LTI causal system question

[emicho]

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 !!

 

[mondunno]

Hi,

Nice question, the impulse response h = 0 for n<0 tells it's a causal system (represented by u in your equation for IR)
for a step input (and the system is at steady state, not excited before this) the system only responds when there is first sample of non zero input signal.

Hence the y = [1-a^(n+1)]/[1-a], is valid for n>0, here the n>0 is one of the condition that "you must" specify as part of y, which is missed out in your equation.

It is true that the y equation yields values for n<0. y(n < 0) must be steady state value of h (for input 0), zero in this case.

 

[emicho]

yes, that's make sense..
thank you very much

 

[mondunno]

please read n > 0 as n >= 0