graph from power spectrum and butterworth filter

[limavolt]

Hi,

I am newbie in labview, I want to see the power spectrum from the samples sound, I want to see if there are some pattern in those samples.

I use spectral measurement, and then I filtered the samples with butterworth filter. For the output I using waveform graphs.

Before filtered the signal, I got in the x axis the graph between 0-22500 and the data mostly in 0-2500.

I have tried to filtered using these values (fs=1000hz, fl=0.125hz, and fh=500hz) and the graphs almost the same but the axis of the graph is between 0-0,5.

questions:

what is the properties of the x axis? why is different but the graphs are the same?

Do I make the mistake with the program?

[ramktamu · Accepted solution]

Hi Limavolt,

You are having problem with Bandpass filter VI, this VI generates only the signal value i.e. DBL array. The signal is a cluster.

solution 1: you have to calculate 'dt' using sampling frequency and use the build waveform VI to build a signal is proper scaling. 

Normally, dt=Sampling frequency/no. of samples

solution 2: replace butterworth filter VI with filter express VI, this will produce signal which does not require scaling.

[johnsold]

Here is an example of the solution 1 of ramktamu.

I generally try to avoid the Express VIs because they hide so much of what they are doing, but in this case they should be fine.

Lynn

[limavolt]

Thank you both of you, you saved my days

I have to learn a lot about this software.

regards

limavolt

[ramktamu]

For the first solution if you consider using build waveform VI. you can calculate dt as,

For low pass and high pass filter.

dt=fs/fc, where fc=cutoff frequency and fs=sampling frequency

and for bandpass and band stop filter.

dt=fs/((flc+fhc)/2)

[johnsold]

Ram,

Where did you get those formulas? 

dt = 1/fs

It does not depend on the filter bandwidth. It is the interval between samples and thus depends only on the sample rate.

Lynn