如何使用快速傅立叶变换计算每分钟的位数

Question

i am getting signals as given bellow this data capture for just 10 second (due to limited space here) now i want to write a code which shows bits per minute and im considering the value 3.00 as a bit

What i did:

I just used a threshold value to print data to count bits per minute

as

threshold  = 2.8     # Threshold to do something if value is more than to that. 
if signal >= 2.8:
    print signal              
    counter  = counter +1 # to count how many times we get value more than 2.8 or (near by 3)
    bits_per_mint = counterx6 # captured for 10 second so converted it to minute 
print bits_per_mint   

data out put in 10 second  

1.7646050347
1.6970572917
1.6774392361
0
3.486762153
1.6310026042
1.6582465278
1.6384114583
1.6501171875
1.6769661458
3.9909898997
0
1.6688020833
1.6627473958
1.6689800347
1.6756423611
1.6579513889
0
1.6809592014
1.6504774306
3.7684857685
1.6463671875
1.67640625
0
1.6509635417
1.6736501736
3.5653423434
1.6581206597
1.6516666667
0
1.6449348958
1.6630338542
1.6605772569
1.6500824653
3.4554564564
0
1.6839409722
1.6495399306
1.6393663194
1.6684244792
------------
--------- so on 

How can i do this using Fast Fourier transform

thanks

Answer 1

Try this and let me know if it helped you. You can use time module to introduce the delay in seconds in an outer while loop and then check for the condition if the signal is greater than the threshold if yes then print it and also append it to a list bits_per_min . Example code shown below.

import time
threshold  = 2.8     # Threshold to do something if value is more than to that. 
seconds = 0
counter = 0

while seconds != 60:
    bits_per_min = []    # this part of code runs for one min duration
    if signal >= threshold:   # to compare if signal is greater than the threshold value
        print signal
        bits_per_min.append(signal)
        counter += 1
    time.sleep(1)
    seconds += 1
print bits_per_min