Exam (elaborations) TEST BANK FOR Fundamentals of Digital Communication By Upamanyu Madhow (Solution Manual)

Problem 2.1: Rather than doing the details of the convolution, we simply sketch the shapes of the waveforms. For a signal s = sc + jss and a filter h = hc + jhs, the convolution y = s ∗ h = (sc ∗ hc − ss ∗ hs) + j(sc ∗ hs + ss ∗ hc) For h(t) = smf (t) = s(−t), rough sketches of Re(y), Im(y) and |y| are shown in Figure 1. Clearly, the maximum occurs at t = 0. = c*hc −ss*hs Re(s* h) sc*hs ss*hc Im(s* h) |s* h| + = + s Figure 1: The convolution of a signal with its matched filter yields at peak at the origin. Problem 2.2: (a) Multiplication in the time domain corresponds to convolution in the frequency domain. The two sinc functions correspond to boxcars in the frequency domain, convolving which gives that S(f) has a trapezoidal shape, as shown in Figure 2. (b) We have u(t) = s(t) cos(100t) = s(t) ej100t + e−j100t 2 ↔ U(f) = S(f − 50) + S(f + 50) 2 The spectrum U(f) is plotted in Figure 2. 1/4 p sinc(t) −1/2 1/2 1 1/2 −1 1 sinc(2t) s(t) = sinc(t) sinc(2t) S(f) 1/2 1/2 3/2 f f CONVOLVE ... ... f f U(f) −51.5 −50 −48.5 48.5 50 51.5 u(t) = s(t) cos(100 t) Figure 2: Solution for Problem 2.2. Problem 2.3: The solution is sketched in Figure 3. (a) We have s(t) = I[−5,5] ∗ I[−5,5]. Since I[−5,5](t) ↔ 10sinc(10f), we have S(f) = 100sinc2(10f). (b) We have u(t) = s(t) sin(1000t) = s(t) ej1000t − e−j100t 2j ↔ U(f) = S(f − 50) − S(f + 50) 2j 0.1 −5 5 −5 5 S(f) = 100 sinc 2 (10f) −10 10 t s(t) 10 1 1 10 sinc(10f) = t * t ... ... −0.2 −0.1 0.1 0.2 f 100 ... ... 500 −500 f Im (U(f)) Re(U(f) = 0 50 Figure 3: Solution for Problem 2.3. Problem 2.4: Part (a) is immediate upon expanding ||s − ar||2. (b) The minimizing value of a is easily found to be amin = hs, ri ||r||2 Substituting this value into J(a), we obtain upon simplification that J(amin) = ||s||2 − hs, ri2 ||r||2 The condition J(amin) ≥ 0 is now seen to be equivalent to the Cauchy-Schwartz inequality. (c) For nonzero s, r, the minimum error J(amin) in approximating s by a multiple of r vanishes if and only if s is a multiple of r. This is therefore the condition for equality in the Cauchy-Scwartz inequality. For s = 0 or r = 0, equality clearly holds. Thus, the condition for equality can be stated in general as: either s is a scalar multiple of r (this includes s = 0 as a special case), or r is a scalar multiple of s (this includes r = 0 as a special case). (d) The unit vector in the direction of r is u = r ||r|| . The best approximation of s as a multiple of r is its projection along u, which is given by ˆs = hs, uiu = hs, r ||r||i r ||r|| and the minimum error is J(amin) = ||s − ˆs||2. Problem 2.5: We have y(t) = (x ∗ h)(t) = Z As( − t0)h(t −  )d = Ahst0 , ht i where st0( ) = s( − t0) and ht( ) = h(t −  ) are functions of  . (Recall that the complex inner product is defined as hu, vi = R uv). (a) Using the Cauchy-Schwartz inequality, we have |y(t)| ≤ |A|||st0||||ht || 2 It is easy to check (simply change variables in the associated integrals) that ||st0 || = ||s|| and ||ht || = ||h||. Using the normalization ||h|| = ||s||, we obtain the desired result that |y(t)| ≤ A||s||2. (b) Equality is attained for t = t0 if ht = ast0 for t = t0 for some scalar a. Since ||st0 || = ||ht || = ||s||, we must have |a| = 1. Thus, we have h(