Region of Convergence, Properties, Stability and Causality of Z-transforms

Contents

What is the Region of Convergence of Z-transform

The S-plane is capable of representing signals such as complex exponentials. When performing Z transform on these signals, we represent these signals on an imaginary and real axis.
The output of these transformations could turn out to meet at a point or go in different directions. The particular signals that tend to meet at a point are the signals that lie in the Region of Convergence(RoC).
The Region of Convergence maps all the values for which the transform converges to a finite value.
Recall the definition of Z-transform. You will remember that the limits of the summation were from -∞ to +∞. However, the Z-transform will exist only for those values of Z, which if put in this series results in a finite value.
In simple words, the ROC is a region in the Z-plane consisting of all the values of Z which make the Z-transform (X(Z)) attain a finite value.
The Region of Convergence is required to determine:
- the stability of a system by examining the transfer function.
- whether the system is causal or non-causal.
- whether the system is finite or infinite.
A stable system is one that produces a bounded output for a bounded input. It can be represented graphically in terms of circles together with a unit circle.

RoC is a representation of where the Z-transform exists. Hence, for a given x[n], the RoC is the range of z for which the output converges. The only condition is that x(z) should not be infinity for any value of z. Mathematically,

$X(z)=\sum _{ n=-\infty }^{ \infty }{ \left| x[n]{ z }^{ -n } \right| } <\infty$

Properties of Region of Convergence of Z-transform

The characteristics of the Region of Convergence are almost analogous to that of the original signal x[n]. Let us look at a few points we should take note of before going about finding out the RoC:

Property 1: The Region of Convergence is a ring or circle in the Z-plane centred about the origin

Explanation: The region of convergence is calculated in the form of the minimum value of Z beyond/within which the transform exists. Geometrically, this relation is possible only with a circle.

Property 2: The Region of Convergence does not contain any pole.

Explanation: Remember the condition that we talked about previously about the RoC not including points that go to infinity, there is a reason. The RoC can only contain Zeros and not Poles. And since Poles are the points where X(z) is infinite, they can’t be included in the RoC.

Property 3: When the Region of Convergence incorporates a unit circle, X(z) converges uniformly

Explanation: For any finite x[n] that includes the unit circle, its ROC exists in the range of z for which its sum converges. The z transform represented by the equation is basically a power series. Power series is a series in which you have a variable x and C[n] which are constants that are added up and represented by the general equation aₙ(x-a)ⁿ. A power series can either converge or diverge. The ROC converges when x[n]z^-n is absolutely summable. Represented mathematically,

$|X(z)|=\sum _{ n=-\infty }^{ \infty }{ \left| x[n]{ z }^{ -n } \right| } =\sum _{ n=-\infty }^{ \infty }{ \left| x[n] \right| } (\left| { z } \right| )^{ -n }$

Property 4: With a finite x[n], your ROC is the entire z plane except when z=0 and z= $\infty$

Explanation: We know that when x[n] is finite, its values are summable. Hence, z cannot be equal to 0 as the summation of finite values will give us a finite value. Similarly, z cannot be equal to infinity as well. Hence, the conclusion that the ROC will be the entire z plane except the aforementioned two values. There are a few exceptions that we will be looking at in the next few properties.

Example: Given x[n] find the RoC

The arrow mark indicates the value for which n=0

Hence, when n=0, x[n]=2

Next, we have to find x(z)

$x(z)=\sum _{ n=-1 }^{ 2 }{ x(n){ z }^{ -n } } =x(-1){ z }^{ 1 }+x(0){ z }^{ 0 }+x(1){ z }^{ -1 }+x(2){ z }^{ -2 }$

if z were equal to 0, the values for which z is in the denominator would tend to infinity.

if z were equal to ∞, the values for which z is in the numerator would tend to infinity.

Hence, the RoC will be the entire z-plane except for when z=0 and z=∞

Property 5: The ROC includes $\infty$ if the signal x[n] is causal

Explanation: Conditions for causality include:

When the RoC includes the region outside the outermost pole of the signal.
When the transfer function is expressed as a ratio of the output to the input of the system, the order of the numerators should be of an order lesser than the order of the denominator.

Example: For the system $H(z)=\frac { z }{ (z-0.2)(z-0.6) }$ , the poles of the transfer function will be z=0.2 and z=0.6

Since both the poles fall inside the unit circle, the system is causal and stable.

Property 6: The ROC includes 0 if the signal x[n] is non-causal

Explanation: When the aforementioned conditions are not met, the ROC would include 0 instead of $\infty$ .

