The sampling rate (fs) and the cutoff frequency (fc) of the Butterworth filter (Point by point mode)

[ThanhDong]

Dear every one,

I want to compare a Butterworth filter (normal mode) and a Butterworth filter (Point by point mode) as low pass filters.

I am wondering about choosing the sampling rate (fs) and the cutoff frequency (fc) of the Butterworth filter (one is normal and the other is a point by point mode). The figure is shown below.

For the normal filter, I choose fs = 1000Hz, fc = 10Hz.

For the PtbyPt filter, I choose fs = 1Hz and fc = 0.01Hz. Here I consider one point of the input of the filter will create one point of the output of the filter. So the fs of the filter must be one calculation per loop at the time T= 1/sampling rate of the input signal (the filter is in the FOR LOOP).

--> It means that the sampling rate of the Butterworth filter is 1Hz at each the time T = 1/1000Hz (1000Hz = sampling rate of the input signal). Of course, for 1Hz of sampling frequency of the filter I choose fc = 0.01(Hz). It corresponds fc/fs = 0.01/1 = 10/1000.

I referenced Mr Henrik's program. The simulation results in two cases are the same.

 

So, fc = 0.01(Hz) and fs = 1Hz for the PtbyPt filter at the time T = 1/1000Hz are correct in principle, aren't they?

Thank you.

Dong

[ARVega]

I see you are having this conversation in the forum post you referenced. 

I am adding this comment so if another user has this same question they will know the should go and see that post. to see Mr. Henrik's answers.

[ThanhDong]

Thanks ARVega .

I also post this conversation because I hope there will many more people who determine.

Dong