How to use envelope function in Matlab

The following Matlab code generate a sound but it contain tak tak like sound which can be easily heard.Can someone removes this noise by using envelope function.

  Fs=44100;
  T=1/Fs;
  M=zeros(1,88);
  for I=7:88
    M(I)=round(36.8*(2^(1/12))^(I-6));
    end
    signal=[];
    FrTm=[50,3;50,3;52,3;54,3;50,3;54,3;52,3;45,3;50,3;50,3;52,3;54,3;50,6;
    49,3;1,3;50,3;50,3;52,3;54,3;55,3;54,3;52,3;50,3;49,3;45,3;47,3;49,3;50,6;
    50,3;1,3;47,5;49,1;47,3;45,3;47,3;49,3;50,3;1,3;45,5;47,1;45,3;43,3;42,6;
    45,3;1,3;47,5;49,1;47,3;45,3;47,3;49,3;50,3;47,3;45,3;50,3;49,3;52,3;50,6;
    50,6];
    for i=1:length(FrTm)
      x=FrTm(i,1);
      y=FrTm(i,2);
      F=M(x);
      time=0:1/Fs:y/18;
      sinewave=*sin(2*pi*10*F*time);

      signal=[signal sinewave];

     end

     stem(signal)
     sound(signal,Fs)

Solution

Here are two possible solutions. They are not perfect since the tak-tak is still here but it is much lower.

Applying a high-pass filter to remove the annoying sound which seems to be low frequency

Fs=44100;
T=1/Fs;
M=zeros(1,88);
for I=7:88
    M(I)=round(36.8*(2^(1/12))^(I-6));
end
signal=[];
FrTm=[50,3;50,3;52,3;54,3;50,3;54,3;52,3;45,3;50,3;50,3;52,3;54,3;50,6;
49,3;1,3;50,3;50,3;52,3;54,3;55,3;54,3;52,3;50,3;49,3;45,3;47,3;49,3;50,6;
50,3;1,3;47,5;49,1;47,3;45,3;47,3;49,3;50,3;1,3;45,5;47,1;45,3;43,3;42,6;
45,3;1,3;47,5;49,1;47,3;45,3;47,3;49,3;50,3;47,3;45,3;50,3;49,3;52,3;50,6;
50,6];
for i=1:length(FrTm)
    x=FrTm(i,1);
    y=FrTm(i,2);
    F=M(x)/4;
    time=0:1/Fs:y/18;
    sinewave=sin(2*pi*10*F*time);

    signal=[signal sinewave];

end

% Creating high-pass filter with passband frequency of 1000 Hz (may be too high)
hpFilt = designfilt('highpassiir','FilterOrder',8, ...
    'PassbandFrequency',1000,'PassbandRipple',0.2, ...
'SampleRate',Fs);
% Filtering sound
signal = filter(hpFilt,signal);

sound(signal,Fs)

Taking the last value from each generated sine wave and finding the corresponding phase for the next sine wave, to make the sinus as continuous as possible

 Fs=44100;
 T=1/Fs;
 M=zeros(1,88);
 for I=7:88
     M(I)=round(36.8*(2^(1/12))^(I-6));
 end
 signal=[];
 FrTm=[50,3;50,3;52,3;54,3;50,3;54,3;52,3;45,3;50,3;50,3;52,3;54,3;50,6;
   49,3;1,3;50,3;50,3;52,3;54,3;55,3;54,3;52,3;50,3;49,3;45,3;47,3;49,3;50,6;
50,3;1,3;47,5;49,1;47,3;45,3;47,3;49,3;50,3;1,3;45,5;47,1;45,3;43,3;42,6;
45,3;1,3;47,5;49,1;47,3;45,3;47,3;49,3;50,3;47,3;45,3;50,3;49,3;52,3;50,6;
50,6];
 previous_end_val = 0;
 for i=1:length(FrTm)
     x=FrTm(i,1);
     y=FrTm(i,2);
     F=M(x)/4;
     time=0:1/Fs:y/18;

     % Calculating required phase to match the end of previous sinewave
     phi = asin(previous_end_val);
     % Generating sinewave with phase phi
     sinewave=sin(2*pi*10*F*time + phi);
     previous_end_val = sinewave(end);

     signal=[signal sinewave];

 end

 sound(signal,Fs)

The first solution is not perfect: if you take too high passband frequency you could not hear the lowest audible frequencies. The second solution is faster and more elegant. Please ask if there is a method you don't understand.