# Root Locus Help (Break in/Away pts)

Discussion in 'Homework Help' started by Whaler, Mar 28, 2011.

1. ### Whaler Thread Starter New Member

Mar 28, 2011
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0
i have a feed-forward tf
G(s) = K(s+2) / s(s+8)(s+1+3j)(s+1-3j)

Poles: 0,-8,-1+j,-1-j (four fintite)
zeros: -2 (Three @ infinity)

Asymptotes:
(-8-1+3j-1-3j)-(-2) / (4-1) = -8/3 = -2.6667
(2k+1)Pi / 4-1 = (Pi/3)(2k+1)
k=0, (Pi/3);
K=1, (Pi);
k=2, (5/3)Pi;

Break In/Away points:
CLTF: K(s+2)/s(s+8)(s+1+3j)(s+1-3j)+k(s+2)=-1
then do i solve for k and take the derivative wrt to s?

jw-axis crossing do i use the denominator from CLTF
s(s+8)(s+1+3j)(s+1-3j)+k(s+2)
and do Routh Hurwitz and manipulate till get row a zero's and figure out what values of k the system is stable and use above row to figure out where it crosses the jw axis?

Any help would be appreciated, been a while and dont remember dealing with imaginaries in the denominator of a feed-forward-loop.

Thanks,

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Unfortunately Matlab doesn't have a feedforward root locus plotter, so we 're going to have to go with math only.

What is CTLF that you mention? The way I see it, the new system will have a TF of:
H=Y/X=G+K
not the one you write right after you first mention CTLF.

Can we clear this before going on?

3. ### Whaler Thread Starter New Member

Mar 28, 2011
6
0
CLTF is my closed-loop transfer function which not sure about.

i figure tf=G/1+GH which is equal to:

denH*numG / denH*denG+numG*numH and h =1 so denH and nomH both equal 1 which is how i got my CLTF.

I have to sketch root locus for feed-forward transfer function (given) in a negative, unity feedback loop.
i thought that you set KGH=-1 solve for k then differentiate, but the imaginaries throw me off when trying to differentiate. i am also not sure if KGH is what i have it set to as CLTF.

Again sorry for not being very helpful but first time with root locus.

Last edited: Mar 28, 2011
4. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Wow, wait a minute. Do you want to do a feedback loop with positive feedback? In this diagram you take the output of G and add it to its input.

In the feedforward mode, you take the input of G and add it directly to its output.

Which of the two you want to do?

5. ### Whaler Thread Starter New Member

Mar 28, 2011
6
0
no negative feedback loop. unfortunately my hw didn't came with a figure, but all my examples/notes are for negative feedback so i am pretty sure he wouldn't switch it on us now.

I am just not sure which equation to use when finding the break away pt's. do i use the given feed-forward tf in negative unity feedback

G(s) = K(s+2) / s(s+8)(s+1+3j)(s+1-3j)

or do i use "my" closed loop tf

K(s+2)/s(s+8)(s+1+3j)(s+1-3j)+k(s+2)

thats where i'm not to sure, then i have the added problem of differentiating a equation with imaginary numbers, which was years ago, but will deal with that when i get to it.

6. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Can you post the homework question as-is, because it seems we have a problem of communication here?

7. ### Whaler Thread Starter New Member

Mar 28, 2011
6
0
sure can, hope this helps.

Sketch the root locus for the following feed-forward transfer functions in a negative, unity feedback loop. Include:
(a) asymptotes
(b) break-away/break-in points
(c) jw-axis crossings
(d) angles of arrival/departure
(e) label the gain, K, at the poles, zeros, break-away/break-in points, and jw-axis crossings
(f) For what range of K are the systems stable?

1.G(s)=K(s+2)/s(s+8)(s+1+3j)(s+1-3j)
2.G(s)=0.75K/s(0.1s^2+1.4s+4.5)

8. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
It looks weird in the place it is, but I don't think that feed-forward here means anything else than that the TF carries the signal forward. As I see it, it is a common unity feedback loop with a feedback gain of K.

If you haven't talked in class about feed-forward control at all, chances are that I am right, so just do your analysis as usual. I will try to do it myself to tomorrow (maybe) and come back to you with the results for comparison.

9. ### Whaler Thread Starter New Member

Mar 28, 2011
6
0
sweet, thanks. I will continue with my efforts and post my solutions to both problems. thanks.

10. ### Whaler Thread Starter New Member

Mar 28, 2011
6
0
for the first one i simplified the equation too:

(s+1+3i)(s+1-3i)=s^2+s-3is+s+1-3i+3is+3i-9i^2
=s^2+2s+10
so then get intial equation of
G(s)=k(s+2)/s(s+8)(s^2+2s+10)
from which i get

poles: (0,-8,-1+3i,-1-3i) four total
zeros: (-2) one finite, so three @ infinity.

σa=[0-8-(1+3i)-(1-3i)-(-2)]/4-1
σa=(-8-1-3i-1+3i+2)/3
σa=-2.6667

θa=60(2k+1)
k=2 add 300, increments of 120

Break-away/in points

set initial equaion to -1 solve for k, then differeneciate wrt to s and solve for s with dirivative set to zero.

k(s+2)/s(s+8)(s^2+2s+10)=-1
k= -[s^4/(s+2)^2] - [10s^3/(s+2)^2] - [26s^2/(s+2)^2] - [80s/(s+2)^2]
which reduces to s^4-24s^3-34s^2-134s-160/(s+2)^2 set to zero and s equals

25.5459, 0
-0.1589, +- 2.2528
-1.228, 0

for jω-axis crossing i use CLTF
k(s+2)/s(s+8)(s^2+2s+10)+k(s+2)
with den of
s^4+10s^3+(k+80)s+2
then thru routh-hurwitz i get
k=160 with crossing at +-4.89898i
so sys stable for k<160 and > than 0.

angles of departure/arrival i am not sure.

hope this looks right, any help would be appreciated.

Last edited: Mar 29, 2011
11. ### ryanw202 New Member

Mar 29, 2011
1
0
yea man your doing everything i would do and btw i believe we are in the same class