Question

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 \(\mathop \sum \limits_{n = 0}^\infty h\left[ n \right] = 4,\) the relationship between α and β is

1. 4α – β = 4
2. 4α + β = 4
3. α = 4β
4. α = -4β

Answer 1

Correct Answer - Option 2 : 4α + β = 4

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)

\(\frac{Y\left( z \right)}{X\left( z \right)}=\frac{\beta }{1-\alpha {{z}^{-2}}}\) 

\( H\left( z \right)=\frac{\beta }{1-\alpha {{z}^{-2}}}\) 

The impulse response can be expressed as

<!--[if gte msEquation 12]>H(z)0∞hnz-n<![endif]--><!--[if !msEquation]--><!--[if gte vml 1]> <![endif]--><!--[if !vml]-->\(H\left( z \right)=\underset{0}{\overset{\infty }{\mathop \sum }}\,h\left[ n \right]{{z}^{-n}}\)

Put z = 1,

<!--[if gte msEquation 12]>H1= 0∞h[n]<![endif]--><!--[if !msEquation]--><!--[if gte vml 1]> <![endif]--><!--[if !vml]-->\(H\left( 1 \right)=~\underset{0}{\overset{\infty }{\mathop \sum }}\,h\left[ n \right]\)

H(1) = 4

\( \frac{\beta }{1-\alpha }=4\) 

\(1-\alpha =\frac{\beta }{4}\)

\(4\alpha +\beta =4\)