Understanding System Stability with Characteristic Polynomial

To determine the stability of a linear system, we analyze its characteristic polynomial. For a system to be stable, all the roots of its characteristic polynomial must lie in the left half of the complex s-plane. This condition is typically checked using the Routh-Hurwitz stability criterion.

Applying the Routh-Hurwitz Criterion

The given characteristic polynomial is:
$P(s) = s^4 + 3s^3 + 5s^2 + 6s + K + 10 = 0$

We construct the Routh array using the coefficients of the polynomial ($a_4=1, a_3=3, a_2=5, a_1=6, a_0=K+10$).

Row	Coefficient 1	Coefficient 2	Coefficient 3
$s^4$	1	5	$K+10$
$s^3$	3	6	0

Now, we calculate the elements for the subsequent rows:

Calculating the $s^2$ Row Coefficients

The first element ($b_1$) of the $s^2$ row is calculated as:

$b_1 = \frac{(3 \times 5) - (1 \times 6)}{3} = \frac{15 - 6}{3} = \frac{9}{3} = 3$

The second element ($b_2$) of the $s^2$ row is:

$b_2 = \frac{(3 \times (K+10)) - (1 \times 0)}{3} = \frac{3(K+10)}{3} = K+10$

The Routh array now looks like this:

Row	Coefficient 1	Coefficient 2	Coefficient 3
$s^4$	1	5	$K+10$
$s^3$	3	6	0
$s^2$	3	$K+10$	0

Calculating the $s^1$ Row Coefficients

The first element ($c_1$) of the $s^1$ row is calculated as:

$c_1 = \frac{(3 \times 6) - (3 \times (K+10))}{3} = \frac{18 - 3K - 30}{3} = \frac{-12 - 3K}{3} = -4 - K$

The Routh array now includes the $s^1$ row:

Row	Coefficient 1	Coefficient 2
$s^4$	1	5
$s^3$	3	6
$s^2$	3	$K+10$
$s^1$	$-4-K$	0

Calculating the $s^0$ Row Coefficient

The first element ($d_1$) of the $s^0$ row is calculated as:

$d_1 = \frac{((-4-K) \times (K+10)) - (3 \times 0)}{(-4-K)} = K+10$

The complete Routh array is:

Row	Coefficient 1	Coefficient 2
$s^4$	1	5
$s^3$	3	6
$s^2$	3	$K+10$
$s^1$	$-4-K$	0
$s^0$	$K+10$	0

Stability Conditions from the First Column

For the system to be stable, all the elements in the first column of the Routh array must be strictly positive.

• $1 > 0$ (Always true)
• $3 > 0$ (Always true)
• $3 > 0$ (Always true)
• $-4-K > 0 \implies -4 > K \implies K < -4$
• $K+10 > 0 \implies K > -10$

Combining these conditions, we get $-10 < K < -4$.