The sequence x(n), illustrated in Fig. 1-8(a), is a linearly decreasing sequence that begins at index n = 0 and ends at index n = 5. The first sequence that is to be sketched, yl(n) = x(4 - n), is found by shifting x(n) by four and time-reversing. | SIGNALS AND SYSTEMS 21 The sequence x n illustrated in Fig. 1 -8 fl is a linearly decreasing sequence that begins at index n 0 and ends at index n 5. The first sequence that is to be sketched yi n x 4 n is found by shifting x n by four and time-reversing. Observe that at index n 4 y n is equal to x 0 . Therefore yi n has a value of 6 at n 4 and decreases linearly to the left decreasing values of n until n 1 beyond which yi n 0. The sequence yt n is shown in Fig. 1-8 Z . Fig. 1-8. Performing signal manipulations. b The second sequence yz n x 2n 3 is formed through the combination of time-shifting and downsampling. Therefore y2 n may be plotted by first shifting x n to the right by three delay as shown in 22 SIGNALS AND SYSTEMS CHAP. 1 Fig. l-8 c . The sequence yz n is then formed by down-sampling by a factor of 2 . keeping only the even index terms as indicated by the solid circles in Fig. l-8 c . A sketch of y2 is shown in Fig. l-8 d . c The third sequence x 8 3 is formed through a combination of time-shifting down-sampling and time-reversal. To sketch y n we begin by plotting x 8 which is formed by shifting x n to the left by eight advance and reversing in time as shown in Fig. 1 -8 e . Then yi n is found by extracting every third sample of x 8 n as indicated by the solid circles which is plotted in Fig. 1 -8 . d Finally y n x n2 2n 1 is formed by a nonlinear transformation of the time variable n. This sequence may be easily sketched by listing how the index n is mapped. First note that if n 4 or n 2 then n2 2n -I- 1 9 and therefore y n 0. For 1 n 3 we have y4 1 y4 3 x 4 2 j4 0 y4 2 x l 5 y4 l x 0 6 The sequence y4 n is sketched in Fig. l-8 g . The notation x n y is used to define the sequence that is formed as follows x n v x n modulo N where n modulo N is the positive integer in the range 0 N 1 that remains after dividing n by N. For example 3 8 3 12 s 4 and 6 4 2. If x n i sin i7r 2 M n make a sketch of a x n 3 and x n - 2 3. a We begin by noting that n