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( h ) x ( n ) - x(n - 2)

Fig. 10-15

( c ) x(2n)

10.43. Find the shaded area in each part of Fig. 10-15. Each heavy dot represents the center of an arc or a circle. (10.15) 10.44. Find the shaded area in each part of Fig. 10-16. Each dot represents the center of an arc or circle. (10.15)

( d ) x ( n )* x ( n - I )

Fig. 10-16

In this case, the acceleration is not constant. Notice that a > 0 when 12t 6 > 0, or t > 1 . This implies (by Theorem 17.3) 2 that the velocity is increasing for t > 1 . 2

2.67 2.68

Locus, in Latin, means location. The plural is loci. A locus of points is the set of points, and only those points, that satisfy given conditions. Thus, the locus of points that are 1 in from a given point P is the set of points 1 in from P. These points lie on a circle with its center at P and a radius of 1 in, and hence this circle is the required locus (Fig. 11-1). Note that we show loci as long-short dashed figures.

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If the DTFT of x ( n ) = ( i ) " u ( n 2) is X(ejW), the sequence that has a DTFT given by Y (ejw) X(ejzo), find = Let x ( n ) be the sequence

If the DTFT of x ( n ) is expressed in terms of its real and imaginary parts as follows,

Fig. 11-1

3 JW X ( e ) - ( 1 - 0.Xe-jw)5

22.1 Describe all the derivatives ( rst, second, etc.) of the following functions: x 1 (b) f (x) = (a) f (x) = x 4 5x 4 x+1

To determine a locus, (1) state what is given and the condition to be satisfied; (2) find several points satisfying the condition which indicate the shape of the locus; then (3) connect the points and describe the locus fully. All geometric constructions require the use of straightedges and compasses. Hence if a locus is to be constructed, such drawing instruments can be used.

x ( e J W= cos3(3w) )

PRINCIPLE 1:

2.71 2.72

The locus of points equidistant from two given points is the perpendicular bisector of the line segment joining the two points (Fig. 11-2).

Let C:-,x(n)

(a) f (x) = x 3 5 (b) f (x) = = f (x) = 3x 2 f (x) = 6x f (4) (x) = 6 f (n) (x) = 0 for n 5 [by the quotient rule] (x + 1)Dx x xDx (x + 1) (x + 1)2 x+1 x 1 (x + 1)(1) x(1) = = = (x + 1) 2 (x + 1)2 (x + 1)2 (x + 1)2 [by the power chain rule]

x : ,

Fig. 11-2

Evaluate the following integral:

PRINCIPLE 2:

x ( n ) = anu(n)

The locus of points equidistant from two given parallel lines is a line parallel to the two lines and midway between them (Fig. 11-3). The locus of points equidistant from the sides of a given angle is the bisector of the angle (Fig. 11-4).

2y(n)- y(n - 2 ) = x ( n Find the frequency response, H (e'").

f (x) = 2(x + 1) 3 Dx (x + 1) 2 = 2(x + 1) 3 (1) = (x + 1)3 +6 f (x) = 2( 3)(x + 1) 4 = (x + 1)4 24 f (4) (x) = 2( 3)( 4)(x + 1) 5 = (x + 1)5 . . . f (n) (x) = 2( 3)( 4)( 5) (n)(x + 1) (n+1) = ( 1)n 1 n! (x + 1)n+1

PRINCIPLE 3:

- 1) +3x(n - 2) +2x(n - 3)

Find the inverse of the system that has a unit sample response h(n) = n(-k)"u(n - 3).

Fig. 11-3

h(n) = hl(n)

PRINCIPLE 4:

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