Signals and Systems

5.2 Properties of the Region of Convergence (ROC)

Imron Rosyadi

Property 1: ROC is a Ring Centered at the Origin

Statement:

The ROC of \(X(z)\) consists of a ring in the \(z\)-plane centered about the origin.

Justification:

The ROC consists of values of \(z = r e^{j\omega}\) for which \(x[n]r^{-n}\) is absolutely summable: \[ \sum_{n=-\infty}^{+\infty}|x[n]| r^{-n}<\infty \]

• Convergence depends only on \(r = |z|\), not on the angle \(\omega\).
• If a value \(z_0\) is in the ROC, then all \(z\) on the circle \(|z|=|z_0|\) are also in the ROC.
• This implies the ROC must be a collection of concentric rings. In fact, it’s always a single ring (or a disk/exterior of a disk).

Figure 10.6 ROC as a ring in the \(z\)-plane.

Note

The inner boundary can extend to the origin, and the outer boundary can extend to infinity.

The first property states that the ROC is always a ring, possibly extending to include the origin or infinity. This comes directly from the definition of convergence, which only depends on the magnitude of \(z\), not its phase angle. If a point \(z_0\) is in the ROC, then any other point on the circle passing through \(z_0\) must also be in the ROC. This creates the characteristic ring shape.

Property 2: The ROC Does Not Contain Any Poles

Statement:

The ROC does not contain any poles.

Justification:

• A pole is a value of \(z\) where \(X(z)\) becomes infinite.
• By definition, for \(X(z)\) to converge, it must be finite.
• Therefore, any value of \(z\) that makes \(X(z)\) infinite (a pole) cannot be part of the ROC.

Warning

The boundaries of the ROC are always defined by poles.

This property is quite intuitive. If the Z-transform is infinite at a certain point, it clearly cannot be said to converge at that point. Thus, poles always lie outside the ROC, typically defining its inner or outer boundaries.

Property 3: Finite Duration Signals

Statement:

If \(x[n]\) is of finite duration, then the ROC is the entire \(z\)-plane, except possibly \(z=0\) and/or \(z=\infty\).

Justification:

A finite-duration sequence has nonzero values only for \(N_1 \le n \le N_2\). \[ X(z)=\sum_{n=N_{1}}^{N_{2}} x[n] z^{-n} \]

• For any finite \(z \ne 0\), each term \(x[n]z^{-n}\) is finite, so \(X(z)\) converges.
• Exclusions:
  • If \(N_1 < 0\) (terms with \(z^k, k>0\)): \(X(z)\) can be unbounded as \(|z| \rightarrow \infty\). ROC excludes \(z=\infty\).
  • If \(N_2 > 0\) (terms with \(z^{-k}, k>0\)): \(X(z)\) can be unbounded as \(|z| \rightarrow 0\). ROC excludes \(z=0\).

Property 3: Finite Duration Signals

Summary of ROC for Finite Duration Signals:

• Finite-length, right-sided (\(N_1 \ge 0\)): ROC includes \(z=\infty\). Excludes \(z=0\) if \(N_2 > 0\). Example: \(x[n]=\delta[n]\) (\(N_1=0, N_2=0\)).
• Finite-length, left-sided (\(N_2 \le 0\)): ROC includes \(z=0\). Excludes \(z=\infty\) if \(N_1 < 0\). Example: \(x[n]=\delta[n+1]\) (\(N_1=-1, N_2=-1\)).
• Finite-length, two-sided (\(N_1 < 0, N_2 > 0\)): ROC excludes \(z=0\) and \(z=\infty\).

For signals that exist for only a finite number of samples, the Z-transform is a finite sum. This means it will almost always converge for all \(z\). The only exceptions are \(z=0\) and \(z=\infty\), depending on whether the signal extends to negative or positive time indices, respectively.

If \(x[n]\) has terms for \(n<0\), it means \(X(z)\) will have positive powers of \(z\), making it blow up at \(z=\infty\). If \(x[n]\) has terms for \(n>0\), it means \(X(z)\) will have negative powers of \(z\), making it blow up at \(z=0\).

