0
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It seems that you can decompose it as such:

$f(n) = a^n u(n) + a^{-n} u(-n-1)$

But I already have issue here,

is it basically saying that $ u(n) + u(-n-1) = 1$?

this is the plot of u(n) and u(-n-1):

enter image description here

if you sum them, wouldn't you just basically combine them? So you get 1 from $- \infty$ to -1, and from 0 to $\infty$

between n = -1 and 0 it's 0

no?

I know that the answer to z-transform $2^k$ is $\frac{z}{z-2}$ but how do you arrive at this?

general formula of z-transform is:

$F(z) = \sum_{k=\infty}^{0} f[k] z^{-k} $

applying formula:

$F(z) = \sum_{k=\infty}^{0} 2^k z^{-k} $

it would look like this, no?

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  • 2
    $\begingroup$ Note that the Z-transform of $2^k$ doesn't exist. What you probably mean is the Z-transform of $2^ku[k]$, where $u[k]$ is the unit step function. In that case it's a matter of applying the formula for the geometric series. $\endgroup$ – Matt L. May 1 '17 at 15:51
  • $\begingroup$ $f[k] = 2^k, h(z) = \frac{1}{1 - 1/2 z^{-1}}$, find output y(t) ... ? $\endgroup$ – Jack May 1 '17 at 23:25
  • $\begingroup$ also, how to do z transform of $3^k u(-k-1)$ ?? or fourier transform? $\endgroup$ – Jack May 1 '17 at 23:49
  • $\begingroup$ @Jack Is $f$ the input to the system $H(z)=\frac{1}{1-1/2z^{-1}}$? If yes, this might help you. $\endgroup$ – msm May 2 '17 at 3:41
  • $\begingroup$ @msm yes, so would the answer be $ 2^k \cdot \frac{1}{1 - 1/2 z^{-1}}$ what do I do next ? $\endgroup$ – Jack May 2 '17 at 23:23

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