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

- anonymous

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

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

\[\theta\] is step function

- anonymous

@phi

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.