Digital Filters
35
The transfer function of this filter is
H(z) =
1/2
1 -
1
2
z
-1
.
(27)
If the filter (26) is fed with a unit impulse at instant 0, the response will be:
y = 0.5, 0.25, 0.125, 0.0625, . . . .
(28)
It is clear that the impulse response is nonzero over an infinitely extended sup-
port, and every sample is obtained by halving the preceding one. Similarly to
what we did for the first-order FIR filter, we analyze the behavior of this fil-
ter using a complex sinusoid having magnitude A and initial phase , i.e. the
signal Ae
j(
0
n+)
. Since the system is linear, we do not loose any generality by
considering unit-magnitude signals (A = 1). Moreover, since the system is time
invariant, we do not loose generality by considering signals having the initial
phase set to zero ( = 0). In a linear and time-invariant system, the steady-
state response to a complex sinusoidal input is a complex sinusoidal output. To
have a confirmation of that, we can consider the reversed form of (26)
x(n) = 2y(n) - y(n - 1) ,
(29)
and replace the output y(n) with a complex sinusoid, thus obtaining
x(n) = 2e
j
0
n
- e
j
0
(n-1)
= (2 - e
-j
0
)y(n) .
(30)
Eq. (30) shows that a sinusoidal output gives a sinusoidal input, and vice versa.
The input sinusoid gets rescaled in magnitude and shifted in phase. Namely,
the output y is a copy of the input multiplied by the complex quantity
1
2-e
-j0
,
which is the value taken by the transfer function (27) at the point z = e
j
0
. The
frequency response is
H() =
1/2
1 -
1
2
e
-j
,
(31)
and there are no simple formulas to express its magnitude and phase, so that we
have to resort to the graphical representation, depicted in fig. 14. This simple
(a)
(b)
0
1
2
3
0
0.2
0.4
0.6
0.8
frequency [rad/sample]
magnitude
0
1
2
3
-0.6
-0.4
-0.2
0
0.2
frequency [rad/sample]
phase [rad]
Figure 14: Frequency response (magnitude (a) and phase (b)) of a one-pole IIR
filter