# The relation between fft(x,8)/8 and fft(x,4)/4

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zohar on 17 May 2011
Dear all matlab users,
I wrote this code
l = 128;
t = 0:l-1;
fs = 32;
f = 32/128 * 17; % f=4.25 Hz
x = sin(2*pi*f/fs*t);
l1 = 128;
l2 = 64 ;
x1 = fft(x,l1)/l1;
x2 = fft(x,l2)/l2 ;
fres1 = fs/l1;
fres2 = fs/l2;
figure(2);
plot([0:length(x1) - 1]*fres1,abs(x1), 'b');
hold('on');
plot([0:length(x2) - 1]*fres2,abs(x2), 'r');
hold('off');
What is the relation between x1 and x2 ?

#### 1 Comment

Arturo Moncada-Torres on 17 May 2011
Much better explained ;)

Daniel Shub on 17 May 2011
The scale factor and the number of frequency components.

#### 1 Comment

zohar on 17 May 2011
Sorry, but i don't understand.
The length of x1 is 128 and the length of x2 is 64.
What is the scale factor and the number of frequency components ?
Thanks

Arturo Moncada-Torres on 17 May 2011
OK, I suggest doing a quick plot:
figure(1);
plot(x);
figure(2);
hold('on');
plot(abs(x1), 'b');
plot(abs(x2), 'r');
hold('off');
You can see that even though it appears that x1 and x2 are different, they are actually the same. This is logical, since you are obtaining the FFT of the same signal. The only difference is that since in x1 you use 128 points and in x2 you use 64 points, the FFT of x1 has a better resolution in the frequency domain (look here for a detailed explanation).
The division between 128 and 64 is just to normalize them and make a fair comparison (try plotting without the division and you will see what I mean).

#### 1 Comment

zohar on 18 May 2011
Arturo Moncada Torres Thanks for the info.
I understand what is frequency resolution , but I want to understand
what is the relation between x1 and x2, I think answer related to dirichlet function ;-(
I rearrange my question.