Finding the -3dB corner frequency of the circuit.

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SUMMARY

The discussion focuses on determining the -3dB corner frequency of a specific circuit using the voltage divider principle. Participants emphasize the importance of deriving the output voltage equation as a function of frequency and applying the formula 3dB = 20 Log10(X) to find the frequency at which the output voltage is reduced to a specified ratio of the input. The initial voltage divider reduces the input voltage by half, which corresponds to a -3dB drop, indicating that the circuit's behavior at different frequencies needs to be analyzed to find the exact -3dB frequency point.

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  • Understanding of voltage dividers and their impact on signal levels
  • Familiarity with the concept of -3dB points in frequency response
  • Knowledge of logarithmic functions and their application in signal analysis
  • Basic circuit analysis skills, particularly in AC signals
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  • Derive the output voltage equation for the circuit as a function of frequency
  • Learn about frequency response and how to calculate -3dB points in circuits
  • Explore the implications of using voltage dividers in AC signal processing
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Electrical engineers, circuit designers, and students studying signal processing who need to understand frequency response and -3dB calculations in circuits.

tarmon.gaidon
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Hey Everyone,

The circuit I am interested in is attached. I was just looking for a little input to help me start this problem. As the title states I am just looking to find the -3dB frequency of this circuit.

Point me in the right direction :)
 

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Write the equation for the output voltage as a function of frequency. Then using that equation, find the frequency that makes the output X times lower than the input where X is given by the equation below.

3dB = 20 Log10(X).
 
The circuit begins with a voltage divider that cuts Vsource in half. That's already more than 3dB down (1/√2 ≈ 0.7071). Perhaps you're looking for where it's 3db down compared to a DC input voltage?
 

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