What is the range of values for stability using the Routh-Hurwitz criterion?

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 3K views
Dustinsfl
Messages
2,217
Reaction score
5

Homework Statement


Find the stability of ##s^3 + 3s^2 + 2(1 + K_p)s + 2K_i = 0##.

Homework Equations


Routh-Hurwitz criterion

The Attempt at a Solution


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## line, 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,
$$
\frac{K_i}{K_p} < \frac{3}{K_p} + 3.
$$
I am supposed to conclude
$$
0 < \frac{K_i}{K_p} < 13.5.
$$
From line ##s^2##, we get greater than zero, but what do I do to go from ##\frac{3}{K_p} + 3## to ##13.5##?
 
Last edited:
Physics news on Phys.org
To be honest I do not know. I went through the routh-Hurwitz criterion and got the same results you did

If you plug in 10 for ki and 1 for kp the system is unstable (i ran it in matlab)
therefore the conclusion that 0<ki/kp<13.5 is incorrect

I recommend confirming that all the information you provided us is correct, and all the information provided to you is correct.
 
donpacino said:
To be honest I do not know. I went through the routh-Hurwitz criterion and got the same results you did

If you plug in 10 for ki and 1 for kp the system is unstable (i ran it in matlab)
therefore the conclusion that 0<ki/kp<13.5 is incorrect

I recommend confirming that all the information you provided us is correct, and all the information provided to you is correct.

Is the information from the book:
I had to use the link since the image appears too big on the site.
http://i.imgur.com/M5Ks5up.jpg?1

M5Ks5up.jpg
 
Last edited:
Must be an error in your book. Your Routh table is correct, so to avoid any sign changes in the first column for ##K_p > 0, K_i > 0##, we must have:
[tex] \begin{align}<br /> 6(1 + K_p) - 2K_i \geq 0 \Leftrightarrow K_i \leq 3(1 + K_p) \quad (1)<br /> \end{align}[/tex]
Although (1) being true is necessary and sufficient for stability, the weaker condition:
[tex] K_i \leq 3K_p \Leftrightarrow \frac{K_i}{K_p} \leq 3[/tex]
is sufficient.