GTU Sem 4 Signals & Systems Summer 2021 PYQ Question 2(c) Solution
🧭 Introduction
In this post, we will compute the convolution sum of two discrete-time signals using the graphical method. This approach gives an intuitive understanding of how two sequences interact through time reversal and shifting.
The given sequences are:
x(n) = { 1 4 3 2 }
↑
h(n) = { 1 3 2 1 }
↑
As shown above, the x(n) sequence has its arrow (origin) at the second position, while h(n) has its arrow at the first position. We’ll use this alignment while performing the convolution step-by-step.
📚 Table of Contents
❓ Question Statement
Determine the convolution sum of two sequences using graphical method:
x(n) = {1, 4, 3, 2}
↑
h(n) = {1, 3, 2, 1}
↑
Note: Use graphical method to evaluate the convolution sum y(n) = x(n) * h(n).
📚 Theory Part Related to the Question
In signal processing, the convolution sum is used to determine the output of a Linear Time-Invariant (LTI) system when the input and system's impulse response are known.
For discrete-time signals, the convolution of two signals x(n) and h(n) is defined as:
y(n) = x(n) * h(n)
= ∑k = -∞∞ x(k) · h(n − k)
Here, y(n) is the output signal, x(k) is the input signal, and h(n - k) is the time-reversed and shifted version of the system's impulse response.
🔍 Graphical Method Steps:
- Time-Reverse: Flip the sequence
h(n)to obtainh(-k). - Shift: Shift
h(-k)bynto geth(n - k). - Multiply & Add: For each value of
n, multiplyx(k)andh(n - k)pointwise and sum the result to gety(n).
This method is especially helpful for short finite sequences, like in this question, where both x(n) and h(n) are of length 4. The resulting output signal y(n) will have a length of:
Length of y(n) = Length of x(n) + Length of h(n) - 1 = 4 + 4 - 1 = 7
Therefore, y(n) will contain 7 values, corresponding to the range of n for which overlap occurs during the shifting and multiplying process.
📘 Solution in Detail
y(n) = x(n) * h(n)
= ∑k = -∞∞ x(k) · h(n − k)
🔹 Given Sequences:
x(n) = { 1 4 3 2 }
↑ (arrow at index 1)
h(n) = { 1 3 2 1 }
↑ (arrow at index 0)
These are finite-duration sequences, so the length of the convolution result will be:
N = L + M − 1 = 4 + 4 − 1 = 7
So the output sequence y(n) will range from y(-1) to y(5).
🔍 Convolution Step-by-Step:
y(-1)= 1·1
= 1
y(0) = 1·3 + 4·1
= 3 + 4
= 7
y(1) = 1·2 + 4·3 + 3·1
= 2 + 12 + 3
= 17
y(2) = 1·1 + 4·2 + 3·3 + 2·1
= 1 + 8 + 9 + 2
= 20
y(3) = 4·1 + 3·2 + 2·3
= 4 + 6 + 6
= 16
y(4) = 3·1 + 2·2
= 3 + 4
= 7
y(5) = 2·1
= 2
📘 Final Result:
y(n) = { 1 7 17 20 16 7 2 }
↑
✅ The arrow is at the second position (index 1), which aligns with the reference of x(n).
📝 Handwritten Solution
The following handwritten steps clearly demonstrate the graphical convolution process for the given sequences from Signals & Systems, Question 2(c) OR, Summer 2021 GTU exam.
📌 Key Takeaways
- Graphical convolution involves flipping, shifting, multiplying, and summing the signals step-by-step.
- The arrow position is crucial for the correct alignment of the signals during convolution.
- Final output sequence: y(n) = {1, 7, 17, 20, 16, 7, 2} with the arrow at the second element.
- This process is useful for understanding LTI system responses in signal processing.
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