How do I plot 5e^(0.5t)sin(2*pi*t)

14 次查看(过去 30 天)
% I have tried to plot 5e^(0.5t)sin(2pi*t) but my graph just show a negative exponential
% But when I compared to demos it's completely wrong
t1 = 0:10;
e = exp((0.5)*t1);
num6 = (5).*e.*sin((2)*pi*t1);
plot(t1, num6, 'c')

采纳的回答

Les Beckham
Les Beckham 2023-2-24
编辑:Les Beckham 2023-2-24
t1 = 0:10;
e = exp((0.5)*t1);
num6 = (5).*e.*sin((2)*pi*t1);
plot(t1, num6, 'c')
grid on
You are plotting your exponential term multiplied by the sine of integer multiples of 2*pi which should be zero. Due to floating point precision issues, it is instead very close to zero and increasing in a negative direction. This causes your plot to look the way it does.
t1 = 0:10;
plot(t1, sin(2*pi*t1), '.')
grid on
If you add points that aren't integer multiples of 2*pi you will see the real behavior of your equation.
t1 = 0:0.01:10;
e = exp(0.5 * t1);
num6 = 5 * e .* sin(2*pi*t1);
plot(t1, num6, 'c')
grid on

更多回答(2 个)

Torsten
Torsten 2023-2-24
移动:Torsten 2023-2-24
Note that sin(2*pi*t) = 0 for t = 0,1,2,3,...,10.
And these are the only points for t you specified.
  1 个评论
Steven Lord
Steven Lord 2023-2-24
You can see this even better if you use sinpi instead of sin.
format longg
t = (0:10).';
y1 = sin(2*pi*t)
y1 = 11×1
1.0e+00 * 0 -2.44929359829471e-16 -4.89858719658941e-16 -7.34788079488412e-16 -9.79717439317883e-16 -1.22464679914735e-15 -1.46957615897682e-15 -1.71450551880629e-15 -1.95943487863577e-15 -2.20436423846524e-15
y2 = sinpi(2*t)
y2 = 11×1
0 0 0 0 0 0 0 0 0 0

请先登录,再进行评论。


Image Analyst
Image Analyst 2023-2-24
You had so few points that you were subsampling the shape away and couldn't see the oscillations. Use more data points. by using linspace. Try it this way:
t1 = linspace(0, 10, 1000);
e = exp(0.5 * t1);
num6 = 5 .* e .* sin(2 * pi * t1);
plot(t1, num6, 'c-', 'LineWidth', 2);
grid on
xlabel('t1')
ylabel('num6')
title('num6 = 5 .* e .* sin(2 * pi * t1)')

类别

Help CenterFile Exchange 中查找有关 Line Plots 的更多信息

产品


版本

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by