IMC-based PID Controller for unstable system

[gfd43tg]

Homework Statement

Homework Equations

The Attempt at a Solution

I'm working on parts (c) and (d) now, but I will show my work for (a) and (b) for completeness.

(a)
Starting with the process transfer function

$$ g_{p} = \frac {-1.43}{s^{2}+3.687s-7.177} $$

The denominator is factorized,

$$ g_{p} = \frac {-1.43}{(s-1.399)(s+5.086)} $$

$$ = \frac {-1.43}{-1.399(\frac {-1}{1.399}s+1)5.086(\frac
{1}{5.086}s+1)} $$

$$ g_{p} = \frac {0.201}{(-0.715s+1)(0.197s+1)} $$

where ##\tau_{1}= -0.715## and ##\tau_{2} = 0.197##. The first time constant makes the process unstable. A singularity occurs at

##s = \frac {1}{\tau_{1}} = \frac {1}{0.715} = 1.399##
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(b)
$$ f(s) = \frac {\gamma s + 1}{(\lambda s + 1)^{n}} $$
The controller must be semi-proper with an unstable system.
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(c)

The invertible part of the process transfer function is found,

$$ \tilde {g_{p-}} = \frac {0.201}{0.197s+1} $$
with a filter
$$ f = \frac {\gamma s+1}{5s+1} $$
$$ q = \tilde {g_{p-}}^{-1}f = \frac {0.197s+1}{0.201} \cdot \frac {\gamma s+1}{5s+1} $$

The controller transfer function is determined,

$$ g_{c} = \frac {q}{1-q \tilde {g_{p-}}} $$

$$ = \frac { \frac {0.197s+1}{0.201} \frac {\gamma s+1}{5s+1}}{1 - \frac
{0.201}{0.197s+1} \cdot \frac {0.197s+1}{0.201} \cdot \frac {\gamma s+1}{5s+1} } $$

$$ = \frac {\frac {0.197s+1}{0.201} \cdot \gamma s+1} {(5- \gamma)s} $$

$$ = 4.975 \bigg [ \frac {0.197 \gamma s^{2}+(0.197+ \gamma)s + 1}{(5- \gamma)s} \bigg ] $$

Which is of the form of an ideal PID controller,

$$ g_{c,PID} = k_{c} \bigg [ \frac {\tau_{I} \tau_{D} s^{2} + \tau_{I}s + 1}{\tau_{I}s} \bigg ] $$
Therefore, for the ##\tau_{I}## term to be equal, ##5- \gamma = 0.197+ \gamma##, so ##\gamma = 2.402##.

$$ g_{c} = 4.975 \bigg [ \frac {0.473s^{2}+2.599s+1}{2.599s} \bigg ] $$
##k_{c} = 4.975##, ##\tau_{I} = 2.599##, ##\tau_{D} = 0.473/2.599 = 0.182##
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(d)
Here is my simulink model

And the inputs to the PID controller

However, my output and manipulated inputs look totally wrong, this controller is not working at all!

So I am wondering if I may have done part (c) incorrectly.

 

[gfd43tg]

I might as well post my work for the next set, which asks to make the controller transfer function semiproper

(e) & (f)
$$ q = \frac {0.197s+1}{0.201} \cdot \frac {\gamma s + 1}{(5s+1)^{2}} $$
So here is my controller transfer function worked out
$$ g_{c} = \frac { \frac {0.197s+1}{0.201} \cdot \frac {\gamma s + 1}{(5s+1)^{2}}}{1 - \frac {0.201}{0.197s+1} \cdot \frac {0.197s+1}{0.201} \cdot \frac {\gamma s+1}{(5s+1)^{2}}} $$

$$ = 4.975 \bigg [ \frac {0.197 \gamma s^{2} + (0.197 + \gamma)s + 1}{25s^{2} + (10-\gamma)s} \bigg ] $$
$$ = 4.975 \bigg [ \frac {0.197 \gamma s^{2} + (0.197 + \gamma)s + 1}{(10- \gamma)s( \frac {25}{10-\gamma}s + 1)} \bigg ] $$
Which is of the form
$$ g_{c} = k_{c} \bigg [ \frac { \tau_{I} \tau_{D} s^{2} + \tau_{I} s + 1}{\tau_{I} s(\tau_{F}s+1)} \bigg ] $$
So ##10 - \gamma = 0.197 + \gamma##, so ##\gamma = 4.902##,
This reduces to
$$ 4.975 \bigg [ \frac {0.966 s^{2} + 5.099s + 1}{5.099s( 4.904s + 1)} \bigg ] $$
Where ##\tau_{F} = 4.904## and ##\tau_{D} = 0.966/5.099 = 0.189##
Then I make a new model in simulink

And my output is still diverging

So it seems neither of these controllers are working

 

[Hesch]

I think you have a sign-error in (c).

k_c must be negative because k_p is negative.

Otherwise you will have a positive feed-back in your control-loop.

 

[gfd43tg]

It means something might have gone wrong in (b)? Because when I take out the -1.399, it makes the numerator positive

 

[Hesch]

It means something might have gone wrong in (b)?

I don't know. I've just sketched a root locus and can see that the rightmost pole ( s = 1.399 ) will cause a root moving to the right if k_c → +∞.
But if you let k_c → -∞, the root will move to the left ( toward the stable area ).

 

[gfd43tg]

I don't think I made a sign error, but it's possible that the ##k_{p}## is the problem statement is wrong. Another classmate had the same issue as me, we have not resolved it.

 

[Hesch]

Well, anyway you can see, that if you follow the circulation path in the loop, the sign will be positive. You are not allowed to change the sign of the transferfunction ( -1.43 ) because it can be due to some inverting amplifier in the hardware, or whatever. But you may change the sign of the PID-controller: You are the one to make that decision.

Try it out, see what happens! Don't make a problem out of that.

 

[gfd43tg]

This is what the grad student responded with in response to the sign of kc and kp being different

Be careful with logic like this. A process gain is the long-time output response of a system. If you have an unstable process, the true long-time response is undefined (infinity). The gain for an unstable process transfer function is no longer meaningful physically so it is not reliable to draw conclusions from the sign.

Also, I changed the sign of the PID controller, and all that happened was a divergence in the opposite direction.

 

[Hesch]

The grad student says: Be careful with logic like this.
Well, then be careful. Know what you are doing.

By inserting a PID-controller in the loop, you will increase the system order by 1 ( 3. order characteristic equation ). I don't have the "tools" to handle that right now, calculating the optimal zero/pole/amplification. But for simplification, try to set PID = -7 ( just a P-controller ).

The characteristic equation for the system will be: ( s + 1.092 )( s + 2.595 ) = 0.
Thus it must be stable.