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How phase angle effect on the sawtooth signal .. if anybody having mathamtical derivation on sawtooth using phase angle.. start frequency, end frequency and phase angle. Thanks & Regards, Sunil


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The product $x(t)y(t)$ of two periodic signals with fundamental periods $T_x$ and $T_y$ is not a periodic signal unless $T_x$ and $T_y$ are rational multiples of one another; that is, $T_x = aT_y$ where $a$ is a rational number. Thus, except when such a relationship holds, $x(t)y(t)$ does not have a Fourier series. When $T_x$ is a rational multiple of $T_y$,...


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You can figure where the peaks would be if the other tone wasn't there by an accurate estimate of the phase. This location will be a rougher estimate in the presence of the other tone. When you add a slanted signal (say a line) to a mode (a peak), it will shift the location of the peak somewhat. Since you are interested in the higher frequency tone, the ...


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Vanilla implementation of each method for image of size m x n and kernel of size k x l will yield: Spatial Domain Convolution - O(mnkl) as for each pixel in the image we do kl multiplications (Additions are discarded). Frequency Domain Convolution - O(mn log(mn) + mn) as the complexity of the FFT is mn log(mn) and we add the multiplication (You could add ...


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Your math is correct, for a delay of 2 seconds, the first null in the PSD should be at 0.25 Hz. I'm guessing the problem is in the code you use to generate the signals and/or the PSD. The sample rates seem awfully low. At a sample rate of 0.25 Hz, your 2 second delay is just half a sample, that means you would have to implement a fractional delay to ...


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The problem is that the formula $$H(f) = \frac{Y(f)}{X(f)},$$ where $H(f)$ is the frequency response, and $X(f)$, $Y(f)$ are the input and output, is valid only for frquencies $f$ where $|X(f)| \neq 0$. Furthermore, if $|X(f_0)| \approx 0$, you're going to run into numerical errors, since $|H(f_0)| \rightarrow \infty$. Your input signal has a bunch of ...


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No, only the amplitude would change. Changing brightness is nothing more than adding (or removing) offset to RGB or luma channel. As it only changes the average value of image, it only changes the DC component of fourier transform.


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If I understand correctly only phase would change not frequency. Is that correct? Assuming "increasing brightness by 10%" just means multiplying each pixel's value with 1.1: No, the phase won't change either. Since the Fourier Transform is a linear operation, you just multiply every point in the spectrum by the same factor. That doesn't affect phase, since ...


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the lomb scargle in matlab, (plomb) returns a frequency vector as the second output. i would be surprised if the routine you are using doesn’t do the same. there is a common tendency for python signal processing libraries to be functionally equivalent to matlab conventions. your plots show symmetry around a “center” frequency which is the same as what a ...


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If $x(t)$ is an action signal like voltage or current, then the square of $x(t)$ is proportional to instantaneous power and the constant of proportionality depends on what kinda animal $x(t)$ is and what load $x(t)$ is connected to. So if $\big|x(t)\big|^2$ is proportional to instantaneous power, so also is the integral of $\big|x(t)\big|^2$ over all time ...


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If you want correct dimensions, the signal $x(t)$ must have dimension $\sqrt{W}$ because then its energy $$E_x=\int_{-\infty}^{\infty}|x(t)|^2dt$$ has the correct dimension $W\cdot s=J$. E.g., if you have a voltage signal, you need to normalize it by some load impedance to get the correct dimension $V/\sqrt{\Omega}=\sqrt{W}$. With the dimension of $x(t)$ ...


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Wouldn't it be much easier to just compare the left-hand side and the right-hand side of $$e^{-j2\pi ft_0}=|H(f)|e^{j\phi(f)}\tag{1}$$ to see that $|H(f)|=1$ and $\phi(f)=-2\pi ft_0$?


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$\varphi(f)=-\arctan(\tan(2{\pi}ft_0)) = -2{\pi}ft_0$ Which is the solution


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without seeing your code, one can only guess. your Fourier Transform doesn’t look like it is uniformly sampled in frequency. There are fewer points covering the frequencies as frequency increases, but the labels imply uniform coverage. I’m guessing that the frequency analysis is of the constant Q type but is being plotted as a linear frequency plot. a ...


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If the size $N$ of the DFT is even, only one "extremal" point (after fftshift) is Nyquist. If $N$ is even, you cannot have an arbitary $c_{-5}$ and $c_5$. They must add to be whatever your $c_{-5}$ term is. If the input to the DFT is purely real and if $N$ is even, consider the Nyquist point, at $c_{N/2}$ to be split in half. One half is the negative-...


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This can be answered simply by considering the definition of the $N$-point DFT: $$X_N[n] = \sum_{k=0}^{N-1} x[k]e^{-j2\pi \frac {n}{N}k }$$ where it's easy to see that the DFT just compares your $N$-point input signal $x[k]$ to a sinusoid of frequency $\frac nN$. Thus, the lowest frequency is always $0$, and the resolution is always $\frac{f_\text{sample}...


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So, your integral with bounds $$X(f) = \int_{a}^b x(t) e^{-i2\pi ft}\,dt$$ could really be written as $$X(f) = \int_{-\infty}^\infty (r_{a,b}(t)\cdot x(t)) e^{-i2\pi ft}\,dt$$ with $r_{a,b}(t) = \begin{cases}1 & a<t\le b\\0&\text{else,}\end{cases}$ i.e. as a "windowed" view at $x(t)$. Luckily, the convolution theorem of the Fourier ...


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All recognition tasks (doesn't even have to be speech recognition) are reductions of a very high-dimensional signal (your speech recording's dimension is the number of audio samples!) to a low-dimensional signal. As such, it is generally advisable to transform the input signal through an easy operation to a representation where the dimensionality can be ...


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The output signal of an LTI system is periodic only if its input signal is periodic as well, your reasoning was right. Now, a signal periodic with a certain period $T_p$ has a line spectrum, i.e., it only contains frequency contents at $n f_0$, $n \in \mathbb{Z}$ where $f_0 = \frac{1}{T_p}$. What it basically means is that your input only excites these ...


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First of all, I think it'd be good to include your result vs the result given by the mentioned library. Here is the code of scipy's ifft. In the docs it states that the function returns a complex array contains y(0), y(1),..., y(n-1) where y(j) = (x * exp(2*pi*sqrt(-1)*j*np.arange(n)/n)).mean(). I suspect that you're trying to write the imaginary ...


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