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State and explain different properties of ROC of z transform.

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 05

Year : DEC 2015

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ROC is the region where Z-transform converges. Z-transform is an infinite power series. This series is not convergent for all values of z. Hence ROC is useful in mentioning z-transform.

Significance of ROC:

1. ROC gives an idea about values of z for which Z-transform can be calculated.

2. ROC can be used to determine causality of the system.

3. ROC can be used to determine stability of the system.

Properties of ROC:

Property 1: The ROC for a finite duration sequence includes entire z-plane except $z = 0$ and/or $|z| = ∞.$

Proof:

Consider the finite duration sequence

$x(n) = \uparrow\{1 2 1 2\} \\ ∴X(z) = 1z^2 + 2z + 1z^0 + 2z^{-1} \\ = z^2 + 2z + 1 + \dfrac 2z$

Here $X(z) = ∞$ for $z = 0$ and $∞.$

This proves first property.

Property 2: ROC does not contain any poles.

Proof:

The Z-transform of $a^n u(n)$ is calculated as

$X(z) = \dfrac 1{1 – az^{-1 }} \\ = \dfrac z{z - a} \space\space\space ROC: |z| \gt |a|$

This function has pole at z = a. Note that ROC is |z| > |a|. This means poles do not lie in ROC. Actually $X(z) = ∞$ at poles by definition of pole.

This proves second property.

Property 3: ROC is the ring in the z-plane centered about origin.

Proof:

Consider

$Z [a^n u(n)] = \dfrac 1{1 – az^{-1 }}\space\space\space\space ROC: |z| \gt |a| \\ Or \\ Z [-a^n u(- n - 1)] = \dfrac 1{1 – az^{-1 }} \space\space\space\space ROC: |z| \lt |a|$

Here observe that |z| is always a circular region (ring) centered about origin.This proves third property.

Property 4: ROC of causal sequence (right hand sided sequence) is of the form |z| > r.

Proof:

Consider right hand sided sequence $a^n u(n).$ Its ROC is |z| > |a|. Thus the ROC of right hand sided sequence is of the form of |z| > r where r is the radius of the circle.

This proves fourth property.

Property 5: ROC of left sided sequence is of the form |z| < r.

Proof:

Consider left sided sequence $-a^n u(- n - 1).$ Its ROC is |z| < |a|. Thus the ROC of left sided sequence is inside the circle of radius r.

This proves fifth property.

Property 6: ROC of two sided sequence is the concentric ring in z-plane.

Proof:

Consider $x(n) = a^n u(n) + b^n u(- n - 1) \\ \text {Now} ,\space Z [a^n u(n)] = \dfrac 1{1 – az^{-1}}\space\space\space\space ROC: |z| \gt |a| \\ And \\ Z [b^n u(- n - 1)] = \dfrac 1{1 – bz^{-1}}\space\space\space\space ROC: |z| \lt |b| \\ ∴X(z) = \dfrac 1{1 – az^{-1}} + \dfrac 1{1 – bz^{-1}}\space\space\space ROC: |z| \gt |a| \space\space \text {and } \space\space |z| \lt |b| \\ i.e. |a| \lt |z| \lt |b| \\ \text {Since ROC is} |a| \lt |z| \lt |b|$

This is the concentric ring in z-plane.

This proves sixth property.