Assigment 4

Problem 1

$$H(z)=\frac{1}{1-a{z}^1}$$

a) For $a=0.9$ we have a pole at $0.9$, which means that low frequencies get amplified a lot → $a=0.9$ is a lowpass filter While for $a=-0.9$ we have a pole at 0.9 which means that high frequencies get amplified a lot. → $a=-0.9$ is a highpass filter

b) The demo proves what i wrote in a)

Problem 2

Given a causal digital filter:

$$H(z)=\frac{1}{\left(1-\frac{1}{2}{z}^{-1}\right)\left(1+\frac{1}{2}{z}^{-1}\right)}$$

a) The inverse filter is given by:

H_{I}(z) &= H^{-1}(z) \\ H_{I}(z) &= \left( 1-\frac{1}{2}z^{-1} \right)\left( 1+\frac{1}{2}z^{-1} \right) \\ H_{I}(z) &= 1-\frac{1}{4}z^{-2} \end{align}

b) As $w\to \infty$ our inverse filter goes towards 1, aka it is finite. It is therefore stable

c) Since the inverse of our inverse filter is stable and causal, it is a minimum-phase filter

d) ?

Problem 3

a) The filter, $A(z)$, has a zero in $z={\alpha }^{-1}$, and a pole in $p=\alpha$. It is therefore an allpass filter.