Signal Processing and Linear Systems Solution Ch.8

Chapter 8 (a) (b) (c) λ (d) λ * 0.6065 0 * 0.5 1-6487 * 0 Fig. 88.2-1 8.2-1 (a) e ^ (-0.3k) = (0.6065) ^ k (b) * e ^ (0.5k) = (1.6487) ^ k (c) epsilon ^ (- j * pi*k) = (epsilon ^ (- j * pi)) ^ k = (- 1) ^ k Figure of A (d) e^ j*k =(e^ j* )^ k = (- 1) ^ k 8.2-2 (8) epsilon ^ (- (1 + j*tau) * k) = (e ^ - 1 * e ^ (- j * pi)) ^ k = (- 1/a) ^ k (b) e ^ (- (1 - j*epsilon) * k) = (e ^ - 1 * e ^ (j*epsilon)) ^ k = (- 1/c) ^ k (c) (1+j pi)k = (e * e ^ (j*pi)) ^ k = (- e) ^ k * (d) * e ^ ((1 - j*pi) * k) = (e * e ^ (- j * tau)) ^ k = (- e) ^ k (e) e ^ (- (1 + j * pi/2) * k) = (e ^ (- i)) ^ k * e ^ (- j * pi/2 * pi/2) = (1/2) ^ k * [cos pi/3 * k - j * sin pi/3 * k](f) * e ^ ((i - j * pi/3) * k) = (e ^ 1) ^ k * e ^ (- j * pi/3 * k) =e^ k |cos pi 3 k-j sin pi 3 k 8.2-3 (a) Periodic because Omega / 2 * pi = 1/4 rational. Using Eq. (8.96), we find that N_{0} = m(2pi / Omega) = m(4) is an integer for the smallest m = 1 . (b) Aperiodic because 0/2 = 1/ (c) Aperiodic because Pi / 2 * pi = 1/4 * pi not rational. (d) Periodic because Omega / 2 * pi = 1/6 rational. 8.2-4 (a) Periodic because n / 2 * pi = 3/10 rational. Using Eq. (8.9b), we find that N_{0} = m(2pi / Omega) = m(10/3) an integer for the smallest m = 3 Hence N_{0} = 10 b) Periodic because Omega / 2 * pi = 2/5 rational. Using the argument in part (a), we obtain N_{0} = 5 ( (a) All the three sinusoids with periods 10. 4, and 5. respectively. Because 2(10) = 5(4) = 4(5) = 20 the period is 20. This is because we can fit exactly 2, 5, and 4 cycles of the three sinusoids 8.2-5 (a) Omega_{f} = 0.8pi, |Omega_{f}| = 0.8pi (b) Omega_{f} = 1.2pi - 2pi = - 0.8pi Omega_{f} |=0.83 (с) Omega_{f} = 6.9 - 2pi = 0.6168 |Omega_{f}| = 0.6168 (d) Omega_{f} = 3.7pi - 4pi = - 0.3pi |Omega_{f}| = 0.3pi (e) Omega_{f} = 22.9pi - 22pi = 0.9pi |Omega_{f}| = 0.9pi 8.2-6 Because 1.4 = 2pi - 0.6pi ,cos(1.4 pi k+ pi 3 )=cos(-0.6 pi k+ pi 3 )=cos(0.6 pi k- pi 3 ). Also Similarly Therefore, cos(0.6pi*k + pi/6) = cos 0.6pi*k * cos(pi/6) - sin 0.6pi*k * sin(pi/6) = (sqrt(3))/2 * cos 0.6pi*k - 1/2 * sin 0.6pi*k coe(0.6pi*k - pi/3) = cos 0.6pi*k * cos(pi/3) + sin 0.6pi*k * sin(pi/3) = 1/2 * cos 0.6pi*k + (sqrt(3))/2 * sin 0pi*k cos(0.6pi*k + pi/6) + sqrt(3) * cos(1.4pi*k + pi/3) = sqrt(3) * cos 0.6pi*k + sin 0.6pi*k = 2cos(0.6pi*k - pi/6) 8.2-7 (a) j(8.2pi*k - theta) - jk*pi*k = e ^ (j(\mathbb{O} * 0.2pi*k + theta)) (b) e ^ (j * 4pi*k - j * 4k) = e ^ (j0k) = 1 e ^ (- j * 1.93k + j * 2pi*k) = e ^ (j4.333k) e ^ (- j * 10, Trkj * 12rk) = e ^ (j * 1.3pi*k) (3) 8.2-8 E_{f} = sum 0 to ∞ (0.8) ^ 2 = sum 0 to ∞ (0.64) ^ 2 = ((0.64) ^ (∞) - (0.64) ^ 6)/(0.64 - 1) = 2.7778 Rest is trivial. The energy of - f * [k] E_{f} = 2.7778 and the energy of cf[k] is c ^ 2 * E_{f} = 2.7778c ^ 2 108 8.2-9 (a) E_{f} = (1) ^ 2 + 2 * (2) ^ 2 + 2 * (1) ^ 2 = 19 b) E_{f} = (3) ^ 