Exam (elaborations) TEST BANK FOR Principles of Communication Systems, Modulation and Noise 5th Edition By R. E. Ziemer and W. H. Tranter (SOLUTION MANUAL)

Problem 2.1 For the single-sided spectra, write the signal in terms of cosines: x(t) = 10cos(4πt + π/8) + 6 sin(8πt + 3π/4) = 10cos(4πt + π/8) + 6 cos(8πt + 3π/4 − π/2) = 10cos(4πt + π/8) + 6 cos(8πt + π/4) For the double-sided spectra, write the signal in terms of complex exponentials using Euler’s theorem: x(t) = 5exp[(4πt + π/8)] + 5 exp[−j(4πt + π/8)] +3 exp[j(8πt + 3π/4)] + 3 exp[−j(8πt + 3π/4)] The two sets of spectra are plotted in Figures 2.1 and 2.2. Problem 2.2 The result is x(t) = 4ej(8πt+π/2) +4e−j(8πt+π/2) + 2ej(4πt−π/4) + 2e−j(4πt−π/4) = 8cos(8πt + π/2) + 4 cos (4πt − π/4) = −8 sin(8πt) + 4cos (4πt − π/4) 1 2 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY f, Hz f, Hz 0 2 4 6 0 2 4 6 10 5 π/4 π/8 Single-sided amplitude Single-sided phase, rad. Figure 2.1: Problem 2.3 (a) Not periodic. (b) Periodic. To find the period, note that 6π 2π = n1f0 and 20π 2π = n2f0 Therefore 10 3 = n2 n1 Hence, take n1 = 3, n2 = 10, and f0 = 1 Hz. (c) Periodic. Using a similar procedure as used in (b), we find that n1 = 2, n2 = 7, and f0 = 1 Hz. (d) Periodic. Using a similar procedure as used in (b), we find that n1 = 2, n2 = 3, n3 = 11, and f0 = 1 Hz. Problem 2.4 (a) The single-sided amplitude spectrum consists of a single line of height 5 at frequency 6 Hz, and the phase spectrum consists of a single line of height -π/6 radians at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 2.5 at frequencies of 6 and -6 Hz, and the double-sided phase spectrum consists of a line of height -π/6 radians at frequency 6 Hz and a line of height π/6 at frequency -6 radians Hz. (b) Write the signal as xb(t) = 3cos(12πt − π/2) + 4 cos(16πt) From this it is seen that the single-sided amplitude spectrum consists of lines of height 3 and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists 2.1. PROBLEM SOLUTIONS 3 f, Hz 0 2 4 6 π/4 π/8 Double-sided phase, rad. f, Hz -6 -4 -2 0 2 4 6 5 Double-sided amplitude -π/8 -π/4 -6 -4 -2 Figure 2.2: 4 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY of a line of height -π/2 radians at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz, respectively, and lines of height 1.5 and 2 at frequencies -6 and -8 Hz, respectively. The double-sided phase spectrum consists of a line of height -π/2 radians at frequency 6 Hz and a line of height π/2 radians at frequency -6 Hz. Problem 2.5 (a) This function has area Area = Z∞ −∞ ²−1 · sin(πt/²) (πt/²) ¸2 dt = Z∞ −∞ · sin(πu) (πu) ¸2 du = 1 A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the central lobe of the function becomes narrower and higher. Thus, in the limit, it approximates a delta function. (b) The area for the function is Area = Z∞ −∞ 1 ² exp(−t/²)u (t) dt = Z∞ 0 exp(−u)du = 1 A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the function becomes narrower and higher. Thus, in the limit, it approximates a delta function. (c) Area = R ² −² 1 ² (1 − |t| /²) dt = R 1 −1 Λ (t) dt = 1. As ² → 0, the function becomes narrower and higher, so it approximates a delta function in the limit. Problem 2.6 (a) 513; (b) 183; (c) 0; (d) 95,583.8; (e) -157.9. Problem 2.7 (a), (c), (e), and (f) are periodic. Their periods are 1 s, 4 s, 3 s, and 2/7 s, respectively. The waveform of part (c) is a periodic train of impulses extending from -∞ to ∞ spaced by 4 s. The waveform of part (a) is a complex sum of sinusoids that repeats (plot). The waveform of part (e) is a doubly-infinite train of square pulses, each of which is one unit high and one unit wide, centered at · · ·, −6, −3, 0, 3, 6, · · ·. Waveform(f) is a raised cosine of minimum and maximum amplitudes 0 and 2, respectively. 2.1.