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The next steps, which are algebraically very tedious, are lo perform a partial fraction expansion of H,,(s),

(c) P(x, y) A Pr( y, x) 5, x)

H (z) =

A dihilation (also called a scaling or an enlargement) is not a rigid motion because it multiplies all lengths by a single scale factor. The image of a figure under a dihilation will always be similar to the original. The dihilation that enlarges everything by a scale factor of k is P(x, y) A Pr(kx, ky).

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0.0105z~-' 0.0168z-~ 0 . 0 0 4 2 ~ -+ 0.0001z-~ ~ 0.0007~ 3.3431~-I 5 . 0 1 5 0 ~ - ~ 4.21532-2 + 2 . 0 7 0 3 ~ -- 0.5593zr5 O.0646zr6 ~

18.17 Performing dihilations Let triangle ABC be formed by A(2, 2), B(4, 2), and C(4, 3). Graph the image of ^ABC under 1 (a) magnification by 2 and (b) scaling by 2.

at and , As a final check on the design, evaluating H ( e J U ) w = 0 . 2 ~ w = 0 . 3 ~we find that

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notation f + g is a function such that ( f + g)(x) = f (x) + g(x) for every x that is in both the domain of f and the domain of g. Similarly, ( f g)(x) = f (x) g(x) for all x common to the domains of f and g.

Therefore, the filter exceeds the passband specifications and comes close to meeting the stopband specifications. If this filter is unacceptable, the design could be modified by decreasing Q, to improve the stopband performance.

See Fig. 18-16.

Using the bilinear transformation to design a Butterworth filter according to the specifications given in Prob. 9.39, we first use the transformation

Fig. 18-16

to map the passband and stopband frequencies of the digital filter to the cutoff frequencies of the analog filter. With Ts = 2, we have

Thus, the slope of the tangent line at (x, 1/x) is 1/x 2 . (b) From part (a), the slope of the tangent line at (2, 1 ) is 2 1 1 1 2 = 2 = 4 x 2 Hence, the slope-intercept equation of the tangent line L has the form 1 y = x+b 4

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(a) Ar(4, 4), Br(8, 4), and Cr(8, 6) (b) As(1, 1), Bs(2, 1), and Cs(2, 1.5)

= tan(0. I n ) == 0.3249

18.18 Naming dihilations Name the transformation that (a) Scales everything 5 times larger (b) Shrinks every length to half size (c) Triples all linear dimensions (d) Depicts everything at 10 scale (e) Dihilates by a scale factor of 12

R, = tan 2 = t a n ( 0 . 1 5 ~L- 0.5095 )

As we found in Prob. 9.39, the required filter order IS N = 6. For the 3-dB cutoff frequency of the analog Butterworth filter, we may choose any frequency in the range

(a) P(x, y) A Pr(5x, 5y)

CHAP. 11]

0.3667 5 R,. 5 0.39 10

1 1 (b) P(x, y) A Pr(2x, 2y)

If we select Q,. =0.3667, the passband specifications will be met exactly, and the stopband specifications will be exceeded. Conversely. if we set R, = 0.3910. the stopband specifications will be met exactly, and the passband specifications exceeded. Picking a frequency between the two extremes will produce an improvement in both bands. Because the stopband deviation is twice that of the deviation in the passband. we will set 0,.= 0.3667 in order to improve the stopband performance. From Table 9-4, we find the coel'ticients in the system function of a sixth-order normalized (52, = I) Butterworth filter to be

(c) P(x, y) A Pr(3x, 3y)

To obtain a Butterworth filter with a cutofl' frequency R, = 0.3666, we perform the low-pass-to-low-pass transformation r

1 1 (d) P(x, y) A Pr(10x, 10y)

FILTER DESIGN This gives Hob) =

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