What is the correct transfer function of a feedback loop if the summing amplifier has different input resistors?

Question

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:

$$H(s)=\frac{H_1(s)}{1+H_1(s)H_2(s)}$$

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?

Answer 1

I can redraw the block diagram as follows:

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

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

Answer 2

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)\$.