emicho
Junior Member level 3
I have a causal LTI system:
h = (a^n)u
and we apply a step function
so by using convolution and little math
we find that
y = [1-a^(n+1)]/[1-a]
my point is :
I read that if I apply a causal input to a causal system that means the output y will be zero for n < 0 ;
but if we substitute some values n<0 in y equation , we didn't that y =0 !!!!
any explanation please !!
h = (a^n)u
and we apply a step function
so by using convolution and little math
we find that
y = [1-a^(n+1)]/[1-a]
my point is :
I read that if I apply a causal input to a causal system that means the output y will be zero for n < 0 ;
but if we substitute some values n<0 in y equation , we didn't that y =0 !!!!
any explanation please !!