How to determine the frequency of each row in a MATLAB spectrogram

Question

I recently learned how to manually compute a spectrogram of an audio signal, which is essentially a matrix. Each column corresponds to a time frame (e.g. 10 ms or something) but I'm asking which row corresponds to which frequency?

See my implementation below:

clc; clear all; close all;
Fs = 44100;
t_max = 3;
T = 1/Fs;

time = 0:T:(t_max-T);
input = chirp(time,1500,1,8000);

window_length_t = 0.01; %10ms window length
window_length_s = round(0.01 * Fs); %window length in samples
if mod(window_length_s,2) == 0
    window_length_s = window_length_s + 1; %make sure we have odd window size
end


signal_framewise = buffer(input , window_length_s , floor(window_length_s/2));
nfft =((window_length_s-1)/2)+1;

out_buffer = zeros(nfft,size(signal_framewise,2));
for jj = 1:size(signal_framewise,2)
    current_frame = signal_framewise(:,jj).*gausswin(window_length_s);
    dtf = fft(current_frame);
    out_buffer(:,jj) = dtf(1:nfft);
end

My theory: The topmost row must be 0.5*Fs (half of the sampling frequency) because of the sampling theorem. All other rows are linear increments to then match to 0.5*Fs. Example: 5 Rows, topmost is 5 Hz, then the first row is 1 Hz, the second is 2 Hz and so forth.

Answer 1

The physical frequency $f$ that a given Discrete Fourier Transform (DFT) frequency bin $k$ corresponds to is $f = \frac{k}{N_{DFT}} \cdot Fs$.

For an $Fs=44100$ Hz, the middle point of a given $N_{DFT}$ point DFT would correspond to $\frac{1}{2} \cdot Fs$, also known as the Nyquist Frequency.

EXCEPT MATLAB, which maps $1$ to $\frac{Fs}{2}$.

For more information, please see here.

Hope this helps.

Answer 2

I don't know what you're doing but MATLAB's spectrogram implementation allows you to get the power spectrum ps, and the exact frequencies f and time instants t at which the spectrogram is computed with the following syntax at the output: [~, f, t, ps], where ps is a $M\times N$ matrix with the following:

• Each $k^\rm{th}$ column is the power spectrum over all frequencies at a fixed ($k^\rm{th}$) time instant. Or $$P_k(f_i), \quad\text{where}\quad i = 1, 2\cdots, M\quad\text{and}\quad k \in \left\{1, 2, \cdots, N\right\}.$$ $P_k(f)$ is a $M$-element vector.

• Each $i^\rm{th}$ row is the power spectrum over time at a fixed ($i^\rm{th}$) frequency value. Or $$P_i(t_k), \quad\text{where}\quad k = 1, 2\cdots, N\quad\text{and}\quad i \in \left\{1, 2, \cdots, M\right\}.$$ $P_i(t)$ is a $N$-element vector.

You don't always have $0$ to $F_s/2$ by default. The frequency range is $0$ to $F_s/2$ for real signal, one side of the spectrum is sufficient because of the Hermitian symmetry of the DFT. And $0$ to $F_s$ for complex signals, where uppper-half end of the frequencies can be shifted to negative frequencies.