Understanding DFT and Its Applications - Week 6 Lecture Highlights
Explore the essential facts of Discrete Fourier Transform (DFT) and its practical application through examples. Learn how to work out DFT for a given data vector and understand the matrix form representation. Dive into signal processing concepts with explanations and visual aids from the online lecture.
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Do you have any question about the video of week-6? We have talked about DFT and its applications
DFT I (Essential facts) N-1 x[n+1]exp(-j2pk X(k)= X(2pk / N)= n) N n=0 Rewritten it in matrix form, assume the data is x[0], x[1], , x[N-1] 1 1 ... 1 X[0] X[1] .. X[N -1] x[0] x[1] .. x[N -1] exp -2p j exp -2p j (N -1) 1 ... N N = ... ... exp -2p j(N -1) 1exp -2p j(N -1) (N -1) N N DFT DATA
Example Let (x[0], x[1], x[2], x[3]) = (1 2 -1 -1) be a data vector of length N=4. Work out its DFT (X[0], X[1], X[2], X[3]) N-1 x[n+1]exp(-j2pk X(k)= X(2pk / N)= n) N n=0 X[0] = 1+2 -1-1=1 X[1] = exp(-0j2pi/4) + 2 exp(-1 j 2 pi/4) -1 exp(-2 j 2 pi/4) -1 exp(-3 j 2 pi/4 = 1 + 2 exp(-j pi /2) (exp(-j pi/2 ))^2 (exp(-j pi/2)) ^3 = 1 + 2 (-j)-(-j)^2-(-j)^3 = 1-2j+1+(-j)=2-3j exp(-j pi /2) = cos( pi/2) j sin(pi/2) = 0 j 1 Any question?
Example Let (x[0], x[1], x[2], x[3]) = (1 2 -1 -1) be a data vector of length N=4. Work out its DFT (X[0], X[1], X[2], X[3]) 3 2pk N X[k]= x[n]exp(-j n) n=0 X[0]=1+2-1-1=1 3 2p1 4 n)= exp(-j2p1 0)+2exp(-j2p1 1)-1exp(-j2p1 2)-12exp(-j2p1 X[1]= x[n]exp(-j 3) 4 4 4 4 n=0 =1+2exp(-jp 2)-exp(-jp 2)-[exp(-jp 22)-exp(-jp 2)]2- [exp(-jp 23) =1+2exp(-jp 2)]3 =1+2(-j)-(-j)2-(-j)3=1-2j+1- j = 2-3j 3 2p2 4 X[2]= x[n]exp(-j n)=1+2(-j)2-((-j)2)2-((-j)3)2=1-2-1+1=-1 n=0 3 2p3 4 X[3]= x[n]exp(-j n)=1+2(-j)3-((-j)2)3-((-j)3)3=1+2j+1+ j = 2+3j n=0
x = [1 2 -1 -1] fft (x)
clear all close all sampling_rate=100;%Hz omega=30; %signal frequecy Hz N=20000; %total number of samples for i=1:N x_sound(i)=cos(2*pi*omega*i/sampling_rate); %signal x(i)=x_sound(i)+2*randn(1,1); %signal+noise axis(i)=sampling_rate*i/N; % for psd time(i)=i/sampling_rate; % for time trace end subplot(1,2,1) plot(time,x); %signal + noise, time trace xlabel('second') ylabel('x') subplot(1,2,2) plot(axis,abs(fft(x)),'r'); % magnitude of signal xlabel('Hz'); ylabel('Amplitude') sound(x_sound)
t=0:0.001:2; % 2 secs @ 1kHz sample rate y=chirp(t,100,1,200,'q'); % Start @ 100Hz, cross 200Hz at t=1sec spectrogram(y,128,120,128,1E3); % Display the spectrogram title('Quadratic Chirp: start at 100Hz and cross 200Hz at t=1sec'); sound(y) y=chirp(t,100,1,200,'q'); % Start @ Fs=5000; filename = 'h.wav'; audiowrite(filename,y,Fs); The first one is in theory, it is ideal. The second one is dealing with our daily life example.
for i=1:T(1,1); zn(i,1)=y(i,1)+.3*randn(1,1); end figure(8) plot(zn); xlabel('time') figure(9) plot(fre_x,abs(fft(zn))) xlabel('Hz') figure(10) spectrogram(zn,128,120,[],Fs); close all clear all load handel; figure(1) plot(y); xlabel('time') T=size(y); for i=1:T(1,1) fre_x(i)=i*Fs/T(1,1); end figure(2) plot(fre_x, abs(fft(y))) xlabel('Hz') figure(3) spectrogram(y,128,120,[],Fs); T1=floor(T(1,1)/2); omega1=1000; for i=1:T(1,1)-T1; z(i,1)=y(i,1)+.3*cos(2*pi*omega 1*i/Fs); end omega2=2000; for i=T-T1+1:T(1,1); z(i,1)=y(i,1)+.3*cos(2*pi*omega 2*i/Fs); end figure(5) plot(z); xlabel('time') figure(6) plot(fre_x,abs(fft(z))) xlabel('Hz') figure(7) spectrogram(z,128,120,[],Fs); ;
Power spectrum of white noise Can you prove that the power spectrum of white noise is flat? = 1 2 exp( ] 1 [ ( | x[k] | n k j m x N 1 2 k = = + X ( k ) X ( 2 k / N ) x [ n 1 ] exp( j n ) N n 0 N 2 k = = + x n j n )) N 0 N 1 2 = N + * ( [ 1 ] exp( m )) N m 0 1 2 k 2 k = 0 n = + + x [ m 1 ] exp( j m ) x [ n 1 ] exp( j n ) N N m Do you have any question here? E x[n]x[m] = 0 if n neq m
N 1 2 k = n = + 2 E | x[k] | E ( x [ n 1 ] exp( j n )) N 0 N 1 2 k = N + * ( x [ m 1 ] exp( j m )) N m 0 1 2 k 2 k = N = + + E x [ m 1 ] exp( j m ) x [ n 1 ] exp( j n ) N N n 0 m 1 = n = + = [ 2 E x n 1 ] N 0
