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A central angle has the same number of degrees as the arc it intercepts. Thus, as shown in Fig. 6-29, a central angle which is a right angle intercepts a 90 arc; a 40 central angle intercepts a 40 arc, and a central angle which is a straight angle intercepts a semicircle of 180 . Since the numerical measures in degrees of both the central angle and its intercepted arc are the same, we may restate the above principle as follows: A central angle is measured by its intercepted arc. The symbol may be used to mean is measured by. (Do not say that the central angle equals its intercepted arc. An angle cannot equal an arc.)

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Given a desired frequency response Hd(eJ"), show that the rectangular window design minimizes the least-squares error

(f ) lim (7x 3 5x 2 + 2x 4) x 2 x 12 x 4 x 4 x 4 + 3x 3 13x 2 27x + 36 (l) lim x 1 x 2 + 3x 4 4 1 (o) lim 2 x 2 x 2 x 4 (i) lim

Fig. 6-29

For this problem, we use Parseval's theorem to express the least-squares error ELs in the time domain:

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An inscribed angle is an angle whose vertex is on the circle and whose sides are chords. An angle inscribed in an arc has its vertex on the arc and its sides passing through the ends of the arc. Thus, j A in Fig. 6-30 is an inscribed angle whose sides are the chords AB and AC. Note that j A intercepts BC and is inscribed in BAC.

If we assume that h(n) is of order N , with h(n) = 0 for n < 0 and n z N ,

Because the last two terms are constants that are not affected by the filler h(n), the least-squares errorELs is minimized by minimizing the first term, which is done by setting h(n) = hd(n) for n = 0, I . . . . N (i.e., using a rectangular window in the window design method).

Fig. 6-30

(h) lim (x [x])

If hd(n) is the unit sample response of an ideal filter, and h ( n ) is an N th-order FIR filter, the least-squares error

A central angle is measured by its intercepted arc. An inscribed angle is measured by one-half its intercepted arc.

fLS= I\

PRINCIPLE 2:

277 -=

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Fig. 6-31

f (x + h) f (x) f (x + h) f (x) and then lim (if the latter exists) for each of the following functions: h h h 0 1 (c) f (x) = 7x + 12 (b) f (x) = (a) f (x) = 3x 2 + 5 x+1 (d) f (x) = x 3 (e) f (x) = x (f ) f (x) = 5x 2 2x + 4

is minimized when h(n) is designed using the rectangular window design method. If ER is the squared error using a rectangular window, find the excess squared emor that results when a Hanning window is used instead of a rectangular window; that is, find an expression for

PRINCIPLE 4:

Using Parseval's theorem. it is more convenient to express the least-squares error in the time domain as follows:

where wH(n) and wx(n) are the Hanning and rectangular windows, respectively. However, the second sum is equal to zero. Therefore, the excess squared error is simply

In the same or congruent circles, inscribed angles having congruent intercepted arcs are congruent. (This is the converse of Principle 3.)

Consider the following specifications for a low-pass filter:

8.8 Assuming that lim f (x) = L and lim g(x) = K, prove rigorously: (a) lim c f (x) = c L, where c is any real number.

PRINCIPLE 5:

Design a linear phase FIR filter to meet these specifications using the window design method. Designing a low-pass filter with the window design method generally produces a filter with ripples of the same amplitude in the passband and stopband. Therefore, because the passband and stopband ripples in the filter specifications are the same, we only need to be concerned about the slopband ripple requirement. A stopband ripple of 6 , = 0.01 corresponds to a stopband attenuation of -40 dB. Therefore. froin Table 9-2 it follows that we may use a Hanning window, which provides an attenuation of approximately 44 dB. The specification on the transition band is that Aw = 0.05rr, or A f = 0.025. Therefore, the required filter order is

Fig. 6-32

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