# The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?

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The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?

(a) 2

(b) 1

(c) 0

(d) -1

I have been asked this question in final exam.

This is a very interesting question from Inverse Z-Transform topic in portion Z-Transform and Digital Filtering of Signals and Systems

by (42.1k points)
Right option is (c) 0

Easy explanation: Given equation = y (n) – 2y (n-1) + y (n-2) = X (n) – X (n-1) has y (n) = 0

For n = 0, y (0)2y (-1) + y (-2) = x (0) – x (-1)

∴ y(0) = x(0) – x(-1)

Or, y (n) = 0 for n<0

For n=1, y (1) = -2y (0) + y (-1) = x (1) – x (0)

Or, y (1) = x (1) – x (0) + 2x (0) – 2x (-1)

Or, y (1) = x (1) + x (0) – 2x (-1)

For n=2, y (2) = x (2) – x (1) + 2y (1) – y (0)

Or, y(2) = x(2) – x(1) + 2x(1) + 2x(0) – 4x(-1) – x(0) + x(-1)

∴y (2) = d (2) + d (1) + d (0) – 3d (-1).

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