I have a control system that we are working on with a forward transfer function of:-

\$ \dfrac{K(2s+3)}{s^2(s^4+2s^3+4s^2+2s+7)}\$

So I've set up the routh table for this, I've found there are 2 sign changes, therefore, two poles. But does a value of k=4 affect the control system? Or any value for that matter?

I can't think why multiplying the transfer function by a constant would affect the stability in any way

  • \$\begingroup\$ I've changed your formulae into MathTex for ease of reading. But the equation seems odd, please check to see if those edits make sense. \$\endgroup\$ May 8, 2013 at 19:29
  • \$\begingroup\$ It should be s^4, not 2^4. I've changed it back again. Thanks for that. Any ideas on my question? \$\endgroup\$ May 8, 2013 at 20:09
  • \$\begingroup\$ Nobody knows how to do this? \$\endgroup\$ May 8, 2013 at 21:24

2 Answers 2


This system has 6 poles and one zero.

fully factored it becomes:

\$ \dfrac{K(2s+3)}{(s-(0.329881+1.27963 i)) (s-(0.329881-1.27963 i)) (s+(1.32988-1.49665 i)) (s+(1.32988+1.49665 i))}\$




\$2^{nd}\$ order at \$s=0\$

\$1^{st} \$ order @ \$s= 0.329881+1.27963 i\$

\$1^{st} \$ order @ \$s= 0.329881-1.27963 i\$

\$1^{st} \$ order @ \$s= -1.32988+1.49665 i\$

\$1^{st} \$ order @ \$s= -1.32988-1.49665 i\$

So this is non-causal, unstable in a strict sense. However, someone who has more recent practise in this area should be able to use the info above and tell you the limits. I would be answering "It isn't stable anyways so the K factor doesn't matter" but then I'm rusty but I did the factoring so I thought I'd put it up here for others to use.

  • \$\begingroup\$ thank you for your detailed answer, I'm intersted in finding out whether a value of 4 for gain would make this unstable or not \$\endgroup\$ May 8, 2013 at 22:30
  • \$\begingroup\$ Well none of the poles lie within the unit circle so it doesn't even meet the criteria for minimal phase either. I't's looking more and more like it isn't a stable system at all. \$\endgroup\$ May 8, 2013 at 23:33
  • \$\begingroup\$ Apparently there are a range of values of K that maintain the stability of this system... \$\endgroup\$ May 9, 2013 at 5:59
  • \$\begingroup\$ You are right about the system being unstable but that is not because of the unit circle, it is for discrete systems. For continuous systems, if all poles are in the left-hand plane, the system is stable. If there is at least one pole at the right-hand side, the system is unstable. \$\endgroup\$ May 9, 2013 at 10:58
  • \$\begingroup\$ Also, the system is causal because the order of the numerator is less the denominator. \$\endgroup\$ May 9, 2013 at 11:09

The position of the poles change with K, that is why it can affect stability.

Your system is unstable because there are poles on the right-hand plane. You can see if you can make the system stable by adjusting K with a root locus plot.

The root locus of your system is like this:

root locus

As you can see the poles on the right hand side will always be on the right-hand side for all K. So, that means the system cannot be stabilized with a proportional controller.

You can also do that with Routh table as you've tried, but I do not remember how that was done. Using root locus is always easier if you have MATLAB (or other root locus plotting software) at hand.


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