EDITOR: B. SOMANATHAN
NAIR
1. CONVERTING H(s) INTO H(z) IN THE
DIGITAL DOMAIN
To convert the analog
expression for H(s) into its equivalent digital expression H(z), we have several
methods. The most popular ones are the impulse-invariant method and the
bilinear transformation method. In this blog, we discuss the impulse-invariant
transformation method.
2. THE
IMPULSE-INVARIANT TRANSFORMATION METHOD
In
the impulse-invariant method, we first find the expression for H(s).
This is then split into first-order equations using the partial-fraction
method. The inverse of these yields the impulse response in time domain. We
then find the z-transform equivalents
of these first-order equations to
obtain H(z). The following example will illustrate the procedure.
Example 1: Consider the first-order transfer function
H(s)
= 1/(s + a) (1)
The
inverse Laplace transform of this is given by:
h(t) = e‒at (2)
Now,
taking the z transform of (2), we get
H(z) = 1/(1‒e‒az‒1 ) (3)
Equation
(3) can be used for the implementation of the filter, as illustrated below.
Example 2: Let us now assume that a = 1. This gives e-1 =
0.368. Then (3) may be written as:
H(z) = 1/(1‒ 0.368z‒1 ) (4)
It can be seen that (4) represents
an infinite-impulse response (IIR) expression and hence it can be easily
implemented by using delay lines or digital computers. Figure 1 shows the
implementation of this. First, we
rewrite (4) in the form
H(z) = Y(z)/X(z) = 1/(1‒ 0.368z‒1 ) (5)
where
Y(z) and X(z) are, respectively, the output and
input functions. Rearranging (5), we get
X(z) = Y(z)(1‒ 0.368z‒1 ) (6)
Taking
the inverse z transform of (6) yields
x(n)
= y(n) ‒ 0.368y(n ‒ 1)
(7)
To
implement the digital filter, we rearrange (7) in the form
y(n) = 0.368y(n ‒ 1) + x(n)
(8)
Equation
(8) can be used to implement the desired filter, as shown in Fig. 1.
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