Search Engine www.edaboard.com

Abs Fft

Add Question

32 Threads found on edaboard.com: Abs Fft
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 (...)
If you change the line: m=abs(y); to m=abs(y2); then the fft plot will go from -Fs/2 to Fs/2. Thus you'll see the DC component in the middle of the plot, and you wanted signal on either side of the DC signal.
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...
Yeah I tried that already but for some reason it's not coming out right. This is my code thus far: Fs = 0.01; % sampling frequency T = 1/Fs; %sample time N = length(data); %length of signal t = (0:N-1)/N; % define time t = t*T; % define time in seconds realData = data(1,:); imagData = data(2,:); FinalData = realData + 1i.*imagData; f
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 (...)