virtins.com

Hello
According to calculations, for a sampling frequency Fs=192KHz and FFT size=2048 samples, the resulted Resolution=192000/2048=93.75Hz, and the FFT segments=93.
In the 2 attached pictures is presented the Bandwidth (-3dB) of the Asus Xonar Essence STX, but the measurement has been obtained in the past with an old version of M.I.
The BW FFT is mine, and I'm pretty sure that Pink Noise was used as a stimulus.
For the colored FFT, which has been obtained from the web, I don't know the stimulus used.
You could see that the number of the FFT segments=2.
All others are correct.
Why this?

Thanks

Fotis

Both screenshots use narrow-band FFT instead of octave-band analysis. Therefore the stimulus used should be white noise or MLS instead of pink noise.

Yes, maybe you are right. After so many years, I forgot many things about MI.
BTW, why in these two FFT captures from the past the FFT segments are 2? It might be 93.

Thanks

In these two graphs, the upper limit of the spectrum analyzer is 96kHz, so the sampling rate should be 192kHz. The resolution is 93.75Hz, so the FFT size should be 2048 (i.e. 192000/2048=93.75Hz). The number of FFT segments is 2, so the Record Length of the oscilloscope should be 4096 (i.e. 4096/2048=2). To set exactly 4096 samples per oscilloscope frame, tick [Setting]>[Display]>"Enable Record Length change via "Point" in Sampling Parameter Toolbar" and then enter "4096" into "Point" in the sampling parameter toolbar (first toolbar from the top).