Region of convergence in Signal Processing - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to compute the Z-transform of a sequence.

Signal Processing

import numpy as np

x = [1, 2, 3]
z = np.array([1, -1j, 1j])
X = sum(x[n] * z[1]n for n in range(len(x)))
print(X)

Drag options to blanks, or click blank then click option'

A**

B*

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '*' instead of '**' causes a multiplication error.

Using '+' or '-' changes the meaning and causes wrong results.

The Z-transform uses powers of z, so we raise z to the power n using '**'.

2fill in blank

medium

Complete the code to check if the magnitude of z is inside the region of convergence (ROC).

Signal Processing

z = 0.5 + 0.5j
roc_radius = 1.0
if abs(z) [1] roc_radius:
    print('Inside ROC')
else:
    print('Outside ROC')

Drag options to blanks, or click blank then click option'

A>

B==

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' reverses the condition and gives wrong results.

Using '==' is too strict and rarely true.

The ROC includes points where the magnitude of z is less than the radius.

3fill in blank

hard

Fix the error in the code to compute the ROC for a causal sequence with pole at p.

Signal Processing

p = 0.8
roc = {z for z in np.linspace(0, 2, 100) if abs(z) [1] abs(p)}
print(len(roc))

Drag options to blanks, or click blank then click option'

A>

B<

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' selects inside the pole, which is incorrect for causal sequences.

Using '<=' or '>=' changes the boundary inclusion incorrectly.

For a causal sequence, the ROC is outside the pole, so |z| > |p|.

4fill in blank

hard

Fill both blanks to create a dictionary of z values and their ROC status for a pole at p.

Signal Processing

p = 0.6
z_values = np.linspace(0, 1.5, 10)
roc_dict = {z: 'inside' if abs(z) [1] abs(p) else 'outside' for z in z_values if z [2] 0}
print(roc_dict)

Drag options to blanks, or click blank then click option'

A<

B>

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' in the first blank reverses the ROC condition.

Using '<=' or '>=' changes the strictness of the condition incorrectly.

We check if |z| < |p| for inside ROC and filter z > 0 to avoid zero or negative values.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps z to magnitude and ROC status for poles p1 and p2.

Signal Processing

p1, p2 = 0.5, 1.2
z_vals = np.linspace(0, 2, 15)
roc_info = {z: (abs(z), 'inside' if abs(z) [1] abs(p1) and abs(z) [2] abs(p2) else 'outside') for z in z_vals if z [3] 0}
print(roc_info)

Drag options to blanks, or click blank then click option'

A>

B<

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Reversing inequalities causes wrong ROC classification.

Not filtering positive z values includes zero or negatives.

The ROC is between poles, so |z| < |p1| and |z| <= |p2|, filtering z > 0.

Practice

1.

What does the Region of Convergence (ROC) represent in signal processing?

easy

A. The frequency range of the signal

B. The set of values where the signal's transform converges

C. The time duration of the signal

D. The amplitude range of the signal

Solution

1. Step 1: Understand the definition of ROC

   The ROC is the set of complex values for which the signal's transform (like Z-transform or Laplace transform) converges.

2. Step 2: Differentiate ROC from other signal properties

   Frequency range, time duration, and amplitude range describe different aspects, not convergence.

3. Final Answer:

   The set of values where the signal's transform converges -> Option B

4. Quick Check:

   ROC = convergence region [OK]

Hint: ROC means where transform sums up properly [OK]

Common Mistakes:

• Confusing ROC with frequency range
• Thinking ROC is about signal duration
• Mixing amplitude with convergence

2.

Which of the following is the correct syntax to denote the ROC of a Z-transform X(z)?

easy

A. ROC = \{z : |z| > r\}

B. ROC = \{z : z = 0\}

C. ROC = \{z : |z| = 0\}

D. ROC = \{z : |z| < 0\}

Solution

1. Step 1: Recall ROC form for causal signals

   For causal signals, ROC is outside a circle of radius r, so |z| > r.

2. Step 2: Check other options for correctness

   Options B, C, and D are invalid because |z| cannot be zero or negative for ROC.

3. Final Answer:

   ROC = \{z : |z| > r\} -> Option A

4. Quick Check:

   ROC syntax uses magnitude inequalities [OK]

Hint: ROC usually involves |z| > or < some radius [OK]

Common Mistakes:

• Using equality instead of inequality for ROC
• Setting ROC radius to zero
• Confusing ROC with single points

3.

Given the Z-transform X(z) = \frac{1}{1 - 0.5z^{-1}}, what is the ROC?

medium

A. |z| < 0.5

B. |z| >= 0.5

C. |z| = 0.5

D. |z| > 0.5

Solution

1. Step 1: Identify the pole location

   The denominator is zero when 1 - 0.5z^{-1} = 0, so z = 0.5.

2. Step 2: Determine ROC for causal sequence

   For a causal sequence, ROC is outside the outermost pole, so |z| > 0.5.

3. Final Answer:

   |z| > 0.5 -> Option D

4. Quick Check:

   ROC outside pole radius [OK]

Hint: ROC is outside the pole radius for causal signals [OK]

Common Mistakes:

• Choosing ROC inside the pole radius
• Confusing equality with inequality
• Ignoring pole location

4.

Find the error in this statement about ROC: "The ROC for a two-sided Z-transform always includes the unit circle |z| = 1."

medium

A. ROC never includes the unit circle

B. ROC always excludes the unit circle

C. ROC may or may not include the unit circle depending on poles

D. ROC is unrelated to the unit circle

Solution

1. Step 1: Understand ROC for two-sided signals

   Two-sided signals have ROC between poles, which may or may not include the unit circle.

2. Step 2: Analyze the statement's generality

   The statement is incorrect because ROC depends on pole locations; it is not guaranteed to include |z|=1.

3. Final Answer:

   ROC may or may not include the unit circle depending on poles -> Option C

4. Quick Check:

   ROC depends on poles, not fixed unit circle [OK]

Hint: ROC inclusion of unit circle depends on poles [OK]

Common Mistakes:

• Assuming ROC always includes unit circle
• Ignoring pole positions
• Confusing two-sided with causal signals

5.

You have a signal with Z-transform X(z) = \frac{z}{z - 2}. Which ROC ensures the signal is stable?

hard

A. |z| < 2

B. |z| = 2

C. |z| > 2

D. |z| >= 0

Solution

1. Step 1: Identify the pole location

   The pole is at z = 2.

2. Step 2: Recall stability condition

   For stability, ROC must include the unit circle |z|=1, so ROC must be inside the pole: |z| < 2.

3. Step 3: Choose ROC that includes unit circle

   Only ROC |z| < 2 includes |z|=1, ensuring stability.

4. Final Answer:

   |z| < 2 -> Option A

5. Quick Check:

   Stable ROC includes unit circle inside poles [OK]

Hint: Stable ROC must include unit circle inside poles [OK]

Common Mistakes:

• Choosing ROC outside pole for stability
• Ignoring unit circle inclusion
• Confusing stability with causality