New answers tagged dft
1
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) ...
1
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 (...
2
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:
$$\sum y[k]^2 = \sum x[n]^2 +...
2
As stated, yes, the variance of zero-mean data can be found by Parseval's theorem.
But that's because the sum of the squares of any data can be found by Parseval's theorem.
So you are correct: Parseval's theorem isn't limited to zero-mean data. I suspect that you were looking in texts about audio or RF, where signals typically are zero mean. In the more ...
2
pwelch() can return a one sided (default) or two sided spectrum. For one sided, the energy of negative frequency is added to the symmetric positive ones.
pwelch() returns a spectral density in something like $W/Hz$ which is actually dependent on the sample rate. By default pwelch() assumes a sample rate of $2\pi$.
Try
pp = pwelch(x, ones(n,1), n-1, n,1,'...
1
Let $\text{DFT}_2 = \text{DFT}(\text{DFT}(...))$. Then,
$$
\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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