Rational transfer function; LTI system
Given the system with Nth order difference equation,
a0 y(n) + a1 y(n-1) + ... + aN y(n-N)
= b0 x(n) + b1 x(n-1) + ... + bM x(n-M), a0 ≠ 0
we can write it in the compact form as follows
We can find transfer function of the system by taking the z-transform on both the sides of equation. We note that in finding the impulse response of a system and, as a result, in finding the transfer function, the system should be initially relaxed ("zero initial conditions"). Hence, if we suppose zero initial conditions, we can use the time-shift and linearity properties to obtain
The impulse response corresponding to it can be found as h(n) = ?-1{H(z)}. The poles of the system transfer function are same as the characteristic values of corresponding difference equation. For the system to be stable, poles should lie within unit circle in z-plane. As a result, for a stable, causal function, the ROC includes unit circle.
The system function, H(z), is the rational function: 
Here N(z) and D(z) stand for the denominator and numerator respectively.
As N(z) and D(z) are polynomials in z, they are expressed in factored form as
Thus H(z) has M finite zeros at z = z1, z2, ..., zM, and N finite poles at z = p1, p2,..., pN, and |N-M| zeros (if N > M) or poles (if N < M) at the origin z = 0. Poles and zeros may also occur at z = ∞ . A pole exists at z = ∞ if H( ∞ ) = ∞ , and a zero exists at z = ∞ if H( ∞ ) = 0. If we count the poles and zeros at z = 0 and z = ∞ as well as the N poles and M zeros, we find that H(z) has exactly the same number of poles and zeros.
By definition the ROC of H(z) should not contain any poles.
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