# Active Filter Derivation

by dillonmhudson
Tags: active, derivation, filter
 P: 50 Can I get some help with this? I don't even know where to begin.
P: 1,089
 Quote by dillonmhudson Can I get some help with this? I don't even know where to begin.
Bandpass filters usually employ both inductors and capictors.

The inductor cuts off the high frequencies, while the capacitor cuts off the low frequencies, ideal for a bandpass filter.
P: 1,089
 Quote by jegues Bandpass filters usually employ both inductors and capictors. The inductor cuts off the high frequencies, while the capacitor cuts off the low frequencies, ideal for a bandpass filter.
Whoops, I just noticed that the the answer is given in the figure. I can see they did not use a capacitor!

I managed to derive the transfer function for this circuit and found the following,

$$T(jw) = \frac{-z}{(jwL + z)(jwL + R)}$$

Now we simply need to select our Z such that the requirement for resonant frequency is obatined.

If we select,

$$z = R_{z}$$

We rearrange the transfer function,

$$\frac{-\frac{1}{R}}{(1 + \frac{jwL}{R_{z}})(1 + \frac{jwL}{R})}$$

This becomes,

$$\frac{-\frac{1}{R}}{(1 + \frac{jw}{w_{L}})(1 + \frac{jw}{w_{H}})}$$

Now we know what wL and wH are we can solve for fL and fH since,

$$f = 2\pi w$$

Then our desired resonant frequency,

$$f_{r} = \sqrt{f_{L} \cdot f_{H}}$$

Simply solve for, $$R_{z}$$.

You should find that, $$R_{z} = \frac{100\pi^{2}L^{2}}{R}$$

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