This problem is in the context of discrete systems and signals.
My goal is to find an equation for y(n) without the summation.
where is the unit step function
So I need to eliminate the unit step functions from the equation by changing the limits of summation:
if then . And so it's eliminated from my summation above.
if then . And it's also eliminated.
So my limits of summation are and . But the limits don't define a finite range! Can anyone help?
Sorry, let me restate the original problem and be more explicit.
Find the convolution of x(n) and h(n).
where
where the convolution is
note the symbol is not multiply, it is the convolution operator. Also y(n) is causal, which means y(n) is only defined for , where n=0,1,2,3....
Now,
Plugging in,
Simplifying,
The argument passed into the unit step must be or else the value of the unit step function is 0. So if I need to change my limits of summation so that the two unit step functions resolve to 1. That is,
if then .
if then .
So the limits on k would be,
if
if
I am bamboozled by the two lower limits on k. How can I do a summation with two lower limits? Would the summation diverge?
Nice solution. Thank you sir!
I was also able to solve it with a bunch more work and sloppy notation... but our solutions agree. I wrote y(n) in piecewise notation.
Are you sure? For large values of n, the numerator has a and the denominator has a ... which would make it converge to 0.