Deriving Active Filters: A Step-by-Step Guide

In summary, The conversation discusses bandpass filters and how they utilize both inductors and capacitors to filter out specific frequencies. The transfer function for a circuit is derived and it is determined that the resonant frequency can be obtained by selecting a specific value for Z. The desired resonant frequency can then be calculated using the values for inductance and capacitance. The conversation also mentions that the value for Rz can be solved for using a specific equation.
  • #1
dillonmhudson
49
0
Untitled-1.jpg


Can I get some help with this? I don't even know where to begin.
 
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  • #2
dillonmhudson said:
Untitled-1.jpg


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.
 
  • #3
jegues said:
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,

[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]
 
Last edited:

1. What is an active filter derivation?

An active filter derivation is the process of designing an electronic circuit that filters out certain frequencies from an input signal using active components such as transistors or op-amps.

2. What are the advantages of using active filters over passive filters?

Active filters have the ability to amplify and shape the filtered signal, resulting in a more precise and customizable output. They also have a high input impedance and low output impedance, making them less susceptible to noise and interference.

3. How do you derive the transfer function for an active filter?

The transfer function for an active filter can be derived using Kirchhoff's laws and basic circuit analysis techniques, such as nodal analysis or mesh analysis. It involves finding the ratio of the output voltage to the input voltage as a function of frequency.

4. What are the different types of active filters?

The different types of active filters include low-pass, high-pass, band-pass, and band-stop filters. They can also be classified as first-order or second-order filters, depending on the number of reactive components in the circuit.

5. What are some common applications of active filters?

Active filters are commonly used in audio signal processing, instrumentation, and communication systems. They can also be found in power supplies, medical equipment, and control systems. Additionally, active filters are often used in combination with other circuits to create more complex filtering functions.

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