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  "body": "\n## Statement\n\nLet $x = (x_n)_{n \\in \\Z}$ be a sequence of complex numbers and let $k$ be an integer.\n\n1. Bilateral form. Let $y_n = x_{n - k}$ for all $n \\in \\Z$. Then $y$ and $x$ have the same [[region of convergence|1P8WX3EGHF]], and on it the [[bilateral Z-transforms|3DBAMXA80B]] satisfy $Y(z) = z^{-k} X(z)$.\n2. Unilateral delay. Suppose $x_n = 0$ for $n \u003c 0$ and $k \\ge 0$. Then the [[unilateral Z-transform|EPG6JPVJE7]] of $(x_{n - k})_{n \\ge 0}$ is $z^{-k} X(z)$.\n3. Unilateral advance. Let $k \\ge 1$ and let $X$ be the unilateral Z-transform of $(x_n)_{n \\ge 0}$. Then the unilateral Z-transform of $(x_{n + k})_{n \\ge 0}$ is\n$$z^{k} \\left( X(z) - \\sum_{j=0}^{k-1} x_j z^{-j} \\right),$$\non the same set of $z$.\n\n## Remarks\n\nEach clause is a change of summation index.\nFor clause 1, $\\sum_n x_{n-k} z^{-n} = z^{-k} \\sum_m x_m z^{-m}$ with $m = n - k$, and the same computation with absolute values shows that the regions agree.\nFor clause 3, $\\sum_{n \\ge 0} x_{n+k} z^{-n} = z^{k} \\sum_{m \\ge k} x_m z^{-m}$, and the sum over $m \\ge k$ is $X(z)$ minus its first $k$ terms.\nExample: for $x_n = 1$ ($n \\ge 0$), $X(z) = z/(z-1)$, and clause 3 with $k = 1$ gives $z (z/(z-1) - 1) = z/(z-1)$, as it must, since the shifted sequence is again constant.\nClause 2 needs $x_n = 0$ for $n \u003c 0$; otherwise the terms $x_{-k}, \\ldots, x_{-1}$ also enter the unilateral transform of the delayed sequence.\n\n## Explanations\n\n### L1 human\n\nDelaying a sequence by one step multiplies its Z-transform by $z^{-1}$.\nAdvancing it by one step multiplies by $z$, but in the one-sided case the first value falls off and must be subtracted.\nThis is why $z^{-1}$ is called the unit delay in signal processing.\n\n### L2 human\n\nClause 3 is the discrete analogue of $\\mathcal{L}\\{f'\\}(s) = s F(s) - f(0)$ for the Laplace transform.\nIt is what makes the Z-transform solve linear difference equations with constant coefficients: each shifted unknown becomes $z^k$ times the transform, plus a polynomial built from the initial values.\nClause 1 shows that a time-invariant system acts on transforms by multiplication.\n",
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