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Match the system open-loop transfer functions given in List-I with the steady-state errors produced a unit ramp input. Select the correct answer using the codes given below the lists:
List-I List-II A. 30/s2 + 6s + 9 1. Zero B. 30/s2 + 6s 2. 0.2 C.
30/s2 + 9s3. 0.3 D. s + 1/s2 4. infinity Codes
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- A B C D
1 2 3 4
- A B C D
4 3 2 1
- A B C D
1 3 2 4
- A B C D
4 2 3 1
- A B C D
Correct Option: B
Steady state error, ess = s → 0 lim sE(s)
where,
= | ||
R(s) | 1 + G(s) H(s) |
or
E(s) = | |
1 + G(s) H(s) |
ess = s → 0lim s. | ![]() | ![]() | |
1 + G(s) H(s) |
given, r(t) = tu(t)
or
R(s) = 1/s2
Now,
(a) When,
G(s) H(s) = | |
s2 + 6s + 9 |
ess =s → 0lim | |
1 + {30 (s2 + 6s + 90)} |
=s → 0lim | = ∞ | |
s2(s2 + 6s + 120) |
(b) When,
G(s) H(s) = 30/s2 + 6s
ess =s → 0lim | |
1 + (30/s2 + 6s) |
=s → 0lim | |
s2(s2 + 6s + 30) |
= 6/30 = 0·2
(c) When,
G(s) H(s) = 30/s2 + 9s
ess =s → 0lim | |
1 + (30/s2 + 9s) |
=s → 0lim | |
s2(s2 + 9s + 30) |
= 9/30 = 0·3
G(s) H(s) = | |
s2 |
ess =s → 0lim | = | = 0 | ||
1 + (s + 1)/s2 | s2 + s +1 |
Hence alternative (D) is the correct choice.