SIGNALS AND SYSTEMS [CHAP. 1 (a) x(r - 2) is sketched in Fig. 1-18(a). (6) x(20 is sketched in Fig. 1-18(b). (c) x(t/2) is sketched in Fig. 1-18(c). (d) X( -t) is sketched in Fig. 1-1Nd). (c) Fig. 1-18 1.2. A discrete-time signal x[n] is shown in Fig. 1-19. Sketch and label each of the following signals. (a) x[n - 21; (b) x[2n]; (c) x[-n]; (d) x[-n + 21 Fig. 1-19 CHAP. 11 SIGNALS AND SYSTEMS (a) x[n - 21 is sketched in Fig. 1-20(a). (b) x[2n] is sketched in Fig. 1-20(b). (c) x[-n] is sketched in Fig. 1-2Nc). (d) x[-n + 21 is sketched in Fig. 1-2Nd). (4 Fig. 1-20 13. Given the continuous-time signal specified by x(t) = (A - It' -1lfll otherwise determine the resultant discrete-time sequence obtained by uniform sampling of x(t) with a sampling interval of (a) 0.25 s, (b) 0.5 s, and (c) 1.0 s. It is easier to take the graphical approach for this problem. The signal x(t) is plotted in Fig. 1-21(a). Figures 1-21(b) to (dl give plots of the resultant sampled sequences obtained for the three specified sampling intervals. (a) T, = 0.25 s. From Fig. 1-21(b) we obtain x[n] = (. . . ,0,0.25,0.5,0.75,1,0.75,0.5,0.25,0,. . .) T (b) T, = 0.5 s. From Fig. 1-21(c) we obtain x[n]= { , 0,0.5,1,0.5,0 , I T SIGNALS AND SYSTEMS [CHAP. 1 (4 Fig. 1-21 (c) T, = 1 s. From Fig. 1-21(d) we obtain x[n] = (. . .,O, 1,O . .) = S[nl 1.4. Using the discrete-time signals x,[n] and x,[n] shown in Fig. 1-22, represent each of the following signals by a graph and by a sequence of numbers. (a) yJn1 =x,[nl +x,[nl; (b) y,[nI= 2x,[nl; (c) y,[nI =xJnIxJnl Fig. 1-22 CHAP. 11 SIGNALS AND SYSTEMS (a) y,[n] is sketched in Fig. 1-23(a). From Fig. 1-23(a) we obtain (b) y2[n] is sketched in Fig. 1-23(b). From Fig. 1-23(b) we obtain (c) yJn] is sketched in Fig. 1-23(c). From Fig. 1-23(c) we obtain (d Fig. 1-23 1.5. Sketch and label the even and odd components of the signals shown in Fig. 1-24. Using Eqs. (1.5) and (1.6), the even and odd components of the signals shown in Fig. 1-24 are sketched in Fig. 1-25. SIGNALS AND SYSTEMS [CHAP. 1 (4 Fig. 1-24 1.6. Find the even and odd components of x(r) = eJ'. Let x,(r) and x,(I) be the even and odd components of ei', respectively. eJ' =x,(I) +x,(I) From Eqs. (1.5) and (1.6) and using Euler's formula, we obtain x,( I) = $(eJr + e-J') = cos I x,,(I) = f(ei'-e-j') =jsint Show that the product of two even signals or of two odd signals is an even signal and that the product of an even and an odd signaI is an odd signal. Let x(t) =xl(t)x2(t). If XJI) and x2(l) are both even, then x(-l) =x,(-I)X,(-t) =xI(I)x2(t) =x(t) and x(t) is even. If x,(t) and x2(t) are both odd, then x(-I) =x,(-I)x,(-I) = -x,(t)[-x2(t)] =x1(t)x2(t) =x(t) and x(t) is even. If x,(t) is even and x2(f) is odd, then and X(I) is odd. Note that in the above proof, variable I represents either a continuous or a discrete variable. CHAP. 11 SIGNALS AND SYSTEMS (4 Fig. 1-25 SIGNALS AND SYSTEMS 1.8. Show that [CHAP. 1 (a) If x(t) and x[n] are even, then (b) If x(t) and x[n] are odd, then x(0) = 0 and x[O] =O k /a ~(r) dr = 0 and x x[n] =O -a n= -k (a) We can write Letting t = -A in the first integral on the right-hand side, we get Since x(t) is even, that is, x( -A) = x(A), we have Hence, Similarly, Letting n = -m in the first term on the right-hand side, we get Since x[n] is even, that is, x[ -m] =x[m], we have Hence, (1.75a) (I. 75b) (6) Since x(t) and x[n] are odd, that is, x( - t) = -x(t) and x[ -n] = -x[n], we have X( -0) = -x(O) and x[-01 = -x[O] CHAP. 11 SIGNALS AND SYSTEMS Hence, Similarly, and in view of Eq. (1.76). 