How can i get the freq plot for 2-D signal i.e. image as we are not knowing that what is max freq component in image.

]]>I’d also like to suggest some simplifications to your code. You don’t need those loops to create your signals x1 and x2.

n = 0:49; x1 = 10*sin(2*pi*n*f1/F1); x2 = 10*sin(2*pi*n*f2/F1);

Also, I suggest that you not use “clear all”. If you want to clear your base workspace variables, just use “clear”.

]]>with my code i cant see overlapping of spectrums??

code:

clear all; f1=10;F1=100;f2=90;F2=100; for n=1:50, x1(n)=10*sin(2*pi*n*f1/F1); fr1(n)=n/25*F1/2; end y1=fft(x1,50); plot(fr1,abs(y1)); xlabel('frequency in Hz'); ylabel('Amplitude'); figure; for n=1:50, x2(n)=10*sin(2*pi*n*f2/F1); fr2(n)=n/25*F1/2; end y2=fft(x2,50); plot(fr2,abs(y2)); xlabel('frequency in Hz');]]>

Its actually simple to relate what you have learnt with that Steve blogged. Here it is how: the two frequencies were 1 and 2pi-1 which is 1 & 5.28 rad/s. Now remember what sampling theorem says – that freq of sampling Fs must be at least twice that of the highest frequency in the signal, right? Steve’s example didn’t honour sampling theorem as Fs was only 1 rad/s (T=1) and hence ended up with aliasing effect.

]]>First let me say I really appreciate and enjoy you explaining the Fourier transforms, wish they would have taught us this in Math in such a detail.

I’m at a loss with the way you described the alaising problem. I have come to know it as a problem, which occurs when the sampling rate of your discreet signal is not at least twice the amount of the highest exisiting frequency. But I can’t see how this fits in with the alaising you discribed above. Would you be so kind and explain that alaising thing next time you got time?

Greetings from Germany

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