As a data acquisition anti-alias filter, an RC Butterworth filter is designed with a cutoff frequency of 100 Hz using a single-stage topology. Determine the attenuation of the filtered analog signal at 10, 50, 75, and 200 Hz

Using the improved Butterworth filter design.

Given that,

Attenuation (dB) = 10log (1 + (f/fc) ^2k)

k is the stage of the filter

For this case it is a single stage

Then k=1.

fc is the cutoff frequency and it is given as 100Hz

fc=100Hz

f is frequency at attenuation

Then taking the frequency one after the other

1. When f=10Hz

Then,

Attenuation (dB) = 10log (1 + (f/fc) ^2k)

k=1, fc=100Hz and f=10Hz

Attenuation (dB) = 10log (1 + (10/100) ^2)

Attenuation (dB) = 10log (1+0.1²)

Attenuation (dB) = 10log (1.01)

Attenuation (dB) = 0.0432dB

2. When f=50Hz

Then,

Attenuation (dB) = 10log (1 + (f/fc) ^2k)

k=1, fc=100Hz and f=50Hz

Attenuation (dB) = 10log (1 + (50/100) ^2)

Attenuation (dB) = 10log (1+0.5²)

Attenuation (dB) = 10log (1.25)

Attenuation (dB) = 0.969dB

3. When f=75Hz

Then,

Attenuation (dB) = 10log (1 + (f/fc) ^2k)

k=1, fc=100Hz and f=75Hz

Attenuation (dB) = 10log (1 + (75/100) ^2)

Attenuation (dB) = 10log (1+0.75²)

Attenuation (dB) = 10log (1.5625)

Attenuation (dB) = 1.938dB

4. When f=200Hz

Then,

Attenuation (dB) = 10log (1 + (f/fc) ^2k)

k=1, fc=100Hz and f=200Hz

Attenuation (dB) = 10log (1 + (200/100) ^2)

Attenuation (dB) = 10log (1+2²)

Attenuation (dB) = 10log (5)

Attenuation (dB) = 6.99dB