Search Results

Search found 10 results on 1 pages for 'ifft'.

Page 1/1 | 1 

  • FFT and IFFT on 3D matrix (Matlab)

    - by SteffenDM
    I have a movie with 70 grayscale frames in MATLAB. I have put them in a 3-D matrix, so the dimensions are X, Y and time. I want to determine the frequencies in the time dimension, so I have to calculate the FFT for every point in the 3rd dimension. This is not a problem but I have to return the images to the original form with ifft. In a normal situation this would be true: X = ifft(fft(X)), but this is not the case it seems in MATLAB when you work with multidimensional data. This is the code I use: for i = 1:length y(:, :, i) = [img1{i, level}]; %# take each picture from an cell array and put it end %# and put it in 3D array y2 = ifft(fft(y, NFFT,3), NFFT, 3); %# NFFT = 128, the 3 is the dimension in which i want %# to calculate the FFT and IFFT y is 480x640x70, so there are 70 images of 640x480 pixels. If I use only fft, y2 is 480x640x128 (this is normal because we want 128 points with NFFT). If I use fft and ifft, y2 is 480x640x128 pixels. This is not normal, the 128 should be 70 again. I tried to do it in just one dimension by using 2 for loops and this works fine. The for loops take to much time, though.

    Read the article

  • Using GNU Octave FFT functions

    - by CFP
    Hello world! I'm playing with octave's fft functions, and I can't really figure out how to scale their output: I use the following (very short) code to approximate a function: function y = f(x) y = x .^ 2; endfunction; X=[-4096:4095]/64; Y = f(X); # plot(X, Y); F = fft(Y); S = [0:2047]/2048; function points = approximate(input, count) size = size(input)(2); fourier = [fft(input)(1:count) zeros(1, size-count)]; points = ifft(fourier); endfunction; Y = f(X); plot(X, Y, X, approximate(Y, 10)); Basically, what it does is take a function, compute the image of an interval, fft-it, then keep a few harmonics, and ifft the result. Yet I get a plot that is vertically compressed (the vertical scale of the output is wrong). Any ideas? Thanks!

    Read the article

  • Can FFT length affect filtering accuracy?

    - by Charles
    Hi, I am designing a fractional delay filter, and my lagrange coefficient of order 5 h(n) have 6 taps in time domain. I have tested to convolute the h(n) with x(n) which is 5000 sampled signal using matlab, and the result seems ok. When I tried to use FFT and IFFT method, the output is totally wrong. Actually my FFT is computed with 8192 data in frequency domain, which is the nearest power of 2 for 5000 signal sample. For the IFFT portion, I convert back the 8192 frequency domain data back to 5000 length data in time domain. So, the problem is, why this thing works in convolution, but not in FFT multiplication. Does converting my 6 taps h(n) to 8192 taps in frequency domain causes this problem? Actually I have tried using overlap-save method, which perform the FFT and multiplication with smaller chunks of x(n) and doing it 5 times separately. The result seems slight better than the previous, and at least I can see the waveform pattern, but still slightly distorted. So, any idea where goes wrong, and what is the solution. Thank you.

    Read the article

  • Extrapolation using fft in octave

    - by CFP
    Using GNU octave, I'm computing a fft over a piece of signal, then eliminating some frequencies, and finally reconstructing the signal. This give me a nice approximation of the signal ; but it doesn't give me a way to extrapolate the data. Suppose basically that I have plotted three periods and a half of f: x -> sin(x) + 0.5*sin(3*x) + 1.2*sin(5*x) and then added a piece of low amplitude, zero-centered random noise. With fft/ifft, I can easily remove most of the noise ; but then how do I extrapolate 3 more periods of my signal data? (other of course that duplicating the signal). The math way is easy : you have a decomposition of your function as an infinite sum of sines/cosines, and you just need to extract a partial sum and apply it anywhere. But I don't quite get the programmatic way... Thanks!

    Read the article

  • Using Cepstrum for PDA

    - by CziX
    Hey, I am currently deleveloping a algorithm to decide wheather or not a frame is voiced or unvoiced. I am trying to use the Cepstrum to discriminate between these two situations. I use MATLAB for my implementation. I have some problems, saying something generally about the frame, but my currently implementation looks like (I'm award of the MATLAB has the function rceps, but this haven't worked for either): ceps = abs(ifft(log10(abs(fft(frame.*window')).^2+eps))); Can anybody give me a small demo, that will convert the frame to the power cepstrum, so a single lollipop at the pitch frequency. For instance use this code to generate the frequency. fs = 8000; timelength = 25e-3; freq = 500; k = 0:1/fs:timelength-(1/fs); s = 0.8*sin(2*pi*freq*k); Thanks.

    Read the article

  • How to generate a lower frequency version of a signal in Matlab?

