How can one calculate the $z$-transform of: $x(n) = \frac{1}{n}$ , where $n \geq 1$? I have searched for table entries, then got stuck while trying to do it with the definition of $z$-transform (summation).
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From what I can gather, by definition, the $z$-transform is the sum $$ \sum_{n = 1}^\infty\frac{z^{-n}}{n} $$ which just so happens to be the power series of $-\ln(1 - 1/z)$, converging as long as $|z| > 1$.
Arthur
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Good Question
$$\frac{u(n-1)}{n} \rightleftharpoons ??$$
We know that Z transform is defined as :
$$X(z) = \sum_{n=-\infty}^{\infty}x(n)z^{-n}$$
$$u(n) \rightleftharpoons \frac{z}{z-1}$$
By the use of time shifting property
$$x(n-1) \rightleftharpoons z^{-1} X(z)$$
$$u(n-1) \rightleftharpoons \frac{1}{z-1}$$
Differentiate the General formula of X(z) with respect to z
$$\frac{dX(z)}{dz} = -\sum_{n=-\infty}^{\infty}nx(n)z^{-n-1}$$
$$\frac{zdX(z)}{dz} = -\sum_{n=-\infty}^{\infty}nx(n)z^{-n}$$
$$nx(n) \rightleftharpoons -\frac{zdX(z)}{dz}$$
Now we want to find out the Z transform of $$\frac{u(n-1)}{n}$$
$$x(n) = \frac{u(n-1)}{n} \rightleftharpoons X(z)$$
$$n x(n) \rightleftharpoons -\frac{zdX(z)}{dz}$$
$$n [\frac{u(n-1)}{n}] \rightleftharpoons -\frac{zdX(z)}{dz}$$
$$u(n-1) \rightleftharpoons -\frac{zdX(z)}{dz}$$
Means
$$-\frac{zdX(z)}{dz} = \frac{1}{z-1}$$
$$\frac{dX(z)}{dz} = -\frac{1}{z(z-1)}$$
$$\frac{dX(z)}{dz} = [\frac{1}{z}] + [\frac{1}{z-1}]$$
$$X(z) = ln(\frac{z}{z-1})$$
Hence
$$\frac{u(n-1)}{n} \rightleftharpoons ln(\frac{z}{z-1})$$
Now we have some interesting series for natural log
$$\frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3z^3} + ...... = ln(\frac{z}{z-1})$$
For z = 2
$$\frac{1}{2} + \frac{1}{2\times2^2} + \frac{1}{3\times2^3} + ...... = ln(2)$$
For $$z = i^2 = -1$$
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{3} + ...... = ln(2)$$
Ashok Saini
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