Example: $H(z)=\frac { 2z+1 }{ { z }^{ 2 }+z-\frac { 5 }{ 16 } }$

Complete the square in the denominator to get the poles $H(z)=\frac { 2z+1 }{ (z+\frac { 5 }{ 4 } )(z-\frac { 1 }{ 4 } ) }$

Hence, the poles are at z=-5/4 and z=1/4

z=1/4 falls within the unit circle but z=-1.25 falls outside the unit circle

Hence, the system is non-causal and unstable.

Property 7: If x[n] is both causal and non-causal, the ROC will be a ring.

Explanation: If the n values fall on both sides of the x-axis, the ROC will be a ring around the radius in the z-plane.

Let us look at an example combining an RSS and an LSS.

Example

x(n)=f(n).u(n)+f(n).u(-n)

where $x(n)\longrightarrow [latex]x(z)=\frac { z }{ z-r }$

The z transform of both components of x(n) would be the same. However, the denominator would change based on the z value. The RSS would point outside the circle, and the LSS would point inside the circle, which means they would cancel out and the RoC would be the ring around the radius at modulus of z equal to r.

Z-plane of both sided sequence — z-plane of BSS

Property 8: The ROC is bounded by poles if x[n] is rational, which means it would extend to $\infty$

Explanation: If the values of x[n] were rational and bounded by poles, the RoC is extended to infinity. The RoC will have combined properties of both poles and fulfill the limits of both.

Conditions: More than one component in x[n] and both are rational, which means it has two poles.

Example: $x(n)={ 4 }^{ n }u(n)-{ 5 }^{ n }u(-n-1)$

The z transform of the individual components will be $\frac { z }{ z-4 }$ and $\frac { z }{ z-5 }$ .

The first component is RSS which means the limits will be |z|>4 and the second component is LSS which means the limits will be |z|<5.

So, combining both the limits RoC will fall in the range 4<|z|<5 as shown by the shaded region in the figure below.

Property 9: For a right sided sequence (RSS) x[n], the ROC will not comprise 0

Explanation: Two conditions, if:

x[n] is entirely positive or what we call a Right Sided Sequence (RSS), as it falls on the right of the x-axis, and
$\left| z \right| =r$ in the RoC

Then all finite values of z for which $\left| z \right| >r$ , will fall in the RoC.

Example: What signal can you think of that is purely a Right-Sided Sequence? A unit step function is all positive, so let's try with that.

For x(n)=f(n).u(n) the z transform will be $x(n)\longrightarrow x(z)=\frac { z }{ z-r }$

For a positive result, the value of z should be more than r to avoid a minus in the denominator. So, the modulus of z should be more than the radius.

Z-plane for a right sided sequence — Z-plane for RSS

Therefore, the RoC lies outside the circle from the pole position.

Property 10: When x[n] is rational, and both poles are RSS, the ROC will be outside the outermost pole.

Explanation: By rational, it means that x[n] has two poles. If both components of x[n] are right-sided, we can say that the RoC will fall in the region outside the outermost pole of the two

Example: $x(n)={ 4 }^{ n }u(n)+{ 5 }^{ n }u(-n-1)$

The z transform of the individual components would be $\frac { z }{ z-4 }$ and $\frac { z }{ z-5 }$

but since both components are RSS, Z will be greater than the respective radii.

The limits will end up being |z|>4 and |z|>5

Now, we need the RoC to satisfy both conditions. |z|>4 will include the values between 4 and 5, not satisfying |z|>5. But |z|>5 will satisfy both the conditions. Hence, choose the limits of the outermost pole, which in this case is |z|>5 (the red arrows).

Property 11: For a left sided sequence(LSS) x[n], the ROC will not comprise $\infty$

Explanation: If x[n] is entirely negative, it would be a Left-Sided Sequence (LSS) as the equation would fall on the left side of the x-axis, and when |z|=r for all finite values of z for which $\left| z \right| <r$ , they will fall in the RoC.

Example: Can you think of any signal that lies solely on the left side of the x-axis? How about the negative of a unit step function?

Hence, let's take x(n)=f(n).u(-n)

the z transform would also be $x(n)\longrightarrow x(z)=\frac { z }{ z-r }$

However, since we want it to be on the left side, we need the denominator to be a negative value. So, the modulus of z should be less than r.

Z-plane of Left Sided Sequence — Z-plane of LSS

Property 12: When x[n] is rational and LSS, ROC will be inside the innermost pole.

