Let-xn - 1- 0- n- 8 0- elsewhere-and hn-1- 0- n- N 0- elsewhere-where- N- 8 is an integer-Given that yn-xn-hn and y4 - 4- y12-0-Then the value of N - -

The signal nbsp; nbsp; y(n) = x(n)*h(n) =∑_k= ∞^∞x(k).h(n k) In this case, this summation reduces to nbsp; y(n) = ∑_k=0^8x(k).h(n k) = ∑_k=0^8h(n k) h ...

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Quantitative Aptitude

Circular Arrangement

Let,xn = 1, ...

Question

Let,

$x (n) = {\begin{matrix} 1, & 0 \leq n \leq 8 0, & e l s e w h e r e \end{matrix} a n d h (n) = {\begin{matrix} 1, & 0 \leq n \leq N 0, & e l s e w h e r e \end{matrix}$
where, $N \leq 8$ is an integer.
Given that $y (n) = x (n) * h (n) a n d y (4) = 4, y (12) = 0.$
Then the value of N = ____.
3

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Solution

The correct option is A 3
The signal $y (n) = x (n) * h (n)$

$= \sum_{k = - \infty}^{\infty} x (k) . h (n - k)$

In this case, this summation reduces to
$y (n) = \sum_{k = 0}^{8} x (k) . h (n - k) = \sum_{k = 0}^{8} h (n - k)$

$h (n) = 0 f o r, n > N [N + 1, s a m p l e s a r e t h e r e]$
$x (n) = 0 f o r, n > 8 [9 s a m p l e s a r e t h e r e]$
$y (n) = 0 f o r, n > (9 + N - 1) [9 + N + 1 - 1 = 9 + N s a m p l e s t h e r e]$
$y (n) = 0 f o r, n > 8 + N$

Given that,
$y (n) = 0 f o r, n = 12$
$∴ 12 > 8 + N$
4 > N
$∴ N < 4$

From here, we may conclude that N can atmost be 3

Given that, y(4) = 4
$y (4) = \sum_{k = 0}^{8} h (4 - k)$
$4 = h (4) + h (3) + h (2) + h (1) + h (0) + h (- 1) + h (- 2) + h (- 3) + h (- 4)$
$4 = h (4) + h (3) + h (2) + h (1) + h (0)$
and this equation satisfies only when,
h(0)=h(1) = h(2) = h(3) = 1
and h(4) = 0
$∴ h (n) = 1, f o r 0 \leq n \leq 3$
$∴$ N = 3

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Let, x(n)={1,0≤n≤80,elsewhereand h(n)={1,0≤n≤N0,elsewhere where, N≤8 is an integer. Given that y(n)=x(n)∗h(n)and y(4)=4,y(12)=0. Then the value of N = ____.3