# Controlled systems - Closed loops

1. Mar 1, 2014

### 215

1. The problem statement, all variables and given/known data[/b]
http://snag.gy/m9Iq0.jpg

2. Relevant equations

3. 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..

2. Mar 1, 2014

### rude man

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. Mar 1, 2014

### 215

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. Mar 1, 2014

### rude man

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. Mar 1, 2014

### 215

I am a bit confused my T(s) looks like this.

6. Mar 1, 2014

### rude man

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. Mar 1, 2014

### 215

I get K /101 + st..

8. Mar 1, 2014

### rude man

\
\
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. Mar 1, 2014

### 215

Well.. the same? I am kinda confused on where you are heading...

Last edited: Mar 2, 2014
10. Mar 1, 2014

### rude man

Try to make your T(s) look like a/(bs + c).

11. Mar 4, 2014

### 215

12. Mar 4, 2014

### rude man

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. Mar 4, 2014

### 215

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

Last edited: Mar 4, 2014
14. Mar 4, 2014

### rude man

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. Mar 4, 2014

### 215

I just realised 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. Mar 4, 2014

### rude man

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. Mar 4, 2014

### 215

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. Mar 4, 2014