Controlled systems - Closed loops

[215]

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..

 

[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.

 

[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.

 

[rude man]

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'.

 

[215]

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

 

[rude man]

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).

 

[215]

I get K /101 + st..

 

[rude man]

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.

 

[215]

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

 

[rude man]

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

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

 

[215]

So something like this..

http://snag.gy/2moUD.jpg

 

[rude man]

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?

 

[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

 

[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?

 

[215]

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.

 

[rude man]

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.

 

[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.

 

[rude man]

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.