Schilling-1120949 book November 12, 2010 9:33
14 Chapter 1 Signal Processing
produce x
q
(k). Careful inspection of Figure 1.11 reveals that at some of the samples there are
noticeable differences between x
q
(k) and x
a
(kT). If rounding is used, then the magnitude of
the error is, at most, q/2.
Most of the analysis in this book will be based on discrete-time signals rather than digital
signals. That is, infinite precision is used to represent the value of the dependent variable.
Finite precision, or finite word length effects, are examined in Chapters 6 and 7 in the context
of digital filter design. When digital filter are implemented in MATLAB using the default
double-precision arithmetic, this corresponds to 64 bits of precision (16 decimal digits). In
most instances this is sufficiently high precision to yield insignificant finite word length effects.
A digital signal x
q
(k) can be modeled as a discrete-time signal x(k) plus random quanti-
Quantization
noise
zation noise, v(k), as follows.
x
q
(k) = x(k) + v(k) (1.2.6)
An effective way to measure the size or strength of the quantization noise is to use average
power defined as the mean, or expected value,ofv
2
(k). Typically, v(k) is modeled as a
Expected value
random variable uniformly distributed over the interval [−q/2, q/2] with probability density
p(x) = 1/q. In this case, the expected value of v
2
(k) is
E[v
2
] =
q/2
−q/2
p(x)x
2
dx
=
1
q
q/2
−q/2
x
2
dx (1.2.7)
Thus, the average power of the quantization noise is proportional to the square of the quanti-
zation level with
E[v
2
] =
q
2
12
(1.2.8)
Example 1.1 Quantization Noise
Suppose the value of a discrete-time signal x(k) is constrained to lie in the interval [−10, 10].
Let x
q
(k) denote a digital version of x(k) using quantization level q, and consider the following
problem. Suppose the average power of the quantization noise, v(k), is to be less than .001.
What is the minimum number of bits that are needed to represent the value of x
q
(k)? The
constraint on the average power of the quantization noise is
E[v
2
] <.001
Thus, from (1.2.3) and (1.2.8), we have
(x
M
− x
m
)
2
12(2
N
)
2
<.001
Recall that the signal range is x
m
=−10 and x
M
= 10. Multiplying both sides by 12, taking
the square root of both sides, and then solving for 2
N
yields
2
N
>
20
√
.012
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