How can I create an impulse wave via triangular pulse

I need to create an impulse wave, through a triangular pulse. I succeded that with the rectangular, but the triangular doesn't seem to work for me.

body of my code:

syms a;
a=0.1;
t=linspace(-0.2,0.2,1000000);
y=zeros(1,1000000);
x=abs(t);
y(x<a)=(1/a)*(1-x);
plot(t,y);
grid;

The error matlab shows me is:

Unable to perform assignment because the left and right sides have a different number of elements.

Error in rectangular_pulse (line 133) (the file name is rectangular pulse)

y(x<a)=(1/a)*(1-x);

Answer 1

[Mathieu NOE]

hello

this solves the assignement problem, but I suspect the shape of the signal is not what you are waiting for

y(x<a)=(1/a)*(1-x(x<a));

[nastia korzoun]

It indeed solves the assignement problem, yet the shape is weird.

thanks anyway!

[Mathieu NOE]

could it be this is what you wanted ?

a=1;
N = 1000000;

y = a - abs(linspace(-a,a,N));
x=(0:N-1);
plot(x,y);
grid;

[nastia korzoun]

The triangle shape is perfect, it just wasn't exactly our assignment.

I just need to fighure out the y axis function, to make the total area of the triangle=1, as in an impulse function.

[Mathieu NOE]

well , with a = 1

the area is

area = 0.5*max(x)*max(y)

I got : area = 499999

NB : this is because the x axis is defined in samples not in seconds (what is the sampling frequency ?)

so you should convert the x axis is seconds, recalculate the area and divide the amplitude of the triangle by this number to get a unitary area

[nastia korzoun]

yeah I started with declaring frequencies and limits, but then the proffesor explained that he wanted to keep things a lot more simple than that.

BUT

I finally figured it out! it was:

y(x<a)=(1/a)*(1-abs((1/a)*t(x<a)));

Thanks again!!!

[Mathieu NOE]

you're welcome