Engineering Finding the -3dB corner frequency of the circuit.

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To find the -3dB corner frequency of the circuit, start by writing the output voltage equation as a function of frequency. The -3dB point corresponds to a reduction in output voltage to a specific ratio, defined by 3dB = 20 Log10(X). The circuit's initial voltage divider already reduces the voltage by more than 3dB, indicating that the focus may be on the frequency where the output is 3dB down from a DC input. Understanding these relationships will guide the analysis of the circuit's frequency response.
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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