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u k() = Ue jkΩ and y k() = Ye jkΩ (6.97)

where Ω is a normalized frequency; typically, in electrical engineering appli-
cations, the real frequency multiplied by the sampling time. Replacing these
expressions in the difference equation, we obtain:

m
m
∑ be − Ω ∑ bz l −
jl
Y = l=0 l = l=0 l ≡
U n − Ω n l − Hz () (6.98)
1 + ∑ ae jl 1 + ∑ az
l
l
l=1 l=1
jΩ
where, by convention, z = e .

Example 6.10
Find the Transfer Function of the following difference equation:

yk() = u k() + 2 yk( − 1 ) − 1 yk( − 2 ) (6.99)
3 3

Solution: By direct substitution into Eq. (6.98), we find:

Hz() = 2 1 1 = z 2 2 1 (6.100)
2
−1
1 − z + z −2 z − z +
3 3 3 3
It is to be noted that the Transfer Function is a ratio of two polynomials. The
zeros of the numerator are called the zeros of the Transfer Function, while the
zeros of the denominator are called its poles. If the coefficients of the differ-
ence equations are real, then by the Fundamental Theorem of Algebra, the
zeros and the poles are either real or are pairs of complex conjugate numbers.
The Transfer Function fully describes any linear system. As will be shown
in linear systems courses, the z-transform of the Transfer Function gives the
weights for the solution of the difference equation, while the values of the
poles of the Transfer Function determine what are called the system modes
of the solution. These are the modes intrinsic to the circuit, and they do not
depend on the specific form of the input function.
Furthermore, it is worth noting that the study of recursive filters, the back-
bone of digital signal processing, can be simply reduced to a study of the
Transfer Function under different configurations. In Applications 2 and 3 that
follow, we briefly illustrate two particular digital filters in wide use.

© 2001 by CRC Press LLC