how can i add Uniformly distributed noise between 44-55 [hz] to a speech signal?

i tried this one but i'm getting an error

Error using randi
Requested 407792x407792 (1239.0GB) array exceeds maximum array size preference. Creation of arrays greater than this limit
may take a long time and cause MATLAB to become unresponsive. See array size limit or preference panel for more information.

clear all;close all;clc
[x,fs,Nbits] = wavread('jennifer.wav');
l=length(x);
y=randi([44,55],l)+x;

Answer 1

[Star Strider]

This creates an (lxl) matrix of random integers between 44 and 55:

y=randi([44,55],l)+x;

That is not what you want to do anyway.

To create uniformly-distributed noise between 44 and 55 Hz, try this:

[x,Fs,Nbits] = wavread('jennifer.wav');

Fn = Fs/2;                                              % Nyquist Frequency (Hz)
Wp = [44   55]/Fn;                                      % Passband Frequencies (Normalised)
Ws = [43   56]/Fn;                                      % Stopband Frequencies (Normalised)
Rp = 10;                                                % Passband Ripple (dB)
Rs = 50;                                                % Stopband Ripple (dB)
[n,Ws] = cheb2ord(Wp,Ws,Rp,Rs);                         % Filter Order
[z,p,k] = cheby2(n,Rs,Ws);                              % Filter Design
[sosbp,gbp] = zp2sos(z,p,k);                            % Convert To Second-Order-Section For Stability

figure(3)
freqz(sosbp, 2^17, Fs)

L=length(x);
noise = rand(1, L);                                     % Normally-Distributed Random Vector
noise_filt = filtfilt(sosbp, gbp, noise);               % Filter Noise Signal
noisy_signal = x + noise_filt;

The filter code works, however I have not used this with your sound file. Note that the output of wavread and audioread scale (or normalise) the output to a maximum amplitude of ±1, so use the rand function, since its output is [0,1]. You may want to reduce the noise amplitude even further to avoid completely ‘swamping’ your audio signal in the noise.

[rivaldo rivaldo]

thank you.

how is this solution-take the original signal of speech and add it to a sum of noisy sin like this:

clear all;close all;clc

t =0:0.01:512;

[x,fs,Nbits] = wavread('jennifer.wav');

nos=sin(2*pi*45*t)+sin(2*pi*46*t)+sin(2*pi*47*t)+sin(2*pi*48*t)+sin(2*pi*49*t)+sin(2*pi*50*t)+sin(2*pi*51*t)+sin(2*pi*52*t)+sin(2*pi*53)+sin(2*pi*54*t)+sin(2*pi*55*t);

y=x+nos;

figure(1);

plot(y);

i want to check it it gives an error:

Error using +

Matrix dimensions must agree

[Star Strider]

My pleasure.

That was not in your original Question!

The error comes from your not including multiplication by ‘t’ in this term:

sin(2*pi*53)

You can create ‘nos’ much more easily by using element-wise vector multiplication:

t = 0:0.01:512;
f = 44:55;
nos = sum(sin(2*pi*f'.*t));

figure(1)
plot(t, nos)
grid
axis([0  0.5    ylim])

[rivaldo rivaldo]

yeah i know that was not my original question.

I'm just trying fegure out what will the most valuebale solution for this and understand it.

still gives an error of dimmensions with the last code:

Error using .*

Matrix dimensions must agree.

[Star Strider]

It worked for me or I’d not have posted it.

Did you copy and paste my code to your script and run it? Note that the ‘f’ vector is transposed (the ‘'’ operator). My code assumes both ‘t’ and ‘f’ are originally row vectors. A slightly different calculation for ‘nos’ that does the same thing is:

nos = sum(sin(2*pi*f(:).*t));

The ‘(:)’ forces ‘f’ to be a column vector (again assuming that ‘t’ is and remains a row vector).

[rivaldo rivaldo]

i dont know why but for some reason the both calculations dont work and i still get this error.

i even tried to run it separately in new script window :/

[Star Strider]

What version of MATLAB are you using?

