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As @Hilmar mentioned I think you get confused between Square wave and Rectangular function. In Wikipedia about Square Wave : A square wave is a non-sinusoidal periodic waveform in which the amplitude alternates at a steady frequency between fixed minimum and maximum values, with the same duration at minimum and maximum. Which its Fourier Transform is only ...


4

The only signal, that really has just one frequency component, is an infinite sine signal. Limiting the signal duration in any way is bound to produce other frequency components, as time limiting can be thought of as multiplying with a rectangle window, which translates to convolution with an si-function in the frequency domain, thus introducing new ...


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I think there is a slight typo in robert bristow-johnson's answer. Should be \begin{align} x(t) &= \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} X(\omega) e^{j \omega t} \, \mathrm{d}\omega\\ \\ &= \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} H(\omega_0) \delta(\omega - \omega_0) e^{j \omega t} \, \mathrm{d}\omega\\ \\ &= \tfrac{1}{2 \pi} H(...


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Let’s be clear on what we will refer to as time delay and phase shift. Due to the common association of individual frequencies as sinusoids many confuse delay and phase shift as being equivalent. However a phase shift in time is simply multiplying a time sample by $e^{j\phi}$ while a time delay is displacing the sample in time such as done with $x(t-\tau)$. ...


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$\DeclareMathOperator{\sgn}{sgn}$ The modulating signal in AM is $$s(t) = C + a(t)\text,$$ where $a(t)$ is the (audio) amplitude, and $C$ is a constant so that $s(t) \ge 0 \;\forall t$. (Otherwise, your audio amplitude would just frequently "switch" the wave's sign, not really modulate the envelope.) That means, $C > - \min_t(s(t))$. Therefore, the ...


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I'd do some small adjustments to your idea (You really nailed them). Assumptions The Signal Model - Signal + Additive White Gaussian Noise (AWGN) Probably we could generalize it more but this is beyond the scope of this question. The DFT of the signal contains Peaks with relatively small roll off This is important as we're almost saying the Signal is a ...


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Depends on what you mean by "square wave". A single rectangular pulse has indeed a sinc spectrum An infinitely repeating series of rectangular pulses has a line spectrum with discrete frequencies


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A square wave is not a Sinc function in the frequency domain, but a sampled Sinc function (Even as a continuous function, the non-zero values are samples of the Sinc function in frequency). An individual rectangular pulse is a continuous Sinc function. The difference is the former is repeating in time. Repetition in one domain relates to sampling in the ...


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There are indeed many peak detection algorithms, and no clear consensus on which ones are "good" or "bad". But for what it's worth, your approach makes sense. Using median or other quantiles to detect sparse signals is common, e.g. the "median clipping" stage in Lasseck (2014), Large-scale identification of birds in audio recordings. In effect, you're ...


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If I compute the Fourier Transform of Z(f), what will I get? If you apply the inverse Fourier Transform, you get the impulse response of the system which is indeed a time domain signal. You can also apply the forward transform and would get the impulse response time reversed and scaled since the forward and inverse transform are quite similar. Which is ...


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I figured it ouy. In order to display DFT image properly they need to be log scaled. I used the log transformation s = c*log(1+r) where r is your normalized input image, and c is a constant. Plot s and you will be able to see the DFT correctly. %Log transform of DFT Image normalized_dftImage_F_uv = F_uv_dftImage/255; c1 = 1; s_LogTransform_F_uv_dftImage = ...


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It is important to understand what the frequency domain means. The bright regions in an image correspond to pixels with a certain value, whether it be color or gray scale. Frequency domain has the interpretation of looking at how fast the values change, not their values. For example, think about an image of a dark blue car against a white background. The ...


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A good stringed instrument music pitch detection/estimation algorithm will do the opposite, e.g. it will not ignore overtones. Instead it will pay attention to the harmonics, specifically the harmonic train and its spacing, as this is a stronger indication of human perceived pitch than spectral content at the fundamental frequency (which could be nearly or ...


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All it means is $$\begin{align} X(\omega) &= H(\omega) \delta(\omega - \omega_0)\\ &= H(\omega_0) \delta(\omega - \omega_0) \\ \end{align}$$ which means, in the time domain $$\begin{align} x(t) &= \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} X(\omega) e^{j \omega t} \, \mathrm{d}\omega\\ \\ &= \frac{1}{2 \pi} \int\limits_{-\infty}^{\...


