Cascading low pass op amps, derive expression for f3db

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SUMMARY

The discussion focuses on deriving the expression for the 3 dB frequency (f3dB) of a cascaded amplifier consisting of two non-identical low-pass stages. Stage 1 has a low-frequency gain of ALF1 and a 3 dB frequency of f1, while Stage 2 has a low-frequency gain of ALF2 and a 3 dB frequency of f2. The polynomial expression for f3dB is formulated as a0 + a1f3dB + a2f23dB + ... = 0, where the parameters a0, a1, and a2 are derived from the gain and frequency characteristics of the individual stages.

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donpacino
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Homework Statement


consider the cascading of two non identical amplifiers stages, each having a low-pass STC frequency response. STage 1 has a low freqency gain of ALF1 and a 3 dB frequency of f1. Stage 2 has a low frequency gain of ALF2 and a 3 dB frequency of f2. Derive a polynomial expression for f3dB, the 3 dB frequency of the cascaded amplifier, in terms of f1 and f2. Express f3dB in the following form.

a0 +a1f3dB+a2f23dB+...=0

find a0,a1,a2...etc

Homework Equations


Av=gain
f=frequency
Av=ALF1/(sqrt(1+(f/f1)))

The Attempt at a Solution


i truly don't know where to start. if anyone knows how to start the problem or has any tips to point me in the right direction it would be greatly appreciated.
 
Last edited:
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It's a while since I've looked at this stuff, so check this over carefully. But -3db point is defined as the frequency where the gain has fallen to \tfrac{1}{\sqrt{2}} of its low-frequency gain. So you need to involve a fraction that looks like this:

3dB\;\; point\;\; \overset{ \mathrm{ \triangle} } {=}\; <br /> \frac{1} {\left |(1+j\frac{{\omega}} {\omega _1})\cdot (1+j\frac{{\omega}}{\omega_2})\right |}\;=\;\frac{1}{\sqrt{2}}

One equation, only one unknown, solve for w. :smile:
 
Last edited:

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