Understanding Feedback System Stability

This solution analyzes the stability of a feedback system described by the characteristic equation:
$s^4 + 2ks^3 + s^2 + 5s + 5 = 0$
We will use the Routh-Hurwitz stability criterion to determine the range of the parameter $k$ for which the system remains stable.

Applying the Routh-Hurwitz Criterion

The Routh-Hurwitz criterion requires constructing a Routh array using the coefficients of the characteristic polynomial. For stability, all entries in the first column of the array must be non-zero and possess the same sign (typically positive).

Routh Array Construction

The given characteristic equation is: $P(s) = 1s^4 + 2ks^3 + 1s^2 + 5s + 5 = 0$ The Routh array is set up as follows:

Calculating Routh Array Elements

Let's compute the necessary elements:

• $s^2$ Row:
  • $b_1 = \frac{(2k)(1) - (1)(5)}{2k} = \frac{2k - 5}{2k}$
  • $b_2 = \frac{(2k)(5) - (1)(0)}{2k} = \frac{10k}{2k} = 5$ (This assumes $k \neq 0$)
  • $b_3 = \frac{(2k)(0) - (1)(0)}{2k} = 0$
• $s^1$ Row:
  • $c_1 = \frac{(b_1)(5) - (2k)(b_2)}{b_1} = \frac{(\frac{2k - 5}{2k})(5) - (2k)(5)}{\frac{2k - 5}{2k}}$ Simplifying this gives: $c_1 = \frac{\frac{10k - 25}{2k} - 10k}{\frac{2k - 5}{2k}} = \frac{10k - 25 - 20k^2}{2k - 5}$
  • $c_2 = \frac{(b_1)(0) - (2k)(b_3)}{b_1} = 0$
  • $c_3 = 0$
• $s^0$ Row:
  • $d_1 = \frac{(c_1)(b_2) - (b_1)(c_2)}{c_1} = \frac{c_1 \cdot 5 - b_1 \cdot 0}{c_1} = 5$ (This assumes $c_1 \neq 0$)
  • $d_2 = 0$
  • $d_3 = 0$

Stability Conditions Analysis

For the feedback system to be stable, all the entries in the first column of the Routh array must be positive. The first column entries are: 1, $2k$, $b_1$, $c_1$, and 5.

1. $s^4$ coefficient: 1. This is always positive ($1 > 0$).
2. $s^3$ coefficient: $2k$. For stability, this must be positive: $2k > 0$, which implies $k > 0$.
3. $s^2$ coefficient ($b_1$): $b_1 = \frac{2k - 5}{2k}$. Since we already established $k > 0$, the denominator $2k$ is positive. For $b_1$ to be positive, the numerator must also be positive: $2k - 5 > 0$, which means $k > \frac{5}{2}$.
4. $s^1$ coefficient ($c_1$): $c_1 = \frac{-20k^2 + 10k - 25}{2k - 5}$.
   • Let's examine the numerator $N = -20k^2 + 10k - 25$. We calculate its discriminant: $\Delta = (10)^2 - 4(-20)(-25) = 100 - 2000 = -1900$. Since the coefficient of $k^2$ (-20) is negative and the discriminant ($\Delta$) is negative, the numerator is always negative for all real values of $k$.
   • For $c_1$ to be positive ($c_1 > 0$), we have the condition $\frac{\text{Negative}}{\text{Denominator}} > 0$. This inequality holds only if the denominator is also negative.
   • The denominator is $D = 2k - 5$. Thus, we require $2k - 5 < 0$, which implies $k < \frac{5}{2}$.
5. $s^0$ coefficient ($d_1$): 5. This is always positive ($5 > 0$).

Conclusion on System Stability

To ensure stability, the parameter $k$ must satisfy all the derived conditions simultaneously:

• From the $s^3$ term analysis: $k > 0$.
• From the $s^2$ term ($b_1$) analysis: $k > \frac{5}{2}$.
• From the $s^1$ term ($c_1$) analysis: $k < \frac{5}{2}$.

There is a contradiction: $k$ cannot be simultaneously greater than $\frac{5}{2}$ and less than $\frac{5}{2}$. This means no value of $k$ can satisfy all the necessary conditions for stability.

Therefore, the feedback system is unstable for all possible values of $k$.