Can an open loop system replace a close loop system

Question

I'm thinking about a mathematical possibility how an open loop system can replace a close loop system so here is my explanation

Let's say I have a plant with transfer function as given in image (1)

But to achieve a desired output with minimum effect of disturbances ,I Make it a close loop with unity feedback and it works very well (image 2)

this time I make a Cascade connection with plant such that overall transfer function is same as close loop transfer function (image 3)

Since transfer function in this case is equal to previous close loop transfer function so for the same input output should be same in both conditions , so we can say an open loop replaced a close loop system?

Problems-

1.but intuitvily it feels that it cannot possible , so there must be something that I'm missing , so where I'm going wrong ?

2. Why close loop system cannot be replaced by open loop system although mathematical conditions are satisfied(as I shown by example)

Answer 1

Too long for a comment.

Think about the physics of the problem. Due to small variations in the components of the plant, the original system may change from \$\frac{1}{s+1}\$ to, say, \$\frac{1}{s+1.2}\$. Then your scheme of cancelling unwanted dynamics won't work.

The situation becomes much much worse when the original system is an unstable system. The original system will get damaged due to its instability while the output of the cascade system hides that fact.

More over, even if the parameters don't change, very very very rarely do we know the exact transfer function of the original system. So this scheme is not suitable for practical applications.

To see why closed loop control is necessary go through the first few chapters of any basic control system text book. They will give more reasons for utilising closed loop control.

Answer 2

But to achieve a desired output with minimum effect of disturbances ,I Make it a close loop with unity feedback and it works very well

Well, you think it works well but it doesn't in reality. If you want a precision control system you need much more gain (and probably integral and maybe some differential) error amplification after the summer (called G below): -

And, if you considered an expanded picture of what a good control loop has to deal with you'd draw this (featuring load effects and "other nasties"): -