Which value of $k$ in $e(k) = 5$ where $E(z)=5z^{-1}+z^{-2}+2z^{-4}$?

Question

I'm new to Laplace transforms so here goes:

From the lookup table we have that $$t^{k-1} = \frac{(k-1)!}{s^k}$$

using that:
$$\begin{align} \scr L\{e(k)\}&=5z^{-1}+z^{-2}+2^{-4}\\ e(k) &= 5\left(\frac{(0-1)!}{k^0}\right)+\frac{(-1-1)!}{k^{-1}}+\frac{(-3-1)!}{k^{-3}}\\ &=-5-2k-24k^3 \end{align}$$

Therefore $k$ would have to be equal to $0$ since $$\begin{align} e(0)&=-5-2\cdot0-24\cdot0^3\\ &=-5\\ \end{align}$$

Now $e(1)=-31$,
$e(2)=-201$, and
$e(3)=-659$

So it must be a sign error or worse. What am I doing wrong?

Answer 1

As much as I can possibly guess at what you are doing (your notation is very confusing), it seems you are trying to find the inverse LT of

$$E(z) = \frac{5}{z} + \frac{1}{z^2} + \frac{2}{z^4}$$

where

$$E(z) = \int_0^{\infty} dt \, e(t) \, e^{-z t}$$

You want to use the easily verified formula

$$\mathcal{L}(t^k) = \frac{k!}{z^{k+1}}$$

It seems you have your signals mixed. You actually want to evaluate (or so I think)

$$\mathcal{L}^{-1}\left ( \frac{1}{z^{k+1}}\right ) = \frac{t^k}{k!}$$

Thus for $k=0$ we have

$$\mathcal{L}^{-1}\left ( \frac{1}{z}\right ) = 1$$

et cetera. The result is

$$e(t) = 5 + t + \frac13 t^3$$