    - by estourodepilha.com
    With a sine input, I tried to modify it's frequency cutting some lower frequencies in the spectrum, shifting the main frequency towards zero. As the signal is not fftshifted I tried to do that by eliminating some samples at the begin and at the end of the fft vector: interval = 1; samplingFrequency = 44100; signalFrequency = 440; sampleDuration = 1 / samplingFrequency; timespan = 1 : sampleDuration : (1 + interval); original = sin(2 * pi * signalFrequency * timespan); fourierTransform = fft(original); frequencyCut = 10; %% Hertz frequencyCut = floor(frequencyCut * (length(pattern) / samplingFrequency) / 4); %% Samples maxFrequency = length(fourierTransform) - (2 * frequencyCut); signal = ifft(fourierTransform(frequencyCut + 1:maxFrequency), 'symmetric'); But it didn't work as expected. I also tried to remove the center part of the spectrum, but it wielded a higher frequency sine wave too. How to make it right?

    Read the article

  • how do i design a high pass filters in matlab without using the builtin function?

    - by noura
    hello everyone, i'm just not sure how to draw the frequency response (H) of the high pass filter? after drawing the frequency response i can get the b coefficient by taking the ifft of (H). so yeah, for a low pass filter, with a cutoff frequency of say pi/2 : the frequency response code will be H = exp(-1*j*w*4).*(((0 <= w) & (w<= pi/2)) | ((2*pi - pi/2 <= w) & (w<=2*pi)); sincr the response is "1" between 0 and pi/2 and between (2*pi - pi/2) and 2*pi. can you help me write H for a high pass filter? thanx in advance.

    Read the article

  • How to remove the boundary effects arising due to zero padding in scipy/numpy fft?

    - by Omkar
    I have made a python code to smoothen a given signal using the Weierstrass transform, which is basically the convolution of a normalised gaussian with a signal. The code is as follows: #Importing relevant libraries from __future__ import division from scipy.signal import fftconvolve import numpy as np def smooth_func(sig, x, t= 0.002): N = len(x) x1 = x[-1] x0 = x[0] # defining a new array y which is symmetric around zero, to make the gaussian symmetric. y = np.linspace(-(x1-x0)/2, (x1-x0)/2, N) #gaussian centered around zero. gaus = np.exp(-y**(2)/t) #using fftconvolve to speed up the convolution; gaus.sum() is the normalization constant. return fftconvolve(sig, gaus/gaus.sum(), mode='same') If I run this code for say a step function, it smoothens the corner, but at the boundary it interprets another corner and smoothens that too, as a result giving unnecessary behaviour at the boundary. I explain this with a figure shown in the link below. Boundary effects This problem does not arise if we directly integrate to find convolution. Hence the problem is not in Weierstrass transform, and hence the problem is in the fftconvolve function of scipy. To understand why this problem arises we first need to understand the working of fftconvolve in scipy. The fftconvolve function basically uses the convolution theorem to speed up the computation. In short it says: convolution(int1,int2)=ifft(fft(int1)*fft(int2)) If we directly apply this theorem we dont get the desired result. To get the desired result we need to take the fft on a array double the size of max(int1,int2). But this leads to the undesired boundary effects. This is because in the fft code, if size(int) is greater than the size(over which to take fft) it zero pads the input and then takes the fft. This zero padding is exactly what is responsible for the undesired boundary effects. Can you suggest a way to remove this boundary effects? I have tried to remove it by a simple trick. After smoothening the function I am compairing the value of the smoothened signal with the original signal near the boundaries and if they dont match I replace the value of the smoothened func with the input signal at that point. It is as follows: i = 0 eps=1e-3 while abs(smooth[i]-sig[i])> eps: #compairing the signals on the left boundary smooth[i] = sig[i] i = i + 1 j = -1 while abs(smooth[j]-sig[j])> eps: # compairing on the right boundary. smooth[j] = sig[j] j = j - 1 There is a problem with this method, because of using an epsilon there are small jumps in the smoothened function, as shown below: jumps in the smooth func Can there be any changes made in the above method to solve this boundary problem?

