Assume a signal $x[n]$ with $N$ samples and its $N$-point FFT $X[k]$. Assume also the simplest power spectral density estimator: $$ S_x[k] = \frac{\left| X[k] \right|^2}{N} $$

With this estimator, the average total power can be evaluated as:

\begin{align} P_x = \sum \limits_{k=0}^{N-1} S_x[k] \Delta F \end{align} where $\Delta F$ is the distance between digital frequency points represented by the FFT and is equal to $\frac{1}{N}$.

At this point, a first question arises, what is the unit of $S_x[k]$? Is it $\frac{V^2}{\text{subcarrier}}$?

In the case of an analog frequency horizontal axis, the total power would be evaluated as: \begin{align} P_x = \sum \limits_{k=0}^{N-1} S_x[k] \Delta f \end{align} where $\Delta f$ is the distance between analog frequency points represented by the FFT and is equal to $\frac{f_s}{N}$ (FFT frequency spacing).

Sequentially, in order to represent the power spectral density over an analog frequency, the PSD estimator has to be modified to: \begin{align} S_x[k]' = \frac{S_x[k]}{f_s} \end{align} where $f_s$ is the sampling frency.

Is this analysis correct?

Here is a MATLAB snippet to verify this:

clear all

N = 4096;   % FFT size
fs = 200e6; % Sampling frequency in Hz

Px = 4;     % Target total power

x = sqrt(Px)*randn(N,1);    % Gaussian noise with power Px

% Verify signal variance
fprintf('Signal total power: %d\n', var(x))

X = fft(x,N);   

Sx = (abs(X).^2)/N;  % Simplest PSD estimator

deltaF = 1/N;        % Digital frequency spacing

plot((0:N-1)*deltaF, 10*log10(Sx))
xlabel('Digital Frequency')
grid on

% Verify total power
Px = sum(Sx*deltaF)

Sx_analog = Sx/fs;

deltaf = fs/N;        % Analog frequency spacing

plot((0:N-1)*deltaf, 10*log10(Sx_analog))
xlabel('Analog Frequency (Hz)')
grid on

% Verify total power
Px_2 = sum(Sx_analog*deltaf)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.