Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Below I've quoted Wikipedia's entry that relates the Z-Transform to the Laplace Transform. The part I don't understand is $z \ \stackrel{\mathrm{def}}{=}\ e^{s T}$; I thought $z$ was actually an element of $\mathbb{C}$ and thus would be $z \ \stackrel{\mathrm{def}}{=}\ Ae^{s T}$ (but then it would be different to the Laplace Transform...). I don't understand why the Z-Transform is not defined as: $$ X(z) = \mathcal{Z}\{x[n]\} = \sum_{n=-\infty}^{\infty} x[n] e^{-\omega n} $$ or something like that.


Z-transform

The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of $$ z \ \stackrel{\mathrm{def}}{=}\ e^{s T} \ $$ where $T = 1/f_s \ $is the sampling period (in units of time e.g., seconds) and $f_s \ $is the sampling rate (in samples per second or hertz)

Let $$ \Delta_T(t) \ \stackrel{\mathrm{def}}{=}\ \sum_{n=0}^{\infty} \delta(t - n T) $$ be a sampling impulse train (also called a Dirac comb) and $$ \begin{align} x_q(t) & \stackrel{\mathrm{def}}{=}\ x(t) \Delta_T(t) = x(t) \sum_{n=0}^{\infty} \delta(t - n T) \\ & = \sum_{n=0}^{\infty} x(n T) \delta(t - n T) = \sum_{n=0}^{\infty} x[n] \delta(t - n T) \end{align} $$ be the continuous-time representation of the sampled x(t) \ $$ x[n] \ \stackrel{\mathrm{def}}{=}\ x(nT) \ $$ are the discrete samples of x(t) The Laplace transform of the sampled signal x_q(t) \ is $$ \begin{align} X_q(s) & = \int_{0^-}^\infty x_q(t) e^{-s t} \,dt \\ & = \int_{0^-}^\infty \sum_{n=0}^\infty x[n] \delta(t - n T) e^{-s t} \, dt \\ & = \sum_{n=0}^\infty x[n] \int_{0^-}^\infty \delta(t - n T) e^{-s t} \, dt \\ & = \sum_{n=0}^\infty x[n] e^{-n s T}. \end{align} $$ This is precisely the definition of the unilateral Z-transform of the discrete function $x[n] \ $. $$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} $$ with the substitution of $z \leftarrow e^{s T} \ $.

Comparing the last two equations, we find the relationship between the unilateral Z-transform and the Laplace transform of the sampled signal: $$ X_q(s) = X(z) \Big|_{z=e^{sT}} $$ The similarity between the Z and Laplace transforms is expanded upon in the theory of time scale calculus.


(Source: http://en.wikipedia.org/wiki/Laplace_transform#Laplace.E2.80.93Stieltjes_transform)

Here: http://en.wikipedia.org/wiki/Z-transform it says that $z \in \mathbb{C}$.

share|improve this question
    
The $s$ in the definition is a complex number also, and the exponential over the complexes is surjective (even if not injective). –  Juan Simões Jun 15 '12 at 13:00
    
Thanks for the reply. I looked up the words surjective and injective but I don't really understand what you mean. Are you saying that since $s$ is a complex number, $e^{st}$ can basically represent any complex number? If it can, then why not stick a $z$ in the Laplace transform instead of $e^{st}$? –  user968243 Jun 15 '12 at 14:21
1  
Exactly. The $z$ or the exponential depend on how you are going to use these series. The exponential is used mostly in the solution of differential equations. If you put a $z$ there, you will see many $\log z$ terms that are not necessary. –  Juan Simões Jun 18 '12 at 23:32
    
@JuanSimões: The exponential is not surjective over $\mathbb{C}$, since $e^z\neq 0$. But on $\mathbb{C}\setminus\{0\}$ it is. –  Mårten W Oct 8 '13 at 20:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.