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

% 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')

Answer 1

[Les Beckham]

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

[Monqiue Parrish]

Thank you that makes more sense now.

[Les Beckham]

You are quite welcome.

Answer 2

[Torsten]

Note that sin(2*pi*t) = 0 for t = 0,1,2,3,...,10.

And these are the only points for t you specified.

[Steven Lord]

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

Answer 3

[Image Analyst]

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)')