Principles of Communication 5Ed R. E. Ziemer, William H. Tranter Solutions Manual

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Transcript of Principles of Communication 5Ed R. E. Ziemer, William H. Tranter Solutions Manual

Chapter 2Signal and Linear System Theory2.1 Problem SolutionsProblem 2.1For the single-sided spectra, write the signal in terms of cosines:x(t) = 10 cos(4t +/8) + 6 sin(8t + 3/4)= 10 cos(4t +/8) + 6 cos(8t + 3/4 /2)= 10 cos(4t +/8) + 6 cos(8t +/4)For the double-sided spectra, write the signal in terms of complex exponentials using Eulerstheorem:x(t) = 5 exp[(4t +/8)] + 5 exp[j(4t +/8)]+3 exp[j(8t + 3/4)] + 3 exp[j(8t + 3/4)]The two sets of spectra are plotted in Figures 2.1 and 2.2.Problem 2.2The result isx(t) = 4ej(8t+/2)+ 4ej(8t+/2)+ 2ej(4t/4)+ 2ej(4t/4)= 8 cos (8t +/2) + 4 cos (4t /4)= 8 sin(8t) + 4 cos (4t /4)1Principles of Communication 5Ed R. E. Zeimer, William H Tranter Solutions Manual2 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORYf, Hz f, Hz0 2 4 6 0 2 4 6105/4/8Single-sided amplitude Single-sided phase, rad.Figure 2.1:Problem 2.3(a) Not periodic.(b) Periodic. To nd the period, note that62 = n1f0 and 202 = n2f0Therefore103 = n2n1Hence, take n1 = 3, n2 = 10, and f0 = 1 Hz.(c) Periodic. Using a similar procedure as used in (b), we nd that n1 = 2, n2 = 7, andf0 = 1 Hz.(d) Periodic. Using a similar procedure as used in (b), we nd 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 6Hz, and the phase spectrum consists of a single line of height -/6 radians at frequency 6Hz. The double-sided amplitude spectrum consists of lines of height 2.5 at frequencies of6 and -6 Hz, and the double-sided phase spectrum consists of a line of height -/6 radiansat frequency 6 Hz and a line of height /6 at frequency -6 radians Hz.(b) Write the signal asxb(t) = 3 cos(12t /2) + 4 cos(16t)From this it is seen that the single-sided amplitude spectrum consists of lines of height 3and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists2.1. PROBLEM SOLUTIONS 3f, Hz 0 2 4 6/4/8Double-sided phase, rad.f, Hz -6 -4 -2 0 2 4 65Double-sided amplitude-/8-/4-6 -4 -2Figure 2.2:4 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORYof a line of height -/2 radians at frequency 6 Hz. The double-sided amplitude spectrumconsists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz, respectively, and lines ofheight 1.5 and 2 at frequencies -6 and -8 Hz, respectively. The double-sided phase spectrumconsists of a line of height -/2 radians at frequency 6 Hz and a line of height /2 radiansat frequency -6 Hz.Problem 2.5(a) This function has areaArea =Z1sin(t/)(t/)2dt=Zsin(u)(u)2du = 1A sketch shows that no matter how small is, the area is still 1. With 0, the centrallobe of the function becomes narrower and higher. Thus, in the limit, it approximates adelta function.(b) The area for the function isArea =Z1 exp(t/)u(t) dt =Z0exp(u)du = 1A sketch shows that no matter how small is, the area is still 1. With 0, the functionbecomes narrower and higher. Thus, in the limit, it approximates a delta function.