barcode lib ssrs A white noise sequence e ( n ) with variance 0; is input to a filter with a system function in Software

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A white noise sequence e ( n ) with variance 0; is input to a filter with a system function
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H (z) =
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+ 2zr2)(1 + 3 z r I ) ( l + z r ' ) (1 + iz-2)(1 + iz-1)
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Find the variance of the output sequence.
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The variance of the output sequence is
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For the given filter, note that
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where the firs1 two terms are allpass filters with
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Therefore, Using Parseval's theorem,
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IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
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[CHAP. 8
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we find Thus. we have
= 720,'
Consider the following cascade of two first-order all-pole filters:
(a) Find the variance of the round-off noise at the output of the cascade for an 8-bit processor with
rounding.
(h) Repeat for the case in which the order of the cascade is reversed.
(a) A model for the round-off noise is shown in the following figure:
where the variance of each noise source is equal to
The system function of the filter is
H(z) =
and the unit sample response is
Note that because e l ( n ) is filtered by h ( n ) , and e2(n) is only filtered by the second filter in the cascade, which has a unit sample response
hdn) =
(:)nw
the output noise, f (n), is
f ( n )=
Therefore, the variance of f (n) is
* h ( n )+ e d n ) * h d n )
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
With
we have Next, we have
Therefore, the variance of the round-off noise at the output of the filter is
which, for an 8-bit processor (B = 7), is
2-28 = 2.8953 - = 0.2413 . 2-l4 = I.4726 . 12
(b) If the order of the cascade is reversed. we have the following network:
The variance of the round-off noise due to e l ( n ) is the same as in part (a),but because the unit sample response of the second system in the cascade is now
h; (n) ( i ) n u ( n ) =
the variance of the noise due to e z ( n ) is
Thus, the variance of the round-off noise at the output of the filter is
a = 1.82860: :
+ 1.33330:
= 3.16190;
which, for an 8-bit processor is
With this structure, the round-off noise is slightly larger.
Consider a linear shift-invariant system with a system function
H (z) =
1-0 . 4 ~ ~ '
( 1 - 0.62-')(I - 0.8~-1)
Suppose that this system is implemented on a 16-bit fixed-point processor and that the sums of products a are accumulated prior to quantization. Let : be the variance of the round-off noise.
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
[CHAP. 8
(a) If the system is implemented in direct form 11, find the variance of the round-off noise at the output of the filter.
(b) Repeat part (a) if the system is implemented in parallel form.
( a ) The direct form I1 implementation of this system is shown in the figure below along with the two round-off noise sources.
Because the sum x(n)
+ 1.4w(n - 1) - 0.48w(n - 2 )
Similarly, because the sum
may be accumulated prior to quantization, the variance of the noise e l ( n )is . : a
may be accumulated prior to quantization, the variance of the noise e2(n)is also me2. With c l ( n )being filtered by the system and with ez(n)being noise that is simply added to the output, the quantization noise at the output of the filter is f ( n ) = h(n) * el ( n ) eAn)
which has a variance equal to
To find the unit sample response of the filter, we expand H ( z ) in a partial fraction expansion as follows: H ( z )= I - 0.4~-I
- O.6zr1)(I - 0.82-I)
-1 1 - 0.62-I
1 - 0.8~-I
Therefore,
h(n) = -(0.6)"u(n)
+ 2(0.8)"u(n)
- 4(0.48)"
lh(n)12 = [-(0.6)"
+ 2(0.8)"]2u(n) [(0.6)'" =
+4(0.8)~"]u(n)
Evaluating the sum of the squares of h(n),we have
Thus, the variance of the output noise is
= 6u:
(b) Using the partial fraction expansion for H ( z ) given in part ( a ) ,the parallel form implementation of this filter is
shown in the following figure:
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
As indicated in the figure, there are two noise sources. The first, cl (n), is filtered with a first-order all-pole filter that has a unit sample response hl(n) = (0.8)"1~(~) and the second, ez(n), is filtered with a first-order all-pole filter that has a unit sample response
Because the output noise is + f ( n ) = cJl(n)* I I I ( ~ )r 2 ( n )* hz(n) the variance of f (n) is
A linear shift-invariant system with a system function of the form
is to be implemented as a cascade of N second-order sections. where each section is realized in either 1 direct form I o r 1 or in their transposed forms. How many different cascaded realizations are possible. Let us assume that each factor in H ( z ) is unique, so that there arc N different second-order polynomials in the numerator and the same number of polynomials in the denominator. In this case. there are N! different pairings of factors in the numerator with factors in the denominator. In addition, for each of these pairings, there are N ! different orderings of these sections. Therefore. there are ( N !)' different pairings and orderings. With four different structures for each section (direct form I, direct form 11, transposed direct form I, and transposed direct form 11), there are a total of 4N(N!)2 different realizations. For a tenth-order system (N = 5). this corresponds to 14,745,600 different structures. This is why general pairing and ordering rules are important.
Let H ( z ) be a pth-order allpass filter with a gain of 1 that is implemented in direct form I1 using a 1 bits. processor with B
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