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HFCC, Hello I have problem with code below

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Michalina Stocka
Michalina Stocka on 19 May 2020
Commented: Rena Berman on 22 Jul 2020 at 18:51
My topic of project is: Analysis of HFCC parameter variability as a function of time for a given signal
I have problem with fft of z(n,t)
Subscripted assignment dimension mismatch.
Error in projekthfcc (line 57)
widmo(t,:) = fft(z(t,:),Nfft);
close all;
clear all;
[s,fp]=audioread('sygnal.wav');
N=length(s);
x=decimate(s,4);
Nx = length(x);
tx = (0:Nx-1)/fp;
%Preemfaza
N = length(x);
y=x(1);
for i=2:N
y(i)=x(i)-0.95*x(i-1);
end
fpr = fp/4;
figure(1)
plot(tx,x);
Ny = length(y);
ty = (0:Ny-1)/fpr;
%t = 0:1/fpr:(length(x)-1)/fpr;
figure(2)
plot(y);
ylabel('Badany sygnal po 4 krotnej decymacji');
xlabel('Czas[s]');
Nr = 25; %frame [ms]
Ns = 10; %step frame [ms]
nr = (Nr*0.001)*fpr; %rame [sample]
ns = (Ns*0.001)*fpr; %step frame [sample]
T = floor((Ny - nr - 1)/ns); %number of frames
Nfft = 2^nextpow2(nr);
N21 = Nfft/2+1;
z1 = zeros(length(0:T-1), length(0:nr-1));
z = zeros(length(0:T-1), length(0:nr-1));
for n = 1:nr
for t = 1:T
z1(t,n) = y(t*ns+n); % framing
z(t,n) = y(t*ns+n) * w(n,nr);% z1 * hamming windows
end
end
widmo = zeros(length(0:T-1), length(0:nr-1));
for t = 1:T
widmo(t,:) = fft(z(t,:),Nfft);
widmo1(t,:) = pow(abs(fft(widmo(t,:),2)));
end
function [ out ] = w( n, Nr )
if n < 1 || n > Nr
out = 0;
return;
end
out = 0.54 - 0.46*cos(2*pi*(n-1)/Nr-1);
end

  2 Comments

Rik
Rik on 1 Jul 2020
Question recovered from Google cache (thanks Michalina Stocka for changing the question title, that makes this easier):
HFCC, Hello I have problem with code below
My topic of project is: Analysis of HFCC parameter variability as a function of time for a given signal
I have problem with fft of z(n,t)
Subscripted assignment dimension mismatch.
Error in projekthfcc (line 57)
widmo(t,:) = fft(z(t,:),Nfft);
close all;
clear all;
[s,fp]=audioread('sygnal.wav');
N=length(s);
x=decimate(s,4);
Nx = length(x);
tx = (0:Nx-1)/fp;
%Preemfaza
N = length(x);
y=x(1);
for i=2:N
y(i)=x(i)-0.95*x(i-1);
end
fpr = fp/4;
figure(1)
plot(tx,x);
Ny = length(y);
ty = (0:Ny-1)/fpr;
%t = 0:1/fpr:(length(x)-1)/fpr;
figure(2)
plot(y);
ylabel('Badany sygnal po 4 krotnej decymacji');
xlabel('Czas[s]');
Nr = 25; %frame [ms]
Ns = 10; %step frame [ms]
nr = (Nr*0.001)*fpr; %rame [sample]
ns = (Ns*0.001)*fpr; %step frame [sample]
T = floor((Ny - nr - 1)/ns); %number of frames
Nfft = 2^nextpow2(nr);
N21 = Nfft/2+1;
z1 = zeros(length(0:T-1), length(0:nr-1));
z = zeros(length(0:T-1), length(0:nr-1));
for n = 1:nr
for t = 1:T
z1(t,n) = y(t*ns+n); % framing
z(t,n) = y(t*ns+n) * w(n,nr);% z1 * hamming windows
end
end
widmo = zeros(length(0:T-1), length(0:nr-1));
for t = 1:T
widmo(t,:) = fft(z(t,:),Nfft);
widmo1(t,:) = pow(abs(fft(widmo(t,:),2)));
end
function [ out ] = w( n, Nr )
if n < 1 || n > Nr
out = 0;
return;
end
out = 0.54 - 0.46*cos(2*pi*(n-1)/Nr-1);
end
Rena Berman
Rena Berman on 22 Jul 2020 at 18:51
(Answers Dev) Restored edit

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Answers (1)

Walter Roberson
Walter Roberson on 19 May 2020
Nfft = 2^nextpow2(nr);
So Nfft will be at least as large as nr, and often larger.
widmo = zeros(length(0:T-1), length(0:nr-1));
length(0:nr-1) is (nr-1) plus 1 for the 0, for a total of nr . So widmo has nr columns.
widmo(t,:) = fft(z(t,:),Nfft);
fft() with Nfft specified tells fft to use Nfft points. We established that Nfft is typically larger than nr, so the fft() would typically produce a vector larger than nr. But you are trying to store it in a row vector that has exactly nr columns.

  3 Comments

Walter Roberson
Walter Roberson on 20 May 2020
I do not understand the point of using length(0:NFFT-1) when that is always going to be the same as NFFT for any non-negative NFFT??
Rik
Rik on 1 Jul 2020
Deleted comment recovered from Google cache:
Thank you for your answer.
I change this part of code:
widmo = zeros(length(0:T-1), length(0:Nfft-1));
for t = 1:T
widmo(t,:) = power(abs(fft(z(t,:),Nfft)),2);
end
and there is no error now, but I'm not sure if this is the correct version. The formula for fft is pictured below.
l - frequency indicator
L - number of spectrum bands
Rik
Rik on 1 Jul 2020
Deleted comment recovered from Google cache:
so "widmo" should be a matrix of what dimensions? and how can I determine the value of "L"? Sorry, maybe there are simple things, but I am a beginner in this ground :/

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