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)enter image description here

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)enter image description here

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

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?


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

  1. Why close loop system cannot be replaced by open loop system although mathematical conditions are satisfied(as I shown by example)
  • \$\begingroup\$ Wikipedia: why closed loop and Why cancellation technique is not used \$\endgroup\$
    – AJN
    Aug 18, 2020 at 15:30
  • 1
    \$\begingroup\$ You can't just stick additional transfer functions after the plant. "Yes boss, my robot is slamming into a brick wall at full speed, but the digital corrections we apply to the sensor readings say that it's exactly where it's supposed to be!" I suppose you can put your new transfer function before the plant, but that's not what's shown in the picture. \$\endgroup\$ Aug 18, 2020 at 17:05

2 Answers 2


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.


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): -

enter image description here

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"): -

enter image description here

  • \$\begingroup\$ Thanks Andy! But about my main concern why I cannot replace close loop system by an open loop system of same transfer function , When both system give same outputs for same input? \$\endgroup\$
    – user215805
    Aug 18, 2020 at 15:11
  • \$\begingroup\$ Because it's unrealistic of a real control system that has to drive against friction, move mass and deal with noise and surges. All of that before we get into plant non-linearity, stiction and dead-band problems. \$\endgroup\$
    – Andy aka
    Aug 18, 2020 at 15:39
  • \$\begingroup\$ Open loop can only control speed or voltage with no loading effects, otherwise load regulation error and no regulation at all except no load inside inside BW unity gain/2 \$\endgroup\$ Aug 18, 2020 at 15:54

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