# Homework Help: Root Loci Problem - Noob need help!

1. Jul 28, 2009

### Kai-Itza

1. The problem statement, all variables and given/known data

3. The attempt at a solution

Hi,

I’ve been studying Root Loci for almost 3 months now, on and off, so I know most of the methods involved with it. I’m still pretty much new to the subject still as I’ve recently hit a problem with this exercise that I’ve been working on for the past 3 weeks, I’ve spent most of the last and this week literally stuck on this particular part, and it’s driving me crazy!

I can’t just leave the thing as I’ll always have the cravings to come back to finish the job, as my curiosity for finding the answer will only grow.

Can someone help me out and end this torture?

Question:

“A position control system has a block diagram as shown above.
The plant is controlled by a Compensator, Gc(s) and variable gain, K
The Specification for the whole system is:-”

1. The Steady State error for a step input is zero
2. the steady state error for a unit ramp input is less than 4
3. the percentage overshoot for a step input is less than 5%
4. the 2% settling time for a step input is less than 10 seconds

Design a compensator and determine K. Use the root locus point design method.

What I’ve done so far…

the Plant is: S²+2S²+2S+1

I figured that the system compensator, Gc = (S+a)/S

As this system must be type 1 for it to be a zero steady state system, where B = 0

This makes the system a P+I type

The percentage overshoot for a step input is less than 5%

Therefore, Overshoot = (e - πζ)/√ (1-ζ³) = 0.05

The line of constant damping:

ζ = Ln(0.5) = -0.6931

The angle can be deprived as:

cosθ = -0.6931

so, cos^-1(0.6931) = 46.1239

The step response of the system does take the form of an exponential rise, upon which a decaying sinusoid takes form.
As the folloing states, the resulting reponse must be within 2% settling time in less than 10 seconds

The 2% settling time for a step input is less than 10 seconds

This is the mathematical term to describe the response of the system shown below:

1-e^-t/T = 1-e^-αt

Where,

α = 1/T
αt = 3.91
t = 10

so, α = 0.391

2. Relevant equations

It is from here on in, that I find myself stuck, as at the moment I’m trying to find ‘a’.

the Angle Criterion is easy enough to follow and everything after finding ‘a’ would be relatively straightforward after that, IMO I find that α looks way too small for my liking, as TBH I’m expecting the result to be much larger, around the α = 1.1+ mark at my best estimate.

This is becoming like less of an exercise and more like chewing iron nails, I know that I’ve been studying the subject for a while but I’m still a noob at it and have lots more to learn by the looks of it. :)

Is there something wrong with my calculations, or have I simply missed out something?

Thanks,

-Kai-Itza-

P.S. If I've missed anything let me know, it took quite a while to type, lol.

2. Jul 29, 2009

### CEL

It seems correct to me. You know the real part of the dominant poles (0.391) and the angle θ, so you can obtain the imaginary part of the poles.
You know that if those poles belong to the root-locus, the sum of the angles from the poles minus the sum of the angles from the zeros of the open-loop function, must be an odd multiple of 180 degrees.
Your only unknown is the zero at -a.

3. Jul 29, 2009

### Kai-Itza

Thanks for the reply, CEL :)

I originally thought that this was right when I checked it, but I thought I’d get it checked by someone else to confirm this. My real concern was the value of ‘a’, as every time I tried to calculate it, it always came out as a slightly different answer, half of which stupidly small.
I honestly think it’s the way that I calculated it, somewhere during the Angle Criterion. I’ll do another draft along with the root locus diagram (with my wonderful skills on Paint) that I’m using for anyone to look at.

To find the value ‘a’, Angle Criterion can be used by locating the pole, P, and the angles of the open loop poles and zeroes on the root locus diagram.

So, P = 0.391 +/- j0.3912tan (46.1239)
= 3.91 +/- j0.4068

‘a’ is placed at random

The Angle Criteria is argG0 = -180
Hence, α1-α2-3α3 = -180

Where α1 is the positive since pole zero and,
α2-3α3 is negative since pole

α2 and α3 can be determined exactly:

α2 = 180-θ = 180-46.1239
α2 = 133.8764

α3 = tan^-1(0.4068/0.6931) = 30.409

α3 = 91.229 – since triple pole.

α1 = -180=133.8764 + 91.229 = 45.106

And I’m looking for the angle opposite to this one, so:

α1 = 180 – 45.106 = 134.894

0.4068/y = tan134.894

so, y = 0.4068/tan(134.894) = -0.4053

Therefore ‘a’ = 0.4068—0.4053

‘a’ = 0.8121

‘a’ does looks like the right value, but I do like to hear another opinion, if any, from someone with a little more experience than me. Is this right?

-Kai-Itza-

4. Jul 29, 2009

### CEL

Where did you get the triple pole at s = -1?
There is one pole at s = -1 and two conjugate poles at $$-\frac{1}{2}\pm j \frac {\sqrt{3}}{2}$$

5. Jul 31, 2009

### Kai-Itza

You must note that the root locus diagram I drew up in my last post was not exactly in scale, that one was a quick 5-min draw up to show what I’ve done so far.

As to your question, CEL, I’ve forgotten to place the transfer function of the plant as per stated from the exercise of which I’m working on and trying to figure out what I’ve missed/ screwed up in my calculations

The Plant described from the exercise is as follows:

Gp = 1/(s + 1)³

(In my earlier posts, I’ve broken it down where my thoughts meandered whilst working through this, my mistake!)

I’ll edit any posts where this has occured

As you can see from Gp, there are:

3 poles at s =-1 (as I got this from the Plant, Gp)

Is this the location of the Pole written in another form? As I can’t see where this had came from, forgive me, but is this another law to Root Locus Design, one that I may have missed/forgotten?
Forgive the questions, as there may be a few holes in my learning. Root Loci is still a new thing to me, I’m still a learner at this :).

-Kai-Itza-

EDIT: Damn, can't edit my past posts on this thread,lol

Last edited: Jul 31, 2009
6. Jul 31, 2009

### CEL

In your original post, you described the denominator of the plant as $$s^3+2s^2+2s+1$$, whose roots are the ones I pointed to.

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