# Making a Square Wave Gradually

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I made a script where I have a square wave audioplayer made from sine waves based off of

for k = 1:2:order

x = x + (sin(k*2*pi.*f.*t))/k;

y((k+1)/2,:) = x;

pause(.01)

end

where f is an array based off the equation 2^x, and t is another array 0:1/fs:T used for sampling. f is used to select frequencies in an exponential manner so that the pitch goes up linearly. This worked very well, but I wanted to try the same thing but flipped around. The script I'm working on takes a certain frequency and creates a waveform that gradually evolves from a sine wave to a square wave. This is what I have so far:

function GradualSquareWave(order,f,T,fs)

tic

t = 0:1/fs:T;

timeDomain = 0:1:(fs * T);

y = zeros(size(t * order));

x = zeros(size(t));

h = waitbar(0,'Initializing Fourier Series');

for k = 1:2:order

x = x + (sin(k*2*pi*f.*t))/k;

y(1 + timeDomain + (((k+1)/2)-1)*fs) = x;

if k/order > .05

waitbar(k/order,h,sprintf('Computing Fourier Series %6.2g%%',(100 * k)/order))

end

pause(.01)

end

elapsedTime = toc;

if elapsedTime < 1

waitbar(1,h,sprintf('Playing Square Wave to Order %d after %6.4g seconds',order,elapsedTime))

end

if elapsedTime >= 1

waitbar(1,h,sprintf('Playing Square Wave to Order %d after%6.4g seconds',order,elapsedTime))

end

player = audioplayer(y,fs);

play(player)

pause(T)

close(h)

It seems that most of it is working just fine and dandy, but I'm not getting the full transition. For each k value, the audioplayer plays a one-second section of the wave form and cuts off at T seconds. If there aren't T sections to play, the last section is extended to T. I have no idea why each section is played for precisely one second. I want each section to last T/((order+1)/2) seconds so that each section of the wave form is played equally throughout time T. If anybody could help me out, I'd really appreciate it.

##### 1 Comment

Walter Roberson
on 22 May 2014

### Accepted Answer

Hugo
on 22 May 2014

Each section is being played for one second because you are setting that in the line

y(1 + timeDomain + (((k+1)/2)-1)*fs) = x;

The key is in the part ((k+1)/2)-1)*fs. When k=1, ((k+1)/2)-1)*fs=0, but when k=3 (which is the next loop in the for statement) ((k+1)/2)-1)*fs=fs. But fs is the number of samples in one second, so the new segment starts one second after the previous one, making the previous one of one second in length.

Notice however that the last segment will have a length equal to T.

To solve this, you should write

y(1 + timeDomain + (((k+1)/2)-1)*fs*T/((order+1)/2)) = x;

This statement will give you a length of T/((order+1)/2) as you requested, but be careful that the number should be an integer.

Just in case, notice that the for loop advances in steps of 2.

Hope this helps.

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