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

Explanation:

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

c)as speed changes, wavelength and frequency change.

everything changes!

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2