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I would say that only 4 does not satisfy the Dirchlet conditions. Has only one discontinuity, is absolutely integrable $\int_{-\infty}^{\infty} e^{-2t}u(t) dt = 1/2$. Is just the function of item 1 made symmetric. No discontinuity and is absolutely integrable. The term $(1 - |t|/2014)$ is not absolutely integrable, but since it is multiplied by $u(t + 2014) ...


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I'll help you out with your first question. The second question is totally unrelated, so it should be asked as a separate question. First of all, it's important to distinguish between the discrete Fourier transform (DFT), and the discrete-time Fourier transform (DTFT). The first one transforms a finite number of (time-domain) samples to a finite number of (...


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This is actually a fun little proof. Let's assume $x[n]$ is mean-free and we create biased signal as $y[n] = x[n]+a$ Taking the Fourier Transforms and looking at the DC part we get $X[0] = 0$ (since it's mean free) and $Y[0] = Na$. For all other frequencies we get $$Y[k] = X[k], k\neq 0$$ Let's sum the squares in the time domain: $$\sum y[k]^2 = \sum x[n]^2 +...


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As stated, yes, the variance of zero-mean data can be found by Parseval's theorem. But that's because the sum of the squares of any data can be found by Parseval's theorem. So you are correct: Parseval's theorem isn't limited to zero-mean data. I suspect that you were looking in texts about audio or RF, where signals typically are zero mean. In the more ...


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pwelch() can return a one sided (default) or two sided spectrum. For one sided, the energy of negative frequency is added to the symmetric positive ones. pwelch() returns a spectral density in something like $W/Hz$ which is actually dependent on the sample rate. By default pwelch() assumes a sample rate of $2\pi$. Try pp = pwelch(x, ones(n,1), n-1, n,1,'...


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Let $\text{DFT}_2 = \text{DFT}(\text{DFT}(...))$. Then, $$ \begin{align} \text{DFT}_2(x[n]) &= N \cdot x[-n] \\ \text{DFT}_3(x[n]) &= N \cdot \text{DFT}(x[-n]) \\ \text{DFT}_4(x[n]) &= N^2 \cdot x[n] \end{align} $$ and thus $$ \text{DFT}_M(x[n]) = N^{\lfloor M/2 \rfloor} \cdot \begin{cases} \sum_{n=0}^{N-1} x[n\cdot (-1)^{\lfloor M/2 \rfloor}] ...


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