How Do Inductors Function in Bandpass Filters Without Capacitors?

[dillonmhudson]

Can I get some help with this? I don't even know where to begin.

 

[jegues]

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.

 

[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}