Controlled systems - Closed loops

In summary, the conversation is about finding the sensitivity of a closed-form at small change using the formula S = ∂ Ln T/∂ Ln G, where T(s) = G_c(s)G(s)/1 + G_c(s)G(s). The discussion revolves around how to solve the formula without G_c(s) and the effect of a change in τ on T(s). The conversation also touches on the relevance of the S parameter, with one person expressing their lack of familiarity with it.
  • #1
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1. Homework Statement [/b]
http://snag.gy/m9Iq0.jpg

Homework Equations





The Attempt at a Solution



I know that the sensistivity of a closed for at small change is given by the formula S = ∂ Ln T/∂ Ln G
And T(s) = G_c(s)G(s)/1 + G_c(s)G(s)
But since i don't have any G_c(s), i don't no how to solve it, and they get the result they get..
 
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  • #2
You did not attach the figure. I'll assume the total forward gain is G Gc and the feedback gain is 1.

So G Gc = 100/(τs+1).
You don't need G and Gc separately. Call G Gc = P (open-loop or "loop" gain) and T = P/(1+P) = closed-loop gain.

You can just compute T(s) from P(s), then get the new time constant τ', then a given dτ results in a corresponding dτ'.

Express as fractional devialtions: dτ/τ → dτ'/τ'. You will find the proportionality dτ'/τ' / dτ/τ
is a constant = K. So a 1% change in τ results in a K % change in τ'.

You will also find that a change in τ does not affect the dc gain of either P (obvious) or T.
 
  • #3
I am not quite sure..
About the picture you are correct.. => http://snag.gy/sceeD.jpg
But i don't understand, what about the (d/ds)G part.. since the sensistivity is a ratio of the T and d derivative.
 
  • #4
215 said:
I am not quite sure..
About the picture you are correct.. => http://snag.gy/sceeD.jpg
But i don't understand, what about the (d/ds)G part.. since the sensistivity is a ratio of the T and d derivative.

You don't have to use the complementary sensitivity formula you cited (S = dT/T / dP/P).
I have never used it so I'd have to go thru it myself. Unless the instructor specifically wants you to use that formula I would proceed as I suggested. It's very simple. The question is just if τ changes by 1%, how much does the new time constant change.

As soon as you compute T(s) you will see that the form of T is the same as the form of P = k/(τs+1). Except there is a new k' and a new τ'.
 
  • #5
I am a bit confused my T(s) looks like this.
I don't see how this should lead to the answer http://snag.gy/XGASC.jpg
 
  • #6
215 said:
I am a bit confused my T(s) looks like this.
I don't see how this should lead to the answer http://snag.gy/XGASC.jpg

Carry ou the product terms in the denominator in your last expression. What do you get?

(I suggest letting k = 100 and using k instead of the numerator until the end).
 
  • #7
I get K /101 + st..
 
  • #8
215 said:
I get K /101 + st..
\
\
Or, K/(K + 1 + sτ), right?

So divide numerator and denominator by (K+1) & what do you get?

BTW don't use t for the original time constant τ. t is always time, a variable.
 
  • #9
Well.. the same? I am kinda confused on where you are heading...
 
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  • #10
215 said:
Well.. the same? I am kinda confused on where you area heading...

Try to make your T(s) look like a/(bs + c).
 
  • #11
So something like this..

http://snag.gy/2moUD.jpg
 
  • #12
Disregard your last line.
Go
T(s) = k/(k+1+τs) from previous line;

= k/(k+1) / (1 + τs/(k+1))

Now you can let k' = k/(k+1)
and τ' = τ/(k+1).

What does your new expression for T(s) look like then?
 
  • #13
T(s) = k'/1+ t'...

I feel pretty confused right now.. I feel like a dummy, which just does what someone say..
Could it be possible to explain how come to do, and how it all is connected?

As far as I've understood i need to calculate, T(s) and then T'(s) ,and G'(s)
And take S = (T'(S) /T(s))/(G'(s)/G(s)) is that correct?

This is my solution for a)..
where is my mistake??

http://snag.gy/O0Z4e.jpg
 
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  • #14
As I said, I don't have the time to learn about stability coefficient S right now.

But: can't you see that your T(s) looks just like your G(s) except for the values of gain and time constant?
So if you change the original time constant in G by say 1 msec., how much does that change the time constant in T? How much does that change the dc gain in T?
 
  • #15
I just realized the answer is written on the back..

I says http://snag.gy/tr5eD.jpg

If I change the time, it's both the same.
 
  • #16
215 said:
I just realized the answer is written on the back..

I says http://snag.gy/tr5eD.jpg

If I change the time, it's both the same.

No, if you change τ by Δτ, τ' changes by Δτ/101. You can see that from τ' = τ/(k+1) = τ/101. And that's what your written answer says also.
 
  • #17
But the answer isn't the same as the one written in my book..

I just took a screenshot of the part of my book which is about the topic

http://snag.gy/WDedY.jpg

Here is the equation, I've been using.
 
  • #18
215 said:
But the answer isn't the same as the one written in my book..

I just took a screenshot of the part of my book which is about the topic

http://snag.gy/WDedY.jpg

Here is the equation, I've been using.

Your question had two parts.

I only addressed the part that asked for the effect on T(s) for a change in τ. And the answer is in τ' = τ/(k+1). So dτ' = dτ/(k+1) and that is the answer they gave also. And the change in k' for a change in τ is zero. That defines the changes in T(s) for a change in τ completely.


I'm not familiar with the S parameter, as I said. I don't have a physical feel for a derivative with respect to s, which is an operator (= d/dt) or complex frequency variable (= σ + jω). This is just because I was never exposed to this parameter, and I have to say I see no point in it, but that's just me. I have no physical feel for taking the derivative of any F(s) with respect to s.

Anyway, to get S just compute S = ∂T/T / ∂G/G. That's what they want.
 

Related to Controlled systems - Closed loops

1. What is a closed loop system?

A closed loop system is a control system where the output of the system is fed back into the input, creating a loop. This allows the system to continuously adjust and maintain its output based on the desired input.

2. How does a closed loop system differ from an open loop system?

In an open loop system, the output is not fed back into the input, so the system cannot make adjustments based on the output. In contrast, a closed loop system can respond to changes in the output and make adjustments to maintain the desired output.

3. What are the advantages of using a closed loop system?

Closed loop systems are more precise and stable compared to open loop systems. They can also compensate for external disturbances and changes in the system, making them more reliable.

4. What are the components of a closed loop system?

A closed loop system typically consists of a controller, a plant, a sensor, and a feedback loop. The controller receives information from the sensor and sends commands to the plant, which produces the desired output. The feedback loop then compares the output to the desired input and makes adjustments as needed.

5. What are some real-world applications of closed loop systems?

Closed loop systems are commonly used in various industries, such as manufacturing, aerospace, and automotive. They are also used in everyday devices like thermostats, automatic doors, and cruise control in cars.

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