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I think you're likely forgetting how the anti-derivative of $h(t)$ affects the gain of the convolution operation. Recall that somewhere in your convolution integral, you'll be taking an integral of the form $\int We^{W\tau} d\tau$. The Chain Rule requires the $W$ in the exponent must appear as a $1/W$ factor after integrating. This factor cancels the $W$ ...

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I think that's because of the DC offset of the signal. A time domain or frequency domain plot provided by you could have been helpful here for the confirmation of your doubt. Following are suggestions 1.try to filter the signals before FFT depending upon the signal band you are interested in 2.Subtract the signal by the mean of the entire signal (which is ...

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Signal $x[n]$ has $N=50$ samples and the filter $h[n]$ has $M=10$ samples. The output $y[n]$ (by linear conv) will have length $L = N+M-1=59$ samples. If you use time domain convolution, for each output sample (except the edges) you will be making $M=10$ real MACs. And this makes the number of total real MACs as $K = L \cdot M = 59 \times 10 = 590$. The ...

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@MBaz solution is neat. Another vision is to use the idea that discrete convolutions turn into polynomial products in the $z$-transform domain: $$[1,1]\ast [1,1] = [1,2,1]$$ is equivalent to $$(1+z^{-1})(1+z^{-1})=(1+2z^{-1}+z^{-2})$$ So separating a long filter into short convolved filters is equivalent to factorizing a polynomial. To ease notations, I'll ...

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First, the circular option relates to the treatment of the borders of the image. Then, standard image kernels can be any $[r,c]$ matrix. If either $r$ or $c$ is equal to $1$, then this is a very flat $2D$ filter, that acts only across one direction: across lines if horizontal, across columns if vertical (with the transpose). Filtering is a linear operation: ...

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First, we define $h_3[n] = h_2[n] \star h_2[n] = [1, 2, 1]$. From this result, we know that $h_1[n]$ must have 5 elements (so that $h[n]$ ends up with 7 elements). Let's define $h_1[n] = [g_1, g_2, g_3, g_4, g_5]$. We can find these as follows, using the definition of discrete convolution. First, we know that $g_1 h_3[0] = h[0] = 1$, so $g_1 = 1$. Then, ...

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Yes, the length of a filter limits the "frequency-selectiveness". If you remember modern physics, that's mathematically very related to the Heisenberg uncertainty principle: something that is temporally short can't be very sharp in the frequency domain, and vice versa. You can find one example of such a constraint in this answer, though the formula only ...

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If I'm not mistaken, a column vector will filter the image across its columns, treating each row independently of the others. Likewise, a row vector will filter across rows, treating all columns the same. edit: Regarding an example - consider the simple image [1,1,1;0,0,0;-1,-1,-1]. It's constant along its rows (i.e., all the columns are the same) and a ...

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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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Could you please share the part of your code where you include Laplacian inside bounding boxes? Many thanks. Regarding to your question, there is a paper reviewing focus methods: Pertuz, Said, Domenec Puig, and Miguel Angel Garcia. "Analysis of focus measure operators for shape-from-focus." Pattern Recognition 46.5 (2013): 1415-1432. Maybe in your case is ...

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For your application, image segmentation would be more useful than bounding boxes that contain also background. Other useful keywords: instance-aware image segmentation, instance segmentation. Figure 1. Instance segmentation example image from Mask R-CNN, by Karol Majek. Bounding boxes are also shown. Examples of implementations using some version of Yolo: ...

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For a metric of the blurredeness, you are using a laplacian which gives you usable limits of the object which are not included in most of the edge of the objects square. If you walk lines of pixels inwards from the edges of the detected zone, omitting the photo edges, when you cross a black zone from your laplacian, it means that you have transitioned inside ...

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Here is what I would try on the source image: Split your image into 5x5 pixel blocks (maybe 3 maybe 7, who knows?) Create output image one fifth (third, seventh, ??) size For each block For each color channel Find best fit plane Measure RMS of (pixel value-plane value) Next Set output pixel to RMS(R,G,B) Next In blurry/plain areas the ...

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