The special case of a finite-duration sequence is given as

\(x\left( n \right) = \left\{ { 2,\begin{array}{*{20}{c}} 4\\ \uparrow \end{array},0,3} \right\}\)

The sequence x(n) into a sum of weighted impulse sequences will be

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  1. 2δ(n + 1) + 4δ(n) + 3δ(n – 2)
  2. 2δ(n) + 4δ(n –1) + 3δ(n – 3)
  3. 2δ(n) + 4δ(n –1) + 3δ(n – 2)
  4. 2δ(n + 1) + 4δ(n) + 3δ(n – 1)

Answer (Detailed Solution Below)

Option 1 : 2δ(n + 1) + 4δ(n) + 3δ(n – 2)
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\(x\left( n \right) = \left\{ { 2,\begin{array}{*{20}{c}} 4\\ \uparrow \end{array},0,3} \right\}\)

The arrow indicates the origin. This can be represented as:

F5 S.B Madhu 28.07.20 D1

In terms of unit impulse sequences, this can be represented as:

x(n) = 2δ(n + 1) + 4δ(n) + 3δ(n – 2)

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