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.

Cheers
Matteo

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”.

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