Exam (elaborations) TEST BANK FOR Principles of Communication Systems,

Exam (elaborations) TEST BANK FOR Principles of Communication Systems, For the single-sided spectra, write the signal in terms of cosines: x(t) = 10 cos(4πt + π/8) + 6 sin(8πt + 3π/4) = 10 cos(4πt + π/8) + 6 cos(8πt + 3π/4 − π/2) = 10 cos(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) = 5 exp[(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) = 8 cos (8πt + π/2) + 4 cos (4πt − π/4) = −8 sin (8πt) + 4 cos (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) = 3 cos(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. PROBLEM SOLUTIONS 5 Problem 2.8 (a) The result is x(t) = Re ¡ ej6πt ¢ + 6 Re ³ ej(12πt−π/2)´ = Re h ej6πt + 6ej(12πt−π/2)i (b) The result is x(t) = 1 2 ej6πt + 1 2 e−j6πt + 3ej(12πt−π/2) + 3e−j(12πt−π/2) (c) The single-sided amplitude spectrum consists of lines of height 1 and 6 at frequencies of 3 and 6 Hz, respectively. The single-sided phase spectrum consists of a line of height −π/2 at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 3, 1/2, 1/2, and 3 at frequencies of −6, −3, 3, and 6 Hz, respectively. The double-sided phase spectrum consists of lines of height π/2 and −π/2 at frequencies of −6 and 6 Hz, respectively. Problem 2.9 (a) Power. Since it is a periodic signal, we obtain P1 = 1 T0 Z T0 0 4 sin2 (8πt + π/4) dt = 1 T0 Z T0 0 2 [1 − cos (16πt + π/2)] dt = 2 W where T0 = 1/8 s is the period. (b) Energy. The energy is E2 = Z ∞ −∞ e−2αt u2(t)dt = Z ∞ 0 e−2αt dt = 1 2α (c) Energy. The energy is E3 = Z ∞ −∞ e2αt u2(−t)dt = Z 0 −∞ e2αt dt = 1 2α (d) Neither energy or power. E4 = lim T→∞ Z T −T dt (α2 + t2) 1/4 = ∞ P4 = 0 since limT→∞ 1 T R T −T dt (α2+t2) 1/4 = 0.(e) Energy. Since it is the sum of x1(t) and x2(t), its energy is the sum of the energies of these two signals, or E5 = 1/α. 6 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (f) Power. Since it is an aperiodic signal (the sine starts at t = 0), we use P6 = lim T→∞ 1 2T Z T 0 sin2 (5πt) dt = lim T→∞ 1 2T Z T 0 1 2 [1 − cos (10πt)] dt = lim T→∞ 1 2T · T 2 − 1 2 sin (20πt) 20π ¸T 0 = 1 4 W Problem 2.10 (a) Power. Since the signal is periodic with period π/ω, we have P = ω π Z π/ω 0 A2 |sin (ωt + θ)| 2 dt = ω π Z π/ω 0 A2 2 {1 − cos [2 (ωt + θ)]} dt = A2 2 (b) Neither. The energy calculation gives E = lim T→∞ Z T −T (Aτ) 2 dt √τ + jt√τ − jt = lim T→∞ Z T −T (Aτ) 2 dt √τ 2 + t2 → ∞ The power calculation gives P = lim T→∞ 1 2T Z T −T (Aτ) 2 dt √ τ 2 + t2 = lim T→∞ (Aτ) 2 2T ln à 1 + p1 + T2/τ 2 −1 + p1 + T2/τ 2 ! = 0 (c) Energy: E = Z ∞ 0 A2t 4 exp (−2t/τ) dt = 3 4 A2τ 5 (use table of integrals) (d) Energy: E = 2 ÃZ τ/2 0 22dt + Z τ τ/2 12dt! = 5τ Problem 2.11 (a) This is a periodic train of “boxcars”, each 3 units in width and centered at multiples of 6: Pa = 1 6 Z 3 −3 Π2 µ t 3 ¶ dt = 