# Steady State Error of a closed loop system

Discussion in 'Homework Help' started by Katie_EE, May 12, 2010.

1. ### Katie_EE Thread Starter New Member

Apr 8, 2010
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Hello,

So for my hw assignment i was given this problem E5.19 Which is attached for some read on my part B) the Steady state error is coming out to be 0. which is wrong. I attached my work for part A and B. I am not sure what is wrong unless my transfer function was wrong to begin with.

If someone could take a look at it and let me know it would be Greatly appreciated!

Thank You!!!!

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2. ### retched AAC Fanatic!

Dec 5, 2009
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You gots purdy hand writin.

Do you use a straight edge for your lines?

Sorry, lemme see if I can trudge through this.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I get E(s)=R(s)*(s^2+3s)/(s^2+(3+Ka)s+Ka)

If R(s)=1/s^2

Would, I think, give SS error as 3/Ka

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You should note that E(s) is not R(s)-Y(s) , rather R(s)-Ka*Y(s)

5. ### Katie_EE Thread Starter New Member

Apr 8, 2010
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wait,

I thought Ess(s), the Steady State Error was = R(s) - Y(s) and the Error Signal E(s) the NON- feedback dependent error signal was R(s)-Ka*Y(s) ?

I guess im a bit confused now about that.

6. ### Katie_EE Thread Starter New Member

Apr 8, 2010
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as for this part i am not entirely sure how you obtained the numerator portion.

and when calculating steady state error

ess = lim (s-> 0) s*Ess(s)

Using the Ess(s) that i have in the attachement (which i totally forgot to include the ss part implying that i was looking for the error signal not the Steady state error....) well anyways when i plug in R(s) = 1/S^2 one S cancles but the other doesnt and the ess = 0 .

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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That's not my understanding - I always consider the error as the resultant output at the feedback summing junction. The goal is generally to reduce the feedback error to zero.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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This was my approach to finding E(s) .....

$E(s)=R(s)-E(s)\frac{(s+1)}{(s^{2}+3s)}Ka$

$E(s)(1+\frac{(s+1)}{(s^{2}+3s)}Ka)=R(s)$

$E(s)(\frac{s^{2}+3s+(s+1)Ka}{s^{2}+3s})=R(s)$

$E(s)=R(s)\frac{(s^{2}+3s)}{(s^{2}+3s+(s+1)Ka)}$

$E(s)=R(s)\frac{(s^{2}+3s)}{(s^{2}+(3+Ka)s+Ka)}$

9. ### Katie_EE Thread Starter New Member

Apr 8, 2010
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Maybe this means the same but im not understanding what you are refering to...

From my teachers lecture notes he tells us that

• Steady state error is the difference between the desired response(input) and the actual response (output) after the transient stage has decayed. (note this shouldn't be confused with Error signal)
For a closed loop:

Ess(s) = R(s) - Y(s)

where Y(s) = T(s)*R(s) T(s)=loop gain/transfer function

therefore Ess(s) = R(s)*[1 - T(s)]
then to calculate ess = lim (s-> 0) s*Ess(s)

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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If R(s)=1/s^2

$SSE=lim_{s->0}[sE(s)]=lim_{s->0}s (\frac{1}{s^2})\frac{(s^2+3s)}{s^2+(3+Ka)s+Ka} = \frac{3}{Ka}$

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I guess the issue over what constitutes error depends upon whether you have unity or non-unity feedback. This is a non-unity feedback case.

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13. ### Katie_EE Thread Starter New Member

Apr 8, 2010
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I do not believe this I with most of it except the initial multiplication. Why are you multiplying the second portion with E(s)??
$E(s)=R(s)-E(s)\frac{(s+1)}{(s^{2}+3s)}Ka t$

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Does this help ???

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15. ### Katie_EE Thread Starter New Member

Apr 8, 2010
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it makes sense BUT its nothing my teacher has ever said or gone over...which is why im doubting it so much...and i have a midterm Monday on this --__--.

Last edited: May 14, 2010
16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Katie_EE likes this.
17. ### Katie_EE Thread Starter New Member

Apr 8, 2010
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Ok this Makes Sense!!! and so does this "You should note that E(s) is not R(s)-Y(s) , rather R(s)-Ka*Y(s)"

HOWEVER these calculations are valid for the Error Signal E(s) but not for the Steady State Error Ess(s) which is equal to Ess(s) = R(s) - Y(s).

There is a difference between the two! at least according to my teachers lecture notes. But im still getting Ess(s) to come out to equal ∞ which is wrong.

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Part (c) of the problem anticipates the existence of a value of Ka which gives a steady state error of zero with a unit step input. The value that satisfies this is Ka=1 or the unity feedback case. In no other circumstance is it possible that Y(s)=R(s) which would be true for your assertion that Ess=R(s)-Y(s). If Ka were 10 then the output Y(s) would be 0.1 with R(s) a unit step input, giving Ess=0.9. But the summing junction output would be zero.

It's clear you won't accept my assertion that the true error in the feedback control system is what comes out of the summing junction. At this stage I can't offer any other advice than I already have. Perhaps you'll get more assistance from other forum members.

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
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One more for the road .... and that's it.

If you do the full analysis for a unit ramp input then the non-transient (steady state) part of the time domain solution for y(t) is

$y(t)=\frac{t}{K_{a}}-\frac{(3+K_{a})}{K_{a}^{2}}$

With r(t)=t [unit ramp] then the error for r(t)-y(t) [with Ka not equal to 1] increases unbounded monotonically - in other words at t=∞ the error is infinite. That's what you calculated.

Last edited: May 16, 2010
20. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Sorry that should be ...

$y(t)=\frac{t}{K_{a}}-\frac{3}{K_{a}^{2}}$