Question:medium

Find the convolution sum of sequences \(x_1[n] = \{1, 2, 3\}\) and \(x_2[n] = \{2, 1, 4\}\).

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N/A
Updated On: Jul 4, 2026
  • \(\{2, 5, 12, 11, 12\}\)
  • \(\{2, 12, 5, 11, 12\}\)
  • \(\{2, 11, 5, 12, 12\}\)
  • \(\{-2, 5, -12, 11, 12\}\)
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The Correct Option is A

Solution and Explanation

Understanding the Concept: The discrete linear convolution of two finite-duration sequences $x_1[n]$ and $x_2[n]$ is defined by the mathematical summation: \[ y[n] = x_1[n] * x_2[n] = \sum_{k=-\infty}^{\infty} x_1[k]x_2[n-k] \] If $x_1[n]$ has a length of $L$ elements and $x_2[n]$ has a length of $M$ elements, the resulting convolved sequence $y[n]$ will contain exactly $L + M - 1$ elements. In this case, both input sequences contain $3$ elements ($L=3, M=3$), so the output sequence will have a length of $3 + 3 - 1 = 5$ elements.

Step 1: Setting up the Matrix Multiplication Method

A reliable way to compute discrete convolution without losing track of terms is the tabular/matrix method. We arrange the elements of sequence $x_1[n]$ along the rows and the elements of sequence $x_2[n]$ along the columns: [h!] {c|ccc} & 2 & 1 & 4
1 & $1 \times 2 = 2$ & $1 \times 1 = 1$ & $1 \times 4 = 4$
2 & $2 \times 2 = 4$ & $2 \times 1 = 2$ & $2 \times 4 = 8$
3 & $3 \times 2 = 6$ & $3 \times 1 = 3$ & $3 \times 4 = 12$

Step 2: Summing along the Diagonals

The values of the output sequence $y[n]$ are found by summing the terms along the diagonal lines running from the top-right to the bottom-left:
First element (\(y[0]\)): Top-left corner element. \[ y[0] = 2 \]
Second element (\(y[1]\)): Sum of the second diagonal terms. \[ y[1] = 4 + 1 = 5 \]
Third element (\(y[2]\)): Sum of the third diagonal terms. \[ y[2] = 6 + 2 + 4 = 12 \]
Fourth element (\(y[3]\)): Sum of the fourth diagonal terms. \[ y[3] = 3 + 8 = 11 \]
Fifth element (\(y[4]\)): Bottom-right corner element. \[ y[4] = 12 \]

Step 3: Compiling the Final Sequence

Combining these calculated diagonal sums gives the final output sequence: \[ y[n] = \{2, 5, 12, 11, 12\} \] This sequence matches choice (A).
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