Plotting one cycle of a wave

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William Chambers
William Chambers 2018-12-18
评论： Walter Roberson 2020-9-20
[sig, fs] = audioread(x);
modfreq = 4;
depth = 100;
a = depth/200;
offset = 1 - a;
len = 1:length(sig);
phasor = a*sawtooth(2*pi*len*(modfreq/fs)) +offset;
I'm generating a wave for a tremolo. I know how to plot the entire wave(at the length of the input signal), however I would like to be able to plot just a single cycle.
I found an answer in a previous question:
fs = 512; % Sampling frequency (samples per second)
dt = 1/fs; % seconds per sample
StopTime = 0.25; % seconds
t = (0:dt:StopTime)'; % seconds
F = 60; % Sine wave frequency (hertz)
data = sin(2*pi*F*t);
plot(t,data)
%%For one cycle get time period
T = 1/F ;
% time step for one time period
tt = 0:dt:T+dt ;
d = sin(2*pi*F*tt) ;
plot(tt,d) ;
However, it generates a second wave at a shorter length. I would like to be able to plot just one cycle of the original wave without creating a second.
Does anyone know how to do this?
Thanks in advance.

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回答（2 个）

Omer Yasin Birey
Omer Yasin Birey 2018-12-18
I believe axis() would work. I wrote an example code with a signal which has cycles.
x = 1:100;
signal = (1-cos(2*pi*0.01*x)).*sin(2*pi*0.15*x);%signal with cycles
plot(signal,'r-')
[peak,locs] = findpeaks(-signal); % Find Minimas
%you can use loop to plot every single cycle. it is just for the first one
firstInd = locs(1);%use a minima and the next one to find cycle limits
lastInd = locs(2);
%limit the plot with the start and end of the both x and y axes
axis([firstInd lastInd -peak(1) peak(2)])
  2 个评论
William Chambers
William Chambers 2018-12-18
Thanks for this, I managed to get it working in my test script, but when I try to implement it within the app designer it doesn't work.
T = 1/modfreq ;
dt = 1/fs;
tt = 0:dt:T+dt ;
plot(app.UIAxes, len, wave)
axis([0 length(tt) 0 1])
I've left out some parts of the code in this snippet, but you should get the idea of what I'm trying to do. When I run the app desginer app, it just plots the unlimited range plot (all cycles) and a random blank graph.
Omer Yasin Birey
Omer Yasin Birey 2018-12-18
编辑：Omer Yasin Birey 2018-12-18
I think length(tt) is the problem here. Because you start from 0 to length(tt), which means the whole range of signal and tt might have a very long sequence. Finding the 2 minimas, left and right, which are bounding a cycle still seems the possible solution to me.
T = 1/modfreq ;
dt = 1/fs;
tt = 0:dt:T+dt ;
[peak,locs] = findpeaks(-signal);
for i = 1:2:length(locs)-2
figure
firstInd = locs(i);
lastInd = locs(i+1);
plot(app.UIAxes, len, wave)
axis([firstInd lastInd -peak(i) peak(i+1)])
end
This code above should plot all the cycles seperately.

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Rohan Basak
Rohan Basak 2020-9-20
编辑：Walter Roberson 2020-9-20
T = 1/modfreq ;
dt = 1/fs;
tt = 0:dt:T+dt ;
[peak,locs] = findpeaks(-signal);
for i = 1:2:length(locs)-2
figure
firstInd = locs(i);
lastInd = locs(i+1);
plot(app.UIAxes, len, wave)
axis([firstInd lastInd -peak(i) peak(i+1)])
end
  1 个评论
Walter Roberson
Walter Roberson 2020-9-20
Rohan Basak:
Why are you create a new figure each time, even though you are plotting on a fixed axes?
Note also that the axis() command you are using is going to apply to the "current" axis, which is going to be a newly generated axes in the newly generated figure.

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