Test: Stability - Electrical Engineering (EE) MCQ

Routh table is as shown in fig. S.6.211

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.

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

Closed loop transfer function

Routh table is as shown in fig. S.6.2.28

2 RHP poles so unstable.

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 

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 bounded and eventually decreases to zero.

3 RHP, 2 LHP poles.

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.

 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.

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.