Root Locus of Negative Feedback System

  • #1
70
6

Homework Statement:

(Included in picture below) Essentially, sketching the root locus for a simple control system with negative feedback

Relevant Equations:

N/A
question.png

From my understanding, the root locus is only concerned with open loop gain. I figured this means you would ignore the negative feedback loop and calculate the root locus from just the plant's function

Workings:
zeros: -1
poles: 0, -2, -2,

relative degree = 2
=> 90-degree asymptotes
meeting point = -3/2

And then sketch using that information

However, in the provided hints:
solution.png

Looking at this, it seems the open loop transfer function is the two functions (plant and controller) multiplied together
We've never covered this in the lectures, but does this mean that for open loop with negative feedback, you'd just include the negative feedback in the main branch?
 

Answers and Replies

  • #2
anorlunda
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From my understanding, the root locus is only concerned with open loop gain. I figured this means you would ignore the negative feedback loop and calculate the root locus from just the plant's function
You vary the open loop gain correct, but the system of equations is for the entire closed loop.
 
  • #3
70
6
You vary the open loop gain correct, but the system of equations is for the entire closed loop.
Thanks for the reply!

So I found the closed loop function to be:

$$\frac {K(0.5s^2 + 1.5s + 1)} {(0.5s^4 + 3s^3 + 6s^2 + (4+K)s + K)}$$

However, the hints for question imply it should be a lot simpler than this

$$ 2K \frac {s+1} {s(s+2)^3} $$

Is there a step after finding the closed loop gain to convert it to the form needed for the root locus analysis?

Workings:
241849


Using the block diagram reduction method for negative feedback:
241848
 
  • #4
70
6
Okay, I think I have it figured out. All of the examples we covered in our notes were unity feedback based. This lead to a misunderstanding on my behalf.

For root locus, you find the closed loop gain.

You then get manipulate it into the form
$$\frac {f(s)} {factor(1+ \frac {h(s)}{g(s)})}$$

This matches the feedback loop equation and so the open loop gain can be considered to be $$\frac {h(s)}{g(s)}$$

I've since figured out the workings for this question:

$$\frac {2K(s+1)} {s(s+2)^3}$$
answer.jpg
 
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