Transfer function H(z) in Signal Processing - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a transfer function H(z) in signal processing?

H(z) is a mathematical representation of a digital filter or system in the z-domain. It relates the input signal to the output signal using complex frequency variable z.

intermediate

How is the transfer function H(z) typically expressed?

H(z) is usually expressed as a ratio of two polynomials in z: H(z) = \( \frac{Y(z)}{X(z)} = \frac{b_0 + b_1 z^{-1} + ... + b_M z^{-M}}{1 + a_1 z^{-1} + ... + a_N z^{-N}} \), where Y(z) is output and X(z) is input.

beginner

What do the coefficients \(b_i\) and \(a_i\) represent in H(z)?

The \(b_i\) coefficients are the feedforward filter coefficients, and the \(a_i\) coefficients are the feedback filter coefficients. They define the filter's behavior.

intermediate

Why is the z-transform used in defining H(z)?

The z-transform converts discrete time signals into a complex frequency domain, making it easier to analyze and design digital filters using algebraic methods.

beginner

What does the magnitude of H(z) tell us about a digital filter?

The magnitude of H(z) shows how much the filter amplifies or attenuates different frequency components of the input signal.

What does the transfer function H(z) relate in a digital system?

A Noise to signal

B Time to frequency

C Amplitude to phase

D Output signal to input signal

H(z) relates the output signal to the input signal in the z-domain.

Which variable is used in the transfer function H(z)?

A z

B f

C t

D s

The variable z is used in the z-transform and transfer function H(z).

In H(z), what do the coefficients \(a_i\) represent?

A Feedforward coefficients

B Feedback coefficients

C Input signal values

D Noise levels

The \(a_i\) coefficients represent feedback filter coefficients.

What is the main benefit of using the z-transform in digital filter design?

A Simplifies algebraic analysis

B Removes noise

C Converts signals to time domain

D Increases signal amplitude

The z-transform simplifies algebraic analysis of discrete signals and filters.

What does the magnitude response of H(z) indicate?

A Phase shift of the filter

B Time delay of the signal

C Amplification or attenuation of frequencies

D Noise reduction level

Magnitude response shows how the filter amplifies or attenuates frequency components.

Explain what a transfer function H(z) is and how it is used in digital signal processing.

Think about how digital filters process signals using algebraic expressions.

Describe the significance of the coefficients in the transfer function H(z) and how they affect the filter's behavior.

Consider how changing coefficients changes the output 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