(c) Area = R 1 (1 |t| /) dt = R11(t) dt = 1. As 0, the function becomes narrowerand 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 spacedby 4 s. The waveform of part (a) is a complex sum of sinusoids that repeats (plot). Thewaveform of part (e) is a doubly-innite train of square pulses, each of which is one unithigh and one unit wide, centered at , 6, 3, 0, 3, 6, . Waveform (f) is a raisedcosine of minimum and maximum amplitudes 0 and 2, respectively.2.1. PROBLEM SOLUTIONS 5Problem 2.8(a) The result isx(t) = Reej6t+ 6 Reej(12t/2) = Rehej6t+ 6ej(12t/2)i(b) The result isx(t) = 12ej6t+ 12ej6t+ 3ej(12t/2)+ 3ej(12t/2)(c) The single-sided amplitude spectrum consists of lines of height 1 and 6 at frequenciesof 3 and 6 Hz, respectively. The single-sided phase spectrum consists of a line of height/2 at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height3, 1/2, 1/2, and 3 at frequencies of 6, 3, 3, and 6 Hz, respectively. The double-sidedphase spectrum consists of lines of height /2 and /2 at frequencies of 6 and 6 Hz,respectively.Problem 2.9(a) Power. Since it is a periodic signal, we obtainP1 = 1T0Z T00 4 sin2(8t +/4) dt = 1T0Z T00 2 [1 cos (16t +/2)] dt = 2 Wwhere T0 = 1/8 s is the period.(b) Energy. The energy isE2 = Z e2tu2(t)dt = Z 0 e2tdt = 12(c) Energy. The energy isE3 = Z e2tu2(t)dt = Z 0e2tdt = 12(d) Neither energy or power.E4 = limTZ TTdt(2 +t2)1/4 = P4 = 0 since limT 1TRTT dt(2+t2)1/4 = 0.(e) Energy. Since it is the sum of x1(t) andx2(t), its energy is the sum of the energies of these two signals, or E5 = 1/.6 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY(f) Power. Since it is an aperiodic signal (the sine starts at t = 0), we useP6 = limT12TZ T0 sin2(5t) dt = limT12TZ T012 [1 cos (10t)] dt= limT12TT2 12 sin(20t)20T0 = 14 WProblem 2.10(a) Power. Since the signal is periodic with period /, we haveP = Z /0 A2|sin(t +)|2dt = Z /0A22 {1 cos [2 (t +)]} dt = A22(b) Neither. The energy calculation givesE = limTZ TT(A)2dt +jt jt = limTZ TT(A)2dt2 +t2 The power calculation givesP = limT12TZ TT(A)2dt2 +t2 = limT(A)22T ln 1 +p1 +T2/21 +p1 +T2/2! = 0(c) Energy:E = Z 0 A2t4exp(2t/) dt = 34A25(use table of integrals)(d) Energy:E = 2Z /20 22dt +Z /2 12dt! = 5Problem 2.11(a) This is a periodic train of boxcars, each 3 units in width and centered at multiples of6:Pa = 16Z 332t3dt = 16Z 1.51.5dt = 12 W2.1. PROBLEM SOLUTIONS 7(b) This is a periodic train of unit-high isoceles triangles, each 4 units wide and centeredat multiples of 5:Pb = 15Z 2.52.52t2dt = 25Z 201 t22dt = 25 231 t2320= 415 W(c) This is a backward train of sawtooths (right triangles with the right angle on the left),each 2 units wide and spaced by 3 units:Pc = 13Z 201 t22dt = 13 231 t2320= 29 W(d) This is a full-wave rectied cosine wave of period 1/5 (the width of each cosine pulse):Pd = 5Z 1/101/10 |cos (5t)|2dt = 2 5Z 1/10012 + 12 cos (10t)dt = 12 WProblem 2.12(a) E = , P = ; (b) E = 5 J, P = 0 W; (c) E = , P = 49 W; (d) E = , P = 2 W.Problem 2.13(a) The energy isE = Z 66 cos2(6t) dt = 2Z 6012 + 12 cos (12t)dt = 6 J(b) The energy isE = Z he|t|/3cos (12t)i2dt = 2Z 0 e2t/312 + 12 cos (24t)dtwhere the last integral follows by the eveness of the integrand of the rst one. Use a tableof dente integrals to obtainE = Z 0 e2t/3dt +Z 0 e2t/3cos (24t) dt = 32 + 2/3(2/3)2+ (24)2Since the result is nite, this is an energy signal.