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  1. Given: KK1 = 99; s = j1 rad/s, the sensitivity of the closed-loop system (shown in the figure) to variation in parameter K is approximately—
    1. 0.01
    2. 0.1
    3. 1.0
    4. 10
Correct Option: B

Given, K K1 = 99
s = j1 it means ω = 1 rad/sec.
We have to calculate sensitivity of closed loop system to variation in parameter K i.e.,

SMK =
∂M/M
∂K/K

=
K
×
∂M
M∂K

or
SMK =
K
∂M
G∂K

Now,
M =
G(s)
=
K/(10s + 1)
1 + G(s)H(s)(1 + KKt)/(10s + 1)

or
M =
K
(10s + 1+ KKt)

∂M
=
(10s + 1 + kkt)(∂/∂K).K - K (∂/∂K)(10s + 1 + kk1)
∂K(10s + 1 + KKt)

or
∂M
=
10s + 1 + kkt - kkt
∂K(10s + 1 + KKt)2

or
∂M
=
10s + 1
∂K(10s + 1 + KKt)2

and SMK =
K
×
(10s + 1)
K/(10s + 1 + kkt)(10s + 1 + kkt)2

or
or SMK =
(10s + 1)
(10s + 1 + kkt)

or SMK =
(10s + 1)
10s + 1 + 99

SMK =
(10s + 1)
10s (s + 10)

or
SMK =
100ω2 + 1
10√102ω2

or
SMK =
100 + 1
10√100 + 1

or SMK = 0·1
Hence alternative (B) is the correct choice.



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