Can the Routh-Hurwitz Criterion Determine the Stability of this Equation?

[Dustinsfl]

Determine the stability for \(s^3 + 3s^2 + 2(1+K_p)s + 2K_i=0\).

By the Routh-Hurwitz stability criterion, we have
\[
\begin{array}{ccc}
s^3 & 1 & 2(1 + K_p)\\
s^2 & 3 & 2K_i\\
s^1 & \frac{6(1 + K_p) - 2K_i}{3} & 0\\
s^0 & 2K_i & 0
\end{array}
\]
From the \(s^3\) line, we have that \(2(1 + K_p) > 0\); therefore, \(K_p > -1\). From the \(s^2\) and \(s^0\) lines, we have that \(2K_i > 0\); thus, \(K_i > 0\). From the \(s^1\) line, we have that \(6K_p - 2K_i + 6 > 0\); therefore, \(\frac{K_i}{K_p} < \frac{3}{K_p} + 3\).
The book says I should obtain:
\[
0<\frac{K_i}{K_p} < 13.5
\]
I get the greater than zero piece. However, I don't see I can go from
\[
\frac{K_i}{K_p} < \frac{3}{K_p} + 3 < 13.5
\]

 

[jvicens]

First, we can simplify the inequality \(\frac{K_i}{K_p} < \frac{3}{K_p} + 3\) by multiplying both sides by \(K_p\), since we know that \(K_p\) is positive (from the \(s^3\) line). This gives us \(K_i < 3 + 3K_p\).
Next, we can use the fact that \(K_p > -1\) to substitute \(K_p = -1 + x\), where \(x\) is a positive number. This gives us \(K_i < 3 - 3x\).
Finally, we can use the fact that \(K_i > 0\) to substitute \(K_i = y\), where \(y\) is a positive number. This gives us \(y < 3 - 3x\), which can be rewritten as \(y < 3(1 - x)\).
Since we know that \(x\) and \(y\) are positive numbers, we can set the upper bound for \(x\) to be \(1\) (since \(1 - x\) will always be positive) and find the corresponding upper bound for \(y\). This gives us \(y < 3(1 - 1) = 0\).
Therefore, we have the inequality \(0 < y < 3 - 3x\), where \(x\) and \(y\) are both positive numbers. To find the maximum value of \(y\), we can set \(x = 0\) (since \(x\) is a positive number) and get \(y < 3\). This gives us the final inequality:
\[0 < \frac{K_i}{K_p} < 3\]
which can be rewritten as \(0 < \frac{K_i}{K_p} < 13.5\).