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Round-off Effects in FIR Digital Filters 585
2. The noise sources are independent of each other—that is,
e i (n) ⊥ e j (n)
for i ≠ j
3. Each noise source is a white noise source with σ 2 e i
=2 −2B /12.
We will now consider the issues of round-off noise and scaling (to prevent
overflow) for the cascade-form realization.
Round-off noise Let the noise impulse response at the output from
the e i (n) node bedenoted by g i (n). Then the length of g i (n) isequal to
(M − 2i). Let q i (n) bethe output noise due to e i (n). Then its power is
given by
M−2i
∑
σq 2 i
= σe 2 i
|g i (n)| 2 = 2−2B
M−2i
∑
|g i (n)| 2 (10.91)
12
0
Since q(n) = ∑ K
i=1 q i(n) weobtain the total noise power as
(
K∑
K
)
σq 2 = σq 2 i
= 2−2B ∑
M−2i
∑
|g i (n)| 2 (10.92)
12
i=1
i=1 n=1
The expression ∑ K ∑ M−2i
i=1 n=1
|g i (n)| 2 shows that the error power depends
on the order of the cascade connections. It has been shown that for the
majority of the orderings the noise power is approximately the same.
Scaling to prevent overflow From Figure 10.28 we note that one
must prevent overflow at each node. Let h k (n) bethe impulse response
at each node k; then we need a scaling constant X max as
1
X max = ∑
max k |hk (n)|
so that |y(n)| ≤1. Clearly, this is a very conservative value. A better
approach is to scale the impulse responses of every section {h i (n)} so
that ∑ |h i | =1for each i. Hence the output of every section is limited
between −1 and +1 if the input x(n) isdistributed over the same interval.
Assuming that x(n) isuniformly distributed over [−1, +1] and is white,
the output signal power is
σ 2 y = σ 2 x
M−1
∑
0
|h(n)| 2 = 1 3
M−1
∑
0
0
|h(n)| 2 (10.93)
where h(n) isthe overall impulse response of the filter. Let ĝ i be the
corresponding scaled impulse responses in (10.92). Now the output SNR
can be computed as
SNR = σ2 y
σ 2 q
( ∑K ∑ M−2i
i=1
∑ M−1
=2 2(B+1) 0
|h(n)| 2
n=1
|ĝ i (n)| 2 ) (10.94)
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