1.9. Show that the complex exponential signal ( t ) = ,j@d is periodic and that its fundamental period is 27r/00. By Eq. (1.7), x(t) will be periodic if ei@dt + TI = eiwd Since eiw~(r + T) = eiqreiq,T we must have eimoT = 1 (1.78) If w, = 0, then x(t) = 1, which is periodic for any value of T. If o0 # 0, Eq. (1.78) holds if 27T ooT=m2r or T=m- m = positive integer a0 Thus, the fundamental period To, the smallest positive T, of x(t) is given by 2r/oo. 1.10. Show that the sinusoidal signal x(t) = cos(w,t + 8) is periodic and that its fundamental period is 27r/wo. The sinusoidal signal x(l) will be periodic if cos[o,(t + T) + 81 = ws(oot + 8) We note that cos[w,(t + T) + 81 = cos[oot + 8 + woT] = cos(oot + 8) SIGNALS AND SYSTEMS [CHAP. 1 27 w0T=m2.rr or T=m- m = positive integer *o Thus. the fundamental period To of x(r) is given by 2.rr/wo. 1.11. Show that the complex exponential sequence x[n] = e~"~" is periodic only if fl0/2.rr is a rational number. By Eq. (1.9), x[n] will be periodic if ,iflo(" +Nl = ,in,,n,i~hp = ,inon or ein~N = 1 Equation (1.79) holds only if floN =m2~ m = positive integer or a0 m -= - = rational number 2.rr N Thus, x[n] is periodic only if R0/27r is a rational number 1.12. Let x(r) be the complex exponential signal with radian frequency wo and fundamental period To = 2.rr/oo. Consider the discrete-time sequence x[n] obtained by uniform sampling of x(t) with sampling interval Ts. That is, x[n] =x(nT,) =eJ"unT. Find the condition on the value of T, so that x[n] is periodic. If x[n] is periodic with fundamental period N,,, then ,iou(n+N,,)T, = ,iw~nT,,iwuN,J', = ejwun-l; Thus, we must have T, m -= - rational number To No Thus x[n] is periodic if the ratio T,/T,, of the sampling interval and the fundamental period of x(t) is a rational number. Note that the above condition is also true for sinusoidal signals x(t) = cos(o,,t + 8). CHAP. 11 SIGNALS AND SYSTEMS 1.13. Consider the sinusoidal signal x(t) = cos 15t Find the value of sampling interval T, such that x[n] = x(nT,) is a periodic sequence. Find the fundamental period of x[n] = x(nT,) if TT = 0.1~ seconds. The fundamental period of x(t) is To = 2*rr/wo = 27/15. By Eq. (1.81), x[n] =x(nTs) is periodic if where m and No are positive integers. Thus, the required value of T, is given by Substituting T, = 0.1~ = ~/10 in Eq. (1.821, we have Thus, x[n] =x(nT,) is periodic. By Eq. (1.82) The smallest positive integer No is obtained with m = 3. Thus, the fundamental period of x[nl = x(0.l~n) is N, = 4. .4. Let x,(t) and x,(t) be periodic signals with fundamental periods T, and T2, respec- tively. Under what conditions is the sum x(t) =x,(t) + x2(t) periodic, and what is the fundamental period of x( t) if it is periodic? Since x,(t) and x,(t) are periodic with fundamental periods TI and T,, respectively, we have xl(t) =x,(t + TI) =x,(t + mT,) m = positive integer x2(t) =x2(t + T2) =x2(f + kT2) k = positive integer Thus, In order for x(t) to be periodic with period T, one needs Thus, we must have mT, = kT2 = T TI k = rational number T2 m In other words, the sum of two periodic signals is periodic only if the ratio of their respective periods can be expressed as a rational number. Then the fundamental period is the least [...].. .SIGNALS AND SYSTEMS [CHAP 1 common multiple of T, and T2, and it is given by Eq (1.84) if the integers m and k are relative prime If the ratio T,/T, is an irrational number, then the signals x,(t) and x,(t) do not have a common period and x(t) cannot be periodic 1.15 Let x,[n] and x2[n] be periodic sequences with fundamental periods N , and N2, respectively Under what... = 4 < m 4 2N+ 1 m Thus, x [ n ] is a power signal BASIC SIGNALS N - C 1 = 2N+1 2' -N CHAP 11 SIGNALS AND SYSTEMS Let Since T T = - t Then by definition (1.18) > 0 and 7 < 0 imply, respectively, that t < 0 and r > 0, we obtain which is shown in Fig 1-26 Fig 1-26 1.22 A continuous-time signal following signals At) is shown in Fig 1-27 Sketch and label each of the ( a ) x ( t ) u ( l - t ) ; ( b ) x... ( 1 1 9 ) and x(r)u(l - t ) is sketched in Fig 1-28(a) ( 6 ) By definitions (1.18) and (1.19) u -( t - 1 = and x ( t ) [ u ( r )- u(t - I ) ] is sketched in Fig 1-28(b) O . 11 SIGNALS AND SYSTEMS (4 Fig. 1-25 SIGNALS AND SYSTEMS 1.8. Show that [CHAP. 1 (a) If x(t) and x[n] are even, then (b) If x(t) and. SIGNALS AND SYSTEMS [CHAP. 1 (4 Fig. 1-24 1.6. Find the even and odd components of x(r) = eJ'. Let x,(r) and x,(I) be the even and