Example 10.5: Finite Duration Signals

1. Unit Impulse: \(x[n] = \delta[n]\)

• \(X(z) = \sum_{n=-\infty}^{+\infty} \delta[n] z^{-n} = \delta[0]z^0 = 1\)

• ROC: Entire \(z\)-plane (includes \(z=0\) and \(z=\infty\)).

  (Here, \(N_1=0, N_2=0\), so no positive or negative powers of \(z\) are present in the sum.)

2. Delayed Impulse: \(x[n] = \delta[n-1]\)

• \(X(z) = \sum_{n=-\infty}^{+\infty} \delta[n-1] z^{-n} = \delta[0]z^{-1} = z^{-1}\)

• ROC: Entire \(z\)-plane, except \(z=0\) (pole at \(z=0\)).

  (Here, \(N_1=1, N_2=1\), only negative powers of \(z\). Excludes \(z=0\).)

3. Advanced Impulse: \(x[n] = \delta[n+1]\)

• \(X(z) = \sum_{n=-\infty}^{+\infty} \delta[n+1] z^{-n} = \delta[0]z^{1} = z\)

• ROC: Entire finite \(z\)-plane, except \(z=\infty\) (pole at \(z=\infty\)).

  (Here, \(N_1=-1, N_2=-1\), only positive powers of \(z\). Excludes \(z=\infty\).)

Example 10.5: Finite Duration Signals

Interactive: Pole/Zero for Delayed/Advanced Impulse

Let’s visualize the “poles” at \(z=0\) and \(z=\infty\).

viewof impulse_type = Inputs.select(["delta[n]", "delta[n-1]", "delta[n+1]"], {label: "Select Impulse Type"});

These simple examples illustrate Property 3. The unit impulse \(\delta[n]\) has a Z-transform of 1, which is finite everywhere, so its ROC is the entire Z-plane. \(\delta[n-1]\) introduces a \(z^{-1}\) term, creating a pole at \(z=0\), thus excluding \(z=0\) from the ROC. Conversely, \(\delta[n+1]\) introduces a \(z\) term, creating a pole at \(z=\infty\), excluding it from the ROC. The interactive plot helps visualize these subtle points.

Property 4: Right-Sided Sequences

Statement:

If \(x[n]\) is a right-sided sequence, and if the circle \(|z|=r_0\) is in the ROC, then all finite values of \(z\) for which \(|z|>r_0\) will also be in the ROC.

Interpretation:

The ROC for a right-sided sequence is an exterior region, extending outward from the outermost pole.

• Causal sequences: If \(x[n]=0\) for \(n<0\), then \(N_1 \ge 0\). The ROC includes \(z=\infty\).
• Non-causal right-sided sequences: If \(N_1 < 0\), the ROC does not include \(z=\infty\).

\[ X(z)=\sum_{n=N_{1}}^{\infty} x[n] z^{-n} \]

Figure 10.8 (a) ROC for right-sided sequence;

A right-sided sequence means \(x[n]\) is zero for \(n\) less than some value \(N_1\). If \(X(z)\) converges for some radius \(r_0\), it will also converge for any radius \(r_1 > r_0\). This is because the term \(r^{-n}\) decays faster for larger \(r\) when \(n\) is positive, and since \(x[n]\) is zero for negative \(n\), there are no terms that would grow unboundedly for increasing \(r\). Therefore, the ROC extends outwards from some boundary.

Property 5: Left-Sided Sequences

Statement:

If \(x[n]\) is a left-sided sequence, and if the circle \(|z|=r_0\) is in the ROC, then all values of \(z\) for which \(0<|z|<r_0\) will also be in the ROC.

Interpretation:

The ROC for a left-sided sequence is an interior region, extending inward from the innermost pole.

• Anti-causal sequences: If \(x[n]=0\) for \(n>0\), then \(N_2 \le 0\). The ROC includes \(z=0\).
• Non-anti-causal left-sided sequences: If \(N_2 > 0\), the ROC does not include \(z=0\).

\[ X(z)=\sum_{n=-\infty}^{N_{2}} x[n] z^{-n} \]

Figure 10.8 (b) ROC for left-sided sequence;

Similarly, for a left-sided sequence (\(x[n]\) is zero for \(n\) greater than some value \(N_2\)), if \(X(z)\) converges for \(r_0\), it will converge for any \(r_1 < r_0\). This is because \(r^{-n}\) grows faster for smaller \(r\) when \(n\) is negative. Since \(x[n]\) is zero for positive \(n\), there are no terms that would grow unboundedly for decreasing \(r\). Hence, the ROC extends inwards from some boundary.

