The corner frequency of the highpass fiter circuit as shown below is approximate

Question

The corner frequency of the highpass fiter circuit as shown below is approximately 1 Hz. Scale the circuit up in frequency by a factor of 10, while keeping the values of the inductors unchanged, given that R1 R2 12, C C1F, C2 0.3 F, L1L0.04 H 0.3 F 0.04 Ho 0.04 H (a) Determine the magnitude and frequency scailing factars Km and Kf Km : Kr Submit Answer Tries 0/3 (b) Determine the corresponding scailed Capactance C or C3) of C, lar Ca) (pF) Submit Anaer Tries /3 (c) Determine the corresponding scailed Capac tance C'2 of C2 (pF) Submit Answer Tries 0y3 (d) Determine the corresponding scailed resistance R'1 (o R2) of R1 ar R2) (k9) Submit Answer Tries 0/3

Explanation / Answer

The specific formula applies only for a first order RC low pass filter. This is derived from its frequency response:

H(j?)=11+j?RC

H(j?)=11+j?RC

The cutoff frequency is defined as the frequency where the amplitude of H(j?)H(j?) is 12?12times the DC amplitude (approximately -3dB, half power point).

|H(j?c)|=112+?2cR2C2???????????=12–??|H(j0)|=12–?

|H(j?c)|=112+?c2R2C2=12?|H(j0)|=12

Solve it for ?c?c (cutoff angular frequency), you'll get 1RC1RC. Divide that by 2?2? and you get the cutoff frequency fcfc.

If you know the frequency response of your filter, you can apply this method (given that the cutoff frequency is defined as above). Obviously, for high-pass filters for example, you calculate with the value for ?????? as opposed to the DC value (always the maximum of the amplitude response, relative to which there is a 3dB decrease in amplitude at the cutoff frequency.)

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