(c) The energy isE = Z {2 [u(t) u(t 7)]}2dt = Z 70 4dt = 28 J8 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORYSince the result is nite, this is an energy signal.(d) Note thatZ tu() d = r (t) = 0, t < 0t, t 0which is called the unit ramp. The energy isE = Z [r (t) 2r (t 10) +r (t 20)]2dt = 2Z 100 t102dt = 203 Jwhere the last integral follows because the integrand is a symmetrical triangle about t = 10.Since the result is nite, this is an energy signal.Problem 2.14(a) Expand the integrand, integrate term by term, and simplify making use of the orthog-onality property of the orthonormal functions.(b) Add and subtract the quantity suggested right above (2.34) and simplify.(c) These are unit-high rectangular pulses of width T/4. They are centered at t =T/8, 3T/8, 5T/8, and 7T/8. Since they are spaced by T/4, they are adjacent to eachother and ll the interval [0, T].(d) Using the expression for the generalized Fourier series coecients, we nd that X1 =1/8, X2 = 3/8, X3 = 5/8, and X4 = 7/8. Also, cn = T/4. Thus, the ramp signal isapproximated bytT = 181 (t) + 382 (t) + 583 (t) + 784 (t) , 0 t Twhere the n (t)s are given in part (c).(e) These are unit-high rectangular pulses of width T/2 and centered at t = T/4 and 3T/4.We nd that X1 = 1/4 and X2 = 3/4.(f) To compute the ISE, we useN = ZT |x(t)|2dt NXn=1cnX2nNote that RT |x(t)|2dt = RT0 (t/T)2dt = T/3. Hence, for (d),ISEd = T3 T4 164 + 964 + 2564 + 4964 = 5.208 103T.For (e), ISEe = T3 T2 116 + 916 = 2.083 102T.2.1. PROBLEM SOLUTIONS 9Problem 2.15(a) The Fourier coecients are (note that the period = 12 20)X1 = X1 = 14; X0 = 12All other coecients are zero.(b) The Fourier coecients for this case areX1 = X1 = 12 (1 +j)All other coecients are zero.(c) The Fourier coecients for this case are (note that the period is 220)X2 = X2 = 18; X1 = X1 = 14; X0 = 14All other coecients are zero.(d) The Fourier coecients for this case areX3 = X3 = 18; X1 = X1 = 38All other coecients are zero.Problem 2.16The expansion interval is T0 = 4 and the Fourier coecients areXn = 14Z 22 2t2ejn(/2)tdt = 24Z 20 2t2cosnt2dtwhich follows by the eveness of the integrand. Let u = nt/2 to obtain the formXn = 2 2n3Z n0 u2cos u du = 16(n)2 (1)nIf n is odd, the Fourier coecients are zero as is evident from the eveness of the functionbeing represented. If n = 0, the integral for the coecients isX0 = 14Z 22 2t2dt = 83The Fourier series is thereforex(t) = 83 + Xn=, n6=0(1)n 16nejn(/2)t10 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORYProblem 2.17Parts (a) through (c) were discussed in the text. For (d), break the integral for x(t) upinto a part for t < 0 and a part for t > 0. Then use the odd half-wave symmetry contition.Problem 2.18This is a matter of integration. Only the solution for part (b) will be given here. Theintegral for the Fourier coecients is (note that the period really is T0/2)Xn = AT0Z T0/20 sin(0t) ejn0tdt= Aejn0t0T0 (1 n2) [jnsin(0t) + cos (0t)]T0/20= A1 +ejn0T0 (1 n2), n 6= 1For n = 1, the integral isX1 = AT0Z T0/20 sin(0t) [cos (jn0t) j sin(jn0t)] dt = jA4 = X1This is the same result as given in Table