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Hello, In the audio processing field, after applying fft on a sound signal, sometimes, engineers use 10*log10(abs(value)) and some others just sqrt(i**2 + j**2) to get the amplitude of frequencies. Why one instead of the other ? Thanks, Larry
Hi, Set the range for 'n' (as we are in discrete time domain), say n = Compute values for the given signal x in the above taken range. For spectral analysis, compute fft using fft(signal, N) and plot the abs value to get magnitude spectrum. Let me know if you need the code.
hi, I have adopt the ADS5560 for sampling, and the sample sequence is expresses as X in matlab. Then I want to calculate the power/energy spectrum in decibel, I have a problem whether to choose 20 or 10 as the multipy coefficient, i.e., 20*log10(abs(fft(X)) or 10*log10(abs(fft(X)) thanks to all. --yakex
I am having a periodic impulse train of N=5 y=ones(1,5) i want to decompose it into complex exponentials by writing a matlab code. Can anybody guide me how to write a matlab code the aswer of code should be same as of builtin command stem(abs(fft(y,5))) thanks a lot
I have written the cepstral method to determine the pitch of a voice signal which is much simpler. =wavread('/home/..specify where your speech signal is located...'); %determination of cepstral coefficients f=fft(x); f1=abs(f); f2=log(f1); f3=ifft(f2); %finding the strongest cepstral coefficient pitch=max(f3); display(pitch); Hope it
Here are the steps which I normally follow: 1. Choose sufficient number of points of output signal. I usually take points, k such that n = (log k) / (log 2), where n is an integer. 2. Take fft of k. 3. Take abs value of the step-2 result. 4. Multiply each element of step 3 by itself. e.g. step4 = step3.^2 5. The max point available fro
plzzzzzz help me i dont now what its wrong in my m file ///////////////////////////////////////// =wavread('NoisySignal.wav'); figure(1) plot(x) Ts=1/fs; y=fft(x); y=Ts*fftshift(y); f=-fs/2:fs/length(y):fs/2-fs/length(y); figure(2) plot(f,abs (y)) x1=.4*(hardlim(f+3050))-.4*(hardlim(f+2950)); figure(3) plot(f,x1)
Hello All, Please solve my problem, I want to find fft plot for a 50 Hz and 5V sinusoidal signal. So I tried the following code.. fm=50; fs=150; N=512; t=0:1/fs:100; y=sin(2*pi*(fs/fm)*t); F=fftshift(abs(fft(y,N))); k=(-N/2:N/2-1)*(fs/N); plot(k,F); grid on; But I tried something but, I never (...)
you are making the following transformation:ifft(abs(fft(waveform2)))instead of: ifft(fft(waveform2))The abs function changes the fft of waveform2...
Try, t=0:Ts:10; y=x1+x2+x3; plot(t,abs(fft(y)));
how to get frequency spectrum of EEG signal using matlab?abs(fft(x)) is not giving good plot
Hi, i want to see the spectrum of the output in a pll. In order to do so, i have put 5 blocks in series: the first is the Zero-Order-Hold (its sampling time is more than twice the main frequency of the output), the second is a Buffer, the third is the fft block(having fft length equal to that of the buffer), the fourth is the abs block and (...)
Hi members, In order to recongnize the differentes harmonics of a function I wrote a matlab script to calculate the DFT. In order to verfy that may script is correct I applyed this function to a simple sinus function. Here is the code Fe=8E9 %Sampling frequency N=8 %8 samples df=Fe/N %step t=(1:N)/Fe %the range f0=1E9 %Frequ
You need to know the sampling frequency, Fs. Then you can plot the spectrum as follows: =wavread('sound.wav'); N=length(x); X=fft(x); f=(0:N/2-1)*Fs/N; plot(f,abs(X(1:N/2)))You might have to adjust the vector sizes, using floor if your number of points is not even, but I hope you get the point...
First of all, you should not use abs for the ifft. Second, is the input data coming from an fft or is it generated by some software? In order to calculate the ifft, you need an array of complex data where the second N/2 samples are conjugate of the first N/2 samples (and flipped). Here is the formula: (...)
Have a good slap: plot (abs(Y1)) ;-) The fft returns an imaginary number (magnitude an phase).
