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