Transfer function H(z) in Signal Processing - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the transfer function H(z) using numerator and denominator coefficients.

Signal Processing

from scipy.signal import TransferFunction
num = [1, -0.5]
den = [1, -0.9]
H = TransferFunction([1], den, dt=1)

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

Aden

B[0.5, 1]

C[1, 0]

Dnum

Attempts:

3 left

💡 Hint

Common Mistakes

Using denominator coefficients instead of numerator.

Passing a wrong list of coefficients.

The numerator coefficients are stored in num, which define the zeros of the transfer function H(z).

2fill in blank

medium

Complete the code to compute the frequency response of H(z) at 512 points.

Signal Processing

from scipy.signal import freqz
w, h = freqz(H.num, H.den, worN=[1])

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

A1024

B128

C512

D256

Attempts:

3 left

💡 Hint

Common Mistakes

Using too few points leading to rough plots.

Using a non-integer or invalid value.

512 points is a common choice for smooth frequency response plots.

3fill in blank

hard

Fix the error in the code to correctly calculate the magnitude response in decibels.

Signal Processing

import numpy as np
magnitude_db = 20 * np.log10([1])

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

Aabs(h)

Bh

Cnp.abs(h)**2

Dnp.log10(h)

Attempts:

3 left

💡 Hint

Common Mistakes

Using h directly without absolute value.

Squaring magnitude before log10 incorrectly.

The magnitude must be the absolute value of h before applying log10.

4fill in blank

hard

Fill both blanks to create a dictionary of frequency response with frequency in Hz and magnitude in dB.

Signal Processing

freq_response = [1]: w * [2] / (2 * np.pi) for w, mag in zip(w, magnitude_db)

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

A{'frequency_hz'

Bfs

C1

Dnp.pi

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong dictionary key names.

Incorrect frequency scaling factor.

The dictionary key is 'frequency_hz' and frequency is scaled by sampling frequency fs.

5fill in blank

hard

Fill all three blanks to filter a signal x using the transfer function H and store the output in y.

Signal Processing

from scipy.signal import lfilter
y = lfilter([1], [2], [3])

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

AH.num

BH.den

Cx

DH

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the transfer function object instead of coefficients.

Swapping numerator and denominator coefficients.

The lfilter function takes numerator coefficients, denominator coefficients, and the input signal.

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