## Libniz 3 years ago $\sum _{k=-\infty }^{\infty } \theta (n-k)\theta (-k-1)a^k$

1. Libniz

$\theta$ is step function

2. Libniz

@phi

3. Libniz

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

4. Libniz

5. Libniz

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

6. psi9epsilon

this is convultion

7. phi

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

8. psi9epsilon

use convolution property

9. psi9epsilon

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

10. phi

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

11. phi

|dw:1347579756548:dw|

12. psi9epsilon

you can split up the basic sequence too, its way easier to do it

13. phi

when n>-1 u(n-k) times u(-k-1) is zero for k≥ -1

14. phi

clear as mud?

15. Libniz

16. phi

The first confusion is u(k) is 0 up to k=0 , then it jumps to 1 but u(-k) is 1 and drops down to 0 at k=0 that is because when we plug in (for example) k=-2, we get u(-(-2))= u(2) =1 (2>0 means we jumped up) Both of these steps are backwards, are at 1 and drop to zero at some point

17. Libniz

I get it how u(-t) is reflected across y axis

18. Libniz

trying to parse this "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"

19. phi

u(-k-1) never changes. It is 1 from -inf to -1, then drops to 0 we multiply this step times u(n-k) if n is bigger than -1 this second step drops to zero after -1 but it does not do anything, because we are multiplying the 2 steps and everything past -1 is 0 (from the first step function). if n is less than -1, it drops to zero before the first step does, and it sets the upper limit where the product of the 2 step functions is 1.

20. phi

the picture is trying to show the two cases. If it helps, think logic, both steps have to be 1

21. Libniz

oh, so u(-k-1) is one constant step function whereas u(n-k) is many different function based on differing value of n so say if n=-2 u(-2-k) is it drop to zero before first step does

22. phi

yes, u(-2-k) drops to zero when -2-k=0 or k=-2 when you multiply against the first step, you get 1 only up to k=-2 (that is, k= n)

23. Libniz

I guess my main confusion is $\sum _{k=-\infty }^{-1} a^k$ n>-1 <---- why do we use this even though n is not used in summation itself

24. phi

n is controlling the upper limit. n bigger than -1 cannot make the limit bigger than -1 , but n< -1 can change the upper limit it n (n being smaller than -1)

25. phi

still puzzling over it?

26. Libniz

yeah, a lot of mental gymnastic, but thanks though

27. phi

maybe this helps the u(-k-1) is 1 from -inf to -1 and zero afterwards so the sum $\sum_{k= -\infty}^{+\infty}u(-k-1)a^k= \sum_{k= -\infty}^{-1}a^k$ the step function does not "do" anything except zero out all k's bigger than -1 on the other hand, the step (n-k) affects the sum $\sum_{k= -\infty}^{+\infty}u(n-k)a^k= \sum_{k= -\infty}^{n}a^k$ by tossing all entries where k>n so the step function is setting the upper limit. when we use both (multiply both together), both have to be true (1). you get 2 cases: n<-1 (where the upper limit is n) and n> -1 (where the upper limit is -1)

28. phi

the next trick if figuring out how to do the summations. For a>1, you can change the limits to positive values and invert a: a^-value= (1/a)^+value now it is a geometric series with a closed form solution.

29. Libniz

^that one did it, much more clear now