Proof of Z-transform of n

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How can I prove the following Z-transform:

$$ Z\{n\} = \frac {z} {(z-1)^2} $$

As a tip, I was told to use the 'Multiplication in time'-property of the Z transform, which is the following:

$$ Z\{nx[n]\} = -z \frac{dX(z)}{dz} $$

But I don't see how I can use this.

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2 Answers

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Use the definition of $z$ transform

$$Z\left\{n\right\} = \sum_{n=0}nz^{-n}=\frac{z} {(z-1)^2}. $$

now you can use the geometric series (see my answer) to find the desired result.

Note: The $z$ transform of a sequence $a_n$ is given by

$$ \sum_{n=0}^{\infty} a_n z^{-n}.$$

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I will show you, how you can prove it easily

$$u(n) \rightleftharpoons X(z) = \frac{z}{z-1}$$

According to the property which you mentioned

$$nu(n) \rightleftharpoons -z \frac{d(X(z)}{dz}$$

$$\frac{dX(z)}{dz} = - \frac{1}{(z-1)^2}$$

$$nu(n) \rightleftharpoons \frac{z}{(z-1)^2}$$

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