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$$\begin{align} E(z)&=\frac{z+1}{(z-1)(z-0.6)}\\ \frac{z+1}{(z-1)(z-0.6)}&=\frac{A}{(z-1)}+\frac{B}{(z-0.6)}\\ z+1&=A(z-0.6)+B(z-1) \end{align}$$

set z=0.6: $$\begin{align} 0.6+1&=A(0)+B(0.6-1)\\ 1.6&=B(-0.4)\\ B&=\frac{1.6}{-0.4}\\ &=-4 \end{align}$$ set z=1: $$\begin{align} 1+1&=A(1-0.6)+B(0)\\ 2&=A(0.4)\\ A&=\frac{2}{0.4}\\ &=5 \end{align}$$ So $$E(z)=\frac{5}{(z-1)}-\frac{4}{(z-0.6)}$$ Please explain to me how we go from that to: $$e(k)=\delta(k-1)(5-4(0.6)^k)$$

How do I use the z-transform tables such as the one found on page 4 of this PDF to solve this problem?

Cheerful Parsnip
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Gineer
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1 Answers1

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In addition to the table, you will need to use the fact that a multiplication with $z^{-1}$ corresponds to a time shift with $-1$ in the time domain. More generally:

$$z^{-k}X(z) \leftrightarrow x[n-k]$$

The first term in $E(z)$ can be written as $$5z^{-1}\frac{z}{z-1}$$

Using time shifting and the first entry in the table gives us that $$5z^{-1}\frac{z}{z-1} \leftrightarrow5u[n-1]$$

The second term can similarly be written as $$4z^{-1}\frac{z}{z-0.6}$$

According to the fourth row in the table $$\frac{z}{z-e^{-aT}} \leftrightarrow e^{-at}, \;t>0$$

Now, remember that T is the time corresponding to one sample. The continuous time $t$ can thus be written $t=nT$, and the transform pair can then be expressed using only $n$ as follows:

$$\frac{z}{z-e^{-\alpha}} \leftrightarrow e^{-\alpha n}$$

and, since this is valid for $t>0$ (causality), the right-hand side can be written $e^{-\alpha n}u[n]$.

In this case, $e^{-\alpha} = 0.6$ and this together with the time scaling property gives $$4z^{-1}\frac{z}{z-0.6} \leftrightarrow 4(0.6)^{n-1}u[n-1] = \frac{20}3(0.6)^nu[n-1] $$

Now we can put together the resulting inverse z transform:

$$e[n] = 5u[n-1] - \frac{20}3(0.6)^nu[n-1]=u[n-1]\left(5-\frac{20}3(0.6)^n\right)$$

I don't know where you get $e(k)=\delta(k-1)(5-4(0.6)^k)$ from, but it can't be right, can it?

Daniel R
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