Unstable 4th order system (control theory)

[Maxim Nol]

Homework Statement

Hello everyone!
To simplify, I have a system whose transfer function looks like this:

G = (s^2-1) / (s^4-s^2-1)

It's unstable and I have to find a way to stabilize it using PID controller and/or lead/lag compensators.

Homework Equations

The Attempt at a Solution

So far I tried a simple PID controller (Kp + Ki/s + Kd*s), a PID controller with low pass filter (Kp + Ki/s + Kd*s/(1+N*Kd*s), a PID controller with multiple derivatives and integrators (P*I^m*D^n) but no luck.
I'm using MATLAB to solve it.

Root locus attached.

Could you give me any hints?
Thank you.

 

[Hesch]

I don't think you can do it with a 1. order cortroller in which there are 3 variables to adjust, with 5 system roots.

You must use a 3. order controller, where you have 7 variables to adjust, with 7 system roots.

Then you can do it for sure.

The 3. order filter must be written:

G = K*(s+a)(s+b)(s+c) / ( ( s+d)(s+e)(s+f) )

Now, calculate the algebraic characteristic equation as for the closed loop transfer function, H1(s) = 0
Express the desired characteristic equation, H2(s) = 0.
By inspection of H1(s) = H2(s) you can setup 7 linear equations.
Solve them and you have found the values of K, a . . f.

Please sketch a root locus and attach it. I would like to see it.

 

[donpacino]

another hint.

take a look at a bode plot (if you have MATLAB this should be easy)

once you look at the bode plot that might help you understand what you need to do to stabilize your system

 

[Maxim Nol]

I don't think you can do it with a 1. order cortroller in which there are 3 variables to adjust, with 5 system roots.

You must use a 3. order controller, where you have 7 variables to adjust, with 7 system roots.

Then you can do it for sure.

The 3. order filter must be written:

G = K*(s+a)(s+b)(s+c) / ( ( s+d)(s+e)(s+f) )

Now, calculate the algebraic characteristic equation as for the closed loop transfer function, H1(s) = 0
Express the desired characteristic equation, H2(s) = 0.
By inspection of H1(s) = H2(s) you can setup 7 linear equations.
Solve them and you have found the values of K, a . . f.

Please sketch a root locus and attach it. I would like to see it.

Thank you very much, that helped. I wish I hadn't lost so much time trying to design a PID.
Just one question: does this look like a valid desired C.E. to you? D=(s+p)^5 (s^2+2*omega*zeta+omega^2). I set pole 'p'=10*omega so that desired natural frequency dominated system dynamics.

donpacino,
That's a bode diagram (I have 2 outputs). What conclusions should I make from that? Sorry, I'm all new to this.

 

[Hesch]

Your plant has the transfer function: G = (s^2-1) / (s^4-s^2-1)
In #2 I have suggested a transfer function like: H = K*(s+a)(s+b)(s+c) / ( ( s+d)(s+e)(s+f) )

The two transfer functions are connected in series, so the overall transfer function as for the open loop will be: GH(s) = G(s) * H(s)

It is tempting to place a zero/pole pair in H(s) matching the righthand pole/zero pair in G(s), to cancel/get rid of them, but that doesn't work in practice because you cannot hit them exactly. Thus you must do as stated in #2:

Now, calculate the algebraic characteristic equation as for the closed loop transfer function, H1(s) = 0
Express the desired characteristic equation, H2(s) = 0.
By inspection of H1(s) = H2(s) you can setup 7 linear equations.
Solve them and you have found the values of K, a . . f.

So having found the 3 zeroes and 3 poles in H(s), you must place the overall 5 zeros and 7 poles in the same root locus.

Now, calculate the characteristic equation (Mason's rule) as for the closed loop transfer function, plot the root locus by varying K from 0 to ?. You should get something like this ( just an example with only 3 curves):

At the calculated K-value, you should see all the 7 curves passing the desired locations, left to the imaginary axis ( stable area ).

Normally you can see by intuition, where about zeroes and poles are to be placed, but that's impossible ( for me ) in a 7. order system.
Likewise I cannot (any longer) solve 7 linear equations, with complex number results, by mental calculations. . . .