Transfer function with Disturbance

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SUMMARY

The discussion focuses on calculating the steady-state error due to a disturbance represented by Td(s) = 1/s, while setting R(s) = 0. The user aims to derive Y(s)/Td(s) using the transfer function Y(s)/R(s) but encounters confusion regarding the disappearance of R(s) in the final expression. The key takeaway is that the transfer function for the disturbance input Td(s) is distinct from that of the reference input R(s), necessitating a different approach for accurate results.

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TheNovice
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Homework Statement


Find the steady-state error due to a disturbance Td(s) = 1/s.

Set R(s) = 0.

if given a system:
Snapshot.jpg


Homework Equations


n/a

The Attempt at a Solution



I need Y(s)/Td(s). To do this I must find Y(s) in terms of the transfer function Y(s)/R(s) which I have obtained. This gives me an expression for Y(s) in terms of G(s), Gc(s), and R(s).

3D%20R%28s%29%20%5Cfrac%7BG%28s%29G_%7Bc%7D%28s%29%7D%7B1%20+%20G%28s%29G_%7Bc%7D%28s%29%7D.gif

Now I must solve for Y(s)/Td(s).

If Td(s) = 1/s, then I should just multiply Y(s) by s and be done with it. However, the book gets a totally different answer.

specifically, I am not sure how R(s) disappears along with the rest of the numerator in the result.

Thanks for your time.
 
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TheNovice said:

Homework Statement


Find the steady-state error due to a disturbance Td(s) = 1/s.
Set R(s) = 0.
if given a system:
View attachment 83444

Homework Equations


n/a

The Attempt at a Solution


I need Y(s)/Td(s). To do this I must find Y(s) in terms of the transfer function Y(s)/R(s) which I have obtained.
.
Why? R(s) = 0!
 
Your transfer function is valid for the R(s) input only.
The function referred to the Td(s) input looks different!
 

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