Question

\[\sum _{k=-\infty }^{\infty } \theta (n-k)\theta (-k-1)a^k\]

Answer 1

\[\theta\] is step function

Answer 2

@phi

Answer 3

I can show you the work , it would be great if you can explain it

Answer 4

I get the n<=-1 part but not the n>-1

Answer 5

this is convultion

Answer 6

I know u(-k-1) is 1 for k= -inf to -1
u(x) is 0 for x<0, and 1 for x≥0

u(n-k) is 1 for k= -inf to +n

Answer 7

use convolution property

Answer 8

you can break up the sequence from
-infinity to 0 and from 0 to infinity
for 0 to infinity we know what are the results of a step function

Answer 9

when you multiply the 2 step functions
the one that "steps down" (because we are doing u(-k) ) first sets the upper limit
n wins for n< -1, and the other wins for n> -1