How do I plot 5e^(0.5t)sin(2*pi*t)
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Monqiue Parrish
am 24 Feb. 2023
Kommentiert: Les Beckham
am 24 Feb. 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
am 24 Feb. 2023
Bearbeitet: Les Beckham
am 24 Feb. 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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Weitere Antworten (2)
Torsten
am 24 Feb. 2023
Verschoben: Torsten
am 24 Feb. 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.
1 Kommentar
Steven Lord
am 24 Feb. 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
am 24 Feb. 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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