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(c) a > 0 when t > 3, and a < 0 when t < 3. (d) Assuming that s increases to the right, positive velocity indicates movement to the right, and negative velocity movement to the left. The object moves right until, at t = 2, it is at s = 20, where it reverses direction. It then moves left until, at t = 4, it is at s = 16, where it reverses direction again. Thereafter, it continues to move right (see Fig. 22-2).

Fig. 11-13

Sampling Theorem: If x , ( t ) is strictly bandlimited,

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The frequency Q0 is called the Nyquist frequency, and the minimum sampling frequency, Q , = 2'20, is called the Nyquist rate.

A point or points which satisfy two conditions may be found by drawing the locus for each condition. The required points are the points of intersection of the two loci.

Because the signals that are found in physical systems will never be strictly bandlimited, an analog antialiasing filter is typically used to filter the signal prior to sampling in order to minimize the amount of energy above the Nyquist frequency and to reduce the amount of aliasing that occurs in the AID converter.

11.4 Locating points that satisfy two conditions On a map locate buried treasure that is 3 ft from a tree (T ) and equidistant from two points (A and B) in Fig. 11-14.

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A quantizer is a nonlinear and noninvertible system that transforms an input sequence x(n) that has a continuous range of amplitudes into a sequence for which each value of x ( n ) assumes one of a finite number of possible values. This operation is denoted by ,W) = Qlx(n)l

Fig. 22-2

Fig. 11-14

+ I decision levels X I , xl, . . . . x ~ +that divide the amplitude range for x(n) into L intervals [

The required loci are (1) the perpendicular bisector of AB and (2) a circle with its center at T and radius 3 ft. As shown, these meet in P1 and P2, which are the locations of the treasure. Note: The diagram shows the two loci intersecting at P1 and P2. However, there are three possible kinds of solutions, depending on the location of T with respect to A and B: 1. The solution has two points if the loci intersect. 2. The solution has one point if the perpendicular bisector is tangent to the circle. 3. The solution has no points if the perpendicular bisector does not meet the circle.

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For an input x(n) that falls within interval l k , quantizer assigns a value within this interval, &, tox(n). This the process is illustrated in Fig. 3-3.

quantization intervals and eight possible quantizer outputs. i r . Quantizers may have quantization levels that are either uniformly or nonuniformly spaced. When the quantization intervals are uniformly spaced,

To prove that a locus satisfies a given condition, it is necessary to prove the locus theorem and its converse or its inverse. Thus to prove that a circle A of radius 2 in is the locus of points 2 in from A, it is necessary to prove either that 1. Any point on circle A is 2 in from A. 2. Any point 2 in from A is on circle A (converse of statement 1). or that 1. Any point on circle A is 2 in from A. 2. Any point not on circle A is not 2 in from A (inverse of statement 1). These statements are easily proved using the principle that a point is outside, on, or inside a circle according as its distance from the center is greater than, equal to, or less than the radius of the circle.

1 7 1 + 2 3 2

A is called the quantization step size or the resolution of the quantizer, and the quantizer is said to be a uniform or linear quantizer.' The number of levels in a quantizer is generally of the form

11.5 Proving a locus theorem Prove that the locus of points equidistant from two given points is the perpendicular bisector of the segment joining the two points.

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