The input-output relationship of a causal stable LTI system is given as

y[n] = α y[n – 2] + β x[n]

If the impulse response h[n] of this system satisfies the condition $\sum_{n = 0}^{\infty} h [n] = 4,$ the relationship between α and β is

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GATE EC 2014 Official Paper: Shift 2

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Answer 1

y[n] = α y[n – 2] + β x[n]

by applying z-transform on both the sides,

Y(z) = αz^-2 Y(z)+ β X(z)

Y(z) [1 - αz^-2] = β X(z)

$\frac{Y (z)}{X (z)} = \frac{β}{1 - α z^{- 2}}$

$H (z) = \frac{β}{1 - α z^{- 2}}$

The impulse response can be expressed as

$H (z) = \underset{0}{\sum^{\infty}} h [n] z^{- n}$

Put z = 1,

$H (1) = \underset{0}{\sum^{\infty}} h [n]$

H(1) = 4

$\frac{β}{1 - α} = 4$

$1 - α = \frac{β}{4}$

$4 α + β = 4$