# Z-Transforms (ZT)

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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.

$3\, \to \, x(n) = {1 \over 2\pi} \int\, X(z)z^n {1 \over j } z^{-1} dz = {1 \over 2\pi j} \int \,X(z) z^{n-1} dz$

$$X(Z) = \sum_{n=- \infty }^{\infty} \,x(n)z^{-n}$$ $$x(n) = {1 \over 2\pi j} \int\, X(z) z^{n-1} dz$$