I'm trying to understand about deciding FFT length(size) and relative factors.
what If I got the streaming adc data as the below,
1624
20850
10314
-15042
-18908
3717
21175
8479
-16411
-17907
5796
21308
6493
-17713
-16662
7687
21241
4468
-18724
-15291
9750
20929
2392
-19649
-13833
11508
...
Q1. My first question is from here, I want to process with FFT but I don't know how do we decide proper FFT length (size)?
Q2. If I just decide to use FFT8, then can I use FFT8? or if I want to use FFT8 then what should I have to consider what I need to know? As I know (I'm slowly studying DSP) FFT8's number of inputs are 8. then FFT8 is working as
First term inputs
1624
20850
10314
-15042 => FFT out
-18908
3717
21175
8479
Second term inputs
-16411
-17907
5796
21308 => FFT out
6493
-17713
-16662
7687
Third term inputs
21241
4468
-18724
-15291 => FFT out
9750
20929
2392
-19649
n terms inputs
-13833
11508
...
is this right?
I've read some articles https://electronics.stackexchange.com/questions/12407/what-is-the-relation-between-fft-length-and-frequency-resolution https://spectraplus.com/DT_help/fft_size.htm But I didn't get it exactly.
So Would you help me to undersatnd how does FFT size decide and calculate process?
Q3, From http://support.ircam.fr/docs/AudioSculpt/3.0/co/FFT%20Size.html, What does 21,53 Hz. mean? As their description below.
Reminder : Bins The FFT size defines the number of bins used for dividing the window into equal strips, or bins. Hence, a bin is a spectrum sample, and defines the frequency resolution of the window.
By default :
N (Bins) = FFT Size/2
FR = Fmax/N(Bins)
For a 44100 sampling rate, we have a 22050 Hz band. With a 1024 FFT size, we divide this band into 512 bins.
FR = 22050/1024 ≃ 21,53 Hz.
But I want to know more about 21,53 Hz. exactly FR.