Suppose I have a circuit below:


simulate this circuit – Schematic created using CircuitLab

I know the the transfer function of a negative feedback loop is:


But this applies only if all resistors are equal in the summing amplifier and the input voltage is inverted. The equation for the summing amplifier is:

$$V_3(s) = \frac{-R_3}{R_1} V_{in}(s) + \frac{-R_3}{R_2} V_2(s)$$

Where \$V_3(s)\$ is the resulting voltage of the summing amplifier, based on its two inputs, \$V_{in}(s)\$ and \$V_2(s)\$.

My question is how to correctly put both formulas together to get the overall transfer function of the above circuit?

  • \$\begingroup\$ There seems to be some anomalies in your question. The negative feedback loop formula involes H1 - this cannot be correct. Also V3 - what or where is that? Ditto V1 and V2. \$\endgroup\$ – Andy aka Nov 19 '14 at 20:48
  • \$\begingroup\$ What I mean is that V3 is the output of the summing amplifier, while V1 and V2 are its inputs. What do you mean the first formula is not correct? This is based on only if all resistors has the same value. \$\endgroup\$ – vxs8122 Nov 19 '14 at 20:52

I can redraw the block diagram as follows:

enter image description here

Where $$H_3(s) = \frac{R_3}{R_2}$$ and $$H_4(s) = \frac{R_3}{R_1}$$

From this, $$\frac{V_o}{V_{in}} = \frac{-H_1(s)\times H_4(s)}{1+H_1(s)H_2(s)H_3(s)}$$


The most simple and most direct solution is based on Black´s famous feedback formula for the closed-loop gain Acl:

\$A_{cl}=H_{f}/(1-H_{r}H_{f})\$ with \$H_{f}=forward \: function (feedback \: loop \: open) \$ and \$H_{r}= feedback \: function \$ (for \$V_{in}=0 \$). The product \$H_{r}H_{f}\$ is called "loop gain" (and is negative for negative feedback).

In your case: \$H_{f}=-(R_{3}/R_{1})H_{1}(s)\$ and \$H_{r}=-(R_{3}/R_{2})H_{2}(s)\$.


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