I want to plot my results individually under each other like a row and not a continues impulse response . Something isn't correct when i try it, Can someone help please?

x = zeros(1, 20000);

x(1 : 4000 : 20000);

xresh = reshapex, 4000, 5);

So every 4000fs i want to plotted graph?

Answer 1

[Star Strider]

I am not certain what you want.

Try this:

x = linspace(0, 20000, 20000)';                             % Create Column Vector
y  = sin(x*pi/500);                                         % Create Data

xresh = reshape(x, 4000, 5);
yresh = reshape(y, 4000, 5);
yresh = bsxfun(@plus, yresh, (0:2:2*(size(yresh,2)-1)));    % Add Offsets

figure
plot(xresh, yresh)
grid

Experiment to get your desired result.

[codefanta]

I want to record a sound and plot the results individually rather than one continouse response.

So after i have created my sound and played and recorded it i then used the funtion "Getaudiodata" I want to plot my results individually.

[Star Strider]

I do not have your sound file, so I cannot respond specifically with respect to it.

Copying some code from the audiorecorder documentation, and appending my code to the end of it:

% Record your voice for 5 seconds.
recObj = audiorecorder;
disp('Start speaking.')
recordblocking(recObj, 2.5);
disp('End of Recording.');

% Play back the recording.
play(recObj);

% Store data in double-precision array.
myRecording = getaudiodata(recObj);

% Plot the waveform.
% plot(myRecording);

Fs = 8000;                                                  % Default Sampling Frequency
x = linspace(0, 20000, 20000)'/Fs;                          % Create Column Vector
y = myRecording;
xresh = reshape(x, 4000, 5);
yresh = reshape(y, 4000, 5);
yresh = bsxfun(@plus, yresh, (0:2:2*(size(yresh,2)-1)));    % Add Offsets

figure
plot(xresh, yresh)
grid

figure
plot(xresh(:,1), yresh)
grid
set(gca, 'YTick',(0:2:2*(size(yresh,2)-1)), 'YTickLabel',(1:5))
xlabel('Time (sec)')
ylabel('Signal Segment')

This runs without error.

‘I want to record a sound and plot the results individually rather than one continouse response.’

I am not certain what you are asking for. My code appears to produce the result you requested, as I understand it. The first plot retains the time offsets, the second plot uses the first column of ‘xresh’ and plots everything as a function of it.

[codefanta]

Top man!

[codefanta]

Hello

Are you able to talk me through how your code works please?

[Star Strider]

Sure!

I only made a few changes to the code you already posted. My only significant addition is the bsxfun (link) call, that adds a different constant to each row of ‘yresh’. This provides the necessary separation between the rows for the plot. The maximum amplitude for double-precision values for sounds is ±1, so separating them by steps of 2 will work regardless of the values in ‘yresh’.

A slightly more efficient version of my bsxfun call is:

yresh = bsxfun(@plus, yresh, (0:(size(yresh,2)-1))*2);      % Add Offsets

See the documentatiion for bsxfun that I linked to for an explanation of how it works.

[codefanta]

Hello

Unfortunatly the link doens't work, it says the IP adrress isnt recognisable?!

also

Would you be able to talk me through why you divided by Fs and why an Fs is nessaccary ?

thanks

[Star Strider]

As always, my pleasure.

You can access the documentation for bsxfun by simply typing:

doc bsxfun

in a script or in your Command Window. I am certain you have the function.

This assignment:

x = linspace(0, 20000, 20000)'/Fs;

that would be more robust if I had defined it as:

x = linspace(0, size(y,1), size(y,1))'/Fs;

simply creates a time vector for

y = myRecording;

It creates a linear vector, and then converts it to relevant times by dividing it by ‘Fs’, the sampling frequency in cycles/time, creating a time vector in units of time/cycle. ‘Fs’ is necessary in order that ‘x’ accurately reflect the time of the signal. It would also be necessary if you wanted to correctly define frequencies of the Fourier transform of the signal, and to design filters to process your signal, if you want to do any of those.

[codefanta]

Thank you so much it all make sense.

one last question, if i want to change the recording time do i need to change my reshapes ?

[Star Strider]

As always, my pleasure.

Yes. You can probably make that dynamic by doing something like:

xresh = reshape(x, fix(x/5), 5);

However, you need to choose a recording tiime and sampling frequency that will produce a value for the length of your recorded sound vector that is an integer multiple of the number of segments you want. The reshape function will fail otherwise.

Answer 2

[codefanta]

yresh = bsxfun ( @plus , yresh, ( 0 : 2 : 2*(size (yresh,2) -1)));

set (gca, 'ytick', (0 : 2 : 2*( size( yresh,2) -1) , 'yticklabel', (1:5))

Would you be able to explain this 2 bits of code from your code as I dont 100% understand what they do?

Thanks

[Star Strider]

Sure.

The first one:

yresh = bsxfun ( @plus , yresh, ( 0 : 2 : 2*(size (yresh,2) -1)));    % Add Offsets

or, more efficiently:

yresh = bsxfun(@plus, yresh, (0:(size(yresh,2)-1))*2);      % Add Offsets

adds an integer offset vector created by ‘(0:(size(yresh,2)-1))*2’ to the rows of ‘yresh’. This creates the y-offsets for each row that are used in the plot.

The second one (more efficiently this set call):

set(gca, 'YTick',((0:(size(yresh,2)-1))*2), 'YTickLabel',(1:5))

creates the y-tick values, showing the segment numbers (1:5). It is necessary to tell set what y-tick values to use with the 'YTick' vector, to modify with the 'YTickLabel' vector.

