Transfer function with Disturbance

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To find the steady-state error due to a disturbance Td(s) = 1/s with R(s) set to 0, the focus is on deriving Y(s)/Td(s). The user attempts to express Y(s) using the transfer function Y(s)/R(s) but is confused about the disappearance of R(s) in the final result. The key point is that the transfer function is valid only for the R(s) input, and the form of the transfer function for the disturbance input Td(s) is different. Understanding this distinction is crucial for correctly solving the problem. Clarification on how to handle the disturbance input is essential for accurate results.
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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