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2

As mentioned, a FT or DFT can be used to analyse a transient signal but, as a really hand-wavy rule of thumb, the closer the basis vectors resemble the signal itself the less additional terms may be needed to accurately characterise it. This can help with the analysis phase if it leads to simplified data. This is typically where e.g. wavelets and lifting ...


2

It depends on your application. The term “transient” means different things to different people. Analyzing power line hiccups is different than looking at dolphin chirps. The start up of a steady state signal can be considered of interest. You also have complex sound scapes where there are transients in the presence of steady state signals where the ...


4

You're almost there; you just need to connect a few dots. Let your $\frac{10}{5 + i 2 \pi f} = G(f)$. Then $g(t) = 10 u(t) e^{-5 t}$. Now we get into that exponent part. Your $h(t) = g(t - t_0)$ is correct, but you're applying it incorrectly. You need to apply it to the part that I've labeled as $g(t)$: $h(t) = g(t - t_0) = 10 u(t - t_0) e^{-5 (t - t_0)}...


4

You are right. Fourier transforms are not useful for analysing transient signals compared to time-domain analysis or even wavelet analysis. Transient signal analsysis is a complex endeavour. However, keep in mind that all the information in the signal is retianed in the Fourier transform, albeit in a quite hard to interpret form. A better tool should be ...


5

Fourier transforms are multi-purpose tools. While they are well-suited to "stationary" signals, there are several ways to use them in different contexts. Think about a $2$-point DFT. Up to a scaling factor (for orthonormality), it consists in the matrix $F_2=\begin{bmatrix}1&1\\1&-1\end{bmatrix}$. It computes something that is essentially (thanks ...


7

If the signal is zero-valued outside the time window being analyzed ("analysis window"), then the discrete Fourier transform (DFT) exactly samples the discrete-time Fourier transform (DTFT) of the signal at harmonic frequencies. DTFT implies no time-domain periodicity. The sampling of DTFT by DFT is evident from definitions of the two transforms, if we apply ...


2

Seven years after this question was firstly raised, I run into this confusion similar to @Nicholas Kinar. Here I would like to provide some "unofficial" and "correctness not fully assured" personal perceptual ideas and explanations. The title of the following statements are exaggerated for better intelligibility. Forward process of STFT is not really meant ...


-1

The DFT of a unit step response is $$U(\omega) = \frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega)$$ Applying the shift property as you did will give: $$\mathcal{F}(u[n] - u[n-1]) = U(\omega) - U(\omega)e^{-j \omega} = \frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega) - [\frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega)]e^{-j \omega}$$ that is $$\mathcal{F}(u[n]...


1

Big thanks to @user753642 for spotting my mistakes over on the stackoverflow network: I was computing the $c_n$ coefficients from $n=0 \dots m$, where m is the number of wave functions in the sum. But by definitions the coefficients look like: $$c_m = \frac{1}{2L}\sum_{n = -\infty}^{\infty}c_n\delta_{n,m}\int_{-L}^Ldx = \frac{1}{2L}\int_{-L}^L f(x) \exp(\...


2

Multiplying a kernel and signal spectrum in Fourier domain lead to a circular convolution and not a linear convolution, so in order to it become linear convolution you must zero pad your signal and kernel before taking the Fourier transform (up to M+N-1 where M is the signal's length and N is the kernels's length). (if you compare the blue and orange signals,...


1

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} = \...


0

Energy signals, i.e., signals with finite energy $$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty$$ have zero power and, consequently, a power spectrum that is equal to zero. They do have an energy density spectrum, which is the squared magnitude of their Fourier transform. You will only get a non-zero power spectrum, according to the definition in your ...


1

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-...


0

The notion of a "frequency bin" is somewhat misleading. In general a sine wave of a single frequency will show up in ALL frequency "bins" of the Time discrete Fourier Transform. It will generate the highest value where $f_k$ is closest to the signal frequency and but the values at any other Fourier frequency are not zero, just smaller. The only exception is ...


1

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 $1/ \lambda ^2$ ...


3

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-...


5

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 ...


0

A real symmetric function has the following property $$f(t) = f(-t)$$ whereas a real anti-symmetric function has the following property $$f(t) = -f(-t)$$ So at $t=0$ you have $$f(0) = -f(0)$$ which implies that $$f(0) = 0$$ A complex conjugate-symmetric function has the following property $$f(t) = f^*(-t)$$ whereas a complex conjugate-antisymmetric ...


0

A real, 1D DFT is fully described by the non-negative frequencies, since real DFTs are symmetric about the 0 frequency bin. The function you linked is only computing the FFT along one dimension, and is only returning the non-negative frequency bins. From the manual page: "Notice how the final element of the fft output is the complex conjugate of the ...


2

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 ...


0

Very often, image display is driven by the initial data range (with inheritance). If the data is of uint8 format (per channel), some software keep subsequent calculations in the same range, so value below $0$ and above $255$ (or $1$) are clipped. Conversion of initial data to float, or signed integer, can help. As well, take care of the scaling induced by ...


1

One simple (simplistic) way to see that your result is correct is to realize that $$\mathcal{F}\big\{\delta(t-nT)\big\}=e^{-j\omega nT}\tag{1}$$ So you could argue that $$\mathcal{F}\left\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\right\}=\sum_{n=-\infty}^{\infty}e^{-j\omega nT}\tag{2}$$ which actually turns out to be correct. The result you're looking for ...


1

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. ...


2

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 ...


2

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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