Property 6: Two-Sided Sequences

Statement:

If \(x[n]\) is two-sided, and if the circle \(|z|=r_0\) is in the ROC, then the ROC will consist of an annular ring in the \(z\)-plane that includes the circle \(|z|=r_0\).

Justification:

A two-sided sequence can be decomposed into a right-sided component and a left-sided component.

• The ROC of the right-sided component is an exterior region (\(|z| > R_1\)).
• The ROC of the left-sided component is an interior region (\(|z| < R_2\)).
• The ROC of the two-sided sequence is the intersection of these two regions: \(R_1 < |z| < R_2\).

Important

For a two-sided sequence, the ROC is always an annular region between two circles.

Figure 10.8 (c) intersection of the ROCs in (a) and (b), representing the ROC for a two-sided sequence that is the sum of the right-sided and the left-sided sequence.

When a signal has values for both positive and negative time indices, it’s a two-sided sequence. We can think of it as a sum of a right-sided and a left-sided signal. The ROC of the combined signal must be the intersection of the individual ROCs. This intersection results in an annular ring, bounded by circles whose radii are determined by the poles of the Z-transform.

Example 10.6: Finite Length Sequence

Signal: \[ x[n]= \begin{cases}a^{n}, & 0 \leq n \leq N-1, a>0 \\ 0, & \text { otherwise }\end{cases} \] Z-Transform: \[ X(z) = \sum_{n=0}^{N-1} (az^{-1})^n = \frac{1-(az^{-1})^N}{1-az^{-1}} \] This can be rewritten as: \[ X(z) = \frac{z^{N}-a^{N}}{z^{N-1}(z-a)} \] Poles and Zeros:

• Poles: \(z=0\) (of order \(N-1\)) and \(z=a\).
• Zeros: Roots of \(z^N - a^N = 0\), which are \(z_k = a e^{j(2\pi k/N)}\) for \(k=0, 1, \dots, N-1\).
• The zero for \(k=0\) is \(z_0 = a\), which cancels the pole at \(z=a\).
• Thus, the only remaining pole is at \(z=0\) (of order \(N-1\)).

ROC:

• Since \(x[n]\) is finite duration (\(N_1=0, N_2=N-1\)), the ROC is the entire \(z\)-plane except possibly \(z=0\) and/or \(z=\infty\).
• Because \(x[n]\) is nonzero for \(n>0\), there are negative powers of \(z\) in \(X(z)\), which means \(X(z)\) is unbounded at \(z=0\).
• Therefore, the ROC is the entire \(z\)-plane, except \(z=0\).

Figure 10.9 Pole-zero pattern for Example 10.6 with \(N=16\) and \(0<a<1\). The region of convergence for this example consists of all values of \(z\) except \(z=0\).

This example demonstrates a finite-length sequence. Even though the initial form of \(X(z)\) might seem to have a pole at \(z=a\), a cancellation occurs with one of the zeros. This leaves only a pole at \(z=0\). Since the signal is right-sided and starts at \(n=0\), the Z-transform will have terms with negative powers of \(z\), making it unbounded at \(z=0\). Hence, the ROC excludes \(z=0\).

Example 10.7: Two-Sided Sequence \(x[n]=b^{|n|}\)

Signal: \(x[n]=b^{|n|}\), with \(b>0\).

This is a two-sided sequence. We can write it as: \[ x[n] = b^n u[n] + b^{-n} u[-n-1] \] Individual Z-Transforms:

1. Right-sided part: \(b^n u[n]\)

   \(X_R(z) = \frac{1}{1-b z^{-1}}\), with ROC \(R_R: |z|>b\).

2. Left-sided part: \(b^{-n} u[-n-1]\)

   \(X_L(z) = \frac{-1}{1-b^{-1} z^{-1}}\), with ROC \(R_L: |z|<1/b\).