See the documentation for Axes Properties (link) to understand what the set call does, and all the ways you can modify axes properties.

[codefanta]

Fantastic! Fully understand the code now.

If i want to continue further and work out a mean/average for myrecording and the plot the fft and Ifft and eventually the phase etc

Would it be a simple case of just continuing with "yresh" ??

[Star Strider]

My apologies for my delay in responding. It took a few minuites to get the Fourier transform code to work correctly with ‘yresh’.

‘Would it be a simple case of just continuing with "yresh" ??’

That depends on what you want to do. If you want to do a Fourier transform of your entire vector before you reshape it, then using ‘yresh’ is not the best approach.

If you want to do the Fourier transform on the segments of the vector created in ‘yresh’, then the frequency vector would have to be changed to accommodate it.

Example:

First, re-name the assignment of the bsxfun call to:

yresh2 = bsxfun(@plus, yresh, (0:(size(yresh,2)-1))*2);    % Add Offsets

and change this section of code to conform to that change:

figure
plot(xresh, yresh2)
grid

figure
plot(xresh(:,1), yresh2)
grid
% set(gca, 'YTick',(0:2:2*(size(yresh,2)-1)), 'YTickLabel',(1:5))
set(gca, 'YTick',((0:(size(yresh2,2)-1))*2), 'YTickLabel',(1:5))
xlabel('Time (sec)')
ylabel('Signal Segment')

then do the Fourier transform of each row of the original ‘yresh’ matrix, and plot the results:

Fn = Fs/2;                                      % Nyquist Frequency
N = size(yresh,1);                              % Vector Lengths
NFFT = 2^nextpow2(N);                           % FFT Length (Optional)
FTyresh = fft(yresh,NFFT,1)/N;                  % Fourier Transform
Fv = linspace(0, 1, fix(size(FTyresh,1)/2))*Fn; % Frequency Vector
Iv = 1:numel(Fv);                               % Index Vector

figure
hold all
for k1 = 1:size(FTyresh,2)
    subplot(size(FTyresh,2),1,k1)
    plot(Fv, abs(FTyresh(Iv,k1))*2)
    title(sprintf('Row %d', k1))
    ylabel('Amplitude')
end
hold off
xlabel('Frequency (Hz)')

That should get you started with the frequency domain analysis of your signal.

The rest of the code is unchanged.

Experiment to get the result you want.

[codefanta]

Thank you again, I will play around with this and ill mesage you soon with more questions !

[Star Strider]

As always, my pleasure!

[codefanta]

Hello

Are you able to explain to me why you used a for loop in the code above?

kind regards

Oliver

[Star Strider]

It makes plotting the subplot calls easier.

It is not absolutely necessary to use the subplot function. A method similar to my earlier code will work for the Fourier transformed signals:

FTyresh2 = bsxfun(@plus, 20*log10(abs(FTyresh(Iv,:)*2)), (0:(size(yresh,2)-1))*100);    % Add Offsets
% FTyresh2 = bsxfun(@plus, 1E+4*(abs(FTyresh(Iv,:)*2)), (0:(size(yresh,2)-1))*100);    % Add Offsets
qq1 = mean(FTyresh2)
dqq1 = diff([0 qq1])
qq2 = ((0:(size(yresh2,2)-1))-mean(FTyresh2(:,1)))

figure
plot(Fv, FTyresh2)
set(gca, 'YTick',min(FTyresh2), 'YTickLabel',(1:5), 'YGrid','on')
xlabel('Frequency (Hz)')
ylabel('Fourier Transform of Signal Segment')

The first ‘FTyresh2’ calculation plots in decibels, the second (commented-out here) plots linearly. Experiment with both of them.

[codefanta]

Again thank you very much

[Star Strider]

As always, my pleasure.

Answer 3

[codefanta]

hello

is it not possible to get my frequency domain and have my results plotted one under each other like figure 1 and 2 before the new code?

[Star Strider]

I am not certain what you are asking.

If you want to plot the Fourier transform results similar to ‘yresh2’, that is certainly possible.

This uses similar code to the ‘yresh2’ calculation, and the previous results from the Fourier transform:

FTyresh2 = bsxfun(@plus, 20*log10(abs(FTyresh(Iv,:)*2)), (0:(size(yresh,2)-1))*100);    % Add Offsets

figure
plot(Fv, FTyresh2)
set(gca, 'YTick',min(FTyresh2), 'YTickLabel',(1:5), 'YGrid','on')
xlabel('Frequency (Hz)')
ylabel('Fourier Transform of Signal Segment')

Plotting the decibel transformation is necessary because the Fourier transformed spectra have significantly smaller amplitudes than the time-domain signals.

Another option is simply to multiply them:

FTyresh2 = bsxfun(@plus, 1E+4*(abs(FTyresh(Iv,:)*2)), (0:(size(yresh,2)-1))*100);    % Add Offsets

The subplot approach is actually better.

[codefanta]

Hello

Sorry I didnt reply to your last comment, It was very helpful though!

I was hoping you could talk me through how this funcion works:

"yresh = bsxfun(@plus, yresh, ( 0 : size(yresh,2 ) -1) ) * 2 ); "

So what does the (yresh, 2), the -1 and the *2 do in the function?

thanks

[Star Strider]

As always, my pleasure.

That bsxfun (link) call multiplies each column of ‘yresh’ by the corresponding element of the ‘( 0 : size(yresh,2 ) -1) ) * 2’ vector. That vector is equivalent to ‘[0, 2, 4, 6, 8]’. The purpose is to separate the columns by multiples of 2. Since the sound vectors are limited to a magnitude of ±1, this guarantees that they will be separated, and not overlap in the plot.