Transfer function H(z) in Signal Processing - Mini Project: Build & Apply

Calculate Transfer Function H(z) from Filter Coefficients

📖 Scenario: You are working with digital filters in signal processing. Filters are described by their coefficients, and the transfer function H(z) shows how the filter changes signals in the frequency domain.Understanding H(z) helps you analyze and design filters for audio, communications, and control systems.

🎯 Goal: Build a Python program that calculates the transfer function H(z) for a digital filter given its numerator and denominator coefficients.You will create the coefficient lists, set a frequency point, compute H(z) at that point, and print the result.

📋 What You'll Learn

Create two lists called numerator and denominator with exact coefficients

Create a variable z representing a complex number on the unit circle

Calculate the transfer function H_z as the ratio of numerator and denominator polynomials evaluated at z

Print the value of H_z

💡 Why This Matters

🌍 Real World

Digital filters are used in audio processing, communications, and control systems to shape signals. Calculating H(z) helps engineers understand how filters affect different frequencies.

💼 Career

Signal processing engineers and data scientists working with time series or sensor data often analyze transfer functions to design and evaluate filters.

1

Create filter coefficient lists

Create a list called numerator with values [0.0675, 0.1349, 0.0675] and a list called denominator with values [1.0, -1.1430, 0.4128].

Signal Processing

# Your code here

Hint

Use square brackets to create lists with the exact numbers given.

2

Set the complex variable z on the unit circle

Create a variable called z and set it to the complex number exp(-1j * pi / 4) using the cmath module.

Signal Processing

numerator = [0.0675, 0.1349, 0.0675]
denominator = [1.0, -1.1430, 0.4128]
import cmath
# Your code here

Hint

Use cmath.exp() and cmath.pi to create the complex number.

3

Calculate the transfer function H(z)

Calculate the numerator polynomial value at z and store it in num_val. Calculate the denominator polynomial value at z and store it in den_val. Then calculate H_z as num_val / den_val.

Signal Processing

numerator = [0.0675, 0.1349, 0.0675]
denominator = [1.0, -1.1430, 0.4128]
import cmath
z = cmath.exp(-1j * cmath.pi / 4)
# Your code here

Hint

Use enumerate() to get index and coefficient, then sum coef * z**(-i) for numerator and denominator.

4

Print the transfer function value

Print the value of H_z.

Signal Processing

numerator = [0.0675, 0.1349, 0.0675]
denominator = [1.0, -1.1430, 0.4128]
import cmath
z = cmath.exp(-1j * cmath.pi / 4)
num_val = sum(coef * z**(-i) for i, coef in enumerate(numerator))
den_val = sum(coef * z**(-i) for i, coef in enumerate(denominator))
H_z = num_val / den_val
# Your code here

Hint

Use print(H_z) to display the transfer function value.

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