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276 Chapter 6 IMPLEMENTATION OF DISCRETE-TIME FILTERS
Im{z}
UC
r
x{n}
z −1
y{n}
0
r
Re{z}
−r 2
z −1
Digital filter in Example 6.24 (a) pole-zero plot, (b) filter realiza-
FIGURE 6.30
tion
(a)
(b)
Solution
The filter has two complex-conjugate poles at
p 1 = re jθ and p 2 = re −jθ = p ∗ 1
For aproper operation as a resonator, the poles must be close to the unit
circle—that is, r ≃ 1 (but r < 1). Then the resonant frequency ω r ≃ θ.
The zero-pole diagram is shown in Figure 6.30 along with the filter realization.
Let r =0.9 and θ = π/3. Then from (6.74),
a 1 = −2r cos θ = −0.9 and a 2 = r 2 =0.81
We now represent a 1 and a 2, each using 3-bit sign-magnitude format
representation—that is,
a k = ± b 1 b 2 b 3 = ± ( b 12 −1 + b 22 −2 + b 32 −3) , k =1, 2
where b j represents the jth bit and represents the binary point. Then for the
closest representation, we must have
â 1 =1111=−0.875 and â 2 =0110=+0.75
Hence |△a 1| =0.025 and |△a 2| =0.06. Consider the sensitivity formula (6.73)
in which
∂p 1
= − p2−1 1
∂a 1 (p 1 − p ∗ 1 ) = −p1
2Im{p = −rejθ
1} 2r (sin θ) = ejπ/3
√ , and
3
∂p 1
= − p2−2 1
∂a 2 (p 1 − p ∗ 1 ) = −1
2Im{p = 1
1} 0.9 √ 3
Using (6.73), we obtain
∣ ∣ |△p 1|≤∣ ∂p1 ∣∣ |△a1| + ∣ ∂p1 ∣∣ |△a2|
∂a 1 ∂a 2
= √ 1 (0.025) + 1
3 0.9 √ (0.06) =0.0529 (6.75)
3
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