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in order to make the most efficient use of a (B 1)-bit binary code word. A 3-bit uniform quantizer in which the quantizer output is rounded to the nearest quantization level is illustrated in Fig. 3-4. With L = 2'" quantization levels and a step size A, the range of the quantizer is

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First prove that any point on the locus satisfies the condition: Given: Points A and B. CD is the ' bisector of AB. To Prove: Any point P on CD is equidistant from A and B; that is, PA > PB. Plan: Prove ^PEA > ^PEB to obtain PA > PB. PROOF:

With rounding, the quantization error e(n) = Qlx(n)l - x(n)

Statements 1. CD is the ' bisector of AB. 2. /PEA > /PEB 3. 4. 5. 6. AE > EB PE > PE ^PEA > ^PEB PA > PB

'ln some applications, such as speech coding, the quantizer levels are adaptive (1.e..they change with time).

1 17 2 6

Reasons 1. Given 2. Perpendiculars form right angles; all right angles are congruent. 3. To bisect is to divide into congruent parts. 4. Reflexive property 5. SAS 6. Corresponding parts of > triangles are >.

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A discrete-time system is a mathematical operator or mapping that transforms one signal (the input) into another signal (the output) by means of a fixed set of rules or operations. The notation T [ - ]is used to represent a general system as shown in Fig. 1-4, in which an input signal x(n) is transformed into an output signal y(n) through the transformation T [ . ] . The input-output properties of a system may be specified in any one of a number of different ways. The relationship between the input and output, for example, may be expressed in terms of a concise mathematical rule or function such as

It is also possible, however, to describe a system in terms of an algorithm that provides a sequence of instructions or operations that is to be applied to the input signal, such as yl(n) = 0.5yl(n - 1) 0.25x(n)

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I suggest you use Google as there is a lot of information on the topic: http://en. lmgtfy.com/?q=ucc+ ean - 128 +barcode+generator[^]. —SA.

Then prove that any point satisfying the condition is on the locus: Given: Any point Q which is equidistant from points A and B (QA > QB). To Prove: Q is on the perpendicular bisector of AB. Plan: Draw QG perpendicular to AB and prove by congruent triangles that QG bisects AB. PROOF:

+ y2(n) = 0.25y2(n - 1) + 0.5x(n) ys(n) = 0.4y3(n - 1) + 0.5x(n) + y(n) = Y I ( ~ ) y2(n) + ydn)

Statements 1. Draw QG ' AB . 2. QA > QB 3. /QGA and /QGB are rt. /s; ^QGA and ^QGB are rt. ^s. 4. QG > QG 5. ^QGA > ^QGB 6. AG > GB 7. QG bisects AB . 8. QG is ' bisector of AB.

17 12

In some cases, a system may conveniently be specified in terms of a table that defines the set of all possible input-output signal pairs of interest.

Reasons 1. Through an external point, a line can be drawn perpendicular to a given line. 2. Given 3. Perpendiculars form right angles; ^s with a rt./ are rt. ^s. 4. Reflexive property 5. Hy-leg 6. Corresponding parts of > triangles are >. 7. To bisect is to divide into two congruent parts 8. A line perpendicular to a segment and bisecting it is its perpendicular bisector.

Fig. 1-4. The representation of a discrete-timesystem as a transformation T [ . ] that maps an input signal x ( n ) into an output signal y(n).

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