I'm currently studying the Z-transform, and I'm having issues in understanding the time shift and differentiation properties, to be precise: calculating a Z-transform explicitly, and obtaining it by using those properties yields different results.

For example, a common Z-transform is the transform of the discrete unit step function: $$ \mathcal{Z}\left\{ u\left[n\right]\right\} =\frac{1}{1-z^{-1}}=\frac{z}{z-1} $$ The time shifting property of the Z-transform states (leaving aside its effects on the ROC for the moment) that $\mathcal{Z}\left\{ x\left[n-k\right]\right\} =z^{-k}\mathcal{Z}\left\{ x\left[n\right]\right\} \quad\forall k\in\mathbb{Z} $, thus it implies that: $$ \mathcal{Z}\left\{ u\left[n-1\right]\right\} =z^{-1}\frac{z}{z-1}=\frac{1}{z-1} $$ Since $\frac{d}{dz}\frac{1}{z-1}=\frac{1}{\left(z-1\right)^{2}}$, the differentiation property, which states that $\mathcal{Z}\left\{ n\, x\left[n\right]\right\} =-z\frac{d\mathcal{Z}\left\{ x\left[n\right]\right\} }{dz}$ gives: $$ \mathcal{Z}\left\{ n\, u\left[n-1\right]\right\} =-z\frac{1}{\left(z-1\right)^{2}}=-\frac{z^{-1}}{\left(1-z^{-1}\right)^{2}} $$ which appears to be wrong, since the correct result should be: $$ \mathcal{Z}\left\{ n\, u\left[n\right]\right\} =\frac{z^{-1}}{\left(1-z^{-1}\right)^{2}} $$ What am I doing wrong? Is there perhaps any case in which those properties cannot be applied?


2 Answers 2




(with a minus sign in front). So you have


Now everything is OK because


because for $n=0$ also the right-hand side vanishes, and for $n>0$ both are equal anyway.

  • $\begingroup$ $nu[n+a]=nu[n]$ is true if and only if $a = \{a \in \mathbb{Z} \mid a \geq -1\}$ $\endgroup$
    – Unknown123
    Jan 20, 2021 at 16:15

Note that $$\frac{d}{dz} \frac{1}{z-1} = -\frac{1}{(z-1)^2}$$ (emphasis on the negative sign on the right-hand side of the equation). I think you missed the negative sign in your above derivation, which would account for the difference in the two results that you got to.


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