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With an ideal low-pass filter that has a cutoff frequency of w,. = 0.325 (the midpoint of the transition band), and a delay of N/2 = 62 so that hd(n) is placed symmetrically within the interval [O, 1241, we have

Fig. 6-33

Therefore, the filter is 0.5 - 0.5 cos -

f (x)

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Note that if we were to use a Hamming or a Blackman window instead of a Hanning window, the stopband and passband ripple requirements would have been exceeded, and the required filter order would have been larger. With a Blackman window, for example, the filter order required to meet the transition band requirement is

PRINCIPLE 7:

W e would like t o filter an analog signal x,(t) with a n analog low-pass filter that has a cutoff frequency f,. = 2 kHz, a transition width Af = 500 Hz, a n d a stopband attenuation o f 50 dB. This filter is t o b e implemented digitally, a s illustrated in the following figure:

Fig. 6-34

4 1 )

Fig. 6-35

y(n)

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[Hints: In part (d), for L > 0, f (x) L = ( f (x) L) f (x) + f (x) + L L = f (x) L |f (x) L| L

An angle formed by a tangent and a chord is measured by one-half its intercepted arc. An angle formed by two intersecting chords is measured by one-half the sum of the intercepted arcs. An angle formed by two secants intersecting outside a circle is measured by one-half the difference of the intercepted arcs. An angle formed by a tangent and a secant intersecting outside a circle is measured by onehalf the difference of the intercepted arcs. An angle formed by two tangents intersecting outside a circle is measured by one-half the difference of the intercepted arcs.

H(ejw)

PRINCIPLE 10: PRINCIPLE 11: PRINCIPLE 12: PRINCIPLE 13:

yo ( t )

Design a digital filter to meet the analog filter specifications with a sampling frequency f, = 10 kHz.

Fig. 6-36

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In part (f ), if f (x) and h(x) lie within the interval (L , L + ), so must g(x).] 8.9 In an epsilon-delta proof of the fact that lim (2 + 5x) = 17, which of the following values of is the largest that can be used,

With a sampling frequency of 10 kHz, the digital filter should have a cutoff frequency w, = 2 n f,./f, = 0.41~ and a transition bandwidth Aw = 2nAf/f, = O.lrr. For a stopband attenuation of 50 dB, we may use a Kaiser window with B = 0.l l02(50 - 8.7) = 4.55 For the length of the window, we have

Note: To find the angle formed by a secant and a chord meeting on the circle, first find the measure of the inscribed angle adjacent to it and then subtract from 180 . Thus if secant AB meets chord CD at C on the circle in Fig. 6-36, to find mj y, first find the measure of inscribed j x. Obtain mj y by subtracting mj x from 180 .

or N = 59. Finally, the unit sample response of the ideal filter that is to be windowed is a low-pass filter with a ~ cutoff frequency w,. = 0 . 4 and a delay N / 2 = 29.5. Therefore,

6.10 Applying principles 1 and 2 (a) In Fig. 6-37(a), if mj y (b) In Fig. 6-37(b), if mj y (c) In Fig. 6-37(c), if mj x 46 , find mj x. 112 , find mj x. 75 , find m y .

Find the Kaiser window parameters, B and N , to designa low-pass filter with acutofffrequency w, = n / 2 , a stopband ripple 6 = 0.002, and a transition bandwidth no 1a.rger than 0.117. ,

Fig. 6-37

With 6, = 0.002,

given (a) (b) 2 (c) 4 (d) 5 (e) 8

and we have = 0.1 102(0r, - 8.7) = 4.99 The window length, N , on the other hand, is determined by the stopband ripple, 6,. and the transition width as follows:

(46 )

Consider the following specifications for a bandpass filter:

150 60

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