Mirror Images when I use FFT analysis
in [Matlab]
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From: TideMan on 27 May 2010 04:23 On May 27, 5:56 pm, "Abdullah " <zabdull...(a)hotmail.com> wrote: > hi , > Strange!! > > this time instead of obtaining a spectrum for the signal of 1.5 Mhz and obtaining 3 Mhz even (2 I get 4 MHz) strange!!. > > I used the suggested script (Tideman) , please help me ?? > > clc; > clear all > Fo=10e6; > To=1/Fo; > t=0:To/50:5*To; > xt=sin(2*pi*150000000*t); > L=length(xt); > figure(1); > subplot(311); > plot(t,xt,'r');grid; > xlabel('Cosine Signal xt = cos(4000*pi*t)'); > > figure(2) > > N = 2^nextpow2(L); % Next power of 2 from length of vdata > X = fft(xt,N); > M =abs(X(2:N/2)); > f=(1:N/2-1)*Fo/N; > > plot(f,M); > > Regards and Thanks There's nothing strange about it. I've already told you where you're wrong (Point 3), but you haven't taken a blind bit of notice, you're continuing to use the freq of the sine wave, not the sampling freq. I don't have the patience to deal with people who ignore advice and continue to do the wrong thing, then complain that things are strange. Also, you don't need to use dyadic numbers for the Matlab FFT, so you can omit this line and just use L: N = 2^nextpow2(L); % Next power of 2 from length of vdata
From: Omer on 27 May 2010 07:24 On May 27, 10:56 am, "Abdullah " <zabdull...(a)hotmail.com> wrote: > hi , > Strange!! > > this time instead of obtaining a spectrum for the signal of 1.5 Mhz and obtaining 3 Mhz even (2 I get 4 MHz) strange!!. > > I used the suggested script (Tideman) , please help me ?? > > clc; > clear all > Fo=10e6; > To=1/Fo; > t=0:To/50:5*To; ** This corresponds that your sampling frequency is 50*Fo = 500MHz > xt=sin(2*pi*150000000*t); ** This is not 1.5 MHz rather it is 150 MHz > L=length(xt); > figure(1); > subplot(311); > plot(t,xt,'r');grid; > xlabel('Cosine Signal xt = cos(4000*pi*t)'); ** Strange Label > > figure(2) > > N = 2^nextpow2(L); % Next power of 2 from length of vdata > X = fft(xt,N); > M =abs(X(2:N/2)); > f=(1:N/2-1)*Fo/N; ** Use correct sampling frequency which is 500MHz in this case and all your problems will be solved. > > plot(f,M); > > Regards and Thanks
From: Abdullah on 27 May 2010 23:49 Thank you very much "Tideman" You know the problem was with this formular t=0:To/50:5*To; it changes the parameters that I put for Fo and frequency of sin signal. I did not realize that until Omer confirms that. Regards & thanks,
From: Abdullah on 27 May 2010 23:52
Omer <omer.rajput(a)gmail.com> wrote in message <e735faa3-8e29-4868-9457-d2eac31f9903(a)q36g2000prg.googlegroups.com>... > On May 27, 10:56 am, "Abdullah " <zabdull...(a)hotmail.com> wrote: > > hi , > > Strange!! > > > > this time instead of obtaining a spectrum for the signal of 1.5 Mhz and obtaining 3 Mhz even (2 I get 4 MHz) strange!!. > > > > I used the suggested script (Tideman) , please help me ?? > > > > clc; > > clear all > > Fo=10e6; > > To=1/Fo; > > > t=0:To/50:5*To; > ** This corresponds that your sampling frequency is 50*Fo = 500MHz > > > xt=sin(2*pi*150000000*t); > ** This is not 1.5 MHz rather it is 150 MHz > > > L=length(xt); > > figure(1); > > subplot(311); > > plot(t,xt,'r');grid; > > > xlabel('Cosine Signal xt = cos(4000*pi*t)'); > ** Strange Label > > > > > figure(2) > > > > N = 2^nextpow2(L); % Next power of 2 from length of vdata > > X = fft(xt,N); > > M =abs(X(2:N/2)); > > > f=(1:N/2-1)*Fo/N; > ** Use correct sampling frequency which is 500MHz in this case and all > your problems will be solved. > > > > > plot(f,M); > > > > Regards and Thanks Thank you very much Omer about reply They are solved!!, as you said. Regards &thanks |