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Z-Transforms (ZT)
Analysis of continuous time LTI systems can be done using z-transforms. It is a powerful mathematical tool to convert differential equations into algebraic equations.
The bilateral (two sided) z-transform of a discrete time signal x(n) is given as
$Z.T[x(n)] = X(Z) = \Sigma_{n = -\infty}^{\infty} x(n)z^{-n} $
The unilateral (one sided) z-transform of a discrete time signal x(n) is given as
$Z.T[x(n)] = X(Z) = \Sigma_{n = 0}^{\infty} x(n)z^{-n} $
Z-transform may exist for some signals for which Discrete Time Fourier Transform (DTFT) does not exist.
Concept of Z-Transform and Inverse Z-Transform
Z-transform of a discrete time signal x(n) can be represented with X(Z), and it is defined as
$X(Z) = \Sigma_{n=- \infty }^ {\infty} x(n)z^{-n} \,...\,...\,(1)$
If $Z = re^{j\omega}$ then equation 1 becomes
$X(re^{j\omega}) = \Sigma_{n=- \infty}^{\infty} x(n)[re^{j \omega} ]^{-n}$
$= \Sigma_{n=- \infty}^{\infty} x(n)[r^{-n} ] e^{-j \omega n}$
$X(re^{j \omega} ) = X(Z) = F.T[x(n)r^{-n}] \,...\,...\,(2) $
The above equation represents the relation between Fourier transform and Z-transform.
$ X(Z) |_{z=e^{j \omega}} = F.T [x(n)]. $
Inverse Z-transform
$X(re^{j \omega}) = F.T[x(n)r^{-n}] $
$x(n)r^{-n} = F.T^{-1}[X(re^{j \omega}]$
$x(n) = r^n\,F.T^{-1}[X(re^{j \omega} )]$
$= r^n {1 \over 2\pi} \int X(re{^j \omega} )e^{j \omega n} d \omega $
$= {1 \over 2\pi} \int X(re{^j \omega} )[re^{j \omega} ]^n d \omega \,...\,...\,(3)$
Substitute $re^{j \omega} = z$.
$dz = jre^{j \omega} d \omega = jz d \omega$
$d \omega = {1 \over j }z^{-1}dz$
Substitute in equation 3.
$ 3\, \to \, x(n) = {1 \over 2\pi} \int\, X(z)z^n {1 \over j } z^{-1} dz = {1 \over 2\pi j} \int \,X(z) z^{n-1} dz $
$$X(Z) = \sum_{n=- \infty }^{\infty} \,x(n)z^{-n}$$ $$x(n) = {1 \over 2\pi j} \int\, X(z) z^{n-1} dz$$