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Theory

The transfer function H(s) from input V_in to output V_out can be expressed as a product

$$H(s) = \frac{V_{out}(s)}{V_{in}(s)} = H_1(s) H_2(s) = (1 + \frac{R_2}{R_1})\frac{s}{s+ 1/(RC)}$$

where

$$H_1(s) = \frac{V_+(s)}{V_{in}(s)} = \frac{s}{s+ 1/(RC)}$$

is the transfer function from V_in to V₊ and

$$H_2(s) = \frac{V_{out}(s)}{V_{+}(s)} = 1 + \frac{R_2}{R_1}$$

is the transfer function from V₊ to V_out.

The cut-off frequency or 3dB point is

$$\omega_{c} = 1/(RC) \ \ {\rm rad/sec}$$

which gives $f_{c} = \omega_{c}/(2\pi)$ Hz and the pass-band gain is $(1 + \frac{R_2}{R_1})$.

Verify all of these assertions and calculations.