How would I plot the root locus for the system below?

• https://app.box.com/s/w97s8mpbkgjwuv78klv5
• When G(s)= -0.125(s+0.452)/(s+1.25)(s^2+0.234s+0.0163)

 

http://ctms.engin.umich.edu/CTMS/index.php?example=Introduction&section=ControlRootLocus

 

How would I plot the root locus for the system below?

• https://app.box.com/s/w97s8mpbkgjwuv78klv5
• When G(s)= -0.125(s+0.452)/(s+1.25)(s^2+0.234s+0.0163)

You are missing the -K1 and the 2/(s+2) from the open loop transfer function.
The correct open loop transfer function, which we will call F(s) is:

\(F(s) = (-K1)(\frac{2}{s+2})(\frac {-0.125(s+0.452)}{(s+1.25)(s^2+0.234s+0.0163)})\)
It has the following features:

1. a pole at s =-2
2. a pole at s = -1.25
3. a zero at s = -0.452
4. a complex conjugate pair at -0.117 ± j 0.0511

 

in order to plot root locus in Matlab, would I use the closed loop tf i.e. F(s)/1+F(s)

 

Branch 1 goes from the pole at s = -2 to the zero at infinity
Branch 2 goes from the pole at s = -1.25 to the zero at s = -0.452
At all values of K these responses look like damped exponential responses, this is a good thing for pilots!
Branches 3 & 4 exit the complex conjugate poles near the origin at a departure angle of ≈ 94.5°. They remain in the left half plane for a while, eventually crossing the jω axis at some gain,a real bad thing, and head off to the zeros at infinity along asymptotes with real axis intercepts ≈ -0.774 at angles of ±60°

I think, but I'm not positive, that this is the so called phugoid oscillation which is a minor irritation to most pilots. Of course if the gain is large enough and the roots cross over the jω axis bad things can happen.
http://en.wikipedia.org/wiki/Phugoid

 

in order to plot root locus in Matlab, would I use the closed loop tf i.e. F(s)/1+F(s)

See if you get the same results that I did.

 

That phugoid oscillation really doesn't sound good! Are the first two branches you mention are these the horizontal branches? If they are, then my root locus matches, few...
Thanks alot for taking the time to help me, it really is appreciated.

 

That phugoid oscillation really doesn't sound good! Are the first two branches you mention are these the horizontal branches? If they are, then my root locus matches, few...
Thanks alot for taking the time to help me, it really is appreciated.

As a pilot I can tell you that it is irritating while trying to fly straight and level at a constant altitude. You have to keep making small corrections. Notice how close the complex conjugate poles are to the jω axis. This implies very little damping at a very low frequency. It is not dangerous unless the value of K is increased so as to put the pair of poles in the right half plane. Even slightly in the right half plane is OK as long as you have enough control response to make the correction. It is not uncommon for combat fighters to have unstable characteristics so they can have superior maneuverability.

Are you in an Aeronautical Engineering program?

 

No mechanical, I've done very little control theory. A short module in 1st year & now this piece of coursework. It seems to have gone from solving simple systems to quite complex theory. At least with most mechanical systems I can kina picture them in my head, but when it comes to 'imaginary axes'???

 

No mechanical, I've done very little control theory. A short module in 1st year & now this piece of coursework. It seems to have gone from solving simple systems to quite complex theory. At least with most mechanical systems I can kina picture them in my head, but when it comes to 'imaginary axes'???

Ah not to worry. The use of complex numbers is just a convenient device to keep track of the fact that electrical signals, like all vector quantities, have both magnitude and phase(aka direction). If it makes you happier, think of the imaginary unit j as a rotation operator. Whenever you multiply something by j, you rotate it 90° counter clockwise. When you divide by j, or equivalently multiply by -j, you rotate things 90° clockwise. Thinking in this way avoids the necessity of grappling with the square root of -1.
Thus:

1. (1+j0)*j = 0 +j1
2. (0+j1)*j = -1 + j0
3. (-1+j0)*j = 0-j1
4. (0-j1)*j = 1+j0

and so proceed ad infinitum

 

Your advice has/instruction has helped greatly, as I've the last few days going round in circles. It was only by chance I ended up with the control assignment, my friend took the FEA report & I was left with this one. Now he's finnished his & I have until the 19th to finnish too

 

Your advice has/instruction has helped greatly, as I've the last few days going round in circles. It was only by chance I ended up with the control assignment, my friend took the FEA report & I was left with this one. Now he's finnished his & I have until the 19th to finnish too

Maybe you should give me a footnote in your final report!

--just kidding

 

If it ever gets finnished....