Test: Stability
15 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Test: Stability
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The forward-path transfer function of a ufb system is
For system to be stable, the range of K is
Routh table is as shown in fig. S.6.211
The open-loop transfer function of a ufb system is
The closed loop system will be stable if the value of K is
Routh table is as shown in fig.
200K > 0 → K > 0, 30K2 - 140K > 0
satisfy this condition.
First combine the parallel loop K/s2
and 2/s giving
Then apply feedback formula with 
and 
and then multiply with s2.
The poles location for this system is shown in fig.The value of K is
Denominator = s3 + s2 + 2s + K Routh table is as shown in fig.
Row of zeros when K = 2,
s2 + 2 = 0, ⇒ s = -1, j√2, - j√2
The forward-path transfer function of a ufb system is T(s) = 
The system is
Closed loop transfer function
Routh table is as shown in fig. S.6.2.28
2 RHP poles so unstable.
The open loop transfer function of a system is as
The range of K for stable system will be
The characteristic equation is 1 + G(s)H(s) = 0
⇒ s(s - 0.2)(s2 + s + 0.6)+K(s + 0.1) = 0
s4 +0.8 s3 +0.4s2 +(K - 0.12)s +0.1K = 0
Routh table is as shown in fig. S.62.29
K > 0, 055 -125K > 0 ⇒ K < 0.44 -125K2 +0.63K -0066 >0
(K - 0.149)(K - 0355) < 0, 0.149 < K < 0.355
The open-loop transfer function of a ufb control system is given by
For the system to be stable the range of K is
Characteristic equation
s(sT1 + 1)(sT2 +1) + K = 0
T1T2s3 + (T1 + T2)s2 + s + K = 0
Routh table is as shown in fig S.6.2.30
If the roots of the have negative real parts, then the response is ____________
If the roots of the have negative real parts then the response is bounded and eventually decreases to zero.
The closed loop transfer function of a system is
The number of poles in RHP and in LHP are
3 RHP, 2 LHP poles.
The closed loop transfer function of a system is
The number of poles in LHP, in RHP, and on jω - axis are
No sign change exist from the s4 row down to the s0 row.
Thus, the even polynomial does not have RHP poles. Therefore because of symmetry all four poles must be on jw -axis.
For the system shown in fig. the number of poles on RHP, LHP, and imaginary axis are
Closed loop transfer function
Routh table is as shown in fig. S.6.2.34
From s4 row down to s0 there is one sign change. So LHP–1 + 1= 2 pole. RHP–1 pole, jw - axis - 2 pole.
If the system is given with the unbounded input then nothing can be clarified for the stability of the system.
For the open loop system of fig. location of poles on RHP, LHP, and an jω - axis are
Routh table is as shown in fig
Them is two sign change from the s4 mw down to the s° row. So two roots are on RHS. Because of symmetry rest two roots must be in LHP. From s6 to s4 there is 1 sign change so 1 on RHP and 1 on LHP.
Total LHP 3 root, RHP 3 root.
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