If you have R2007a or later, this should work:

arg = bsxfun(@times, f(:), t);
nos = sum(sin(2*pi*arg));

Otherwise, use a loop to calculate ‘arg’:

for k1 = 1:length(f)
    arg(k1,:) = f(k1) * t;
end

If that doesn’t work, I’m out of ideas.

All of my code posted in my Answer here works.

[rivaldo rivaldo]

i have R2015A, the code with arg and nos finaly working, woohoo!

thank you very much!

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

[rivaldo rivaldo]

accepted. last one, when you run this code does it get any errors?

clear all;close all;clc

[x,Fs,Nbits] = wavread('jennifer.wav');

t = 0:0.01:512;

f = 45:55;

arg = bsxfun(@times, f(:), t);

nos = sum(sin(2*pi*arg));

y=x+nos;

figure(1);

plot(y);

when i adding x to nos its again gives an error

Error using +

Matrix dimensions must agree.

x its the speech signal.

[Star Strider]

Thank you.

The time vectors have to match. (I can’t run your code because I don’t have your ‘jennifer.wav’ file.)

Try this to calculate ‘t’:

[x,Fs,Nbits] = wavread('jennifer.wav');
L = size(x,1);                                          % The ‘x’ Matrix Should Be (Lx2)
t = linspace(0, (L-1)/Fs, L);

You will need to create ‘nos’ as a (Lx1) vector because ‘x’ will be (Lx2):

y = x + nos(:);

MATLAB automatically expands ‘nos(:)’ to be (Lx2). so that should work.

[Walter Roberson]

wavread() is gone as of R2016a, so we can tell that the user is using something before R2016b and so does not have implicit expansion.

[Walter Roberson]

I do not see any "uniform distributed noise" in this: noise is random, whereas this code appears to be just adding signal of known frequency.

There are multiple channels to the sound; it is not clear from the question whether the same noise signal is to be added to each.

It is not clear to me whether the noise is intended to be at exactly 45, 46, 47, etc Hz, or if the noise is intended to be in the range of 45 to 55 Hz.

I would suggest that you should be using white noise, then do a pass-band filter between 45 and 55 Hz, and add that to the signal... like Star Strider showed.

[rivaldo rivaldo]

how can i solve the matrix dim error in the last command?

if true
  Error in secondwork (line 36)
noisy_signal = x + noise_filt;
end

in this code

if true
clear all;close all;clc
[x,Fs,Nbits] = wavread('jennifer.wav');
Fn = Fs/2;                                              % Nyquist Frequency (Hz)
Wp = [44   55]/Fn;                                      % Passband Frequencies (Normalised)
Ws = [43   56]/Fn;                                      % Stopband Frequencies (Normalised)
Rp = 10;                                                % Passband Ripple (dB)
Rs = 50;                                                % Stopband Ripple (dB)
[n,Ws] = cheb2ord(Wp,Ws,Rp,Rs);                         % Filter Order
[z,p,k] = cheby2(n,Rs,Ws);                              % Filter Design
[sosbp,gbp] = zp2sos(z,p,k);                            % Convert To Second-Order-Section For Stability
figure(3)
freqz(sosbp, 2^17, Fs)
L=length(x);
noise = rand(1, L);                                     % Normally-Distributed Random Vector
noise_filt = filtfilt(sosbp, gbp, noise);               % Filter Noise Signal
noisy_signal = x + noise_filt;
end

[Star Strider]

Assuming that ‘x’ is (Nx2), one way would be:

noisy_signal = x + [noise_filt(:)  noise_filt(:)];

[rivaldo rivaldo]

Thank you.

My last mission is to implement this noisy signal with 3 different SNR:

-10dB

0dB

+10dB

I tought may be to create function that calcultates the noisy signal when the input is those 3 different snr.

what do you say?

[Star Strider]

How are you supposed to calculate SNR?

My code gives the amplitudes of the signal and noise. You may need to calculate the RMS amplitudes of both the signal and noise, then solve for the required noise amplitude in terms of the desired SNR with respect to the amplitude of your signal.