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-1

I appreciate your work, my friend. There is one more beautiful and general way of doing the same that is using Eigenvalue Decomposition technique.

1

If you have a function $f \left[ m, n \right] \in \mathbb{R}^{M \times N}$ then the DFT of those functions is an orthogonal basis of functions in $\mathbb{C}^{M \times N}$. So if we have: $$F \left[ k , l \right] = \frac{1}{\sqrt{M N}} \sum_{m = 0}^{M - 1} \sum_{n = 0}^{N - 1} f \left[ m, n \right] {e}^{-j 2 \pi \left( \frac{k}{M} m + \frac{l}{N} n \... 1 No it's not correct. Derive it like this : 1-) 2N-point DFT of x[n] is: X_{2N}[k] = \sum_{n=0}^{2N-1} x[n] e^{-j \frac{2\pi}{2N} k n } 2-) Y[k] = X_{2N}[2k+1] be the odd-indexed samples of X_{2N}[k] 3-) We are looking for y[n] = \text{N-point IDFT}\{ Y[k]\}. 4-) Elaborate on step-2 and step-1 to see that Y[k] = \text{N-point DFT}\{ x[n] e^{-... 1 I would follow a signal block diagram based solution for this problem. First as suggested in the comments, it's very helpful to investigate a few values of y[n] and x[n] for some n :$$ \begin{align} y[0] &= 0 ~~~,~~~ y[1] = x[1] \\ y[2] &= 0 ~~~,~~~ y[3] = x[2] \\ y[4] &= 0 ~~~,~~~ y[5] = x[3] \\ y[6] &= 0 ~~~,~~~ y[7] = x[4] \\ \...

1

Note that if you upsample a sequence $x[n]$ by a factor $M$ you have $$y[Mn]=x[n]\tag{1}$$ The DFT of $x[n]$ is $$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}nk}\tag{2}$$ The DFT length of $y[n]$ is $MN$, so the DFT of $y[n]$ is $$Y[k]=\sum_{n=0}^{MN-1}y[n]e^{-j\frac{2\pi}{MN}nk}\tag{3}$$ Since only every $M^{th}$ sample of $y[n]$ is non-zero, $(3)$ ...

3

Since your signal isn't sampled uniformly some strange things might happen when you apply FFT and look at the results. What you should do is estimate the Uniform DFT of the Non Uniform Time Series. One easy way to do it is use the reference code and analysis I posted on the question - Frequency Analysis of a Signal Without a Constant Sampling Frequency (...

2

The DFT Matrix for Non Uniform Time Samples Series Problem Statement We have a signal $x \left( t \right)$ defined on the interval $\left[ {T}_{1}, {T}_{2} \right]$. Assume we have $N$ samples of it given by $\left\{ x \left( {t}_{i} \right) \right\}_{i = 0}^{N - 1}$. The samples time ${t}_{i}$ is arbitrary and not necessarily uniform. We're ...

0

There is no requirement, it is a matter of choice. Some conceptual clarity is required. A sequence of sampled points does not have a derivative. Derivatives come from continuous functions. If you treat the inverse DFT definition as a continuous function, the result is a continuous interpolation function that will pass through all your sampled points. ...

2

My guess would be that the DC peak is part of the transient response, thus it decreases to zero over time T as the oscillating parts of your signal continue. But I don't know what your signal is so that's a guess

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