Impulse Response h(n) of Difference Equation | GTU Sem 4 Signals and Systems Summer 2021 Q5(c)

GTU Sem 4 | Signals and Systems | Summer 2021 | Question 5(c) Solution

📚 Table of Contents

• 🔹 Introduction
• 🔹 Question Statement
• 🔹 Theory Related to the Question
• 🔹 Solution in Detail
• 🔹 Handwritten Solution Images
• 🔹 Key Takeaways
• 🔹 Final Words
• 🔹 Related Posts

📖 Introduction

This blog post provides a detailed and structured solution to one of the frequently asked GTU questions on the subject of Signals and Systems (3141005). In this question, we are asked to determine the impulse response h(n) for a discrete-time LTI system described by a second-order linear difference equation. This falls under Unit 4 of the syllabus.

📘 Question Statement

Determine the impulse response h(n) for the system described by the difference equation:
y(n) – (3/4)y(n - 1) + (1/8)y(n - 2) = x(n)

📚 Theory Related to the Question

To find the impulse response h(n) of a linear time-invariant (LTI) system, we analyze the system’s output when the input is a unit impulse function, denoted as x(n) = δ(n). The output for this input is, by definition, the system's impulse response.

The general approach involves:

• Substituting x(n) = δ(n) into the given difference equation.
• Taking the Z-transform of both sides of the equation with initial rest conditions.
• Solving for H(z) = Y(z)/X(z) and then applying the inverse Z-transform (preferably using the partial fraction method).

🔍 Concept Refresher – Impulse Response in LTI Systems:

• The impulse response h(n) represents the system output when the input is δ(n).
• We use the Z-transform to convert the difference equation into algebraic form.
• The inverse Z-transform (via partial fractions) helps retrieve h(n) in the time domain.

🛠️ Detailed Solution Using Z-Transform & Partial Fractions

We are given the second-order linear constant-coefficient difference equation:

y(n) – (3/4)y(n - 1) + (1/8)y(n - 2) = x(n)

To determine the impulse response h(n), we assume the input x(n) = δ(n) and solve the equation under initial rest conditions (i.e., y(n) = 0 for n < 0).

💡 Important: We solve for impulse response using the Z-transform method and apply partial fraction expansion on H(z)/z instead of H(z).

📌 Step 1: Take the Z-Transform

Apply the Z-transform to both sides:

Y(z) – (3/4)z⁻¹Y(z) + (1/8)z⁻²Y(z) = X(z)

Factor out Y(z) on the left-hand side:

Y(z) [1 – (3/4)z⁻¹ + (1/8)z⁻²] = X(z)

Substitute X(z) = 1 (Z-transform of δ(n)) and multiply the numerator and denominator by z²:

Y(z) = z² / [z² – (3/4)z + 1/8]

📌 Step 2: Derive H(z) and Compute H(z)/z

Let H(z) be the system transfer function:

H(z) = Y(z)/X(z) = z² / [z² – (3/4)z + 1/8]

We apply partial fractions to H(z)/z:

H(z)/z = z / [(z – 1/2)(z – 1/4)]

📌 Step 3: Perform Partial Fraction Expansion

Let’s write:

z / [(z – 1/2)(z – 1/4)] = A / (z – 1/2) + B / (z – 1/4)

Multiply both sides by the denominator:

z = A(z – 1/4) + B(z – 1/2)

Now solve for A and B:

• Let z = 1/2 ⇒ 1/2 = A(1/4) ⇒ A = 2
• Let z = 1/4 ⇒ 1/4 = B(–1/4) ⇒ B = –1

So we have:

H(z)/z = 2 / (z – 1/2) – 1 / (z – 1/4)

H(z) = 2.z / (z – 1/2) – z / (z – 1/4)

📌 Step 4: Apply Inverse Z-Transform

We now apply the inverse Z-transform using the identity:

🧠 Identity Used:
Z⁻¹ { z / (z – a) } = aⁿ u(n)

Thus, the inverse Z-transform of H(z)/z gives:

h(n) = 2(1/2)ⁿ – (1/4)ⁿ , n ≥ 0

🚫 Common Mistakes to Avoid

• Not converting H(z) to H(z)/z before applying partial fractions.
• Incorrect factorization of the quadratic denominator.
• Sign errors while solving A and B in partial fractions.
• Skipping initial rest conditions.

🧾 Step-by-Step Summary Table

Step	Description
1	Apply Z-transform to the difference equation.
2	Derive H(z) = Y(z)/X(z) and compute H(z)/z.
3	Use partial fractions: H(z)/z = 2/(z – 1/2) – 1/(z – 1/4)
4	Apply inverse Z-transform using known identities.
5	Final result: h(n) = 2(1/2)ⁿ – (1/4)ⁿ for n ≥ 0

🖼️ Handwritten Solution Images

Below are the handwritten solution images that detail the step-by-step solution to the given system's impulse response using Z-transform and partial fractions.

🔑 Key Takeaways

• The Z-transform provides a powerful method to solve difference equations systematically.
• Partial fraction decomposition is essential for simplifying complex rational functions in Z-domain.
• The impulse response h(n) can be derived by applying the inverse Z-transform to the system's transfer function H(z)/z.
• Common mistakes include improper factorization, sign errors in partial fractions, and neglecting initial rest conditions.

💬 Final Words

By solving this problem, we learned how to apply Z-transform and partial fraction decomposition to determine the impulse response of a system. These methods are foundational for analyzing discrete-time systems in signal processing.

🗨️ Share Your Thoughts

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