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Libniz
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\[\sum _{k=\infty }^{\infty } \theta (nk)\theta (k1)a^k\]
 2 years ago
 2 years ago
Libniz Group Title
\[\sum _{k=\infty }^{\infty } \theta (nk)\theta (k1)a^k\]
 2 years ago
 2 years ago

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

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

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

psi9epsilon Group TitleBest ResponseYou've already chosen the best response.0
this is convultion
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I 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
 2 years ago

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

psi9epsilon Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
dw:1347579756548:dw
 2 years ago

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

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

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

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

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

Libniz Group TitleBest ResponseYou've already chosen the best response.0
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"
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
u(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.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
the picture is trying to show the two cases. If it helps, think logic, both steps have to be 1
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
oh, 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
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes, 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)
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
still puzzling over it?
 2 years ago

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

phi Group TitleBest ResponseYou've already chosen the best response.1
maybe 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)
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
^that one did it, much more clear now
 2 years ago
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