Question

Determine the stability of closed loop control system whose characteristic equation is s5+3s4+5s3+4s2+s+3=0 stable marginally stable unstable marginally unstable

Answer 1

Answer:

The system of equation is always Stable  .

Step-by-step explanation:

Given as :

The characteristic equation

CE = $S^{5}+ 3 S^{4}+5 S^{3}+4S^{2}+S+3$

Now, for determining the stability

Apply Routh Array method .

$S^{n}$  $a_0$  $a_2$  $a_4$  .......

$S^{n-1}$ $a_1$  $a_3$  $a_5$  .......

$S^{n-2}$  $b_1$  $b_2$  $b_3$  ......

where $b_1$  = $\dfrac{a_1 a_2 - a_0 a_3}{a_1}$

          $b_2$  = $\dfrac{a_1 a_4 - a_0 a_5}{a_1}$

Now, Applying this in given characteristic equation

$S^{5}$  $1$  $5$  $1$  .......

$S^{4}$ $3$  $4$  3  .......

$S^{3}$  ( $\dfrac{3\times 5-1\times 3}{1}$ ) = 12  $\dfrac{1\times 1-1\times 3}{1}$ = - 2 $b_3$  ......

$S_4$  0  0  .......

here in the first column , the sign is always positive

So, The system of equation is always Stable

Hence, The system of equation is always Stable  . Answer