How do I change this system with function handles to linear equations to plot discretely?

Hi, I don't know how to implement a unit function and still plot discretely. Also, I don't really understand what the "double" does in my code but somehow I need it. Please advise - TIA.

u = @(n) double(n>=0); %converts symbolic u to array of n
uu = @(n) 1*(n>=0); %unit step function
x = @(n) u(n-2)-u(n-4);
n=[-2,12];
h = @(n) uu(n).*(sin(n).*exp(-n));
fplot(x,[-2,12]); %plots within x-limits
hold on
fplot(h,[-2,12]);
grid on
y = @(n) x(n).*h(n);
fplot(y,[-2,12]);

Answer 1

[Walter Roberson]

u = @(n) double(n>=0); %converts symbolic u to array of n
uu = @(n) 1*(n>=0); %unit step function
x = @(n) u(n-2)-u(n-4);
n=[-2,12];
h = @(n) uu(n).*(sin(n).*exp(-n));

T = linspace(-2,12);

stem(T, x(T));  %plots within x-limits
hold on

stem(T, h(T));

grid on
y = @(n) x(n).*h(n);
stem(T, y(T));

ylim([-.1 1.1])

[balla243]

Thank you!

Can you please explain how "double" and plotting against a "linspace" makes this script work?

[Walter Roberson]

The double() is not needed.

fplot() plots against the given range automatically, chosing plotting points according to how bumpy the function is. It does not plot discretely.

stem() plots discretely, but it needs to be told which points to plot.

u = @(n) (n>=0); %converts symbolic u to array of n
uu = @(n) 1*(n>=0); %unit step function
x = @(n) u(n-2)-u(n-4);
n=[-2,12];
h = @(n) uu(n).*(sin(n).*exp(-n));

T = linspace(-2,12);

stem(T, x(T));  %plots within x-limits
hold on

stem(T, h(T));

grid on
y = @(n) x(n).*h(n);
stem(T, y(T));

ylim([-.1 1.1])