How do I plot 5e^(0.5t)sin(2*pi*t)
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Monqiue Parrish
le 24 Fév 2023
Commenté : Les Beckham
le 24 Fév 2023
% 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')
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Les Beckham
le 24 Fév 2023
Modifié(e) : Les Beckham
le 24 Fév 2023
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
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Plus de réponses (2)
Torsten
le 24 Fév 2023
Déplacé(e) : Torsten
le 24 Fév 2023
Note that sin(2*pi*t) = 0 for t = 0,1,2,3,...,10.
And these are the only points for t you specified.
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Steven Lord
le 24 Fév 2023
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
le 24 Fév 2023
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)')
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