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

- anonymous

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

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- schrodinger

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- anonymous

\[\theta\] is step function

- anonymous

@phi

- anonymous

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

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## More answers

- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

this is convultion

- 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

- anonymous

use convolution property

- anonymous

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

- 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

- phi

|dw:1347579756548:dw|

- anonymous

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

- phi

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

- phi

clear as mud?

- anonymous

lol, I am reading all your posts over , will ask questions later

- 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

- anonymous

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

- anonymous

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"

- 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.

- phi

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

- anonymous

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

- 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)

- anonymous

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

- 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)

- phi

still puzzling over it?

- anonymous

yeah, a lot of mental gymnastic, but thanks though

- 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)

- 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.

- anonymous

^that one did it, much more clear now

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