Closed loop transfer function problem

Question

Let's say we have the basic feedback system of the picture enter image description here

We know that the closed loop transfer function is given by $$T(s)=\frac{G(s)}{1+G(s)H(s)}$$ Now, the diagram above is equivalent to an open loop diagram with this transfer function . enter image description here

Here's my problem . In the first diagram it's clearly a feedback system while in the second there is no feedback . How is this possible if these are equivalent ? Can a system have feedback or not depending on the perspective? I'm confused .

• Janka: The feedback is in the 1/(1±G(s)H(s)) part. Without feedback, it would just be G(s). Set H(s)=1 to see the effect of a plain, direct feedback.
• Andy aka: Can you tell if your car engine is a V8 just by driving it?

Answer 1

It's eminently possible for an open loop system to have exactly the same TF as a closed loop system: \$ \frac {1}{1+s}\$ could be the TF of a simple series RC circuit with R=1, C=1, or it could be a unity feedback closed loop system with an integrator, \$\frac{1}{s}\,ドル in the forward path. Or it could be something completely different. It's impossible to tell, from the TF itself, what the primitive structure of the system is.

Equally, putting a Laplace transform into a single box does not mean that the system is open loop. It could be the final abstraction of a primitive closed loop block diagram.

Also, using G for the forward path and H for the feedback path of a CLTF is purely convention. Those letters are not sacrosanct.

Answer 2

Take the simple example of an op-amp with dominant pole compensation. Open loop it has the approximate transfer function of Ao/(1+s*tau).

Put resistive feedback around it and you can get an amplifier with whatever gain you want, and a single pole roll off at the gain-bandwidth divided by the gain. (Of course within GBW limits etc.)

Now of course it's possible to design an amplifier WITHOUT feedback that has a certain gain and a single poll rolloff at a given point.

So you have two systems: One with feedback and one without that have identical transfer functions but different implementations.