2079 Asoj · 3 marks

Find x[2n−3] for a given sequence

Short answer

Substitute 2n−3 into indices: only odd indices of x[n] survive, yielding y[n]=x[2n−3]. Nonzero values: y[0]=0, y[1]=2, y[2]=−2, and y[n]=0 elsewhere.

For x[n] = {1, 2, 0, −2, 1} (origin at the underlined 0, i.e. x[0] = 0), find x[2n−3].

Given $x[n] = \{1, 2, \underline{0}, -2, 1\}$ :

$n$	−2	−1	0	1	2
$x[n]$	1	2	0	−2	1

Let $y[n] = x[2n-3]$ . For each integer $n$ , the argument $m = 2n-3$ is always odd, so only the odd-index samples $x[-1]$ and $x[1]$ can appear (in downsampling, $x[-2], x[0], x[2]$ are lost).

$n$	0	1	2	3
$2n-3$	−3	−1	1	3
$y[n] = x[2n-3]$	0	2	−2	0

For all other $n$ , $2n-3$ is outside $-2 \ldots 2$ , so $y[n] = 0$ .

Answer:

y[n] = x[2n-3] = \{\underline{0}, 2, -2\} = 2\delta[n-1] - 2\delta[n-2]

(Method: first shift, $x[n-3]$ moves the sequence 3 steps right; then scale by 2, keeping only samples at even positions of the shifted sequence and halving their indices.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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