
A command that appears in the Spectrum menu if you select one or more Sound objects. It turns the selected Sound into a Spectrum by an overall spectral analysis, a Fourier transform.
For the Fourier transform, the Praatdefined time domain of the Sound is ignored. Instead, its time domain is considered to run from t=0 to t=T, where t=0 is supposed to be aligned with the first sample, and T is the total duration of the samples, i.e. NΔt, where N is the number of samples and Δet is the sampling period. Thus, the last sample lies at t=T–Δt.
For a sound x(t), defined for all times t in the domain (0, T), the complex spectrum X(f) for any frequency f is the forward Fourier transform of x(t), with a negative exponent:
X(f) = ∫_{0}^{T} x(t) e^{2πift} dt 
If the Sound is expressed in Pascal (Pa), the Spectrum is expressed in Pa·s, or Pa/Hz. Since a Spectrum object can only contain a finite number of frequency samples, it is only computed for frequencies that are multiples of Δf = 1/T. The number of those frequencies is determined by the number of samples N of the sound.
If N is odd, there will be N frequency samples. For instance, if the sound has 20,457 samples, the spectrum will be computed at the frequencies 10,228Δf, 10,227Δf, ..., –Δf, 0, +Δf, ..., +10,227Δf, +10,228Δf. If we suppose that a frequency sample represents a frequency bin with a width of Δf, we see that the frequency samples span adjacent frequency ranges, e.g. the first sample runs from 10,228.5Δf to 10,227.5Δf, the second from 10,227.5Δf to 10,226.5Δf. Together, the frequency samples span the frequency domain of the spectrum, which runs from F to +F, where F = 10,228.5Δf. We can see that this frequency equals one half of the sampling frequency of the original sound: F = 10,228.5Δf = 10,228.5/T = 10,228.5/(20,457Δt) = 0.5/Δt. This is the socalled Nyquist frequency.
If N is even, there will be N+1 frequency samples. For instance, if the sound has 32,768 samples, the spectrum will be computed at the frequencies 16,384Δf, 16,383Δf, ..., Δf, 0, +Δf, ..., +16,383Δf, +16,384Δf. Again, the frequency samples span adjacent frequency ranges, but the first and last samples are only half as wide as the rest, i.e. the first sample runs from 16,384Δf to 16,383.5Δf, the second from 16,383.5Δf to 16,382.5Δf, and the last from +16,383.5Δf to +16,384Δf. Together, the frequency samples again span the frequency domain of the spectrum, which runs from –F to +F, where F = 16,384Δf = 0.5/Δt, the Nyquist frequency.
In a Spectrum object, Praat stores the real and imaginary parts of the complex spectrum separately. The real part is equal to the cosine transform:
re X(f) = ∫_{0}^{T} x(t) cos (2πft) dt 
The imaginary part is equal to the reverse of the sine transform:
im X(f) = – ∫_{0}^{T} x(t) sin (2πft) dt 
The complex spectrum can be reconstructed from the real and imaginary part as follows:
X(f) = re X(f) + i im X(f) 
Since the cosine is a symmetric function of t and the sine is an antisymmetric function of t, the complex spectrum for a negative frequency is the complex conjugate of the complex spectrum for the corresponding positive frequency:
X(f) = re X(f) + i im X(f) = re X(f)  i im X(f) = X^{*}(f) 
For purposes of storage, therefore, the negative frequencies are superfluous. For this reason, the Spectrum object stores re X(f) and im X(f) only for frequencies f = 0, Δf, 2Δf... In the case of a sound with 20,457 samples, the Spectrum object contains the real part of X(0) (its imaginary part is always zero), and the real and imaginary parts of X(f) for frequencies from Δf to 10,228Δf, which makes in total 1+2·10,228 = 20,457 real values. In the case of a sound with 32,768 samples, the Spectrum object contains the real parts of X(0) and X(16,384Δf) (their imaginary parts are always zero), and the real and imaginary parts of X(f) for frequencies from Δf to 16,383Δf, which makes in total 2+2·16,383 = 32,768 real values.
Since the negative frequencies have been removed, the frequency domain now runs from 0 to F. This means that the first frequency bin is now only 0.5Δf wide (i.e. as wide as the last bin for evenN spectra), which has consequences for computations of energies.
If you perform Spectrum: To Sound on the resulting Spectrum object, a Sound is created that is equal to the original Sound (or to the original Sound with appended zeroes).
The frequency integral over the squared Spectrum equals the time integral over the squared Sound:
∫_{F}^{+F} X(f)^{2} df = ∫_{0}^{T} x(t)^{2} dt 
This is called Parceval's theorem.
© ppgb, November 23, 2004