Common Z-transform pairs in Signal Processing - Step-by-Step Execution

Concept Flow - Common Z-transform pairs

Start with time-domain sequence x[n

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Apply Z-transform formula: X(z) = Σ x[n

↓

Simplify sum to get X(z)

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Identify common pair from table

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Use pair for analysis or inverse transform

The flow shows how a time-domain sequence is converted to its Z-transform using the formula, then matched to common pairs for easy analysis.

Execution Sample

Signal Processing

x[n] = a^n u[n]
X(z) = Σ (a^n) z^(-n) from n=0 to ∞
X(z) = 1 / (1 - a z^{-1}) for |z| > |a|

This code calculates the Z-transform of the sequence a^n times the unit step, resulting in a common transform pair.

Execution Table

Step	Expression	Action	Result	Condition
1	x[n] = a^n u[n]	Define sequence	Sequence ready for transform	n ≥ 0
2	X(z) = Σ x[n] z^{-n}	Apply Z-transform sum	Sum expression formed	n=0 to ∞
3	Σ (a^n) z^{-n}	Substitute x[n]	Σ (a z^{-1})^n	n=0 to ∞
4	Sum of geometric series	Simplify sum	1 / (1 - a z^{-1})	|z| > |a|
5	Identify pair	Match to common Z-transform pair	Confirmed pair	Valid region of convergence
6	-	End	Z-transform found	-

💡 Sum converges only if |z| > |a|, otherwise transform does not exist.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
x[n]	undefined	a^n u[n]	a^n u[n]	a^n u[n]	a^n u[n]
X(z)	undefined	Σ a^n z^{-n}	Σ (a z^{-1})^n	1 / (1 - a z^{-1})	1 / (1 - a z^{-1})

Key Moments - 3 Insights

Why do we need the condition |z| > |a| for the Z-transform to exist?

Because the sum is a geometric series that converges only if the common ratio |a z^{-1}| < 1, which means |z| > |a|. This is shown in step 4 of the execution_table.

Why do we start the sum from n=0 and not from negative n?

Because the sequence x[n] includes the unit step u[n], which is zero for n < 0, so terms before n=0 do not contribute. This is clear in step 1 and step 2.

How do we simplify the sum Σ (a z^{-1})^n to a closed form?

We recognize it as a geometric series with ratio r = a z^{-1}, and use the formula sum = 1/(1-r) when |r|<1, as shown in step 4.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the simplified form of the Z-transform?

A Σ a^n z^n

B a / (1 - z^{-1})

C 1 / (1 - a z^{-1})

D 1 - a z^{-1}

💡 Hint

Refer to the 'Result' column at step 4 in the execution_table.

At which step does the condition |z| > |a| become important?

A Step 4

B Step 1

C Step 2

D Step 5

💡 Hint

Check the 'Condition' column in the execution_table where convergence is discussed.

If the sequence was x[n] = a^n without the unit step u[n], how would the sum limits change?

A Sum from n=0 to ∞

B Sum from n=-∞ to ∞

C Sum from n=1 to ∞

D Sum from n=-∞ to 0

💡 Hint

Without the unit step, the sequence exists for all n, so sum limits cover all integers.

Concept Snapshot

Z-transform converts discrete time sequences x[n] into X(z) using X(z) = Σ x[n] z^{-n}.
Common pairs like x[n] = a^n u[n] transform to X(z) = 1/(1 - a z^{-1}) for |z| > |a|.
The region of convergence ensures the sum converges.
Use these pairs to analyze signals and systems easily.

Full Transcript

The Z-transform takes a sequence x[n] and converts it into a function X(z) by summing x[n] times z to the negative n power. For example, the sequence a^n times the unit step u[n] has a Z-transform of 1 divided by (1 minus a times z to the negative one). This sum converges only if the magnitude of z is greater than the magnitude of a. We start summing from n equals zero because the unit step is zero for negative n. Recognizing the sum as a geometric series helps simplify it to a closed form. These common pairs help us analyze signals and systems in the z-domain efficiently.

Practice

1. Which of the following is the Z-transform of the sequence x[n] = a^n u[n], where u[n] is the unit step?

easy

A. X(z) = z / (z - a), |z| < |a|

B. X(z) = 1 / (1 - a z^{-1}), |z| > |a|

C. X(z) = (1 - a z^{-1}) / z, |z| > |a|

D. X(z) = a / (1 - z^{-1}), |z| > 1

Solution

1. Step 1: Recall the definition of the Z-transform for geometric sequences

   The Z-transform of a^n u[n] is a standard pair given by the sum of a geometric series: \( \sum_{n=0}^\infty a^n z^{-n} = \frac{1}{1 - a z^{-1}} \) for \( |z| > |a| \).

2. Step 2: Match the formula with the options

   X(z) = 1 / (1 - a z^{-1}), |z| > |a| matches the formula exactly with the correct region of convergence.

