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How to find a single period from a graph?

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Yasmine Sellwood
Yasmine Sellwood 2020 年 8 月 20 日
編集済み: Utkarsh Belwal 2020 年 8 月 24 日
Can somone please help me? I've produced the code below and I now need to find a single period from Signal 3 and plot it on a new graph with an appropriate time vector, tPeriod and save it to a new variable additive_noise.
Thank you very much for your time :)
%% 1.0 Plot the received speech singal
T = 5; %period
sample = 44100*20; % samples
period = 4; %required number of periods
SinglePeriod = linspace (0, T, 44100*5 + 1); %single period
SinglePeriod(end) = []; %time vetor
t = linspace(0, T*period, sample+1); %time vector required
t(end) = [];
noiseSound = repmat(noiseSound, 1, period);
plot(t, noiseSound, 'b');
title('Received speech signal')
xlabel('Time (s)')
ylabel('Amplitude')
filtered = zeros(size(noiseSound));
%% 1.1 Identifying the two interfering waveforms
s1 = (SinglePeriod/A); %signal one
Signal1 = repmat(s1, 1, period);
figure
subplot(3,1,1)
plot(t, Signal1, 'b')
title('Signal 1')
xlabel('Time (s)')
ylabel('Amplitude')
hold on
s2 = exp(-(SinglePeriod-B)/4); %signal 2
Signal2 = repmat(s2, 1, period);
subplot(3,1,2)
plot(t, Signal2, 'g')
title('Signal 2')
xlabel('Time (s)')
ylabel('Amplitude')
s3 = zeros(1, period);
s3(SinglePeriod >= 0 & SinglePeriod < 2.5) = ((A*SinglePeriod(SinglePeriod >= 0 & SinglePeriod < 2.5))/4);
s3(SinglePeriod >= 2.5 & SinglePeriod < 5) = (((-A*SinglePeriod(SinglePeriod >= 2.5 & SinglePeriod < 5))+(5*A))/4);
Signal3 = repmat(s3, [1 period]); %signal 3
subplot(3,1,3)
plot(t, Signal3, 'r')
title('Signal 3')
xlabel('Time (s)')
ylabel('Amplitude')
hold off
%% 2.0 Generate the noise waveforms - a single period
additive_noise_first =
  1 件のコメント
Utkarsh Belwal
Utkarsh Belwal 2020 年 8 月 24 日
編集済み: Utkarsh Belwal 2020 年 8 月 24 日
Can you define the variables noiseSound and A as without defining them code will give an error.

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