° as ω → ∞.
- An all pass network imparts only—
-
View Hint View Answer Discuss in Forum
Transfer function of an all pass network,
= 1 – s 1 + s = 1 – jω 1 + jω, ∠ φ = tan– 1 (ω) – tan– 1 ω
As ω → ∞, ∠ φ → – 180°
However, if the transfer function= 1 - s = 1 -jω 1 + s 1 – jω
∠ φ = tan–1 ω – tan–1 (– ω) = 2 tan–1 ω,
as ω → ∞, ∠ φ → + 180°.
Thus, an all pass transfer function impart ± 180° phase shift to the input. Hence alternative .Correct Option: D
Transfer function of an all pass network,
= 1 – s 1 + s = 1 – jω 1 + jω, ∠ φ = tan– 1 (ω) – tan– 1 ω
As ω → ∞, ∠ φ → – 180°
However, if the transfer function= 1 - s = 1 -jω 1 + s 1 – jω
∠ φ = tan–1 ω – tan–1 (– ω) = 2 tan–1 ω,
as ω → ∞, ∠ φ → + 180°.
Thus, an all pass transfer function impart ± 180° phase shift to the input. Hence alternative .