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Let apand 52, be the desired passband and stopband cutoff frequencies of an analog low-pass filter, and let 6, and 6, be the passband and stopband ripples. Show that the order of the Butterworth filter required to meet these s~ecifications is log d N Z log k with the 3-dB cutoff frequency a, being any value within the range

Write Then, g(x) g(x + h) g(x) = = =

Find x and y in each part of Fig. 6-62 (t and t9 are tangents).

Because I H,(jQ)l is monotonically decreasing, the maximum error in the passband and stopband occurs at the band edges, Q, and a,, respectively. Therefore, we want

(6.15)

("n)

(note that the inequality is reversed because log k < 0). Because the right side of this equation will not generally be an integer, the order N is taken to be the smallest integer larger than (logd)/(log k ) . Finally, once the order N is fixed, it follows from Eqs. (9.20) and (9.21) that a,, may be any value in the range

Fig. 6-62

[by Theorem 30.4]

a a,, tip,and ,

(6.16)

where is an Nth-order Chebyshev polynomial. Over the passband, In1 < Q,, the magnitude of the frequency response oscillates between 1 and ( l c2)-'I2. Therefore, the ripple amplitude, 6,. is

Fig. 6-63

6, = 1

200 102

= (1 - a,)-'

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By the mean-value theorem for integrals, the last integral is equal to hf (x ) for some x between x and x + h. Hence, g(x + h) g(x) = f (x ) h g(x + h) g(x) = lim f (x ) = Dx (g(x)) = lim h h 0 h 0

which we want to be less than or equal to 8::

3:1:2

cosh-'(I I d ) C O S ~ - I (QJR,)

X c X mb 3mX X X (h) mjB if ma 75 and mX 2mb c X X 5mb (i) m/ X if AB ' BC and ma C (j) mX if mjA:mjB:mjC 5:4:3 c

- cosh-'( l / d ) COS~-~(I,X)

(6.16)

Now as h 0, x + h x, and so x x (since x* lies between x and x + h). Since f is continuous,

find Ha(s)H,(-s).

If ABCP is an inscribed quadrilateral, PD a tangent, and AF a secant in Fig. 6-64, find (a) mj1 if ma

= 0. I ,

X 42 X (h) ma if AC is a diameter and mj5 35 X (i) mb if AC ' BP and mj2 57 (j) mX if AC and BP are diameters and mj5 c X X (k) md if mj1 95 and mb 95

where TN(x)is an Nth-order Chebyshev polynomial that is defined recursively as follows,

with To(x) = 1 and Tl(x) = x. Therefore, to find the third-order Chebyshev polynomial, we first find T2(.u) as follows, T2(x)= 2xTl(x) - To(x) = 2x2 - I and then we have T3(x) = 2xT2(x) - Tl(x) = 4x3 - 2x - x = x(4x2 - 3) Thus, the denominator polynomial in I Hu(jR)12is

lim f (x ) = f (x)

Fig. 6-64

we make the substitution R = s / j in 1H,(jC2)l2as follows:

Find x and y in each part of Fig. 6-65 (t and t9 are tangents).

398 9.28

(6.16)

Show that the bilinear transformation maps the jQ-axis in the s-plane onto the unit circle, lzl = 1, and maps the left-half s-plane. Re(s) < 0 insidc the unit circle, l z ( < 1.

31.6 (Change of Variables) Consider a f (x) dx. Let x = g(u), where, as x varies from a to b, u increases or decreases from c to d. [See Fig. 31-1; in effect, we rule out g (u) = 0 in [c, d].] Show that

Fig. 6-65

To investigate the characteristics of the bilinear transformation, let z = reJ"' and s = a transformation may then be written as

='(T ,

Find x and y in each part of Fig. 6-66.

Therefore,

(6.16)

and Note that if r < I, then o < 0, and if r 1, then 0 z 0. Consequently, the left-half s-plane is mapped inside the unit circle. and the right-half s-plane is mapped outside the unit circle. If 1, = I, then a = 0, and

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