Combined Z-Transform:

\(X(z) = X_R(z) + X_L(z) = \frac{1}{1-b z^{-1}} - \frac{1}{1-b^{-1} z^{-1}}\) \[ X(z) = \frac{(1-b^{-1}z^{-1}) - (1-bz^{-1})}{(1-bz^{-1})(1-b^{-1}z^{-1})} = \frac{(b-b^{-1})z^{-1}}{(1-bz^{-1})(1-b^{-1}z^{-1})} \] \[ X(z) = \frac{(b^2-1)/b \cdot z}{(z-b)(z-1/b)} \]

Example 10.7: Two-Sided Sequence \(x[n]=b^{|n|}\)

ROC Analysis:

• For \(X(z)\) to converge, both \(X_R(z)\) and \(X_L(z)\) must converge.
• This means the ROC is the intersection of \(R_R\) and \(R_L\).

Case 1: \(b>1\)

• \(R_R: |z|>b\)
• \(R_L: |z|<1/b\)
• Since \(b>1\), then \(1/b<1\). So, \(b > 1/b\).
• The regions \(|z|>b\) and \(|z|<1/b\) do not overlap. No common ROC. Therefore, for \(b>1\), \(x[n]=b^{|n|}\) does not have a Z-transform.

Case 2: \(0<b<1\)

• \(R_R: |z|>b\)
• \(R_L: |z|<1/b\)
• Since \(0<b<1\), then \(1/b>1\). So, \(b < 1/b\).
• The regions overlap. Common ROC: \(b < |z| < 1/b\). This is an annular region.

Poles are at \(z=b\) and \(z=1/b\). The ROC is between these two poles.

This example is a classic illustration of a two-sided sequence and how the ROC of its components determines the overall ROC. We break \(b^{|n|}\) into a right-sided and a left-sided part.

The crucial part is analyzing the intersection of their ROCs. If \(b>1\), the ROCs do not overlap, meaning the signal \(b^{|n|}\) does not have a Z-transform. This is because the signal grows too fast for both positive and negative \(n\).

However, if \(0<b<1\), the ROCs overlap, forming an annular region. This means the Z-transform exists for this range of \(b\).

Interactive: ROC for \(x[n]=b^{|n|}\)

Let’s visualize the ROC for \(x[n]=b^{|n|}\).

viewof b_val = Inputs.range([0.1, 2.0], {value: 0.5, step: 0.1, label: "Value of 'b'"});

Use this interactive plot to see how the parameter ‘b’ affects the poles and the ROC. When ‘b’ is less than 1, you’ll see an annular ROC, and the unit circle will be within this ROC, implying the DTFT exists.

However, as ‘b’ increases to 1 or greater, the two poles move closer or cross each other. When \(b \ge 1\), the ROCs of the individual components (\(|z|>b\) and \(|z|<1/b\)) no longer overlap, and thus the Z-transform for \(b^{|n|}\) does not exist. This illustrates a critical point: not all signals have a Z-transform.

Properties 7, 8, 9: Rational Z-Transforms

Property 7: ROC Bounded by Poles

If \(X(z)\) is rational, its ROC is bounded by poles or extends to infinity.

Property 8: Rational Right-Sided Sequences

If \(X(z)\) is rational and \(x[n]\) is right-sided:

• The ROC is the region outside the outermost pole.
• If \(x[n]\) is causal (\(x[n]=0\) for \(n<0\)), the ROC also includes \(z=\infty\).

Property 9: Rational Left-Sided Sequences

If \(X(z)\) is rational and \(x[n]\) is left-sided:

• The ROC is the region inside the innermost nonzero pole.
• If \(x[n]\) is anti-causal (\(x[n]=0\) for \(n>0\)), the ROC also includes \(z=0\).

These properties summarize how the ROC behaves for rational Z-transforms, which are very common in practical engineering problems. They essentially state that for rational transforms, the poles define the boundaries of the ROC, and the type of sequence (right-sided, left-sided, two-sided) dictates whether the ROC is an exterior, interior, or annular region, respectively.

Example 10.8: Multiple ROCs for a Single \(X(z)\)

Given Z-Transform: \[ X(z)=\frac{1}{\left(1-\frac{1}{3} z^{-1}\right)\left(1-2 z^{-1}\right)} \] Rewrite in powers of \(z\): \[ X(z)=\frac{z^2}{\left(z-\frac{1}{3}\right)(z-2)} \] Poles: \(z=1/3\) and \(z=2\).

Zeros: \(z=0\) (of order 2).

