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I am studying z transform and I couldn't get how to derive these two formulas

$$Z[ nu_{n}] =-z \frac{d}{dz} Z(u_{n}) $$

$$Z\left[ \frac{1}{n} u_{n}\right] =- \int_0^z z^{-1}Z(u_{n}) $$

These formula are somewhat similar to the corresponding ones of Laplace transforms which also involves derivatives and integrals. Any idea on how to prove it?

Thanks.

bertozzijr
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user471651
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  • would you mind telling me where did you get the second property from (a book or webpage maybe)? Also, is the integral in $dn$ or $dz$? The derivative one is pretty straightforward to prove, but the integral one seems a bit trickier... – bertozzijr Oct 21 '18 at 03:43

1 Answers1

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Look, the first one is pretty straightforward to prove:

For a discrete time domain function $f(k)$, its Z-Transform is defined as:

$$\mathcal{Z}\{f(k)\}=\sum_{k=-\infty}^{+\infty}f(k)z^{-k}$$

We can differentiate on both sides:

\begin{align} \frac{d}{dz}\mathcal{Z}\{f(k)\}&=\frac{d}{dz}\sum_{k=-\infty}^{+\infty}f(k)z^{-k}\\ &=\sum_{k=-\infty}^{+\infty}\frac{d}{dz}f(k)z^{-k}\\ &=\sum_{k=-\infty}^{+\infty}f(k)\frac{d}{dz}z^{-k}\\ &=\sum_{k=-\infty}^{+\infty}-kf(k)z^{-k-1}\\ &=-z^{-1}\sum_{k=-\infty}^{+\infty}kf(k)z^{-k}\\ &=-z^{-1}\mathcal{Z}\{kf(k)\}\\ \end{align}

So, finally:

$$\mathcal{Z}\{kf(k)\}=-z\frac{d}{dz}\mathcal{Z}\{f(k)\}$$

The integral property, on the other hand, is a little harder to prove with such a direct method, but let's go:

\begin{align} \int\mathcal{Z}\{f(k)\}dz&=\int\left[\sum_{k=-\infty}^{+\infty}f(k)z^{-k}\right]dz\\ &=\sum_{k=-\infty}^{+\infty}\int f(k)z^{-k}dz\\ &=\sum_{k=-\infty}^{+\infty}f(k)\int z^{-k}dz\\ &=\sum_{k=-\infty}^{+\infty}\frac{f(k)}{1-k}z^{1-k}\\ &=-z\sum_{k=-\infty}^{+\infty}\frac{f(k)}{k-1}z^{-k}\\ \end{align}

Defining $p=k-1$, then $k=p+1$ and the sum can be rewritten as:

\begin{align} &=-z\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p-1}\\ &=-\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p}\\ \end{align}

Applying the convolution theorem:

\begin{align} -\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p}&=-\left[\sum_{p=-\infty}^{+\infty}f(p+1)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\ \end{align}

Now, applying the time-shift property of the Z-transform:

\begin{align} -\left[\sum_{p=-\infty}^{+\infty}f(p+1)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]&=-\left[z\sum_{p=-\infty}^{+\infty}f(p)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\ &=-z\left[\sum_{p=-\infty}^{+\infty}f(p)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\ &=-z\left[\sum_{p=-\infty}^{+\infty}\frac{f(p)}{p}z^{-p}\right]\\ \end{align}

Now, reverting the substitution of $p=k-1$ and $k=p+1$, we get:

\begin{align} &=-z\left[\sum_{k=-\infty}^{+\infty}\frac{f(k-1)}{k-1}z^{-k+1}\right]\\ &=-z\left[z\sum_{k=-\infty}^{+\infty}\frac{f(k-1)}{k-1}z^{-k}\right]\\ &=-z\left[\sum_{k=-\infty}^{+\infty}\frac{f(k)}{k}z^{-k}\right]\\ &=-z\mathcal{Z}\left\{\frac{f(k)}{k}\right\} \end{align}

Which means that:

$$\int\mathcal{Z}\{f(k)\}dz=-z\mathcal{Z}\left\{\frac{f(k)}{k}\right\}$$

And, finally:

$$\mathcal{Z}\left\{\frac{f(k)}{k}\right\}=-\frac{1}{z}\int\mathcal{Z}\{f(k)\}dz$$

Which actually differs from the property you posted on your question... If the integral is in $z$, then it makes a difference to have $z^{-1}$ inside or outside the integral.

Hope I was able to help.

bertozzijr
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