This is a theoretical question. I saw this circuit somewhere:
System (a) has transfer function: \$H(s)= V_{out}(s) / V_{in}(s)\$
And it stated that when system (a) is put into system (b) in the exact way that is shown, then circuit (b) has transfer function: \$ H_{o}(s) = 1 / H(s) = V_{in}(s) / V_{out}(s) \$
I tried to prove that but unfortunately i can't figure it out. I want to rigorously prove that this is true. Can someone help me? Thanks in advance!
EDIT
I found the answer. It's trivial but i guess i'll post it here.
In circuit \$(β)\$ the voltage of the inverting input of the ideal opamp is the same as in the non inverting input, because of virtual ground, therefore it's \$V_-(s)=V_+(s)=V_{in}(s)\$. But the voltage on the inverting input also equals to \$H(s)V_{out}(s)\$ (the transfer function of system \$(α)\$ times whatever it has in it's input, which in system \$(β)\$ that is \$V_{out}(s)\$. The two previous equations give us that\$:H_{o}(s)=\$ voltage on it's output/voltage on it's input =\$V_{out}(s)/(H(s)V_{out}(s))=1/H(s) \$ which is what i wanted to prove.
Best Answer
The classical feedback formula (H. Black) is
Ho=Ao/(1+AoH)=1/(1/Ao+H) with Ao=open-loop gain and H=feedback function. Of course, for Ao approaching infinity we have Ho=1/H.
Example: If the feedback path consists of a resistive voltage divider R1/(R1+R2) we get the classical closed-loop gain expression for a non-inverting opamp stage Acl=1+R2/R1.