GTU Sem 4 Signals & Systems Summer 2021 PYQ Question 2(b) Solution: Natural Response of the System
📘 Table of Contents
Introduction:
In this blog post, we will solve a difference equation and find the natural response of a discrete-time system using the Z-transform technique. The system is described by the following difference equation:
y(n) - 1.5y(n-1) + 0.5y(n-2) = x(n)
With initial conditions y(-1) = 1 and y(-2) = 0, we will find the solution when the input x(n) = 0 (since we are looking for the natural response).
Question Statement:
The problem is to find the natural response of the system defined by the following difference equation:
y(n) - 1.5y(n-1) + 0.5y(n-2) = x(n)
With the initial conditions:
- y(-1) = 1
- y(-2) = 0
Natural Response using Z-Transform – Signals & Systems Summer 2021 Question 2(b)
We are interested in finding the natural response, which occurs when the input x(n) = 0.
In this case, the difference equation y(n) - 1.5y(n-1) + 0.5y(n-2) = x(n) represents a second-order linear recurrence relation for the system's output, where we will use the Z-transform to find the natural response.
Theory Part Related to the Question:
To solve for the natural response, we will use the Z-transform method. This technique transforms a difference equation into an algebraic equation in the Z-domain. The steps are:
- Take the Z-transform of both sides of the equation.
- Solve for the transfer function in the Z-domain.
- Use inverse Z-transforms to find the solution in the time domain.
The Z-transform of a discrete-time signal y(n) is:
Z{y(n)} = Y(z)
The Z-transforms of the shifted terms y(n-1) and y(n-2) are:
Z{y(n-1)} = z-1Y(z) + y(-1)
Z{y(n-2)} = z-2Y(z) + z-1y(-1) + y(-2)
For a natural response, x(n) = 0 means no external input is applied.
Solution in Detail:
We are given the difference equation:
y(n) - 1.5y(n−1) + 0.5y(n−2) = x(n)
With initial conditions:
- y(−1) = 1
- y(−2) = 0
Since we are asked to find the natural response, we take x(n) = 0.
Step 1: Apply Z-transform
Taking Z-transform on both sides of the equation:
Y(z) - 1.5[z⁻¹Y(z) + y(−1)] + 0.5[z⁻²Y(z) + z⁻¹y(−1) + y(−2)] = 0
Step 2: Substitute initial values
Y(z) - 1.5[z⁻¹Y(z) + 1] + 0.5[z⁻²Y(z) + z⁻¹(1) + 0] = 0
Y(z) - 1.5z⁻¹Y(z) - 1.5 + 0.5z⁻²Y(z) + 0.5z⁻¹ = 0
Step 3: Group Y(z) terms
Y(z)[1 - 1.5z⁻¹ + 0.5z⁻²] = 1.5 - 0.5z⁻¹
Step 4: Multiply both sides by z² to eliminate negative exponents
Y(z)[z² - 1.5z + 0.5] = 1.5z² - 0.5z
Step 5: Factor the denominator
z² − 1.5z + 0.5 = (z − 1)(z − 0.5)
Y(z) = z(1.5z − 0.5) / [(z − 1)(z − 0.5)]
Step 6: Partial Fraction Expansion
Let’s write:
Y(z)/z = (1.5z − 0.5) / [(z − 1)(z − 0.5)] = A / (z − 1) + B / (z − 0.5)
Multiply both sides by the denominator:
1.5z − 0.5 = A(z − 0.5) + B(z − 1)
Step 7: Solve for A and B
Let z = 1:
1.5(1) − 0.5 = A(0.5) ⇒ A = 2
Let z = 0.5:
1.5(0.5) − 0.5 = B(−0.5) ⇒ B = −0.5
Step 8: Take Inverse Z-transform
Using the standard Z-transform pair:
Z⁻¹ { z / (z − a) } = aⁿ·u(n)
Therefore:
y(n) = 2·(1)ⁿ − 0.5·(0.5)ⁿ, for n ≥ 0
Final Answer:
Handwritten Solution Images:
Key Takeaways:
- This solution focuses on the natural response of a discrete-time LTI system governed by a second-order linear difference equation.
- We assumed x(n) = 0 as per the natural response condition, simplifying the Z-transform analysis.
- Initial conditions y(−1) = 1 and y(−2) = 0 were used effectively during transformation.
- Z-transform and partial fraction expansion techniques played crucial roles in solving this question analytically.
- The final result was found as:
y(n) = 2 − 0.5·(0.5)ⁿ, for n ≥ 0 - Understanding this problem enhances your ability to solve typical GTU Signals & Systems PYQ problems using Z-transforms.
Final Words:
In this post, we successfully solved the GTU Sem 4 Signals & Systems Summer 2021 Question 2(b), which asked us to find the natural response of a system described by a difference equation using the Z-transform method. By applying the right steps — from transforming the equation, substituting initial conditions, and performing partial fraction expansion to taking the inverse Z-transform — we arrived at a clean and clear final result.
Understanding problems like this is crucial for mastering the concept of system response analysis in Signals & Systems. If you’re preparing for GTU exams or simply revising PYQs, make sure to practice more questions involving the Z-transform technique and natural responses.
If you found this solution helpful, do check out the other Signals & Systems GTU PYQ solutions linked below. Let us know in the comments if you have any doubts or need step-by-step solutions for other questions.
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