Electronic – Op Amp gain less than 1

Bringing the resistances out

Starting from your final equation, $$\frac{V_o}{V_i} = \frac{1}{\frac{R_1}{R_f}-j\frac{f_c}{f}}\tag{1}$$ where \$f_c =\dfrac{1}{2\pi R_f C}\$.

Let \$f_1 = \dfrac{1}{2\pi R_1 C}\$ then, $$f_c = \frac{R_1}{R_f}\times f_1$$ Applying this in \$(1)\$, $$\frac{V_o}{V_i} = \frac{1}{\frac{R_1}{R_f}-j\frac{f_1}{f}\times \frac{R_1}{R_f}}$$ $$\left|\frac{V_o}{V_i}\right| = \left|\frac{R_f}{R_1}\times\frac{1}{1-j\frac{f_1}{f}}\right|$$ $$= \frac{R_f}{R_1}\frac{1}{\sqrt{1+\frac{f_1^2}{f^2}}}$$

Alternate method

You should have started this way. Especially when you had the final answer with you. :)

$$ \frac {V_o}{V_i} = \frac {R_f}{R_1 - \frac {j}{\omega C}}$$ $$= \frac{R_f}{R_1}\frac{1}{1-j\frac{1}{\omega R_1 C}}$$ $$= \frac{R_f}{R_1}\frac{1}{1-j\frac{f_1}{f}}$$

Taking absolute value results in the required result.

It looks like both filter sections are designed with the -3dB point at the same frequency, or very close together, so this filter is doing what it should.

In the crossover region, both sections contribute to the output, so it is higher than either alone. The slight peax at the crossover frequency would be 3dB if both signals were in phase(so they added coherently), so presumably they aren't. EDIT : apparently the small separation between -3dB points, rather than phase, accounts for this peak being less than 3dB.

For a classic design without that bulge, read up on the Linkwitz-Riley crossover, commonly used in loudspeakers where you want HPF and LPF outputs to sum to unity.

I don't know what you were expecting but if you wanted a notch you'd have to separate the -3dB frequencies, then the depth of notch will depend how far apart they are, and it won't be a deep notch.

If you wanted a deep notch, one approach is the Twin-T filter which can be made as narrow as you want.

Or start by specifying the frequency, notch width (at -3dB), and notch depth you want, and research filter design techniques to meet that specification.