If h(n) is the impulse response of an LTI system T and h1(n) is the impulse response of the inverse system T-1, then we know that h(n)*h1(n)=δ(n)=>[h(n)*h1(n)]*x(n)=x(n).

Given H(z)= 1-5/2z-1-3/2z-2

For an IIR filter whose system function is defined as H(z)=\\(\\frac{B(z)}{A(z)}\\) to be said a mixed phase and if the system is stable and causal, then the poles are inside the unit circle and some, but not all of the zeros are outside the unit circle.

Given H(z)=1-z-1-6z-2

Given h(n)= δ(n)-aδ(n-1)

For an IIR filter whose system function is defined as H(z)=\\(\\frac{B(z)}{A(z)}\\) to be said a minimum phase,

If we know the output of a system y(n) of a system and if we can determine the input x(n) of the system uniquely, then the system is said to be invertible. That is there should be one-to-one correspondence between the input and output signals.

The system function of given system is H(z)=\\(1-\\frac{1}{2} z^{-1}\\)

Given impulse response is h(n)=(1/2)nu(n)

Given H(z)=6+z-1-z-2