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=== Z-transform === {{further|Z-transform#Relationship to Laplace transform}} The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of <math display=block> z \stackrel{\mathrm{def} }{ {}={} } e^{sT} ,</math> where {{math|1=''T'' = 1/''f<sub>s</sub>''}} is the [[sampling interval]] (in units of time e.g., seconds) and {{math|''f<sub>s</sub>''}} is the [[sampling rate]] (in [[samples per second]] or [[hertz]]). Let <math display=block> \Delta_T(t) \ \stackrel{\mathrm{def}}{=}\ \sum_{n=0}^{\infty} \delta(t - n T) </math> be a sampling impulse train (also called a [[Dirac comb]]) and <math display=block>\begin{align} x_q(t) &\stackrel{\mathrm{def} }{ {}={} } x(t) \Delta_T(t) = x(t) \sum_{n=0}^{\infty} \delta(t - n T) \\ &= \sum_{n=0}^{\infty} x(n T) \delta(t - n T) = \sum_{n=0}^{\infty} x[n] \delta(t - n T) \end{align}</math> be the sampled representation of the continuous-time {{math|''x''(''t'')}} <math display=block> x[n] \stackrel{\mathrm{def} }{ {}={} } x(nT) ~.</math> The Laplace transform of the sampled signal {{math|''x''<sub>''q''</sub>(''t'') }} is <math display=block>\begin{align} X_q(s) &= \int_{0^-}^\infty x_q(t) e^{-s t} \,dt \\ &= \int_{0^-}^\infty \sum_{n=0}^\infty x[n] \delta(t - n T) e^{-s t} \, dt \\ &= \sum_{n=0}^\infty x[n] \int_{0^-}^\infty \delta(t - n T) e^{-s t} \, dt \\ &= \sum_{n=0}^\infty x[n] e^{-n s T}~. \end{align}</math> This is the precise definition of the unilateral Z-transform of the discrete function {{math|''x''[''n'']}} <math display=block> X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} </math> with the substitution of {{math|''z'' β ''e''<sup>''sT''</sup>}}. Comparing the last two equations, we find the relationship between the unilateral Z-transform and the Laplace transform of the sampled signal, <math display=block>X_q(s) = X(z) \Big|_{z=e^{sT}}.</math> The similarity between the Z- and Laplace transforms is expanded upon in the theory of [[time scale calculus]].
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