Sound spectrogram

Question

I have made an app that paints FFT to the screen realtime (from mic). Time on x-axis, frequency on y-axis and the color of the pixel represents the amplitude (pretty much a vanilla FFT spectrogram).

My problem is that even though I can see a pattern from the music there is also a lot of noise. Googling it I see people applying a logarithmic calculation to the amplitude. Should I be doing this? And if so, what would the formula look like? (I'm using C#, but I can translate the math into code so any sample is ok.)

I can bypass this problem by applying a color scheme showing lower values as darker colors. I'm just not sure if the audio is correctly represented without a logarithmic calculation on it.

Answer 1

Representation of the amplitude on a logarithmic scale approximates the sensitivity of the human auditory system, and therefore gives you a better representation of what you hear, as compared to a non-logarithmic scale. Mathematically, all you have to do is:

Alog = 20*log10 (abs (A))

Where A is the amplitude of the FFT data, and Alog is the output. the factor of 20 is just a convention and has no effect on the image, which you probably scale anyway to a color-scheme.

EDIT

Explanation regarding the 20 factor: The dB (decibel) unit is a logarithmic unit measuring ratios: it represents a scale on which the distance between 100 and 10, is the same as between 1000 and 100 (since they have the same ratio: 1000/100 = 100/10). If you measure it in dB you get:

10*log10 (1000/100) = 10*log10 (100/10) = 10

The factor of 10 is because deci means tenth, which means 1 Bel is 10 deciBels, (like 1 kilogram is 1000 grams)

Since the human auditory system is also (approximately) measuring ratios, it makes sense to measure sound level on a logarithmic scale, i.e measure the ratio of sound level to some reference value. Since the level of a sound is associated with the power (in Watts) of the sound wave, you actually measure the ratio of powers P/Pref. Also, the power is proportional to the amplitude squared, so all in all you get:

10*log10 (P/Pref) = 10*log10 (A^2 / Aref^2) = 20*log10 (A/Aref)

by the log rules. That's the origin of the 20 factor - remember that in the computer the audio is represented by the instantaneous amplitude of the sound wave.

Answer 2

Viewing your spectrogram logarithmically is, indeed, a better way to look a audio signals. Keep in mind also, that you need good resolution in both time-direction and frequency-direction. If you have too few bins in one or the other, it might look strange.

Another important point is that viewing your STFT on a log scale is not a denoising technique. What you see as "noise" might be actual noise or it could be things like harmonics, transients, spectral leakage and other things that are expected to be there. Depending on your application, if you need to do short-time analysis of a signal, a wavelet transform might be more suitable for your needs. It takes away certain disadvantages of the STFT such as constant window size.