Plotting and returning handle to line?

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Matt Amador
Matt Amador 2017 年 11 月 17 日
コメント済み: Matt Amador 2017 年 11 月 17 日
Hi there, i'm supposed to write a program that plots the function x(t) = cos(2*pi*t - theta) between the limits -1 > t > 1, and saves the handle of the resulting line. The angle theta is supposed to initially be 0 radians, and then, I need to re-plot line over and over with (theta = pi/10 rad), (theta = 2*pi/10 rad), (theta = 3*pi/10rad) and so forth up to (theta = 2*pi rad). I'm not sure how this is supposed to turn out, so this is what I have so far below. Can anyone help?
clc
clear all
for t_steps = 10 : 10 : 100
t = linspace(-1, 1, t_steps);
theta = 0; % or pi/10 or 2*pi
hlot = cos(2*pi*t - theta);
set(hlot, 'XData', t);
drawnow;
pause(1);
end
for t_steps = 10 : 10 : 100
t = linspace(-1, 1, t_steps)
theta = pi/10; % or pi/10 or 2*pi
xt = cos(2*pi*t - theta);
set(hlot, 'YDATA', t);
drawnow;
pause(1);
end
  4 件のコメント
KALYAN ACHARJYA
KALYAN ACHARJYA 2017 年 11 月 17 日
編集済み: KALYAN ACHARJYA 2017 年 11 月 17 日
I have given the answer as per your question, reduce the increase the step size in t, and check again. Pls, vote the answer.
t=1:.5:10; % .1 replace with .5
Matt Amador
Matt Amador 2017 年 11 月 17 日
Okay, now I got it. Thanks!

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KALYAN ACHARJYA
KALYAN ACHARJYA 2017 年 11 月 17 日
clc;clear all;
t=1:.5:10; % reduce the steps, now more clear graphs visible
for n=0:100;
theta=n*pi/10;
hlot=cos(2*pi*t-theta);
plot(hlot);
hold on;
end
hold off;
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