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Thus, the slope of the tangent line at (x, f (x)) is 6x 6. (b) From part (a), the slope of the tangent line at (0, 4) is 6x 6 = 6(0) 6 = 0 6 = 6. Hence, the slope-intercept equation has the form y = 6x + b. Since the line passes through (0, 4), the y-intercept b is 4. Thus, the equation is y = 6x + 4. (c) We want the graph of y = 3x 2 6x + 4. Complete the square: y = 3 x 2 2x + 4 3 = 3 (x 1)2 + 1 3 = 3(x 1)2 + 1

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Therefore, the system function is H0(s) = s Q,.

Fig. 1-9

H (z) =

Fig. 1-10

0.3249 0.3249(1 + Z-I) 0.2452(1 + z-I) 1 - z-' 1 - 0.5095~-I (1 - 2-I) O.3249(1 z-I) 0.3249 1 z-'

The graph (see Fig. 11-6) is obtained by moving the graph of y = 3x 2 one unit to the right [obtaining the graph of y = 3(x 1)2 ] and then raising that graph one unit upward.

The size of /B in Fig. 1-10 would not be changed if its sides AB and BC were made larger or smaller. No matter how large or small a clock is, the angle formed by its hands at 3 o clock measures 90 , as shown in Figs. 1-11 and 1-12.

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( a ) Determine the locations of the poles and zeros of H ( z ) if the filter is designed using the bilinear transformation with T, = 2.

Fig. 1-11

(b) Repeat part (a)for the impulse invariance technique, again with T, = 2.

z2 = I-h To find the zeros, it is necessary to combine the two terms in the system function over a common denominator. Doing this. we have

Fig. 1-12

Finding the roots of the numerator may be facilitated by noting that H,(s) has a zero at s = co,which gets mapped to z = - 1 with the bilinear transformation. Therefore. one of the factors of the numerator is ( 1 z-I). Dividing the numerator by this factor, we obtain the second factor, which is [(2 - a - b) - 2z-'1. Thus, H ( ; ) has zeros at

11.5 The normal line to a curve at a point P is de ned to be the line through P perpendicular to the tangent line at P. Find the slope-intercept equation of the normal line to the parabola y = x 2 at the point ( 1 , 1 ). 2 4

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Angles that measure less than 1 are usually represented as fractions or decimals. For example, one360 thousandth of the way around a circle is either 1000 or 0.36 . In some fields, such as navigation and astronomy, small angles are measured in minutes and seconds. One degree is comprised of 60 minutes, written 1 60 . A minute is 60 seconds, written 1 60 . In this 21 36 1296 360 notation, one-thousandth of a circle is 21 36 because 60 3600 3600 1000 .

z,, = i(e2"

Fig. 1-13

(a) If this filter was designed using impulse invariance with T, = 2, find the system function, H&), of an analog filter that could have been the analog filter prototype. Is your answer unique

Fig. 1-14

(b) Repeat part (a) assuming that the bilinear transformation was used with T, = 2.

By Problem 11.1, the tangent line has slope 2( 1 ) = 1. Therefore, by Theorem 4.2, the slope of the normal line is 1, 2 and the slope-intercept equation of the normal line will have the form y = x + b. When x = 1 , y = x 2 = ( 1 )2 = 1 , 2 2 4 whence, 1 1 = 4 2 +b or b= 3 4

(a) Because H ( z ) is expanded in a partial fraction expansion, the poles at z = a k in H ( z ) are mapped from poles in Ha(s)according to the mapping ( y k = esl

Fig. 1-15

Therefore, if T, = 2,

Thus, in Fig. 1-16, m(st. /B) 180 . Note that the sides of a straight angle lie in the same straight line. But do not confuse a straight angle with a straight line!

Because the mapping from the s-plane to the z-plane is not one to one, this answer is not unique. Specifically, note that we may also write

Fig. 1-16

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