Control systems miscellaneous
- The transfer function Eo (s)/Ei (s) of the circuit is—
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From given figure
E0(s) = Ls · Ei(s) R + Ls or E0(s) = Ls Ei(s) R + Ls or E0(s) = s Ei(s) s + (R/L) 
Correct Option: D
From given figure
E0(s) = Ls · Ei(s) R + Ls or E0(s) = Ls Ei(s) R + Ls or E0(s) = s Ei(s) s + (R/L) 
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.G (s) = K s(1 + sT)
This system is operated in closedloop with unity feedback. The closed-loop system is—
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Given that G(s) = K and H (s) = 1. s (1 + sT)
C.E. 1 + G(s) H(s) = 0⇒ 1 + K .1 = 0 s (1 + sT)
⇒ s + s2T + K = 0
⇒ s2T + s + K = 0
R.H.C. of the above C.E.
s2 T K
s1 1
s0 K
for the system to be stable T, K > 0. Therefore, the system is conditionally stable.
Hence alternative (D) is the correct answer.Correct Option: D
Given that G(s) = K and H (s) = 1. s (1 + sT)
C.E. 1 + G(s) H(s) = 0⇒ 1 + K .1 = 0 s (1 + sT)
⇒ s + s2T + K = 0
⇒ s2T + s + K = 0
R.H.C. of the above C.E.
s2 T K
s1 1
s0 K
for the system to be stable T, K > 0. Therefore, the system is conditionally stable.
Hence alternative (D) is the correct answer.
- The transfer function of a closed-loop system is
Fk(s) = K(s + 3) 1 + s – s2 + s3
. This system is—
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Given closed loop system.
Fk(s) = K(s + 3) 1 + s – s2 + s3
C.E. = 1 + s – s2 + s3 = 0
since the coefficient of s2 is negative, therefore the given system is unstable.Correct Option: B
Given closed loop system.
Fk(s) = K(s + 3) 1 + s – s2 + s3
C.E. = 1 + s – s2 + s3 = 0
since the coefficient of s2 is negative, therefore the given system is unstable.
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G (s) = 1 – s s(s + 2)
The system with this transfer function is operated in closed-loop with unity feedback. The closedloop system is—
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Given, G(s) = 1– s and H(s) = 1 s(s + 2)
C.E. ⇒ 1 + G (s) H (s)=0⇒ 1 + 1 – s .1 = 0 s(s + 2)
⇒ s2 + 2s + 1 – s = 0
⇒ s2 + 2s + 1 – s = 0
⇒ s2 + s + 1 = 0
R.H.C. for C.E. = s2 + s + 1 = 0
s2 1 1
s1 1 0
s0 1
All the elements in the first column are positive.
Hence the closed loop system will be stable system.Correct Option: A
Given, G(s) = 1– s and H(s) = 1 s(s + 2)
C.E. ⇒ 1 + G (s) H (s)=0⇒ 1 + 1 – s .1 = 0 s(s + 2)
⇒ s2 + 2s + 1 – s = 0
⇒ s2 + 2s + 1 – s = 0
⇒ s2 + s + 1 = 0
R.H.C. for C.E. = s2 + s + 1 = 0
s2 1 1
s1 1 0
s0 1
All the elements in the first column are positive.
Hence the closed loop system will be stable system.
- The transfer function is
.K (s + 1)(s + 2)(s + 3)
The break point lies between—
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The C.E. is
(s + 1) (s + 2) (s + 3) + K = 0
K = – (s2 + 3s + 2) (s + 3)
or
K = – (s3 + 3s2 + 2s + 3s2 + 9s + 6)
K = – (s3 + 6s2 + 11s + 6)
dK/ds = – (3s2 + 12s + 11)
put
dK/ds = 0
3s2 + 12s + 11 = 0s = – 12 ±√122 – 4 × 3 × 11 = – 12 ± √12 2 × 3 6
which is lies between – 1 and – 2.Correct Option: B
The C.E. is
(s + 1) (s + 2) (s + 3) + K = 0
K = – (s2 + 3s + 2) (s + 3)
or
K = – (s3 + 3s2 + 2s + 3s2 + 9s + 6)
K = – (s3 + 6s2 + 11s + 6)
dK/ds = – (3s2 + 12s + 11)
put
dK/ds = 0
3s2 + 12s + 11 = 0s = – 12 ±√122 – 4 × 3 × 11 = – 12 ± √12 2 × 3 6
which is lies between – 1 and – 2.
