fft

问题 How can I use the fourier transform to find out the frequency components which are responsible for the texture on the surface? Then I have to remove them to have a smooth surface without texture. This is the image. Thanks 回答1: If you use Fourier Transform and analyze frequency components, then removing high frequency components of the image gives a close effect of low pass filter. However, it does not seem natural since you also manipulate the phase of the image. As people suggest, I also

问题 I use the matlab software. To my question. I have a audio signal, on which i am applying a STFT. I take a segment (46 ms, specifially chosen) out of my signal y(audio signal) and use a FFT on it. Then i go to the next segment, until to end of my audio signal. My WAV-File is 10.8526 seconds long. If I have a sample frequency of 44100Hz, this means my y is 10.8526*fs = 478599.66 which is shown in the workspace as 478 6000 x2 double . The length of my fft is 2048 . My signal are differentiated

问题 Are there known algorithms which will take a big integer with n digits encoded in one base/radix and convert it to another arbitrary base? (Let's say from base 7 to base 19.) n can be really big, like more than 100 000 digits, so I am looking for something better than O( n 2 ) run time. I have seen some algorithms that can multiply two huge integers using the Fast Fourier Transform (FFT), with the theoretical complexity of O( n log n ), where n is the number of digits, so I wonder if

问题 Here is my m-file for Fourier series plot: clear clc syms n a0=input('Enter coefficient a0: '); an=input('Enter coefficient an: '); bn=input('Enter coefficient bn: '); a=input('Enter lower boundary: '); b=input('Enter upper boundary: '); t=linspace(a,b,10000); suma=0; for n=1:10 %% n could be any number, bigger n - better approximation suma=suma+(subs(an,'n',n).*cos(2.*n.*pi.*t./(b-a))+subs(bn,'n',n).*sin(2.*n.*pi.*t./(b-a))); end series=a0+suma; plot(t,series) grid Problem is, it is so slow!

问题 After taking the Discrete Fourier Transform of some samples with scipy.fftpack.fft() and plotting the magnitude of these I notice that it doesn't equal the amplitude of the original signal. Is there a relationship between the two? Is there a way to compute the amplitude of the original signal from the Fourier coefficients without reversing the transform? Here's an example of sin wave with amplitude 7.0 and fft amplitude 3.5 from numpy import sin, linspace, pi from pylab import plot, show,

问题 I want to analyze a signal I get from my microphone port by using portaudio and fftwpp . For that I followed the explanation provided here. My questions concerning that are now: There it is stated that I should chunk a window out of the incoming data. My data is already chunked, after I am only recording for a short time, and afterwards process it. Thus I am assuming that a rectangular window is already applied to my data. Is that correct? Now I am getting 200k data points, should I directly

问题 I'm trying to plot a simple signal in fourier domain using Matlab. It's not plotting the correct signal. Here is my code: clc; clear all; close all; x=1:0.001:10; f1=sin(2*pi*10*x); f2=sin(2*pi*15*x); f3=sin(2*pi*30*x); f=f1+f2+f3; plot(2*pi*x,fft(f1)); figure plot(x,fft(f1)); I've expected a peak at 10 since the frequency is 10. But it is giving a peak at some other point Here are the two plot images: This is the image for plot(x,fft(f1)) This is the image for plot(2*pi*x,fft(f1)) It is not

问题 From the question: How do I obtain the frequencies of each value in an FFT? I have a similar question. I understand the answers to the previous question, but I would like further clarification on frequency. Is frequency the same as the index? Let's go for an example: let's assume we have an array (1X200) of data in MATLAB. When you apply 'abs(fft)' for that array it gives the same size array as the result (1X200). So, does this mean this array contains magnitude? Does this mean the indices of

问题 According to the convolution theorem, a convolution in the time domain is a product in the fft domain. With correct zero-padding, it works: % convolution in time domain a = [1 2 3]; b = [4 5 6]; c = conv(a,b); a_padded=[a 0 0]; b_padded=[b 0 0]; c_bis=ifft(fft(a_padded).*fft(b_padded)); % we do find c_bis=c However, this theorem is suposed to work the other way around as well, a product in the time domain is a convolution in the fft domain. I dont get this part: d = a.*b; D=conv(fft(a_padded)

问题 I am trying to create an application for calculating coefficients for a graphic equalizer FIR filter. I am doing some prototyping in Matlab but I have some problems. I have started with the following Matlab code: % binamps vector holds 2^13 = 8192 bins of desired amplitude values for frequencies in range 0.001 .. 22050 Hz (half of samplerate 44100 Hz) % it looks just fine, when I use Matlab plot() function % now I get ifft n = size(binamps,1); iff = ifft(binamps, n); coeffs = real(iff); %