3. Final Answer:

   X(z) = 1 / (1 - a z^{-1}), |z| > |a| -> Option B

4. Quick Check:

   Geometric sequence Z-transform = 1/(1 - a z^{-1}) [OK]

Hint: Geometric sequences transform to 1/(1 - a z^{-1}) [OK]

Common Mistakes:

• Confusing region of convergence
• Mixing numerator and denominator terms
• Using wrong inequality for |z|

2. Which of the following is the correct Z-transform expression for the unit impulse sequence δ[n]?

easy

A. X(z) = z / (z - 1)

B. X(z) = z^{-1}

C. X(z) = 1 / (1 - z^{-1})

D. X(z) = 1

Solution

1. Step 1: Understand the unit impulse sequence

   The unit impulse δ[n] is 1 at n=0 and 0 elsewhere.

2. Step 2: Calculate the Z-transform

   By definition, \( X(z) = \sum_{n=-\infty}^\infty δ[n] z^{-n} = z^{0} = 1 \).

3. Final Answer:

   X(z) = 1 -> Option D

4. Quick Check:

   Impulse Z-transform = 1 [OK]

Hint: Impulse transforms to 1 always [OK]

Common Mistakes:

• Confusing impulse with step function
• Adding z terms unnecessarily
• Forgetting impulse is zero except at n=0

3. What is the Z-transform of the sequence x[n] = n u[n]?

medium

A. X(z) = z^{-1} / (1 - z^{-1})^2, |z| > 1

B. X(z) = 1 / (1 - z^{-1})^2, |z| > 1

C. X(z) = z / (z - 1), |z| > 1

D. X(z) = 1 / (1 - z)^2, |z| > 1

Solution

1. Step 1: Recall the Z-transform of the sequence n u[n]

   The sequence n u[n] has Z-transform \( X(z) = \frac{z^{-1}}{(1 - z^{-1})^2} \) with region of convergence \( |z| > 1 \).

2. Step 2: Match the formula with the options

   X(z) = z^{-1} / (1 - z^{-1})^2, |z| > 1 matches the formula exactly including the region of convergence.

3. Final Answer:

   X(z) = z^{-1} / (1 - z^{-1})^2, |z| > 1 -> Option A

4. Quick Check:

   Linear sequence Z-transform = z^{-1}/(1 - z^{-1})^2 [OK]

Hint: Linear sequences transform to z^{-1}/(1 - z^{-1})^2 [OK]

Common Mistakes:

• Confusing powers of z
• Using (z - 1) instead of (1 - z^{-1})
• Ignoring region of convergence

4. Identify the error in the following Z-transform expression for x[n] = a^n u[n]: X(z) = 1 / (1 - z a^{-1})

medium

A. The variable z and a are swapped in the denominator

B. The region of convergence is missing

C. The exponent on a should be negative

D. The formula is correct as is

Solution

1. Step 1: Recall the correct Z-transform formula for a^n u[n]

   The correct formula is \( X(z) = \frac{1}{1 - a z^{-1}} \), not \( 1/(1 - z a^{-1}) \).

2. Step 2: Analyze the given expression

   The given expression swaps z and a in the denominator, which changes the meaning and is incorrect.

3. Final Answer:

   The variable z and a are swapped in the denominator -> Option A

4. Quick Check:

   Correct formula uses a z^{-1}, not z a^{-1} [OK]

Hint: Check order of variables in denominator carefully [OK]

Common Mistakes:

• Swapping z and a in denominator
• Ignoring negative powers of z
• Missing region of convergence

5. Given the sequence x[n] = (1/2)^n u[n] + 3 δ[n], what is the combined Z-transform X(z)?

hard

A. X(z) = 3 * z / (z - 1/2), |z| > 1/2

B. X(z) = 3 z^{-1} + 1 / (1 - (1/2) z^{-1}), |z| > 1/2

C. X(z) = 3 + 1 / (1 - (1/2) z^{-1}), |z| > 1/2

D. X(z) = 3 + 1 / (1 - 2 z^{-1}), |z| > 2

Solution

1. Step 1: Find Z-transform of each sequence part

   The Z-transform of (1/2)^n u[n] is \( \frac{1}{1 - (1/2) z^{-1}} \) with \( |z| > 1/2 \). The Z-transform of 3 δ[n] is 3.

2. Step 2: Add the transforms for combined sequence

   Since Z-transform is linear, add both: \( X(z) = 3 + \frac{1}{1 - (1/2) z^{-1}} \).

3. Final Answer:

   X(z) = 3 + 1 / (1 - (1/2) z^{-1}), |z| > 1/2 -> Option C

4. Quick Check:

   Sum of transforms = 3 + 1/(1 - 0.5 z^{-1}) [OK]

Hint: Sum transforms for combined sequences linearly [OK]

Common Mistakes:

• Multiplying instead of adding transforms
• Incorrect region of convergence
• Wrong coefficient placement