2 + 2 * (2) ^ 2 + 2 * (1) ^ 2 = 19 (c) E_{f} = 2 * (3) ^ 2 + 2 * (6) ^ 2 + 2 * (9) ^ 2 = 252 d E_{f} = 2 * (2) ^ 2 + 2 * (4) ^ 2 = 40 8.2-10 (a) aligned P f =lim N infty 1 2N+1 sum =-N ^ N (1)^ 2k =1\\ P_{f} = lim N -> ∞ sum N = N to N (1) ^ 2 = 0.5 aligned (b) Py = lim (1) 1 N (c) P_{f} = lim N -> ∞ 1/(2N + 1) * sum 0 to N (1) ^ 2 = 0.5 (4) P f =|im N...as 1/(2.7 + 1) * sum 0 to N (- 1) ^ (2k) =0.5| (e) N 0 = mathfrak F = mathfrak F =6,P f = 1 8 sum 0 ^ 5 (cos( pi 3 k+ pi 2 ))^ 2 = 1 hbar [((sqrt(2))/2) ^ 2 + 0 ^ 2 + (- (sqrt(3))/2) ^ 2 + (- (sqrt(3))/2) ^ 2 + 0 ^ 2 + ((sqrt(3))/2) ^ 2] = 0.5 8.2-11 (a) P_{f} = 1/6 * [(3) ^ 2 + 2 * (2) ^ 2 + 2 * (1) ^ 2] = 16 6 ( b) P_{f} = 1/12 * [2 * (1) ^ 2 + 2 * (2) ^ 2 + 2 * (3) ^ 2 = 7/3] 8.2-12 (a) In the following discussion. (T*pi)/N_{0} = Omega_{0} Because D e^ j11 0 X is periodic with period No P_{f} = 1/N_{0} * sum k = 0 to N_{0} - 1 |D * e ^ (I*C_{0}*k)| ^ 2 = 1/N_{0} * sum k = 0 to N_{0} - 1 |D| ^ 2 = |D| ^ 2 (b) P f = 1 N 0 sum k=0 ^ N 0 -1 | sum r=0 ^ N 0 -1 D r e^ j*G 0 k |^ 2 = 1 N 0 sum k=0 ^ N 0 -1 [ sum r=0 ^ N 0 -1 D r e^ i Omega 0 k sum m,80 ^ N 0 -1 D r 0 ^ * e ^ (- j * pi*i*Omega_{0}*k) ] Interchanging the order of summation yields P f = 1 N 0 sum r=0 ^ N g -1 sum r=0 ^ N g -1 D rq ^ * [sum k = 0 to N_{g} - 1 e ^ (f(r - m) * U_{0}*k)] From Eq. (3.43) in Appendix 5.1, the sum inside the parenthesis is No when r = m and is zero otherwise. Hence P_{f} = sum r = 0 to N_{9} - 1 |D_{r}| ^ 2 8.3-1 In the present case. mathcal F * = 1 / T = 2 MHz. Therefore Eq. (8.21) yields 2 * 10 ^ 6 = 0 + 2 * 10 ^ 6 Hence, the reduced frequency is 0 (dc). 8.3-2 For the sinusoid 10cos(11pi*t + pi/8) the samples sinusoid, obtained by replacing t with kT = 0.1k is 10cos(11/10 * pi*k + pi/6) = 10cos[(2pi - (9pi)/10) * k + pi/6] = 10cos(- 9/10 * pi*k + pi/6) = 10ccs(9/10 * pi*k - pi/6) For the sinusoid 5 cos(29pi*t - pi/6) the samples sinusoid (obtained by replacing & with kT = 0.1k ) 5cos(29/10 * pi*k - pi/6) = 5cos[(2pi + (9pi)/10) * k - pi/5] = 5cos(9/10 * pi*k - pi/6) 8.3-3 (a) In this case, T = 1/4000 Hence, the sampled signal, which is obtained by replacing t with kT = k / 4000 is 10cos pi/2 * k + sqrt(2) * sin (3pi)/4 * k + 2cos((5pi)/4 * k + pi/4) = 10cos pi/2 * k + sqrt(2) * sin (3pi)/4 * k + 2cos((- 3pi)/4 * k + pi/4) = 10cos pi/2 * k + sqrt(2) * sin (3pi)/4 * k + 2cos((3pi)/4 * k - pi/4) We can combine the last 2 terms by using a suitable trigonometric identity as follows: sqrt(2) * sin (3pi)/4 * k + 2cos((3pi)/4 * k - pi/4) = sqrt(2) * sin (3pi)/4 * k + 2cos (3pi)/4 * k * cos(pi/4) + 2sin (3pi)/4 * k * sin(pi/4) = sqrt(2) * sin (3pi)/4 * k + sqrt(2) * cos (3pi)/4 * k + sqrt(2) * sin (3pi)/4 * k = 2sqrt(2) * sin (3pi)/4 * k + sqrt(2) * cos (3pi)/4 * k = sqrt(10) * cos((3pi)/4 * k - 1.107) The frequency 5pi / 4 has been reduced to 3pi / 4 this indicates aliasing 109