3

It's convention, they're equivalent: $$ \exp{\left(j2 \pi \frac{N}{2}n/N \right)} = \exp{\left(j2\pi \frac{-N}{2}n/N\right)} \\ \Rightarrow e^{j\pi n} = e^{-j \pi n} \Rightarrow \cos(\pi n) = \cos(-\pi n)=(-1)^n,\ j\sin(\pi n) = j\sin(-\pi n) = 0 $$ MATLAB and Numpy go $[-N/2, ..., N/2-1]$, which is unfortunate for analytic representations (+ freqs only). ...


3

I guess this probably just a mistake in your analysis code. Quantization noise is white and for a noise signal it's uncorrelated to the original signal, so spectrum of the original noise doesn't matter. I did repeat your steps and saw exactly what I expected: The quantization noise is white and the spectrum of the quantized signal follows the original signal ...


2

What may explain why your spectrum is noisy is that you are computing it using a single burst of data. You will have to smooth it by averaging successive spectrums applied on your measurements. The number of samples that shall be used for the averaging have to be tuned depending on the speed of variation of your phenomena. After having done this, the other ...


2

You don't have to perform a full FFT. Take the definition of DFT: $$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}kn}$$ What you need is one element of the sequence $X[k]$. For example, suppose the sampling rate is $f_s$, and the signal has a length of $N$, so the frequency resolution is $\Delta f=f_s/N$, and $k$ corresponding to 1000 Hz equals to $1000/\...


1

Typically, you use microcontrollers to process signals when you need to do it real-time. When you process a pre-recorded signal, like an audio file, you filter waveforms of these recordings in applications that you run on computers like desktops: you can process a recording "in the bulk", putting the entire length of the recording into an input ...


1

Given $ \left\{ x \left[ n \right] \right\}_{n \in M} $ where $ M $ is the set of indices given for the samples of $ x \left[ n \right] $. The trivial solution (Which it would be great to have a faster more efficient solution is what I'm looking for) would be: $$ \arg \min_{y} \frac{1}{2} \left\| \hat{F}^{T} y - x \right\|_{2}^{2} $$ Where $ \hat{F} $ is ...


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