# z-transform of $2^k$

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): 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?

• 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. – Matt L. May 1 '17 at 15:51
• $f[k] = 2^k, h(z) = \frac{1}{1 - 1/2 z^{-1}}$, find output y(t) ... ? – Jack May 1 '17 at 23:25
• also, how to do z transform of $3^k u(-k-1)$ ?? or fourier transform? – Jack May 1 '17 at 23:49
• @Jack Is $f$ the input to the system $H(z)=\frac{1}{1-1/2z^{-1}}$? If yes, this might help you. – msm May 2 '17 at 3:41
• @msm yes, so would the answer be $2^k \cdot \frac{1}{1 - 1/2 z^{-1}}$ what do I do next ? – Jack May 2 '17 at 23:23