# Root locus sketching

## Homework Statement

Re-arrange the open-loop transfer function and sketch the root locus for the system for variation of the parameter 'a' from 0 to infinity.

## Homework Equations

G(s)H(s) = 10(s+a) / [s(s+1)(s+8)]

## The Attempt at a Solution

my problem is that i dont know how to re-arrange the transfer function into the familiar form of KF(s). any help would be appreciated. Thanks!

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uart
Hi wh88. Can you tell me what must be the angle (and magnitude) of G(s)H(s) for the point "s" to be a pole of the closed loop TF?

sorry i made a mistake in the opening post. it should be 'variation of the parameter a'. i missed out the 'a'.

to uart, i believe its -180 degs and magnitude 1

uart
sorry i made a mistake in the opening post. it should be 'variation of the parameter a'. i missed out the 'a'.
That's ok, I guessed that

i believe its -180 degs and magnitude 1
Yep that's correct (180 degrees + any multiple of 360). The key to the root locus method is to find points in the complex plane where the sum of the angles on the numerator differs by 180 degrees (+360n) from the sum of angles on the denominator.

Start by plotting all the poles and zeros (of GH) on the complex plane and recognize that the angle that each pole (or zero) contributes to GH is equal to the angle of the vector drawn from the pole (or zero) to the test point "s". It's definitely a process that requires some practice before you become proficient.

That's ok, I guessed that

Yep that's correct (180 degrees + any multiple of 360). The key to the root locus method is to find points in the complex plane where the sum of the angles on the numerator differs by 180 degrees (+360n) from the sum of angles on the denominator.

Start by plotting all the poles and zeros (of GH) on the complex plane and recognize that the angle that each pole (or zero) contributes to GH is equal to the angle of the vector drawn from the pole (or zero) to the test point "s". It's definitely a process that requires some practice before you become proficient.
Thanks for your help, uart. My understanding is that before plotting the poles and zeros on the complex plane, we have to get the open loop TF arranged in the form KF(s) if we want to find out how variation of K from 0 to infinity affects the system.
So in this case, wont we have to re-arrange the TF to the form aF(s)?

Hi wh88,

while uart has very kindly answered your question if you are doing a course in linear control you MUST (along with anyone else interested in this topic) look at the series of lectures by prof Madan Gopal on utube. I think root locus starts around lec 33.

Likely you will find these vids of more use than any text or lecture you might otherwise sit in on. Definitely worth you time. extremely well presented!

uart
Thanks for your help, uart. My understanding is that before plotting the poles and zeros on the complex plane, we have to get the open loop TF arranged in the form KF(s) if we want to find out how variation of K from 0 to infinity affects the system.
So in this case, wont we have to re-arrange the TF to the form aF(s)?
Ok you're right, this case is quite different to a traditional root locus question where the parameter under variation is the loop gain. It's been a long while since I've had to do root locus so I'm pretty rusty, but I think that in the traditional case that every point in the complex plane that satisfies the angle condition automatically satisfies magnitude condition for some value of of loop gain. In this case however, where the parameter under variation is the position of a zero, then it's definitely not true that magnitude condition is automatically satisfied, so this seems like a much harder problem (compared to the standard type root locus problem).

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uart
wh88, is there any further context to this problem. Do you know if you're meant to try and solve it numerically or are you pretty sure that you're meant to be solving it graphically using the angle and magnitude conditions of root locus?

wh88, is there any further context to this problem. Do you know if you're meant to try and solve it numerically or are you pretty sure that you're meant to be solving it graphically using the angle and magnitude conditions of root locus?
uart, the question asks us to re-arrange the open loop transfer function and sketch the root locus for the system as a varies from 0 to infinty. I doubt numerical methods are required. Its the re-arrangement thats troubling me.

uart
uart, the question asks us to re-arrange the open loop transfer function and sketch the root locus for the system as a varies from 0 to infinty. I doubt numerical methods are required. Its the re-arrangement thats troubling me.
Ok I can see what they want you to do now.

You have to find where G(s) H(s) = -1. So split it across the numerator as :

$$\frac{10s +10a}{s (s+1) (s+8) } = \frac{10s}{s (s+1) (s+8) } + \frac{10a}{s (s+1) (s+8) }$$

Using the above you can rearrange GH=-1 to :

$$\frac{10a}{s (s+1) (s+8) } =- \left(1 + \frac{10s}{s (s+1) (s+8) } \right)$$

Now put the RHS to a common denominator and factorize the resulting numerator (it factorizes pretty easily). Finally mult/divide both sides in the obvious way so as to re-establish "-1" on the RHS.

You end up with an equation in the traditional root locus form, where the variable is proportional to the loop gain.

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Ok I can see what they want you to do now.

You have to find where G(s) H(s) = -1. So split it across the numerator as :

$$\frac{10s +10a}{s (s+1) (s+8) } = \frac{10s}{s (s+1) (s+8) } + \frac{10a}{s (s+1) (s+8) }$$

Using the above you can rearrange GH=-1 to :

$$\frac{10a}{s (s+1) (s+8) } =- \left(1 + \frac{10s}{s (s+1) (s+8) } \right)$$

Now put the RHS to a common denominator and factorize the resulting numerator (it factorizes pretty easily). Finally mult/divide both sides in the obvious way so as to re-establish "-1" on the RHS.

You end up with an equation in the traditional root locus form, where the variable is proportional to the loop gain.
i finally got it. Thank you very much uart!