Creating a 1st Order High-Pass Filter Using Bilinear Transform

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The discussion focuses on creating a first-order high-pass filter using the bilinear transform with a sampling frequency of 150 Hz and a cutoff frequency of 50 Hz. Participants clarify the correct formulation of the transfer function and the importance of including the sampling frequency in the bilinear transform. There is confusion regarding the passband gain and the corner frequency, with suggestions that the calculations may not yield the desired results. Additionally, concerns are raised about the effectiveness of the filter design due to the low sampling frequency relative to the cutoff frequency. Overall, the conversation emphasizes the need for accurate calculations and considerations in filter design.
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With bilinear transform (s=z-1/z+1) determine the differential equation for a 1st order discrete HP filter fith samplingfrequency f_s=150 Hz and lower cutoff frequency f_c=50 Hz. Also set the passpandgain to 1.


This is how i did:

Discrete variable: Ω=2π/3

Equivalent cuttof frequency for analog filter: ω_a=sqrt(3)

Analog HP-filter transfer function: H(s)=(s/s+ω_a)

Then use the bilinear transform om H(s).

how do i get passbandgain = 1?

If they mean H(z)=H(1)=1 by that, then H(z)=0, so that can't be right.

Because my answer didnt seem correct. Help please. :D
 
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beyondlight,
How did you get corner frequency ω_a=sqrt(3)? Should be ω_a=2πf = 2π(50) right?
How did you get discrete variable Ω=2π/3?, your independent variables are z and s right?

You are missing your sampling frequency in your bilinear transform. It should be: s=150(z-1)/(z+1).

By the way with such a low sample freq compared to corner frequency (150Hz and 50Hz respectively) this will be a very bad discrete time realization of your first order filter (see attached image, red is ideal first order, blue is your discrete time realization, vertical is dB).
Does this help?
 

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