Since the algebraic expression for \(X(z)\) is fixed, the pole-zero pattern is fixed. However, different ROCs can be associated with this pattern, leading to different time-domain signals \(x[n]\).

Example 10.8: Multiple ROCs for a Single \(X(z)\)

Possible ROCs:

1. Right-Sided Sequence:
   • ROC is outside the outermost pole.
   • \(|z| > 2\).
   • Corresponds to \(x[n] = (1/3)^n u[n] - (2)^n u[n]\).
2. Left-Sided Sequence:
   • ROC is inside the innermost nonzero pole.
   • \(|z| < 1/3\).
   • Corresponds to \(x[n] = -(1/3)^n u[-n-1] + (2)^n u[-n-1]\).
3. Two-Sided Sequence:
   • ROC is an annular region between the poles.
   • \(1/3 < |z| < 2\).
   • Corresponds to \(x[n] = -(1/3)^n u[-n-1] - (2)^n u[n]\).

This example beautifully demonstrates the critical importance of the ROC. The same algebraic expression for \(X(z)\) can represent three entirely different signals in the time domain, depending solely on the specified ROC.

This is why you must always specify the ROC when defining a Z-transform! It’s not just a mathematical detail; it carries essential information about the nature of the signal.

Example 10.8: Visualizing the Multiple ROCs

Pole-Zero Pattern for \(X(z)=\frac{z^2}{(z-1/3)(z-2)}\)

1. Right-Sided Sequence (ROC: \(|z|>2\))

(a) ROC for right-sided sequence

Here, we see the shared pole-zero pattern. The poles are at \(1/3\) and \(2\), and there’s a double zero at the origin.

For the right-sided case, the ROC extends outwards from the largest pole (at \(z=2\)). This is typical for causal or right-sided signals.

Example 10.8: Visualizing the Multiple ROCs (Cont.)

2. Left-Sided Sequence (ROC: \(|z|<1/3\))

(b) ROC for left-sided sequence

3. Two-Sided Sequence (ROC: \(1/3 < |z| < 2\))

(c) intersection of the ROCs in (a) and (b), representing the ROC for a two-sided sequence that is the sum of the right-sided and the left-sided sequence

For the left-sided case, the ROC is inside the smallest pole (at \(z=1/3\)). This is characteristic of anti-causal or left-sided signals.

Finally, for the two-sided case, the ROC is the annular region between the two poles. This is the only scenario where the unit circle (which has radius 1, and \(1/3 < 1 < 2\)) is included in the ROC, meaning only this two-sided sequence has a converging Discrete-Time Fourier Transform. This clearly shows how the ROC provides crucial information beyond just the algebraic expression of \(X(z)\).

Summary of ROC Properties

Key Properties:

1. Ring-shaped: ROC is always an annular region centered at the origin.
2. No Poles: The ROC never contains any poles. Boundaries are defined by poles.
3. Finite Duration: ROC is entire \(z\)-plane, possibly excluding \(z=0\) and/or \(z=\infty\).
4. Right-Sided: ROC is an exterior region, extending outward from the outermost pole.
5. Left-Sided: ROC is an interior region, extending inward from the innermost nonzero pole.
6. Two-Sided: ROC is an annular region between two poles.
7. Rational \(X(z)\): ROC is bounded by poles or extends to infinity.

Summary of ROC Properties

Implications for System Analysis:

• Causality: For a causal LTI system, the ROC of its transfer function \(H(z)\) must be an exterior region, extending to infinity.
• Stability: For a stable LTI system, the ROC of \(H(z)\) must include the unit circle.
• Uniqueness: The Z-transform and its ROC uniquely determine the time-domain signal.
• DTFT Existence: The Discrete-Time Fourier Transform (DTFT) exists if and only if the ROC of the Z-transform includes the unit circle.

Important

Understanding the ROC is fundamental to analyzing discrete-time signals and systems, particularly for determining causality, stability, and the existence of the Fourier Transform.

This slide provides a concise summary of all the ROC properties discussed. Emphasize that these properties are not just theoretical constructs but have direct, practical implications for characterizing signals and, more importantly, for understanding the behavior of discrete-time LTI systems. The relationship between ROC and system properties like causality and stability is a recurring theme in Signals and Systems.