Quarter amplitude response method: PID settings

[James FC]

Homework Statement

see attached, question 8a and b are what I'm attempting. I've also attached the VERY brief course notes that I'm required to work with.
I know Tc= 8 seconds from the graph (0.13minutes)

Gain set to 4. is PB therefore 100/4=25%?

I'm struggling to calculate the ratio of successive amplitudes. If I figured this out I'd be able to calculate lnX/2π (see relevant equations below)

Homework Equations

Ti=Tc/2√1+Δ^2

Td=Tc/8√1+Δ^2

Tc=period of oscillation in minutes.
Ti=Integral action time in minutes.
Td=Derivative action time in minutes.
Δ= lnX/2π (where X is the ratio of amplitude expressed as 1:X

The Attempt at a Solution

Any help would be greatly appreciated.

 

[Tom.G]

In the image you posted as 'PID', what is the maximum amplitude of the first positive peak; of the second positive peak?
In the image you posted as 'PID', how do the amplitudes of the first two peaks compare to each other?

At the bottom of the first page of 'Course notes', the last two lines of text define the ratio of the peaks from FIG. 4.
And the top of the second page defines 'X'.

Cheers,
Tom

 

[David J]

is the ratio 1:2 ? The first peak (call it a1) is 4 units, the second (call it a2) is 2 units but then the 3rd (call it a3) is less than 1 unit and the 4th (call it a4) is less than 0.5 unit. Does this tell us that the ratio of 1:4 does not always apply ?

 

[Tom.G]

I'm struggling to calculate the ratio of successive amplitudes.

is the ratio 1:2 ?

Yes.

Does this tell us that the ratio of 1:4 does not always apply ?

The course notes tell you what method to apply when you adjust the gain for a 1:4 ratio.
And the 2nd page of the course notes tell you the corrections to apply when the gains is adjusted for other than 1:4.

 

[David J]

So as per post #1 the gain is given as 4 which means the %PB=$\frac{100}{4}=25\%$

Tc will be 8 seconds or 0.13 minutes

Ratio is 1:2 which means $\bigtriangleup=\frac{lnX}{2\pi}$ so $\frac{ln2}{6.2832}$= 0.1103178 so $\bigtriangleup^2$ must be 0.01217

The gain we are given is the proportional only gain. Kc, which is 4. We need the working gain, Kp which is $0.6Kc$

The lessons tell me that I should have 3 settings:-

Gain Kp (which will be 0.6Kc)

Integral action time

Derivative action time

so the gain will be 0.6Kc

If I use the equation $Ti=\frac{Tc}{2{\sqrt{1+\bigtriangleup^2}}}$ then this should give me the approx setting time for the integral action time in minutes

If I use the equation $Td=\frac{Tc}{8{\sqrt{1+\bigtriangleup^2}}}$ then this should give me the approx setting time for the derivative action time in minutes

Is this more or less the correct way to do this?

 

[Tom.G]

Ratio is 1:2 which means △=lnX2π\bigtriangleup=\frac{lnX}{2\pi} so ln26.2832\frac{ln2}{6.2832}= 0.1103178 so △2\bigtriangleup^2 must be 0.01217

Our numbers agree.

The gain we are given is the proportional only gain. Kc, which is 4. We need the working gain, Kp which is 0.6Kc

Please document the '0.6' value. Where did it come from?

EDIT:
One of the posters supplied the following from the course being taken at 'Teesside University Open Learning':
Proportional + Integral + Derivative Controllers
For a P + I + D controller the applied settings should be:

Gain Kp = 0.6Kc

Integral Action Time Ti = Tc/2.0 minutes

Derivative Action Time Td = Tc/8.0 minutes

End EDIT

Is this more or less the correct way to do this?

Yes.
But please continue showing your results as there is a further item in the 1st post that needs attention.

(I wonder what happened to the OP, @James FC. He has been online but has not continued in this thread.)

Cheers,
Tom

 

[David J]

$Ti=\frac {Tc}{2{\sqrt{1+\bigtriangleup^2}}}$ so $Ti=\frac {0.13}{2{\sqrt{1+0.1103178^2}}}$ so $Ti=\frac {0.13}{2.01213}=0.06461$ (Integral action time in minutes)

$Td=\frac {Tc}{8{\sqrt{1+\bigtriangleup^2}}}$ so $Ti=\frac {0.13}{8{\sqrt{1+0.1103178^2}}}$ so $Ti=\frac {0.13}{8.04853}=0.01615$ (Derivative action time in minutes)

The gain Kp = 0.6 Kc so gain = 0.6 X 4 = 2.4

I think the above are correct?

