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The widespread use of the unilateral Laplace transform reflects the fact that in practice we often deal with causal systems and signals that have a defined starting time (usually chosen as $t_0=0$). The Fourier transform is mainly used for analyzing ideal signals and systems, such as ideal filters (e.g., low pass, high pass, etc.) and ideal signals such as ...


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Essentially, your code does not respect the inherent Hermitian symmetry of the output of the FFT. Here, your signal is odd-sized $2K+1$. Hence, this FFT yields a complex vector of coefficients $d$ (real) and $a_k$ (generally complex), arranged as: $$ \left[d,a_1,a_2,\ldots,a_K,\overline{a_K},\ldots,,\overline{a_2},\overline{a_1} \right]$$ If you want to ...


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For sinusoids that are not exactly integer periodic in the FFT length, an FFT measures the phase at a circular discontinuity. And that discontinuity flips direction as frequency changes from slightly below to slightly above an exact integer periodic-in-aperture frequency. This is part of the effect of the default rectangular windowing of any finite length ...


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if $$\text{FFT}\{[a, b, c, d]\} = [p, q, r, s]$$ then $$\text{FFT}\{[a, b, c, d, a, b, c, d]\} = [p, 0, q, 0, r, 0, s, 0]$$ or $$=[2p, 0, 2q, 0, 2r, 0, 2s, 0]\text,$$ depending on normalization. Edit: Using this property, we can make FFT nearly 2x faster on this specialized data. But the complexity is still $O(n \log n)$. I mean, big O notation doesn'...


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First of all beware of the difference between image sharpening and highpass filtering. The former is actually a high frequency amplified image, while the latter removes the low frequency and completely eliminates the DC component. So assuming that you indeed wanted to remove low frequency content by appling a highpass filter, then your output will be a ...


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if $v(t)$ is an electromotive force (a voltage expressed in units of volt) that is a function of time (let's say expressed in units of second), then the Fourier Transform (or the Laplace Transform): $$ \mathscr{F} \Big \{ v(t) \Big \} \triangleq V(i \omega) = \int\limits_{-\infty}^{\infty} v(t) e^{-i \omega t} \, \mathrm{d}t $$ will have units of volt-...


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How can I justify the expression of G′(ν)? That's easy enough. Denominator is the sum of the signal energy $|X(\omega)|^2$ and the noise energy $\lambda ^2$. If the signal energy is significantly larger, then the whole expression simplifies to $G(\omega)$. If the noise is larger, we can't do a anything useful with the information and the $1/ \lambda ^2$ ...


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You define PSD proportional to $|X(\omega)|^2$, whose unit is $V^2 \cdot s^2$, where $X(\omega)$ is the Fourier transform of the input signal $x(t)$ in units of Volts. But this quantity is not a PSD but ESD (Energy Spectral Density) used for energy signals instead. You shall define PSD for periodic signals with a normalization by time: $$ \text{PSD_x} = \...


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What makes you think it's wrong? You're just not done yet. In your final expression, if $\omega n T $ is a multiple of $2\pi$, you'll sum "infinitely many ones" which gives you "infinite", whereas for other values, you're going to some numbers "equally spaced" over the complex unit circle, which gives you zero. This is exactly how a train of deltas behaves. ...


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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 \...


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Upon closer inspection, this doesn't even seem to be a rotation: $R_n$ being the output sample corresponding to the input sample $A_n$, we can understand their equation $$R_n = A_n - \Delta_n \tag1$$ as the statement that we add sequence $(-\Delta_n)_{n=1,\ldots,k}$ to the input signal $a$ and get the output signal $r$. I'll call that sequence $s$, $S_n ...


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First of all, as you said the sampling rate is probably $12$ Hz, rather than $12$ kHz, and perhaps they want to demonstrate an aliasing example. Given a bandlimited continuous-time periodic signal $$x(t) = \cos(16\pi t + \phi)$$ the samples taken at the rate $F_s = 12$ Hz will be denoted as $x[n]$ and will be obtained by via $x[n] = x(t_n)$ with $t_n = n ...


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Your first formula in your question is generally wrong, that's why you can't prove it. The correct formula is $$\sum_{n=-\infty}^{\infty}x[n]y^*[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})Y^*(e^{j\omega})d\omega\tag{1}$$ which is just Parseval's theorem. If $x[n]$ and $y[n]$ are real-valued, $(1)$ can be written as $$\sum_{n=-\infty}^{\infty}x[n]y[n]...


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Dr. Manuel Kuehner, You are close. You need to take the square root of the linear values squared. $$P_{\mbox{total_linear}}=\sqrt{p_1^2+p_2^2+...}$$ $$P_{\mbox{total_dB}}=20 log_{10}\left( P_{\mbox{total_linear/20E-6}} \right)$$ FYI: I wrote a MATLAB function to do exactly as you request. It is here Looking to read? See page 16 of this book: https://...


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I have churn over this question for a very long time. I never liked how answers to this question are again presented in mathematical form. Nevertheless, i continued looking at the same problem over and over again and whatever i have understood is represented below. In this way i understood why frequency axis of PSD still represents the frequency of a time-...


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