A community for students.
Here's the question you clicked on:
 0 viewing
Libniz
 2 years ago
\[\sum _{k=\infty }^{\infty } \theta (nk)\theta (k1)a^k\]
Libniz
 2 years ago
\[\sum _{k=\infty }^{\infty } \theta (nk)\theta (k1)a^k\]

This Question is Closed

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0\[\theta\] is step function

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0I can show you the work , it would be great if you can explain it

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0I get the n<=1 part but not the n>1

phi
 2 years ago
Best ResponseYou've already chosen the best response.1I know u(k1) is 1 for k= inf to 1 u(x) is 0 for x<0, and 1 for x≥0 u(nk) is 1 for k= inf to +n

psi9epsilon
 2 years ago
Best ResponseYou've already chosen the best response.0use convolution property

psi9epsilon
 2 years ago
Best ResponseYou've already chosen the best response.0you 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
 2 years ago
Best ResponseYou've already chosen the best response.1when 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

psi9epsilon
 2 years ago
Best ResponseYou've already chosen the best response.0you can split up the basic sequence too, its way easier to do it

phi
 2 years ago
Best ResponseYou've already chosen the best response.1when n>1 u(nk) times u(k1) is zero for k≥ 1

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0lol, I am reading all your posts over , will ask questions later

phi
 2 years ago
Best ResponseYou've already chosen the best response.1The 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

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0I get it how u(t) is reflected across y axis

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0trying 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
 2 years ago
Best ResponseYou've already chosen the best response.1u(k1) never changes. It is 1 from inf to 1, then drops to 0 we multiply this step times u(nk) 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
 2 years ago
Best ResponseYou've already chosen the best response.1the picture is trying to show the two cases. If it helps, think logic, both steps have to be 1

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0oh, so u(k1) is one constant step function whereas u(nk) is many different function based on differing value of n so say if n=2 u(2k) is it drop to zero before first step does

phi
 2 years ago
Best ResponseYou've already chosen the best response.1yes, u(2k) drops to zero when 2k=0 or k=2 when you multiply against the first step, you get 1 only up to k=2 (that is, k= n)

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.1n 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)

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, a lot of mental gymnastic, but thanks though

phi
 2 years ago
Best ResponseYou've already chosen the best response.1maybe this helps the u(k1) is 1 from inf to 1 and zero afterwards so the sum \[\sum_{k= \infty}^{+\infty}u(k1)a^k= \sum_{k= \infty}^{1}a^k\] the step function does not "do" anything except zero out all k's bigger than 1 on the other hand, the step (nk) affects the sum \[\sum_{k= \infty}^{+\infty}u(nk)a^k= \sum_{k= \infty}^{n}a^k\] 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
 2 years ago
Best ResponseYou've already chosen the best response.1the 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.

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0^that one did it, much more clear now
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.