Transfer function H(z) in Signal Processing - Step-by-Step Execution

Concept Flow - Transfer function H(z)

Start: Define input signal x[n

↓

Apply Z-transform to x[n

↓

Define system difference equation

↓

Take Z-transform of system equation

↓

Express output Y(z) in terms of X(z)

↓

Define Transfer function H(z) = Y(z)/X(z)

↓

Analyze H(z) for system behavior

The flow shows how an input signal is transformed using Z-transform, then the system's difference equation is converted to Z-domain to define the transfer function H(z), which relates output to input.

Execution Sample

Signal Processing

import numpy as np
z = 0.9 + 0.1j
b = [1, -0.5]
a = [1, -0.8]
H_z = np.polyval(b, z) / np.polyval(a, z)
print(H_z)

Calculate H(z) at a specific z value using numerator and denominator polynomials.

Execution Table

Step	Action	Input	Computation	Result
1	Define numerator coefficients b	[1, -0.5]	Store as polynomial numerator	b = [1, -0.5]
2	Define denominator coefficients a	[1, -0.8]	Store as polynomial denominator	a = [1, -0.8]
3	Choose z value	z = 0.9 + 0.1j	Use for evaluation	z = 0.9 + 0.1j
4	Evaluate numerator polynomial at z	b, z	1*(z^1) + (-0.5)	num = 0.9+0.1j - 0.5 = 0.4+0.1j
5	Evaluate denominator polynomial at z	a, z	1*(z^1) + (-0.8)	den = 0.9+0.1j - 0.8 = 0.1+0.1j
6	Calculate H(z) = num/den	num, den	(0.4+0.1j)/(0.1+0.1j)	H(z) = 2.5 - 1.5j
7	Output H(z)	H(z)	Print result	2.5 - 1.5j

💡 Completed evaluation of H(z) at chosen z value

Variable Tracker

Variable	Start	After Step 4	After Step 5	After Step 6	Final
b	[1, -0.5]	[1, -0.5]	[1, -0.5]	[1, -0.5]	[1, -0.5]
a	[1, -0.8]	[1, -0.8]	[1, -0.8]	[1, -0.8]	[1, -0.8]
z	undefined	0.9+0.1j	0.9+0.1j	0.9+0.1j	0.9+0.1j
num	undefined	0.4+0.1j	0.4+0.1j	0.4+0.1j	0.4+0.1j
den	undefined	undefined	0.1+0.1j	0.1+0.1j	0.1+0.1j
H(z)	undefined	undefined	undefined	2.5-1.5j	2.5-1.5j

Key Moments - 3 Insights

Why do we evaluate polynomials at a complex number z?

Because H(z) is defined in the complex z-plane, evaluating numerator and denominator polynomials at z gives the system's frequency response at that point (see steps 4 and 5 in execution_table).

Why is H(z) a ratio of two polynomials?

The system's difference equation transforms into a ratio of polynomials in z after Z-transform, representing output over input (step 6 shows division of numerator by denominator).

What does the value of H(z) tell us?

It tells how the system modifies the input signal at frequency corresponding to z; the complex value shows magnitude and phase (step 7 outputs this result).

Visual Quiz - 3 Questions

Look at the execution_table step 4, what is the numerator polynomial evaluated at z?

A 2.5 - 1.5j

B 0.1 + 0.1j

C 0.4 + 0.1j

D 1 - 0.5

💡 Hint

Check the 'Result' column at step 4 in execution_table

At which step does the denominator polynomial get evaluated at z?

A Step 5

B Step 6

C Step 3

D Step 7

💡 Hint

Look for 'Evaluate denominator polynomial' in the 'Action' column

If z was changed to 1 + 0j, how would the value of H(z) change at step 6?

A H(z) would be (1 + 0.1j)/(0.1 + 0.1j)

B H(z) would be (1 - 0.5)/(1 - 0.8) = 0.5/0.2 = 2.5

C H(z) would be zero

D H(z) would be undefined

💡 Hint

Evaluate numerator and denominator polynomials at z=1+0j using steps 4 and 5 logic

Concept Snapshot

Transfer function H(z) relates output Y(z) to input X(z) in Z-domain.
H(z) = Y(z)/X(z) = B(z)/A(z), ratio of numerator and denominator polynomials.
Evaluate polynomials at complex z to find system response.
Used to analyze discrete-time system behavior and stability.

Full Transcript

The transfer function H(z) is a key concept in signal processing that relates the output and input of a system in the Z-domain. We start by defining the input signal and applying the Z-transform to get X(z). The system's difference equation is also transformed to Z-domain, resulting in a ratio of polynomials B(z) and A(z). This ratio, H(z) = B(z)/A(z), is the transfer function. To analyze the system, we evaluate these polynomials at a complex number z, which corresponds to a frequency point. The example code shows how to calculate H(z) by evaluating numerator and denominator polynomials at a chosen z value and dividing them. The execution table traces each step, showing how variables change and how the final H(z) value is computed. Understanding this process helps in analyzing system frequency response and stability.

Practice

1. What does the transfer function H(z) represent in signal processing?

easy

A. The amplitude of the output signal

B. The time delay of a signal

C. The frequency of the input signal

D. The ratio of output to input signals in the z-domain

Solution

1. Step 1: Understand the definition of transfer function

   The transfer function H(z) is defined as the ratio of the output signal to the input signal in the z-domain.

2. Step 2: Identify the correct description

   Among the options, only The ratio of output to input signals in the z-domain correctly describes H(z) as this ratio.

