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I am calculating spectrogram of a audio file of 36 second using the following code snippet:

window = np.hamming(window_width)
length = signal.size
count = 0
for i in range(0, length-window_width, window_overlap):
    cur_frame = signal[i:i+window_width]
    window_frame = cur_frame*window
    spec = sp.fft(window_frame)
    if(i == 0):
        width = int(length/window_overlap)
        height = int(spec.size)
        specgram = np.empty([height, width])
        energy = np.empty(width)
    energy[count] = getEnergy(spec)
    for j in range(height):
        val = spec[j]       
        specgram[j, count] = math.sqrt(math.pow(val.real, 2)+math.pow(val.imag, 2))
    count = count+1
print abs(specgram).max()

I get a ndarray of shape (4000L, 1008L)

but when i calculate spectrogram using Pylab:

Pxx, freqs, bins, im=pylab.specgram(sound_info, Fs=frame_rate)

I get ndarray of dimension: (129L, 125999L).

Actually I need to get the spectrogram which is the dimension of (frequency,nframes)

Thank you in advance.

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    $\begingroup$ What is the question exactly? Figure out what kind of frame rate (= overlap) and resolution (= FFT size) you want, and pass the appropriate parameters to specgram or modify your code accordingly. $\endgroup$ Jul 31, 2013 at 10:19

1 Answer 1

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How @Pichenettes said, we need more information to help properly

I'll take a guess:

If your Frame Rate are 44100, you will have 1587600 samples for 36 second of audio (36*44100), for the calculation example:

Pxx, freqs, bins, im=pylab.specgram(sound_info, Fs=frame_rate)

pylab.specgram seems use FFT Size = 256 and Overlap = 128 (50%) its give me the follow dimensional array:

floor(1587600 / 128) = 12403

(128, 12403)

You need set properly FFT Size and Overlap in your souce code.

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