2

No, at least not the way it's written. $U_1(\omega)$ contains two frequency peaks (10Hz and 0Hz) and $U_2(\omega)$ only one at $\omega _2$. $G(\omega)$ can change the relative height of the peaks but the overall but not create new ones. $A_2$ and $U_2(\omega)$ can vary the height and position of the peak but not the number of peaks.


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I am very thankful to Dilip-sarwate and Gilles, who took their precious time to understand my problem and guide me. So, Now I'm going to write the correct solution to my question. Which is as follows: $$ ℱ[x(t)g(t)] = \int_{t=-\infty}^{\infty}[x(t)g(t)]e^{-j\omega t}dt $$ As we know : $$ x(t) = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)e^{j\alpha ...


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$\Omega$ and $\omega$ are frequencies (in radians). Usually, $\omega$ is used, but if you need to deal with continuous-time as well as discrete-time systems, it's common to use one for discrete time and one for continuous time in order to distinguish the two. There is no real standard as to which one is which. I've seen $\Omega$ used for continuous time as ...


1

BOTTOM LINE UP FRONT: I think the exponential decay growth in $\left<|x(t)|^2\right>$ can be shown in the frequency domain only if the "boundary terms" are nonzero when we compute the Fourier transform of $dx(t)$ from the original SDE. I provide only a start in the work below. Since these processes seem like they could possibly be complex-...


1

Frequency domain analysis has much broader application (more numerous to list) than just analyzing sinusoidal components of a signal. Frequency domain analysis appears as a mathematical tool whenever the equivalent operations in the time domain can be simplified, and vice versa. For example, convolution in one domain is multiplication in the other which can ...


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If you zero out most or all of the harmonics (you set them all to -inf in your cap), then the HPS algorithm can’t find them.


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For any function $f(x)$ that is continuous at $x=x_0$ the following holds: $$f(x)\delta(x-x_0)=f(x_0)\delta(x-x_0)\tag{1}$$ So the result is a Dirac impulse at $x=x_0$ scaled by $f(x_0)$.


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The first term in $e^2_p$ is just the variation of the estimates around the true value. The second term is due to the problem that happens at the end of the periodic region. Sometimes, the noise is enough to move the estimate from $-\pi+\alpha$ to $\pi - \beta$. For example, the orange plot in the figure below represents the variation of the estimates for ...


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If I understood that right, you are trying to minimize the infinity norm over a discrete set of points. If that is the case you set over which you are trying to minimize is not convex and this is very much a NP hard problem, unless the set is very small, so that we can rigorously loop over it. Those are my thoughts on this.


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What kind of windowing/overlap scheme do you use? Did you look into cepstral processing, seeing as your bass should produce a harmonic series? Perhaps dynamic programming to pick a sensible «time-pitch-contour»? Fundamental pitch tracking is a long standing challenge. There should be many hints in the litterature.


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These terms refer to the general duality properties of these two domains. It should be obvious that higher frequency components mean faster changes in time and lower rise/fall times. In most natural systems, some of these fundamental properties oppose each other: Shorter rise time means more overshoot and higher settling times and vice versa; lower ...


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No it's not possible... [YES it's possible as the code below shows] The discrete Fourier transform of an image is a decomposition of it into an orthogonal set of periodic sine (complex exponential) waves. Intensity domain information is best dealt with spatial domain processing, which can be nonlinear type of operations without a (useful or simple) ...


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This is a very complicated problems and I don't think there exists a one-size-fits-all solution. You can try Matlab's $invfreqz()$ and see if it works for your purposes https://www.mathworks.com/help/signal/ref/invfreqz.html In general this is a error minimization problem but the actual data and the way you set the your error function and the search ...


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This serves as potential starting points only, on how to transpose classical signal processing tools to specific kinds of binary data. First, remember that standard spectrograms, fed as inputs to classifiers, can be seen as means to extract attributes from non-stationary data. There is apparently little need for inversion, so we are interested in analysis ...


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Of course you could. But I cannot find a useful manner. Because of duality, product or convolution in the image domain can be re-expressed as a convolution or product in the frequency domain. Also because if you apply it locally, components of a frequency transformation generally produce a value akin to intensity. In the extreme and trivial case, a one-...


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It is null-null when the signal is bandpass and zero-null when the signal is baseband. So you will always take the positive frequencies.


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