What Is Z-Transform and Inverse Z-Transform In Easy Way

If you are Searching for the z transform and Inverse Z-Transform then here is the Easy Way to understand What Is Z-Transform and Inverse Z-Transform with an explanation.

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)=Σ∞n=−∞x(n)z−n$Z.T[x(n)]=X(Z)={\mathrm{\Sigma }}_{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n)z^{-n}$

The unilateral (one sided) z-transform of a discrete time signal x(n) is given as

Z.T[x(n)]=X(Z)=Σ∞n=0x(n)z−n$Z.T[x(n)]=X(Z)={\mathrm{\Sigma }}_{n=0}^{\mathrm{\infty }}x(n)z^{-n}$

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)=Σ∞n=−∞x(n)z−n......(1)$X(Z)={\mathrm{\Sigma }}_{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n)z^{-n}......(1)$

If Z=rejω$Z=r{e}^{j\omega }$ then equation 1 becomes

X(rejω)=Σ∞n=−∞x(n)[rejω]−n$X(r{e}^{j\omega })={\mathrm{\Sigma }}_{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n)[r{e}^{j\omega }{]}^{-n}$

=Σ∞n=−∞x(n)[r−n]e−jωn$={\mathrm{\Sigma }}_{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n)[r^{-n}]e^{-j\omega n}$

X(rejω)=X(Z)=F.T[x(n)r−n]......(2)$X(r{e}^{j\omega })=X(Z)=F.T[x(n)r^{-n}]......(2)$

The above equation represents the relation between Fourier transform and Z-transform.

X(Z)|z=ejω=F.T[x(n)].$X(Z){|}_{z=e^{j\omega }}=F.T[x(n)].$
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Z-Transforms Properties

Z-Transform has following properties:

Linearity Property

If x(n)⟷Z.TX(Z)$x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z)$

and y(n)⟷Z.TY(Z)$y(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}Y(Z)$

Then linearity property states that

ax(n)+by(n)⟷Z.TaX(Z)+bY(Z)$ax(n)+by(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}aX(Z)+bY(Z)$

Time Shifting Property

If x(n)⟷Z.TX(Z)$x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z)$

Then Time shifting property states that

x(n−m)⟷Z.Tz−mX(Z)$x(n-m)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}{z}^{-m}X(Z)$

Multiplication by Exponential Sequence Property

If x(n)⟷Z.TX(Z)$x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z)$

Then multiplication by an exponential sequence property states that

an.x(n)⟷Z.TX(Z/a)$a^n.x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z/a)$

Time Reversal Property

If x(n)⟷Z.TX(Z)$x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z)$

Then time reversal property states that

x(−n)⟷Z.TX(1/Z)$x(-n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(1/Z)$

Differentiation in Z-Domain OR Multiplication by n Property

If x(n)⟷Z.TX(Z)$x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z)$

Then multiplication by n or differentiation in z-domain property states that

nkx(n)⟷Z.T[−1]kzkdkX(Z)dZK$n^{k}x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}[-1{]}^k{z}^k\frac{d^{k}X(Z)}{d{Z}^K}$

Convolution Property

If x(n)⟷Z.TX(Z)$x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z)$

and y(n)⟷Z.TY(Z)$y(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}Y(Z)$

Then convolution property states that

x(n)∗y(n)⟷Z.TX(Z).Y(Z)$x(n)\ast y(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z).Y(Z)$

Correlation Property

If x(n)⟷Z.TX(Z)$x(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z)$

and y(n)⟷Z.TY(Z)$y(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}Y(Z)$

Then correlation property states that

x(n)⊗y(n)⟷Z.TX(Z).Y(Z−1)$x(n)\otimes y(n)\stackrel{\mathrm{Z}.\mathrm{T}}{⟷}X(Z).Y(Z^{-1})$

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Inverse Z-transform

X(rejω)=F.T[x(n)r−n]$X(r{e}^{j\omega })=F.T[x(n)r^{-n}]$

x(n)r−n=F.T−1[X(rejω]$x(n)r^{-n}=F.T^{-1}[X(r{e}^{j\omega }]$

x(n)=rnF.T−1[X(rejω)]$x(n)=r^{n}F.T^{-1}[X(r{e}^{j\omega })]$

=rn12π∫X(rejω)ejωndω$=r^n\frac{1}{2\pi }\int X(re{}^j\omega )e^{j\omega n}d\omega$

=12π∫X(rejω)[rejω]ndω......(3)$=\frac{1}{2\pi }\int X(re{}^j\omega )[r{e}^{j\omega }{]}^{n}d\omega ......(3)$

Substitute rejω=z$r{e}^{j\omega }=z$.

dz=jrejωdω=jzdω$dz=jr{e}^{j\omega }d\omega =jzd\omega$

dω=1jz−1dz$d\omega =\frac{1}{j}{z}^{-1}dz$

Substitute in equation 3.

3→x(n)=12π∫X(z)zn1jz−1dz=12πj∫X(z)zn−1dz

X(Z)=∑n=−∞∞x(n)z−n$$X(Z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n)z^{-n}$$

x(n)=12πj∫X(z)zn−1dz