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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

Digital Signal Processing :: 11-18-2013 20:24 :: LarryJ :: Replies: **1** :: Views: **626**

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.

Digital Signal Processing :: 02-25-2013 05:45 :: MikuIyer :: Replies: **1** :: Views: **997**

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

Digital Signal Processing :: 09-10-2011 03:09 :: yakex :: Replies: **3** :: Views: **742**

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

Digital Signal Processing :: 04-26-2011 14:06 :: moonnightingale :: Replies: **0** :: Views: **1233**

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=i**fft**(f2);
%finding the strongest cepstral coefficient
pitch=max(f3);
display(pitch);
Hope it

Digital Signal Processing :: 04-01-2011 04:25 :: pushpanjali sharma :: Replies: **2** :: Views: **1183**

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

Analog Circuit Design :: 02-01-2011 03:53 :: vigyanjain :: Replies: **0** :: Views: **2672**

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***fft**shift(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)

Digital Signal Processing :: 01-01-2011 12:14 :: dorome :: Replies: **0** :: Views: **623**

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=**fft**shift(**abs**(**fft**(y,N)));
k=(-N/2:N/2-1)*(fs/N);
plot(k,F);
grid on;
But I tried something but, I never (...)

PC Programming and Interfacing :: 12-22-2010 19:37 :: ravitejaelx :: Replies: **0** :: Views: **1528**

you are making the following transformation:i**fft**(**abs**(**fft**(waveform2)))instead of: i**fft**(**fft**(waveform2))The **abs** function changes the **fft** of waveform2...

Digital Signal Processing :: 12-07-2010 01:00 :: JoannesPaulus :: Replies: **9** :: Views: **19695**

Try,
t=0:Ts:10;
y=x1+x2+x3;
plot(t,**abs**(**fft**(y)));

Software Problems, Hints and Reviews :: 11-16-2010 09:45 :: anmol :: Replies: **2** :: Views: **1431**

how to get frequency spectrum of EEG signal using matlab?**abs**(**fft**(x)) is not giving good plot

Digital Signal Processing :: 10-20-2010 04:46 :: thasneem1 :: Replies: **5** :: Views: **2579**

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 (...)

Digital Signal Processing :: 07-04-2010 12:08 :: elcink :: Replies: **4** :: Views: **8317**

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.

Digital communication :: 07-01-2010 22:36 :: RBB :: Replies: **9** :: Views: **2029**

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...

Digital Signal Processing :: 06-17-2010 03:09 :: JoannesPaulus :: Replies: **3** :: Views: **962**

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

Digital Signal Processing :: 05-06-2010 01:24 :: virginiatech :: Replies: **3** :: Views: **5158**

Have a good slap: plot (**abs**(Y1)) ;-)
The **fft** returns an imaginary number (magnitude an phase).

Digital Signal Processing :: 04-26-2010 05:04 :: JoannesPaulus :: Replies: **2** :: Views: **1262**

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) );

Analog Circuit Design :: 02-17-2010 11:15 :: hur :: Replies: **1** :: Views: **2509**

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...

Digital Signal Processing :: 01-29-2010 14:09 :: amgc :: Replies: **4** :: Views: **3621**

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,N**fft**);
plot(f,**abs**(**fft**(Data,N**fft**)));
grid on;
Is this what you are looking for??
HTH,
-- Ashwini

Digital Signal Processing :: 12-11-2009 09:47 :: ashwini1 :: Replies: **2** :: Views: **1636**

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 i**fft** to recontruct the image field (x2) at different z distance values.
I create the amplitude image by using command
Intensity = **abs**(X).^2; where (...)

Digital Signal Processing :: 11-12-2008 11:03 :: niw :: Replies: **0** :: Views: **3225**

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