N = 2048; cycles = 67; fs = 1000; fx = fs*cycles/N; LSB = 2/2^10; %generate signal, quantize and take fft x = cos(2*pi*fx/fs*); x = round(x/LSB)*LSB; s = abs(fft(x)); s = s(1:end/2)/N*2; % calculate SNR f = /N; sigbin = 1 + cycles; noise = ; snr =10*log10( s(sigbin)^2/sum(noise.^2) );
I got the samples of various arrhythmias and i took the fft of it using matlab.. Its found to be different... The dominent frequencies are found to be varying for various arrhythemia which i understood while plotting its abs value... Is it possible to extract those dominent frequencies??? that is frquency with max value...
x=5sin(2*pi*50*t)+3sin(2*pi*100*t) W=hann(length(x)); Data=x.*W; figure; f=(sampling_freq)*linspace(0,1,Nfft); plot(f,abs(fft(Data,Nfft))); grid on; Is this what you are looking for?? HTH, -- Ashwini
Hi, I am trying to obtain phase and amplitude images by using Fresnel tranform. In my matlab program, I call the saved image (acquired by a CCD), use fft and ifft to recontruct the image field (x2) at different z distance values. I create the amplitude image by using command Intensity = abs(X).^2; where (...)
a simple script segment to do fft for a tone plus additive white noise. nbpts = 32768; fs = 1e3*nbpts; nbprds = 7; fi = fs/nbpts*nbprds; amplitude = 1; pnoise = 1e-6; noise = random('norm',0,sqrt(pnoise),1,nbpts); signal = amplitude*sin(2*pi*fi/fs*) + noise; Sxx = abs(fft(signal)./nbpts).^2; Snn =
try plotting abs(fft(signal))... you will get the frequency spectrum of the signal... this will clearly show you the side bands , carrier frequency... etc...
>> freq = fft(impulseTrain, Len); >> stem(, abs(freq)) it shows a peak at 0 frequency, that result is correct. what happens is the fft just gives u the first window, its made to give only the non repeating part... actually every discrete signal is a periodic in frequency domain. so the actual fft that u obtained(which is
you can see fft of x as y=fft(x,512); df=fs/512; m=fftshift(abs(y)); m=m.^2; m1=20*log10(m); p=unwrap(angle(y)); f = -(fs/2)+df:df:(fs/2); %f=(0:length(y)-1)*fs/length(y); figure; plot(f,m1);title('magnitude')
hi, please help me in these statements (1) , ( 2 ), ( 3 ) only , if anyone knows this........... N = 512; dt = 1/fs; %sampling time T = 1; t = 0:dt:1; y=textread('......txt'); Y = fft(y,N); Pyy = 2*abs(Y) / min(N,T/dt); ( 1 ) mag1
I assume you want MATLAB to compute the fft. Try typing a column or row of numbers into Excel, save it to "xdata.xls", and then run this MATLAB command: plot(abs(fft(xlsread('xdata.xls')))); MATLAB provides other methods for accessing Excel data, including DDE transfers to/from the Excel window. Try searching MATLAB help for "excel".
It sometimes depends on who's fft function you are using. Here is a MATLAB example. Notice the fs/N and 2/N. If you are unfamiliar with MATLAB, the abs() function returns the magnitude of a complex number. N = 64; % number of points fs = 4000; % sample rate f1 = 750; % signal 1 frequency f2 = 1000; % signal 2 f
Hi why dont include your code, it is difficult to know what you are really doing. In any case it should be something like this fftshift(abs(fft(x))) where x is has the values of your signal if you think this info is useful click on "helped me" for statistics Sal
I create a sine-wave in matlab and do fft fbin = 11; n = 512; u = 1/2 * sin(2*pi*fbin/n*); w = hann(n); w1 = norm(w,1); U = fft(w.*u)/(w1); % scale for sine-wave psd = abs(U).^2; pow = sum(psd); but I get the wrong answer, pow supposed to be 0.125 which is 0.5^2/2. The reason is that there are two extra non-ze
Unless some adjustment in frequency axis, you can start trying the code below. >> w1=5; >> t=0:1/7:8; >> ws=7; >> x=sin(w1*t); >> Xw=abs(fftshift(fft(x))); >> Ws=*ws; >> plot(Ws,Xw)
Hi Yes, the fft results contain both amplitude and phase information. To plot only the amplitude try abs( fft( ... ) ); Another useful MATLAB function for psd plotting is pwelch. Regards
little matlab demo for fractional fft...enjoy N=64; t=2*pi/N*;c=1; for k=-N/2:N/2 B(c,:)=1/sqrt(N+1)*exp(-j*k*t);c=c+1;end; f=exp(-t.^2); f=f-min(f); f=f'; figure(1);clf;set(gcf,'Renderer','zbuffer'); for n=0:0.1:2.0 Bn=B^n;fn=abs(Bn*f);plot(fn,'b');hold on;plot(f,'r--');hold off; text(10,1.7,sprintf('%0.2f',n));