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Find the roots of the following polynomial by use of the root locus method.

3s⁴ + 10s³ + 21s² + 24s + 30 = 0 (The root locus plot has not been given.)

Can you please help me with this?

This is a question from Control Systems engineering by M Gopal and I J Nagrath 5th edition.

And this is not a homework question or anything related to assignment. It's a sheer attempt to increase my knowledge on root locus method

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closed as unclear what you're asking by Eugene Sh., Dave Tweed Dec 11 '18 at 20:12

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ Tried anything so far? \$\endgroup\$ – Eugene Sh. Dec 11 '18 at 16:11
  • \$\begingroup\$ Homework questions with no attempt at a solution get closed pretty quickly here. Show what you've already tried. \$\endgroup\$ – Hearth Dec 11 '18 at 17:06
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    \$\begingroup\$ Nishanth, your edit to the question doesn't add important information that is missing from your question. You've been requested to show what you've tried so far. Even if this isn't for homework, it is in the form of homework. It is an exercise that you have not shown work for. Thus how do we know where you are stuck? "Can you help me with this" is completely generic. Help us to help you. If you want effort in an answer, put effort in your question. \$\endgroup\$ – Bort Dec 12 '18 at 5:38
  • \$\begingroup\$ I've now studied Root Locus intensively the last days and come to the conclusion that your question makes no sense. - The Root Locus method appears to be about sketching where poles and zeros move as you adjust some parameter on one of the terms in your polynomial. - For getting the roots first so you got a starting point you would either use Newton's method, like I did, or use Durand - Kerner method, or some other method. \$\endgroup\$ – Harry Svensson Dec 15 '18 at 2:59
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You could use Newton's method. Yeah I know, it's not the "root locus method", but hey, it works.


Let \$F(s)=3s^4+10s^3+21s^2+24s+30\$

Clean it up by dividing everything by 3 so our largest s is alone

Let \$f(s)=\dfrac{F(s)}{3}=s^4+\frac{10}{3}s^3+7s^2+8s+10\$

then \$f'(s)=4s^3+10s^2+14s+8\$

Now we have everything we need to start converging onto the values that makes the equation go to 0.

\$s_{n+1} = s-\dfrac{f(s)}{f'(s)}= s-\dfrac{s^4+\frac{10}{3}s^3+7s^2+8s+10}{4s^3+10s^2+14s+8}\$

Set s to some good initial value and evaluate the equation above until you get a value that stops changing. In other words, you've find a value that makes \$f(s)=0\$.

If you start with:

  • \$s = +i\$, then \$s\$ converges to \$-0.059 + 1.525i\$
  • \$s =-i\$, then \$s\$ converges to \$-0.059 - 1.525i\$

So we have a root pair at \$-0.059\pm1.525i\$


Now, let's find the other two.

So we know that \$f(s)\$ is on the following form because it is a 4th power. $$f(s)=(s+0.0589+1.5255i)(s+0.0589-1.5255i)(s+\text{something})(s+\text{something})$$

So let's make a new function called \$f_2(s)\$ which will be equal to: $$f_2(s)=\dfrac{f(s)}{(s+0.059 + 1.525i)(s+0.059 - 1.525i)}$$

This will cancel our first two roots so Newton's method will converge onto the other two roots. So now we need to find the derivative of the above equation, lovely. Or we just make an approximation and divide by some "large number, say 10. The only difference between dividing by a constant and dividing by the derivative is that it will converge much slower. Let's also reduce the order of \$f'(s)\$ by 2, because the two poles add an order of 2. And remove the terms that has an s to a negative power.

To make it crystal clear what the "optimal" equation would be, it would be:
\$s_{n+1}=s-\dfrac{f_2(s)}{f'_2(s)}\$

But instead the approximation below is used because I'm slightly lazy, and it will converge anyways.

\$s_{n+1} = s-\dfrac{f_2(s)}{10f'(s^{-2})}= s-\dfrac{(s^4+\frac{10}{3}s^3+7s^2+8s+10)}{10(s+0.059 + 1.525i)(s+0.059 - 1.525i)(4s+10)}\$

If you start with:

  • \$s = +i\$, then \$s\$ converges to \$-1.608 + 1.306i\$
  • \$s =-i\$, then \$s\$ converges to \$-1.608 - 1.306i\$

So we have a root pair at \$-1.608 \pm1.306i\$


So we can conclude that \$F(s)\$ is equal to:

$$\small F(s)=3(s+0.059+1.525i)(s+0.059-1.525i)(s+1.608+1.306i)(s+1.608-1.306i)$$

Don't forget the factor 3 for \$s^4\$.


Now, time to verify it. Entering the following code into octave/matlab

h=zpk([(-0.058899+1.525541i) (-0.058899-1.525541i) (-1.608+1.306i) (-1.608-1.306i)],[],3)

returns

Transfer function 'h' from input 'u1' to output ...

 y1:  3 s^4 + 10 s^3 + 21 s^2 + 24 s + 30.01

Continuous-time model.

Which means that it actually worked, to my surprise.

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    \$\begingroup\$ It's engineering, but not as we know it. +1 for the good clear (math) answer. \$\endgroup\$ – Andy aka Dec 11 '18 at 18:52
  • \$\begingroup\$ Inb4 OP's professor takes off for not doing it the right way, +1 for your numerical methods :) \$\endgroup\$ – KingDuken Dec 11 '18 at 19:59
  • \$\begingroup\$ Thank you for this method. However it doesn't answer my question \$\endgroup\$ – Jarvis Dec 12 '18 at 1:19
  • \$\begingroup\$ @NishanthARao Yeah I know, but you didn't show any attempt at solving it on your own, so I gave an answer on purpose that wouldn't solve it the way you wanted, the same as you purposely did not show any attempts at solving it on your own. - But I'll be the bigger man and at least show you the way. \$\endgroup\$ – Harry Svensson Dec 12 '18 at 14:31

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