    Read the article

  • DSP - Problems using the inverse Fast Fourier Transform

    - by Trap
    I've been playing around a little with the Exocortex implementation of the FFT, but I'm having some problems. First, after calculating the inverse FFT of an unchanged frequency spectrum obtained by a previous forward FFT, one would expect to get the original signal back, but this is not the case. I had to figure out that I needed to scale the FFT output to about 1 / fftLength to get the amplitudes ok. Why is this? Second, whenever I modify the amplitudes of the frequency bins before calling the iFFT the signal gets distorted at low frequencies. However, this does not happen if I attenuate all the bins by the same factor. Let me put a very simplified example of the output buffer of a 4-sample FFT: // Bin 0 (DC) FFTOut[0] = 0.0000610351563 FFTOut[1] = 0.0 // Bin 1 FFTOut[2] = 0.000331878662 FFTOut[3] = 0.000629425049 // Central bin FFTOut[4] = -0.0000381469727 FFTOut[5] = 0.0 // Bin 3, this is a negative frequency bin. FFTOut[6] = 0.000331878662 FFTOut[7] = -0.000629425049 The output is composed of pairs of floats, each representing the real and imaginay parts of a single bin. So, bin 0 (array indexes 0, 1) would represent the real and imaginary parts of the DC frequency. As you can see, bins 1 and 3 both have the same values, (except for the sign of the Im part), so I guess these are the negative frequency values, and finally indexes (4, 5) would be the central frequency bin. To attenuate the frequency bin 1 this is what I do: // Attenuate the 'positive' bin FFTOut[2] *= 0.5; FFTOut[3] *= 0.5; // Attenuate its corresponding negative bin. FFTOut[6] *= 0.5; FFTOut[7] *= 0.5; For the actual tests I'm using a 1024-length FFT and I always provide all the samples so no 0-padding is needed. // Attenuate var halfSize = fftWindowLength / 2; float leftFreq = 0f; float rightFreq = 22050f; for( var c = 1; c < halfSize; c++ ) { var freq = c * (44100d / halfSize); // Calc. positive and negative frequency locations. var k = c * 2; var nk = (fftWindowLength - c) * 2; // This kind of attenuation corresponds to a high-pass filter. // The attenuation at the transition band is linearly applied, could // this be the cause of the distortion of low frequencies? var attn = (freq < leftFreq) ? 0 : (freq < rightFreq) ? ((freq - leftFreq) / (rightFreq - leftFreq)) : 1; mFFTOut[ k ] *= (float)attn; mFFTOut[ k + 1 ] *= (float)attn; mFFTOut[ nk ] *= (float)attn; mFFTOut[ nk + 1 ] *= (float)attn; } Obviously I'm doing something wrong but can't figure out what or where.

    Read the article

  • DSP - Filtering in the frequency domain via FFT

    - by Trap
    I've been playing around a little with the Exocortex implementation of the FFT, but I'm having some problems. Whenever I modify the amplitudes of the frequency bins before calling the iFFT the resulting signal contains some clicks and pops, especially when low frequencies are present in the signal (like drums or basses). However, this does not happen if I attenuate all the bins by the same factor. Let me put an example of the output buffer of a 4-sample FFT: // Bin 0 (DC) FFTOut[0] = 0.0000610351563 FFTOut[1] = 0.0 // Bin 1 FFTOut[2] = 0.000331878662 FFTOut[3] = 0.000629425049 // Bin 2 FFTOut[4] = -0.0000381469727 FFTOut[5] = 0.0 // Bin 3, this is the first and only negative frequency bin. FFTOut[6] = 0.000331878662 FFTOut[7] = -0.000629425049 The output is composed of pairs of floats, each representing the real and imaginay parts of a single bin. So, bin 0 (array indexes 0, 1) would represent the real and imaginary parts of the DC frequency. As you can see, bins 1 and 3 both have the same values, (except for the sign of the Im part), so I guess bin 3 is the first negative frequency, and finally indexes (4, 5) would be the last positive frequency bin. Then to attenuate the frequency bin 1 this is what I do: // Attenuate the 'positive' bin FFTOut[2] *= 0.5; FFTOut[3] *= 0.5; // Attenuate its corresponding negative bin. FFTOut[6] *= 0.5; FFTOut[7] *= 0.5; For the actual tests I'm using a 1024-length FFT and I always provide all the samples so no 0-padding is needed. // Attenuate var halfSize = fftWindowLength / 2; float leftFreq = 0f; float rightFreq = 22050f; for( var c = 1; c < halfSize; c++ ) { var freq = c * (44100d / halfSize); // Calc. positive and negative frequency indexes. var k = c * 2; var nk = (fftWindowLength - c) * 2; // This kind of attenuation corresponds to a high-pass filter. // The attenuation at the transition band is linearly applied, could // this be the cause of the distortion of low frequencies? var attn = (freq < leftFreq) ? 0 : (freq < rightFreq) ? ((freq - leftFreq) / (rightFreq - leftFreq)) : 1; // Attenuate positive and negative bins. mFFTOut[ k ] *= (float)attn; mFFTOut[ k + 1 ] *= (float)attn; mFFTOut[ nk ] *= (float)attn; mFFTOut[ nk + 1 ] *= (float)attn; } Obviously I'm doing something wrong but can't figure out what. I don't want to use the FFT output as a means to generate a set of FIR coefficients since I'm trying to implement a very basic dynamic equalizer. What's the correct way to filter in the frequency domain? what I'm missing? Thanks in advance.

    Read the article

1