If the function H1(z) = (1 + 1.5z– 1

If the function
H₁(z) = (1 + 1.5z^{– 1} – z^{– 2}) and H₁(z) = (z² + 1.5z – 1) then—

1. the poles and zeros of the functions will be the same
2. the poles of the functions will be identical but not zeros
3. the zeros of the functions will be identical but not the poles
4. neither the poles nor the zeros of the two functions will be identical

Given H₁(z) = 1 + 1·5z^{– 1} – z^{– 1}…(A)
and H₂(z) = z² + 1·5z – 1 …(B)
Equation (A) can be written as

H₁(z) =	z² + 1·5z – 1
	z²

poles at z = 0, 0
zeros are z² + 1·5z – 1 = 0
or 2z² + 3z – 2 = 0
or 2z² + 4z – z – 2 = 0
2z(z + 2) – 1(z + 2) = 0
(2z – 1) (z + 2) = 0

z =	1	and – 2
	2

However, in equation (B)
H₂(z) = z² + 1·5z – 1
There is no poles
zeros are at z² + 1·5z – 1 = 0

or z =	1	and – 2
	2

Hence, the alternative (C) is the correct choice.