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Active Filter Derivation 
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#1
Apr2911, 04:29 PM

P: 50

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


#2
Apr2911, 05:07 PM

P: 1,089

The inductor cuts off the high frequencies, while the capacitor cuts off the low frequencies, ideal for a bandpass filter. 


#3
Apr2911, 05:15 PM

P: 1,089

I managed to derive the transfer function for this circuit and found the following, [tex]T(jw) = \frac{z}{(jwL + z)(jwL + R)}[/tex] Now we simply need to select our Z such that the requirement for resonant frequency is obatined. If we select, [tex]z = R_{z}[/tex] We rearrange the transfer function, [tex]\frac{\frac{1}{R}}{(1 + \frac{jwL}{R_{z}})(1 + \frac{jwL}{R})}[/tex] This becomes, [tex]\frac{\frac{1}{R}}{(1 + \frac{jw}{w_{L}})(1 + \frac{jw}{w_{H}})}[/tex] Now we know what wL and wH are we can solve for fL and fH since, [tex]f = 2\pi w[/tex] Then our desired resonant frequency, [tex]f_{r} = \sqrt{f_{L} \cdot f_{H}}[/tex] Simply solve for, [tex]R_{z}[/tex]. You should find that, [tex]R_{z} = \frac{100\pi^{2}L^{2}}{R}[/tex] 


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