The only issue I can see from post #1 is question B to be addressed now? or is there something I am missing, we have to reassess the entire situation (I think) because the gain has been increased from 4 to 6 and we are now using the ultimate cycle method instead of the Quarter Amplitude Response Method

 

[James FC]

see attached what I submitted.

 

[James FC]

your answers are correct

 

[Tom.G]

Both of you need to review the 'Course notes.pdf' file attached to the first post in this thread. Pay particular attention to the definition of 'T_C' on page 9, and its further definition on pg 8.

 

[David J]

period of oscillation "in minutes" ?

 

[Tom.G]

period of oscillation "in minutes" ?

Yes.
What is the period of one cycle in the attachment 'PID.pdf'?
What number are you two using for Tc in the calculations?

 

[David J]

it looks like 16 seconds for a full cycle so it should have been 0.267 minutes

 

[David J]

So it should look like this

$Ti=\frac {Tc}{2{\sqrt{1+\bigtriangleup^2}}}$ so $Ti=\frac {0.267}{2{\sqrt{1+0.1103178^2}}}$ so $Ti=\frac {0.267}{2.01213}=0.1327$ (Integral action time in minutes)

$Td=\frac {Tc}{8{\sqrt{1+\bigtriangleup^2}}}$ so $Ti=\frac {0.267}{8{\sqrt{1+0.1103178^2}}}$ so $Ti=\frac {0.267}{8.04853}=0.0332$ (Derivative action time in minutes)

The gain Kp = 0.6 Kc so gain = 0.6 X 4 = 2.4

 

[Tom.G]

@James FC, do you agree with the results from @David J?
If not, could you explain why?

If you agree, do you have the time to assist David J with Part b? I will continue to monitor this thread and will post when needed, but thought I would give you the opportunity to help a fellow student.

Tom.g

 

[David J]

The answers using the "Ultimate Cycle Method" appear to be the same as the answers using the "Quarter Amplitude Response Method" apart from the gain which is slightly higher

$gain Kp = 0.6Kc = 3.6$

$Ti = \frac{Tc}{2}$ = $\frac{16}{2}$ = 8 seconds or 0.1333 minutes

$Td = \frac{Tc}{8}$ = $\frac{16}{8}$ = 2 seconds or 0.0333 minutes

This is too much of a coincidence so I am suspecting the answers are correct

 

[Tom.G]

David, for part 'a', I get a different value for Kp.
See pg 9 of' 'Course notes.pdf' attached to the first post.

 

[David J]

$\%PBp =\frac{\%PBc}{0.5+2.27\bigtriangleup}$

$\%PBp =\frac{25}{0.5+(2.27*0.11032)}$

$\%PBp =\frac{25}{0.5+0.2504}$

$\%PBp =\frac{25}{0.7504}$

$\frac{25}{0.7504} = 33.314$

$gain = \frac{100}{33.3146} = 3.002$

So the gain in "part a" should be 3.002?

 

[Tom.G]

@James FC, I agree with your part 'a', haven't seen your part 'b'.
@David J, I agree with both parts 'a' and 'b'.

For extra credit:
Could you use the Quarter Cycle equations in the Ultimate Cycle tuning? Why?

Cheers,
Tom

 

[David J]

I think you will be able to use the Quarter cycle equations in the ultimate cycle tuning. The gain in the Quarter Cycle is lower than the ultimate cycle so would this mean that the quarter cycle would give a more stable output ??

 

[Tom.G]

Mmm... perhaps. I'm having trouble following all three documents at once. But I am a little concerned that the gains are different with the same time constants. So either we are all wrong or the course material has a problem... or perhaps it is a safety factor for the Δ correction being approximate. (and I'm too lazy to work out all of them)

Here is what I had in mind. I noticed on pg 8 of 'Course notes.pdf' that the denominator of the formulas for Ti and Td have Δ in them, with Δ defined in the last equation as a function of X.

In the Ultimate Cycle method, X=1. When X is 1, Δ becomes 0, and the Quarter Amplitude and Ultimate Cycle formulas for Ti and Td are identical.

I see the terms involving Δ in Quarter Cycle as being 'corrections' to Ultimate Cycle.

If you run across the reason for the gain difference, please post back here to help us all learn!

Cheers,
Tom

 

[David J]

I will submit my answers as they are for now and see what response I get back from the uni when they are marked. They usually point out any errors. I still have 3 other questions to complete so I will leave this one for now. Thanks for your help with this one

 

[David J]

Hello again, just an update, I submitted this and it came back as marked correct. There were no comments other than `correct` so just a thanks for your help with this. I did not start this thread so I cannot mark it as solved but if anyone reading this needs help post a question.

thanks again