This question was previously asked in

UPRVUNL AE EC 2016 Official Paper

Option 2 : y[n] + 4y[n - 1] + 7y[n - 2] = -x[n] – 4x[n - 1]

Hindi Subject Test 1

4474

10 Questions
10 Marks
10 Mins

Given:

x[n] = [1, 4, 7]

y(n) = [-1, -4]

X(z) = z + 4z^{-1} + 7z^{-2}

Y(z) = -1 – 4z^{-1}

We know that,

\(H(z)=\frac{Y(z)}{X(z)}\)

\(H(z)=\frac{-1-4z^{-1}}{1+4z^{-1}+7z^{-2}}\)

Y(z) [1 + 4z^{-1} + 7z^{-2}] = X(z) [-1 – yz^{-1}]

Y(z) + 4z^{-1} Y(z) + 7z^{-2} Y(z) = -X(z) – 4 × (z) z^{-1}

Taking z inverse transformation,

y(n) + 4 y(n – 1) + 7 y(n – 2) = -x(n) – 4x(n - 1)