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32 = 9 = 3. (b) ( 3)2 = 9 = 3. (c) By (1.4), x 2 = |x|2 ; hence, since |x| 0, x 2 = |x|. (d) By part (c), x 2 = |x|, but |x| = x is false when x < 0. For example, ( 3)2 = 9 = 3 = 3.

where T&) is a kth-order Chebyshev polynomial [see Eq. (9.9)]. Therefore, Eq. (9.4) may be written as

Fig. 1-7

Thus, A(ejw)is an Lth-order polynomial in coso. With Ad(ei") a desired amplitude, and W(eiw)a positive weighting function, let

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(b) Since semicircle ADB contains 180 , AD contains 180 (c) Since semicircle ACB contains 180 , BC contains 180

be a weighted approximation error. The equiripple filter design problem thus involves finding the coefficients a ( k ) that minimize the maximum absolute value of E(ejU)over a set of frequencies, F,

An angle is the figure formed by two rays with a common end point. The rays are the sides of the angle, while the end point is itsSvertex. The symbol for angle is / or ]; the plural is . S Thus, AB and AC are the sides of the angle shown in Fig. 1-8(a), and A is its vertex.

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By (1.9), |3x + 2| < 1 is equivalent to 1 < 3x +2 < 1. Subtracting 2, we obtain the equivalent relation 3 < 3x < 1. This is equivalent, upon division by 3, to 1 < x < 1 . 3

For example, to design a low-pass filter, the set F will be the frequencies in the passband. [O, w,,], and the n], illustrated in Fig. 9-4. The transition band, (w,, o,), is a don't care region, and it is not as stopband,

Don't Care Fig. 9-4. The set R in the equiripple filter design problem, consisting of the passband [0, w,] and the stopband [op. , ] . The transition band ( o , , w,) is a don't care region. o

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An angle may be named in any of the following ways: 1. With the vertex letter, if there is only one angle having this vertex, as /B in Fig. 1-8(b). 2. With a small letter or a number placed between the sides of the angle and near the vertex, as /a or /1 in Fig. 1-8(c). 3. With three capital letters, such that the vertex letter is between two others, one from each side of the angle. In Fig. 1-8(d), /E may be named /DEG or /GED.

considered in the minimization of the weighted error. The solution to this optimization problem is given in the alternation theorem. which is as follows:

Fig. 1-8

to be the unique function that minimizes the maximum value of the weighted error I E(eJW)I over the set 3is that the E(eJW) have at least L 2 alternations. That is to say, there must be at least L 2 extremalfiequencies.

By (1.9), 2 < 5 3x < 2. Subtracting 5, 7 < 3x < 3. Dividing by 3, 7 > x > 1. 3 algebra review Multiplying or dividing both sides of an inequality by a negative number reverses the inequality: if a < b and c < 0, then ac > bc. To see this, notice that a < b implies b a > 0. Hence, (b a)c < 0, since the product of a positive number and a negative number is negative. So bc ac < 0, or bc < ac.

Fig. 18-8

Thus, the alternation theorem states that the optimum filter is equiripple. Although the alternation theorem specifies the minimum number of extremal frequencies (or ripples) that the optimum filter must have, it may 3 extremal frequencies. A low-pass have more. For example, a low-pass filter may have either L 2 or filter with L 3 extrema is called an extraripplefilter. From the alternation theorem, it follows that

Only (b), (e), and (f) have reflectional symmetry. Note that when (d) is flipped, it will look like Fig. 18-9, which is different from the original in that the upper-left-hand crossing is horizontal instead of vertical.

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