Explanation: If the z-transform of x[n] is rational, that is it has more than one pole, and the finite values of x[n] lie on the left side of the x-axis. The RoC will fall in the region inside the innermost pole of the two poles of x[n].

Example: $x(n)={ 4 }^{ n }u(n)+{ 5 }^{ n }u(-n-1)$

The z transform of the individual components would be $\frac { z }{ z-4 }$ and $\frac { z }{ z-5 }$

but since both components are RSS, Z will be greater than the respective radii.

The limits will end up being |z|>4 and |z|>5

Again, RoC should satisfy both conditions, if we take |z|<5, for the values between 4 and 5,|z|<4 will not be satisfied. However, |z|<4 covers both conditions. Hence, choose the limits of the innermost pole, which in this case is |z|<4.

Steps to find RoC and Z-transform

Here is a basic outline as to how to approach an RoC problem.

Step 1: Identify the point at the origin.

Step 2: Find out X(z) with the equation $X(z)=\sum _{ n=-\infty }^{ \infty }{ \left| x[n]{ z }^{ -n } \right| } <\infty$ for the limits determined from x[n].

Step 3: Identify whether the value of X(z) goes to infinity at any point (especially when z=0 and z=∞).

Step 4: Your RoC will be the entire z plane except for the region that you figured out in step 3.

Stability of an LTI system (What it is and conditions)

An LTI system is said to be stable if, for an input that is bounded, the output of the system is also bounded for all values of n.

$|x[n]|<{ n }_{ 1 }\quad$ for which $|y[n]|<\infty$

y[n] is commonly known as the convolution of x[n] and the impulse response, h[n]. This can also be represented as the summation expression shown below:

$y[n]=x[n]*h[n]=\sum _{ m=-\infty }^{ \infty }{ h[m]x[n-m] }$ this should be a finite value

hence, the terms inside the summation should not lead to infinity which means we can come to the conclusion that

$\sum _{ m=-\infty }^{ \infty }{ |h[m]|<\infty }$

Thus,

For the impulse response to be finite, we need to ensure that h[m] is absolutely integrable. This would ensure that the system will be stable. Hence, the bottom line is that we need an absolutely summable impulse response.

Energy signals are absolutely integrable

This stability of a system can also be determined using the RoC by fulfilling a couple of conditions

Conditions:

The system's transfer function H(z) should include the unit circle.
Also, for a causal LTI system, all the poles should lie within the unit circle. Read on to find out more about the causality of an LTI system.

BIBO stability of an LTI system

Assuming all the initial conditions of an LTI system are zero when the system output is bounded for each and every bounded input, it is referred to as a BIBO (Bounded Input Bounded Output System). By bounded, we mean the absolute value of a signal does not exceed some finite constant.

For a continuous output y(t)

$y(t)=\int _{ -\infty }^{ \infty }{ h(\lambda )x(t-\lambda )\quad d\lambda }$

Therefore, for a bounded input $\left| x(t-\lambda ) \right| \le N$

y(t) will also be bounded like below

$\left| y(t) \right| \le N\int _{ -\infty }^{ \infty }{ h(\lambda )d\lambda } \le M$

Hence, the conclusion that we can make from these expressions is

$\int _{ -\infty }^{ \infty }{ \left| h(\lambda ) \right| } d\lambda <\infty$

With this information, let us determine the BIBO stability and/or causality of the following system:

Example: h(t)=u(t+1)

this response can be represented by the graph below

From the graph, we can see clearly that h(t)=0 for all t<0, hence it is not causal. Now let's take the integral to get an idea about the stability.

$\int _{ -\infty }^{ \infty }{ u(\lambda +1)\quad d\lambda =\int _{ -1 }^{ \infty }{ d\lambda =\infty +1 } }$

The output is not bounded as it goes to infinity, hence, the system is not BIBO stable as well.

Causality of an LTI system (What it is and conditions)

If you can recall what we discussed in the previous post about causality, it means the output of the LTI system depends on the present and past input values of x[n]. If there is a future value, it is automatically not causal.

We know that

$y[n]=x[n]*h[n]=\sum _{ m=-\infty }^{ \infty }{ h[m]x[n-m] } =\sum _{ m=0 }^{ \infty }{ h[m]x[n-m] }$

when the output of y[n] is 0, for a causal function, n should be less than 0 or 0 itself.

Therefore, we can conclude that for n<0 h[m] will be 0.

Conditions:

Causality is satisfied when the RoC falls in the region outside the outermost pole of the signal.
When the transfer function is expressed as a ratio of the output to the input of the system, the order of the numerators should be of an order lesser than the order of the denominator.