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I'm not understanding the comments. Of course you can do this. It is simply a matter of understanding what a DFT means, how to calculate DFT bin values, and how to interpret those bin values as continuous fourier series coefficients. First off, the plane you are looking at is the complex plane. Your points are a set of $N$ discrete samples. Each sample ...

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I was able to solve the problem using Hint 2 mentioned by Hilmar. Basically consider the FFT operation Z=FI, where I is a function whose alternate sample are zero. Then obtain the ifft of I to get the time domain sequence i(n) . The output sequence y(n) = x(n) convolution i(n). The sequence i(n) has only two non zero samples at n=0 and n=N/2. The output y(n)...

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Hint 1: try to solve the symmetric problem: what happens in the frequency domain if zeri out every second sample in the time domain? Hint 2: try to express the operation "zero out the odd samples" as a set of multiplications, convolutions and/or add and subtractions with certain signals

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You can use expansion analysis to deduce what happens when a sequence is zero staffed in between its samples. Let $x[n]$ be your original length $N$ sequence. Then its expansion by $2$ yields the new sequence $y[n]$ of length $2N$ : $$y[n] = \begin{cases} { x[n/2] ~~~,~~~n = 2m \\ ~~~~~ 0 ~~~,~~~ \text{otherwise} } \end{cases}$$ Then the DTFT relation ...

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the DFT bins for stationary noise processes tend toward independence as the separation increases. A white noise process is often simulated by using independent bin values. The Toeplitz structure of covariance matrices comes from assuming stationary. One might use the assumption of short term stationarity for noise like signals or low SNR signals in ...

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Whether that decorrelation happens depends on the signals you put in – it's not a general property of the DFT! Especially, when you model your signal as sum of narrowband signals, then you'll find that due to the DFT of a narrowband signal being concentrated on one or very few bins, and practically 0 elsewhere, the decorrelation simply stems from the fact ...

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Signal $x[n]$ has $N=50$ samples and the filter $h[n]$ has $M=10$ samples. The output $y[n]$ (by linear conv) will have length $L = N+M-1=59$ samples. If you use time domain convolution, for each output sample (except the edges) you will be making $M=10$ real MACs. And this makes the number of total real MACs as $K = L \cdot M = 59 \times 10 = 590$. The ...

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First of all, I think it'd be good to include your result vs the result given by the mentioned library. Here is the code of scipy's ifft. In the docs it states that the function returns a complex array contains y(0), y(1),..., y(n-1) where y(j) = (x * exp(2*pi*sqrt(-1)*j*np.arange(n)/n)).mean(). I suspect that you're trying to write the imaginary ...

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