$y(n)=0.9y(n-1)+bx(n)$ i) Determine b such that |H(0) |=1 ii) Determine the frequency at which |H(ω) |=$\frac{1}{√2}$

iii) Identify the filter type based on passband.

We have,

$y(n)=0.9y(n-1)+bx(n)$

Taking z transform

$Y(z)=0.9Y(z) z^{-1}+bX(z)$

$Y(z)-9z^{-1} Y(z)=bX(z)$

$H(z)=\frac{(Y(z))}{(X(z))}=\frac{b}{(1-0.9z^{-1}}$

$Put z=e^{jω}$

$H(ω)=\frac{b}{(1-0.9e^{-jω} )}$

i) Put ω=0

$H(0)=\frac{b}{(1-0.9)}=1$

b=0.1

ii) We have,

$H(ω)=\frac{0.1}{(1-0.9e^{-jω} )}$

$=\frac{0.1}{(1-0.9(cos⁡ω-j sin⁡ω)}$

$H(ω)=\frac{0.1}{((1-0.9 cos⁡ω) )}$

By plotting magnitude response, it is observe that,

$ω=\frac{π}{2}$

iii) Identify the type by pass band