# Tag Info

In my understanding you need a little more information. You need to know the minimum and maximum frequency and number of coefficients in the MFCC representation. If the coefficients are linearly spaced in the mel scale, $m(f)$ The first coefficient is $m(f_1)$ the last coefficient is $m(f_N)$, and the $i-$th coefficient is $$m(f_i) = m(f_1) + \frac{i-1}{N-1}... 0 The Discrete Fourier Transform (DFT, what the FFT algorithm computes) has the origin in the top-left corner. It relates a time-domain signal sampled at n = 0..N-1, and a frequency-domain signal sampled at k = 0..N-1. k is assumed periodic, such that k = N is the same as k = N. The same periodicity can be assumed for n, which is why the convolution computed ... 3 I wonder why the negative part and positive part of X axis is not symmetrical? The data is symmetrical. The problem is that you have an odd FFT length and hence your frequency grid does NOT include the Nyquist frequency and you construct the frequency vector incorrectly. It should be f1 = -(Fs/2-dF/2):dF:(Fs/2-dF/2); 0 The structure of the FFT coefficients can be found in an answer to the post How is the frequency vector being derived?. Basically, they are arranged as: DC / (N-1) AC / (NC) / [(N-1) AC flipped and conjugate] with: DC: the average, DC-component or 0-frequency (real) AC: the set of generic coefficients with increasing positive frequency index (complex in ... 3 why does this power measurement (which is an instant function of time) It's NOT a instant function of time. For example instantaneous power of a sine wave is$$ p(t) = x(t)^2 = A \cdot cos^2(t) = \frac{A^2}{2}(1+cos(2t)) You have to apply some amount of time averaging to get rid of the $cos(2t)$ component and that's what all spectral analyzer do. : if I ...