Signal Processing and Linear Systems Solution Ch.1

Chapter 1 1.1-1 For parts (a) and (b) (a) Ef = integral from 0 to 2pi of (sin^2 t) dt = integral from 0 to 2pi of ((1 - cos 2t) / 2) dt = (1/2)t - (1/4)sin 2t evaluated from 0 to 2pi = pi + 0 = pi (c) Ef = integral from 0 to 2pi of (sin^4 t) dt = integral from 0 to 2pi of (((1 - cos 2t) / 2)^2) dt = (1/4) integral from 0 to 2pi of (1 - 2cos 2t + cos^2 2t) dt = (1/4) integral from 0 to 2pi of (1 - 2cos 2t + ((1 + cos 4t) / 2)) dt = (1/4) integral from 0 to 2pi of (3/2 - 2cos 2t + (1/2)cos 4t) dt = (1/4) [ (3/2)t - sin 2t + (1/8)sin 4t ] evaluated from 0 to 2pi = (1/4) [3pi - 0 + 0] = 3pi/4 (d) Ef = integral from 0 to 2pi of ((2sin t)^2) dt = 4 integral from 0 to 2pi of (sin^2 t) dt = 4 [ (1/2)t - (1/4)sin 2t ] evaluated from 0 to 2pi = 4[pi + 0] = 4pi 1.1-2 Ef = integral from 0 to 1 of ((2t^3)^2) dt = integral from 0 to 1 of (4t^6) dt = (4/7)t^7 evaluated from 0 to 1 = 4/7 Efs = integral from 0 to infinity of ((-t)^2 e^t) dt = integral from 0 to infinity of (t^2 e^t) dt = [(-1)^2 e^t] (This seems incorrect for convergence) Efs = integral from 0 to 1 of ((-1)^2 t^3) dt = integral from 0 to 1 of (t^3) dt = (1/4)t^4 evaluated from 0 to 1 = 1/4 Efs = integral from 0 to 1 of ((t-1)^2) dt = integral from 0 to 1 of (t^2 - 2t + 1) dt = (1/3)t^3 - t^2 + t evaluated from 0 to 1 = 1/3 - 1 + 1 = 1/3 Efs = integral from 0 to 1 of (x^2) dx = (1/3)x^3 evaluated from 0 to 1 = 1/3 1.1-3 (a) Ex = integral from 0 to 1 of (1^2) dt = 1 Ey = integral from 0 to 1 of (1^2) dt = 1 Ey = integral from 0 to 1 of ((-1)^2) dt = 1 Ey = integral from 0 to 1 of (2^2) dt = 4 Ex+y = integral from 0 to 1 of (2^2) dt = 4 Therefore Ex+y = Ex + Ey. (b) Ex = integral from 0 to 1 of (1)^2 dt + integral from 1 to 2 of (1)^2 dt = 1+1=2 Ey = integral from 0 to pi/2 of (1)^2 dt + integral from pi/2 to pi of (-1)^2 dt = pi/2 + pi/2 = pi Ey = integral from 0 to pi/2 of (1)^2 dt + integral from pi/2 to 3pi/2 of (-1)^2 dt + integral from 3pi/2 to 2pi of (1)^2 dt = pi/2 + pi + pi/2 = 2pi Ex+y = integral from 0 to pi/2 of (2)^2 dt + integral from pi/2 to pi of (0)^2 dt + integral from pi to 3pi/2 of (0)^2 dt + integral from 3pi/2 to 2pi of (2)^2 dt = (pi/2) * 4 + (pi/2) * 4 = 4pi Similarly, we can show that Ex-y = 4pi. Therefore Ex+y = Ex + Ey. (c) Ex = integral from 0 to pi/4 of (1)^2 dt = pi/4 Ey = integral from 0 to pi/4 of (1)^2 dt = pi/4 Ey = integral from 0 to pi/4 of (-1)^2 dt = pi/4 Ew = integral from 0 to 1 of (1)^2 dt = 1 Ex+y = integral from 0 to pi/4 of (2)^2 dt = pi Ex-y = integral from 0 to pi/4 of (0)^2 dt = 0 Ey-x = integral from 0 to pi/4 of (0)^2 dt = 0 Eu+v = integral from 0 to pi/4 of (2)^2 dt + integral from pi/4 to pi/2 of (-2)^2 dt = pi + pi = 2pi Eu-v = integral from 0 to pi/4 of (0)^2 dt + integral from pi/4 to pi/2 of (-2 - (-2))^2 dt = integral from pi/4 to pi/2 of (0)^2 dt = 0 (This calculation seems incorrect, should be integral from pi/4 to pi/2 of (-4)^2 dt = 16(pi/2 - pi/4) = 4pi) Eu+v = integral from 0 to pi/4 of (2)^2 dt = pi (This value is from earlier in the line, the other 2pi value is more consistent with the other similar problem). Ey = integral from 0 to pi/4 of (-2)^2 dt = pi Ez = integral from 0 to pi/4 of (2)^2 dt = pi Ev = integral from 0 to pi/4 of (2)^2 dt = pi Therefore, in general Ex+y is not equal to Ex + Ey. 1.1-4 (a) P3f = (1/4) integral from -2 to 2 of ((-t)^2) dt = (1/4) integral from -2 to 2 of (t^2) dt = (1/4) [ (1/3)t^3 ] evaluated from -2 to 2 = (1/4) (8/3 - (-8/3)) = (1/4) * (16/3) = 4/3 Pf-f = (1/4) integral from -2 to 2 of ((-t)^2) dt = 4/3 (Same as above, looks like it's a repetition of the problem) (b) P3f = (1/4) integral from 0 to 4 of ((2t^3)^2) dt = (1/4) integral from 0 to 4 of (4t^6) dt = integral from 0 to 4 of (t^6) dt = (1/7)t^7 evaluated from 0 to 4 = (4^7)/7 = 16384/7 Pf-f = (1/4) integral from 0 to 4 of ((-t)^2) dt = (1/4) integral from 0 to 4 of (t^2) dt = (1/4) [ (1/3)t^3 ] evaluated from 0 to 4 = (1/4) * (64/3) = 16/3 Pf-f = (1/4) integral from 0 to 4 of ((t^3)^2) dt = (1/4) integral from 0 to 4 of (t^6) dt = (1/4) [ (1/7)t^7 ] evaluated from 0 to 4 = (4^7)/28 = 16384/28 = 4096/7 Sign change of a signal does not affect its power. Multiplication of a signal by a constant c increases the power by a factor c^2. 1.1-5 P_f = lim T->infinity (1/T) integral from -T/2 to T/2 of (f(t)f*(t)) dt = lim T->infinity (1/T) integral from -T/2 to T/2 of (|f(t)|^2) dt P_f = summation from k = -infinity to infinity of (|D_k|^2) The integrals of the cross-product terms (when k is not equal to r) are finite because the integrands are periodic signals (made up of sinusoids). These terms, when divided by T -> infinity, yield zero. The remaining terms (k = r) yield P_f = lim T->infinity (1/T) integral from -T/2 to T/2 of (summation from k = -infinity to infinity of (D_k D_k^*)) dt = summation from k = -infinity to infinity of (|D_k|^2) 1.1-6 (a) Power of a sinusoid of amplitude C is C^2/2 [Eq. (1.5a)] regardless of its frequency (omega is not equal to 0) and phase. Therefore in this case P = (10)^2/2 = 50. (b) Power of a sum of sinusoids is equal to the sum of the powers of the sinusoids [Eq. (1.5b)]. Therefore, in this case P = (10)^2/2 + (5)^2/2 = 50 + 12.5 = 62.5. (Note: in the image, it says 175, which seems to be a typo or different calculation for 10^2 + 5^2 = 125, not 50 + 12.5 = 62.5. It probably refers to (10)^2 + 5^2 + 7.5^2 for the next part.) (c) (10 + 2sin 3t) cos 10t = 10cos 10t + 2sin 3t cos 10t. Hence from Eq. (1.5b) P = (10)^2/2 + 2^2/2 + (sqrt(3))^2/2 + 1/2 = 50 + 2 + 1.5 + 0.5 = 54. (d) 10cos 5t cos 10t = 5(cos 5t + cos 15t). Hence from Eq. (1.5b) P = (5)^2/2 + (5)^2/2 = 12.5 + 12.5 = 25. (e) 10sin 5t cos 10t = 5(sin 15t - sin 5t). Hence from Eq. (1.5b) P = (5)^2/2 + (-5)^2/2 = 25. (f) integral from t_0 to t_0+T of (c(t) - e^(t^2)) dt + c(t) + integral from -infinity to infinity of (e^t + (1/T)) dt. Using the result in Prob. 1.1-5, we obtain P = (1/4) + (1/4) = 1/2. 1.3-1 fa(t) = f1(t) - f2(t) + f1(t - 1) fb(t) = f1(t) - 0.5f1(t+1) fc(t) = f1(t - 0.5) + f1(t + 0.5) The signal f3(t) can be obtained by (i) delaying f(t) by 1 second (replace t with t-1), (ii) then time-expanding by a factor 2 (replace t with t/2), (iii) then multiply with 1.5. Thus f3(t) = 1.5f(t/2 - 1). 1.3-2 All the signals are shown in Fig. S1.3-2 (Diagrams with waveforms f(t), f(-t), f(t-delta), f(t/a), f(t-a), f(t/2), f(t/2 - 1)) Fig. S1.3-2 1.3-3 All the signals are shown in Fig. S1.3-3 (Diagrams with waveforms f(t-4), f(-t-2), f(t-2), f(2t-4), f(t-2), f(t-2))