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1)-bit numbers is rounded to B I bits before any additions are performed, find the variance of the round-off noise at the output of the filter.
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(b) Repeat part (a) for the case in which sums of products are accumulated prior to quantization.
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IMPLEMENTATION OF DISCRETE-TIME SYSTEMS (a) The system function for a pth-order allpass filter has the form
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[CHAP. 8
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H (z) =
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+ a ( p - l)z-I + . . . + z-" 1 + a(l)z-' + . . . + a(p)z-P
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With a direct form I1 implementation of this system, rounding each product to 5 any additions, we have the round-off noise model shown in the following figure:
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where the variance of each noise source is
2-28
u2 =
Note, however, that this noise model may be simplified as illustrated in the following figure:
where
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS Assuming that each noise source is uncorrelated with the others, the variances of e,(n) and eh(n)are u,Z=,,u; Because the output noise is
~ ; = ~ u ;
f (n) = e d n ) * h ( n ) + eh(n)
where h ( n ) is the unit sample response of the allpass filter, the variance of f ( n ) is
Equivalently, we may write this using Parseval's theorem as follows:
Because H(e1") is an allpass filter with I H (elW)I= I, the variance of the output noise is a = uZ : ,
+ 02
= 2pu,'
(b) If the products are accumulated prior to quantization, the variances of e,(n) and eh(n) in the noise model given in part (a)will be reduced by a factor of p:
Therefore, the variance of the output noise becomes
In the figure below are direct form I1 and transposed direct form I1 realizations of the first-order system
(a) Direct form 11.
( b ) Transposed direct form 11.
Assume that both systems are implemented using ( B 1)-bit fixed-point arithmetic and that all products are rounded to B 1 bits before any additions are performed.
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
[CHAP. 8
(a) Using a linear noise model for the round-off noise, find the variance of the round-off noise at the output of the direct form I1 filter.
(b) Repeat part (a)for the transposed direct form I1 filter.
(c) How would the variance of the output noise change if the sums of products were accumulated prior to quantization
( a ) The linear noise model for round-off noise in the direct form I1 implementation is shown in the figure below.
The variance of the noise eo(n) is a and the variance of el( n ) is 2 4 , where :
Because the output noise is
f ( n )=
+ e d n ) * h(n)
where h ( n ) is the unit sample response of the filter, the variance of the output noise is
With it follows that the unit sample response is
Therefore,
and the output noise variance is
( b ) For the transposed direct form I1 implementation, the noise model is as follows,
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
: a where the variance of eo(n) is , and the variance of e l ( n ) is 2u:. Note that because neither noise source is filtered by the zeros of the system, an equivalent model for the generation of the filtered noise f ( n ) is as shown in the figure below,
where e ( n ) = en@) e l ( n - I). Because e&) and e l ( n ) are uncorrelated, the variance of e ( n ) is 30:. and the variance of the output noise is
(c) If we accumulate the sums of products prior to quantization, for the direct form I1 implementation, the variance of the noise e l ( n ) would be a,' instead of 2u:, and everything else remains the same. For the transposed direct form I1 structure, on the other hand, the variance of el ( n ) would be 0 instead of 2u:, which implies that the ; variance of e ( n ) would be 2u:,
A sixth-order filter with a system function
H (z) =
(1 - 1.6 cos(lr/4)z-I
+ zp2)(I + ~ - ' ) ~ ( 12cos(lr/6)z-' + zp2) + 0 . 6 4 ~ - ~ ) (+ 1.6 cos(lr/4)z-I + 0.64zr2)(1 - 1.8 cos(rr/6)z-' + 0 . 8 1 r 2 ) 1
is to be implemented as a cascade of second-order sections. Considering only the effects of round-off noise, determine what the best pole-zero pairing is, and the best ordering of the second-order sections.
Note that all six zeros of this system lie on the unit circle, with two at z = -1, a complex pair at z = fj, and a complex pair at z = eiJ"I6. The poles, on the other hand, are at z = 0.8e*J"I4, z = 0 . 8 e * ~ ~and z = 0.9e*J"16. ~J~, A pole-zero diagram showing each of these poles and zeros is given below.
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