3. Final Answer:

   The ratio of output to input signals in the z-domain -> Option D

4. Quick Check:

   Transfer function = output/input ratio [OK]

Hint: Remember: H(z) = output signal ÷ input signal [OK]

Common Mistakes:

• Confusing H(z) with time delay
• Thinking H(z) is just frequency
• Mixing amplitude with transfer function

2. Which of the following is the correct general form of a digital filter transfer function H(z)?

easy

A. H(z) = (b0 + b1z + b2z^2) / (1 + a1z + a2z^2)

B. H(z) = (b0 + b1z^{-1} + b2z^{-2}) * (1 + a1z^{-1} + a2z^{-2})

C. H(z) = (b0 + b1z^{-1} + b2z^{-2}) / (1 + a1z^{-1} + a2z^{-2})

D. H(z) = (b0z + b1z^2 + b2z^3) / (1 + a1z + a2z^2)

Solution

1. Step 1: Recall the standard form of digital filter transfer function

   The transfer function is usually expressed with negative powers of z, representing delays: numerator and denominator polynomials in z^{-1}.

2. Step 2: Compare options with the standard form

   H(z) = (b0 + b1z^{-1} + b2z^{-2}) / (1 + a1z^{-1} + a2z^{-2}) correctly shows numerator and denominator with negative powers of z, matching the standard form.

3. Final Answer:

   H(z) = (b0 + b1z^{-1} + b2z^{-2}) / (1 + a1z^{-1} + a2z^{-2}) -> Option C

4. Quick Check:

   Negative powers of z indicate delays [OK]

Hint: Look for negative powers of z in numerator and denominator [OK]

Common Mistakes:

• Using positive powers of z incorrectly
• Multiplying numerator and denominator instead of dividing
• Forgetting the leading 1 in denominator

3. Given the transfer function H(z) = (1 - 0.5z^{-1}) / (1 - 0.9z^{-1} + 0.15z^{-2}), what is the value of H(z) at z = 1?

medium

A. 2.0

B. 1.0

C. 0.5

D. 0.0

Solution

1. Step 1: Substitute z = 1 into H(z)

   Calculate numerator: 1 - 0.5 * 1^{-1} = 1 - 0.5 = 0.5
   Calculate denominator: 1 - 0.9 * 1^{-1} + 0.15 * 1^{-2} = 1 - 0.9 + 0.15 = 0.25

2. Step 2: Compute H(1) = numerator / denominator

   H(1) = 0.5 / 0.25 = 2.0 exactly.

3. Final Answer:

   2.0 -> Option A

4. Quick Check:

   H(1) = 2.0 [OK]

Hint: Plug z=1, compute numerator and denominator separately [OK]

Common Mistakes:

• Forgetting to use z^{-1} as 1/z
• Incorrect arithmetic in numerator or denominator
• Choosing zero instead of computed value

4. Identify the error in the following transfer function expression:
H(z) = (1 + 0.3z^{-1} - 0.2z^{-2}) / (1 + 0.5z + 0.1z^{-2})

medium

A. Numerator has negative powers of z which is invalid

B. Denominator has a positive power of z which is incorrect

C. Transfer function must not have more than two terms in numerator

D. Denominator should be multiplied, not divided

Solution

1. Step 1: Examine powers of z in numerator and denominator

   The numerator has negative powers of z, which is correct for delays.
   The denominator has a positive power term 0.5z, which is not standard and incorrect.

2. Step 2: Confirm correct form of denominator

   The denominator should have terms with negative powers of z (delays), not positive powers (advances). So the positive power term is an error.

3. Final Answer:

   Denominator has a positive power of z which is incorrect -> Option B

4. Quick Check:

   Denominator powers must be negative [OK]

Hint: Check denominator powers: must be negative or zero [OK]

Common Mistakes:

• Allowing positive powers of z in denominator
• Confusing numerator and denominator powers
• Thinking number of terms limits validity

5. You have a transfer function H(z) = (1 - 0.7z^{-1}) / (1 - 1.2z^{-1} + 0.45z^{-2}). Which statement best describes the system's stability based on the poles of H(z)?

hard

A. The system is stable because all poles lie inside the unit circle

B. The system is unstable because at least one pole lies outside the unit circle

C. The system is marginally stable because poles lie on the unit circle

D. Stability cannot be determined without zeros

Solution

1. Step 1: Find the poles by solving denominator polynomial

   The denominator is 1 - 1.2z^{-1} + 0.45z^{-2} = 0.
   Rewrite as z^2 - 1.2z + 0.45 = 0 (multiply both sides by z^2).

2. Step 2: Calculate roots of the quadratic

   Using quadratic formula:
   z = [1.2 ± sqrt(1.44 - 1.8)] / 2
   Discriminant = 1.44 - 1.8 = -0.36 (complex roots)
   Roots: z = 0.6 ± 0.3i

3. Step 3: Calculate magnitude of poles

   Magnitude = sqrt(0.6^2 + 0.3^2) = sqrt(0.36 + 0.09) = sqrt(0.45) ≈ 0.67 < 1

4. Step 4: Re-examine coefficients for stability

   Actually, the denominator polynomial is in z^{-1}, so poles are roots of denominator polynomial in z.
   But the coefficients suggest poles inside unit circle, so system is stable.

5. Step 5: Check option correctness

   Options say system is unstable if poles outside unit circle.
   Our calculation shows poles inside unit circle, so system is stable.

6. Final Answer:

   The system is stable because all poles lie inside the unit circle -> Option A

7. Quick Check:

   Poles magnitude < 1 means stable [OK]

Hint: Calculate pole magnitudes; inside unit circle means stable [OK]

Common Mistakes:

• Confusing zeros with poles for stability
• Ignoring complex roots magnitude
• Assuming zeros affect stability