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Round-off Effects in IIR Digital Filters 577
where ε 1 (n) and ε 2 (n) are the relative errors in the corresponding quantizers.
The exact analysis even for the 1st-order case is tedious; hence we
make a few practically reasonable approximations. If the absolute values
of the errors are small, then we have ŷ(n − 1) ≈ y(n − 1) and ĝ(n) ≈ y(n);
hence from (10.68a) we obtain
e 1 (n) ≈ αε 1 (n) y(n − 1)
e 2 (n) ≈ ε 2 (n) y(n)
(10.69a)
(10.69b)
Furthermore, we make the following assumption about the noise sources:
1. ε 1 (n) and ε 2 (n) are white noise sources.
2. ε 1 (n) and ε 2 (n) are uncorrelated with each other.
3. ε 1 (n) and ε 2 (n) are uncorrelated with the input x(n).
4. ε 1 (n) and ε 2 (n) are uniformly distributed between −2 −B and 2 −B .
Let x(n) beazero-mean, stationary random sequence. Then y(n) is
also a zero-mean, stationary sequence. Hence from (10.69)
σ 2 e 1
= |α| 2 σ 2 ε 1
σ 2 y (10.70a)
σ 2 e 2
= σ 2 ε 2
σ 2 y (10.70b)
Let the error in the output due e 1 (n) beq 1 (n) and that due to e 2 (n) be
q 2 (n). Let h 1 (n) and h 2 (n) bethe corresponding noise impulse responses.
Note that h 1 (n) =h 2 (n) =h(n) =α n u(n). Then the total error q(n) is
with
where
q(n) =q 1 (n)+q 2 (n) (10.71)
σ 2 q = σ 2 q 1
+ σ 2 q 2
(10.72)
∑
∞ ∑
∞
σq 2 1
= σe 2 1
|h 1 (n)| 2 and σq 2 2
= σe 2 2
|h 2 (n)| 2 (10.73)
0
Hence using (10.72), (10.73), and (10.70),
σq 2 = ( σe 2 1
+ σe 2 ) ( ) ( )
1
2
1 −|α| 2 = σy
2 1 (|α| 2
1 −|α| 2 σε 2 1
+ σε 2 )
2
0
(10.74)
Using σε 2 1
= σε 2 2
=2 −2B /3, we obtain
( )( )
2
σq 2 = σy
2 −2B 1+|α|
2
3 1 −|α| 2
Therefore
SNR = σ2 y
σ 2 q
=3 ( 2 2B) ( 1 −|α| 2
1+|α| 2 )
(10.75)
(10.76)
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