Question

The characteristic equation of system is given by s4+20 s3+15s2+2s+k=0. determine range of k for system to stable.

Answer 1

Answer:

k <1.49

Explanation:

S*4         1        15        K

S³          20       2

S²           14.9     K

S            ((29.8-20K)/14.9)

S*0         1

The system to be Stable,there should be no sign changes,  i.e

((29.8-20K)/14.9) > 0

29.8>20K

K<1.49

Answer 2

Answer:

$0 < K < 1.49$

Explanation:

For stability,

all the elements of the 1st column of the Routh table must be positive,

i.e. $K > 0,$

and $29.8 − 20$

$k 14.9 > 0 29.8−20 k14.9 > 0 K < 1.49$

Hence, For the stability range is, $0 < K < 1.49$

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