# Tag Info

## Hot answers tagged convolution

4

Simply keep the first 1049 samples of the IFFT result and throw the rest away. You don't have to do a fftshift y1 = conv(s,h); y2 = ifft(fft(s, 2048) .* fft(h, 2048)); figure; plot(y1); hold on; plot(y2(1:length(y1)), '--'); BTW, compared to linear convolution, it is more efficient to use FFT when s and h have similar lengths. In your case uniformly ...

4

You may follow the answers to the following questions which implements the paper you linked above: Kernel Convolution in Frequency Domain - Cyclic Padding (Exact same paper). 2D Frequency Domain Convolution Using FFT (Convolution Theorem). Applying 2D Image Convolution in Frequency Domain with Replicate Border Conditions in MATLAB. Replicate MATLAB's conv2()...

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(Poster and I have iterated on this in a chat here: https://chat.stackexchange.com/rooms/131792/isi ) The OP mentions that the filter should not impact the signal since the signal is within the passband- this would only be true if this was a linear phase filter, otherwise we must also give consideration to the phase response and distortion that can cause. ...

1

Am I right, I just need to take convolution of the amplitude response and the pulse or I'm mistaken? Yes, you are mistaken. The amplitude response alone is not sufficient: the transfer function of a first order lowpass filter is complex: $$H(\omega) = \frac{1}{1+j\omega RC}$$ In order to implement this numerically, you need to sample it at a certain ...

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Sometimes (very rare), circular convolution may be more complexity efficient when compared to FFT. Let us say we circ-convolve to two arrays with lengths of $N$ and $K$, respectively. Recall that circular convolution can be performed with $N\times K$ operations. On the other hand, to perform circular convolution with the help of FFT, you need to run the FFT ...

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The issue is probably because of the start-up transients. You can possibly fix this using the zi parameter on the filter function: The other option might be to use the filtfilt method which attempts to choose the initial conditions: If I take that approach, then the resulting filterings are shown below. The original sinusoid is in blue, the filter version ...

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I created a MATLAB function which is basically conv2() applied in Frequency Domain: function [ mO ] = ImageConvFrequencyDomain( mI, mH, convShape ) % ----------------------------------------------------------------------------------------------- % % [ mO ] = ImageConvFrequencyDomain( mI, mH, convShape ) % Applies Image Convolution in the Frequency Domain. % ...

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Worth looking into the log Fourier transform (and FFTLog), which I know nothing about except that its abstract reads exactly like what you seek: We present an exact and analytical expression for the Fourier transform of a function that has been sampled logarithmically Note, any such method necessarily imposes a prior. That is, convolution demands ...

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I did want to note that I figured out what fractional slow component was. Following the references Ultradian Oscillations of Insulin Secretion in Humans Chantal Simon and Gabrielle Brandenberger Ref 19 ----------------------- Direct measurement of pulsatile insulin secretion from the portal vein in human subjects S H Song 1 , S S McIntyre, H Shah, J D ...

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