what is difference between fft(x) & fft(x,n)

clc; close all; clear all;

fc=1000; fs=15000; f1=1/fs; t=[0:f1:1-f1]; x=sin(2*pi*fc*t); figure; plot(t(1:100),x(1:100)); xlabel('time'); ylabel('amplitude'); title('Sine wave of Frequency 1 KHz');

fc=2000; f1=1/fs; t=[0:f1:1-f1]; y=sin(2*pi*fc*t); figure; plot(t(1:100),y(1:100)); xlabel('time'); ylabel('amplitude'); title('Sine wave of Frequency 2 KHz');

z=x+y; figure; plot(t(1:100),z(1:100)); xlabel('time'); ylabel('amplitude'); title('Mixed signal');

z1=fft(z,4096); len=length(z1); zdft = (1/len)*z1; freq=0:(fs/len):fs/2-(fs/len); figure; plot(freq,abs(zdft(1:len/2))) xlabel('Frequency'); ylabel('amplitude'); title('Single sided Frequency Spectrum');

In this code at z1 if we use fft(x) i am getting 2Hz width at 2K frequency in frequency spectrum,But if i use fft(x,4096) i am getting width of 20-30 Hz at 2K frequency....Why is this so , Can any one explain the reason.

[Rick Rosson]

Please format your code. Thanks.

Answer 1

[Dishant Arora]

fft(x) computes discrete fourier transform having as many samples as the original signal x whereas fft(x,n) will have n points in dft. Usually n is taken as integral power of 2 to fasten the computation. In order to recover original signal from fft, number of samples in fft must be greater than or equal to number of samples in original signal . As n increases resolution in frequency domain increases i.e the distance between two consecutive frequency bins decreases .

Answer 2

[Rick Rosson]

Compare:

   >> dF_1 = fs/length(z)

with:

   >> dF_2 = fs/length(z1)

[Kiki]

Hi

I had a similar question. My codes are as follows.

Fs = 1000;            % Sampling frequency
T = 1/Fs;             % Sampling period
L = 1000;             % Length of signal
t = (0:L-1)*T;        % Time vector
S = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
n=L;
Y2 = fft(S,n);
P22 = abs(Y2/n);
P11 = P22(1:n/2+1);
P11(2:end-1) = 2*P11(2:end-1);
f2 = Fs*(0:(n/2))/n;
figure(4)
plot(f2,P11)
title('1000 Point DFT Single-Sided Amplitude Spectrum of S(t)')
xlabel('f (Hz)')
ylabel('|P11(f)|')

n=4096;
Y2 = fft(S,n);
P22 = abs(Y2/n);
P11 = P22(1:n/2+1);
P11(2:end-1) = 2*P11(2:end-1);
f2 = Fs*(0:(n/2))/n;
figure(5)
plot(f2,P11)
title('1024 Point DFT Single-Sided Amplitude Spectrum of S(t)')
xlabel('f (Hz)')
ylabel('|P11(f)|')

n=1000 gives me a perfect spectrum with narrowest main lobe and 0 side lobes, while n=1024 gives me a wider main lobe and some side lobes. I thought bigger N would give us better frequency solution, but why would it add some "noise" to it?

Thank you very much! Kiki