How do I plot 5e^(0.5t)sin(2*pi*t)
13 ビュー (過去 30 日間)
古いコメントを表示
% 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')
0 件のコメント
採用された回答
Les Beckham
2023 年 2 月 24 日
編集済み: Les Beckham
2023 年 2 月 24 日
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
2 件のコメント
その他の回答 (2 件)
Torsten
2023 年 2 月 24 日
移動済み: Torsten
2023 年 2 月 24 日
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 件のコメント
Steven Lord
2023 年 2 月 24 日
You can see this even better if you use sinpi instead of sin.
format longg
t = (0:10).';
y1 = sin(2*pi*t)
y2 = sinpi(2*t)
Image Analyst
2023 年 2 月 24 日
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)')
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Directed Graphs についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!