IIR filter design - Continuous and Discrete Time Signals and Systems - page 730

16.3.1 Mapping between the s-plane and the z-plane

For k = 1, Eq. (16.23) can be represented in the following form:

= 1 + s z

1 − s .

(16.25)

Substituting s = σ + j ω into Eq. (16.25), we obtain

= 1 + σ

+ j ω

− j ω ,

z (16.26)

1 − σ

with an absolute value given by

(1 + σ ) 2 + ω 2
(1 − σ ) 2 + ω 2 .

z = (16.27)

By substituting different values of s = σ + j ω corresponding to the right-half,
left-half, and imaginary axes of the s-plane in Eq. (16.27), we derive the fol-
lowing observations.

Left-half s-plane ( σ< 0) For σ < 0, we observe that the value of the
denominator (1 − σ ) 2 + ω 2 in Eq. (16.27) exceeds the value of the numerator
(1 + σ ) 2 + ω 2 , resulting in z < 1. In other words, the bilinear transformation
maps the left-half of the s-plane to the interior of the unit circle within the
z-plane.

Right-half s-plane ( Ω< 0) For σ> 0, the value of the numerator (1 + σ ) 2 +
ω 2 in Eq. (16.27) exceeds the value of the denominator (1 − σ ) 2 + ω 2 , resulting
in z > 1 . Consequently, the bilinear transformation maps the right-half of the
s-plane to the exterior of the unit circle within the z-plane.

Imaginary axis ( σ = 0) For σ = 0, the denominator and numerator in Eq.

(16.27) are equal, resulting in z = 1. The bilinear transformation maps the
imaginary axis of the s-plane onto the unit circle within the z-plane.

Note that the mapping in Eq. (16.25) is a one-to-one mapping, which means
that no two points in the s-plane will map to the same point in the z-plane, and
vice versa.

16.3.2 IIR filter design using bilinear transformation

The steps involved in designing IIR filters using the bilinear transformation are
as follows.

Step 1 Using Eq. (16.24), ω = k tan( Ω / 2), transform the specifications of the
digital filter from the DT frequency ( Ω ) domain to the CT frequency ( ω ) domain.
For convenience, we choose k = 1.