1 6 Z 1.5 −1.5 dt = 1 2 W 2.1. PROBLEM SOLUTIONS 7 (b) This is a periodic train of unit-high isoceles triangles, each 4 units wide and centered at multiples of 5: Pb = 1 5 Z 2.5 −2.5 Λ2 µ t 2 ¶ dt = 2 5 Z 2 0 µ 1 − t 2 ¶2 dt = −2 5 2 3 µ 1 − t 2 ¶3 ¯ ¯ ¯ ¯ ¯ 2 0 = 4 15 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 = 1 3 Z 2 0 µ 1 − t 2 ¶2 dt = −1 3 2 3 µ 1 − t 2 ¶3 ¯ ¯ ¯ ¯ ¯ 2 0 = 2 9 W (d) This is a full-wave rectified cosine wave of period 1/5 (the width of each cosine pulse): Pd = 5 Z 1/10 −1/10 |cos (5πt)| 2 dt = 2 × 5 Z 1/10 0 · 1 2 + 1 2 cos (10πt) ¸ dt = 1 2 W Problem 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 is E = Z 6 −6 cos2 (6πt) dt = 2 Z 6 0 · 1 2 + 1 2 cos (12πt) ¸ dt = 6 J (b) The energy is E = Z ∞ −∞ h e−|t|/3 cos (12πt) i2 dt = 2 Z ∞ 0 e−2t/3 · 1 2 + 1 2 cos (24πt) ¸ dt where the last integral follows by the eveness of the integrand of the first one. Use a table of definte integrals to obtain E = Z ∞ 0 e−2t/3dt + Z ∞ 0 e−2t/3 cos (24πt) dt = 3 2 + 2/3 (2/3)2 + (24π) 2 Since the result is finite, this is an energy signal. (c) The energy is E = Z ∞ −∞ {2 [u (t) − u (t − 7)]}2 dt = Z 7 0 4dt = 28 J 8 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Since the result is finite, this is an energy signal. (d) Note that Z t −∞ u (λ) dλ = r (t) = ½ 0, t< 0 t, t ≥ 0 which is called the unit ramp. The energy is E = Z ∞ −∞ [r (t) − 2r (t − 10) + r (t − 20)]2 dt = 2 Z 10 0 µ t 10¶2 dt = 20 3 J where the last integral follows because the integrand is a symmetrical triangle about t = 10. Since the result is finite, this is an energy signal. Problem 2.14 (a) Expand the integrand, integrate term by term, and simplify making use of the orthogonality 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 each other and fill the interval [0, T]. (d) Using the expression for the generalized Fourier series coefficients, we find that X1 = 1/8, X2 = 3/8, X3 = 5/8, and X4 = 7/8. Also, cn = T /4. Thus, the ramp signal is approximated by t T = 1 8 φ1 (t) + 3 8 φ2 (t) + 5 8 φ3 (t) + 7 8 φ4 (t), 0 ≤ t ≤ T where the φn (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 find that X1 = 1/4 and X2 = 3/4. (f) To compute the ISE, we use ²N = Z T |x (t)| 2 dt −X N n=1 cn ¯ ¯X2 n ¯ ¯ Note that R T |x (t)| 2 dt = R T 0 (t/T) 2 dt = T /3. Hence, for (d), ISEd = T 3 − T 4 ¡ 1 64 + 9 64 + 25 64 + 49 64 ¢ = 5.208 × 10−3T. For (e), ISEe = T 3 − T 2 ¡ 1 16 + 9 16 ¢ = 2.083 × 10−2T. 2.1. PROBLEM SOLUTIONS 9 Problem 2.15 (a) The Fourier coefficients are (note that the period = 1 2 2π ω0 ) X−1 = X1 = 1 4 ; X0 = 1 2 All other coefficients are zero. (b) The Fourier coefficients for this case are X−1 = X∗ 1 = 1 2 (1 + j) All other coefficients are zero. (c) The Fourier coefficients for this case are (note that the period is 2π 2ω0 ) X−2 = X2 = 1 8 ; X−1 = X1 = 1 4 ; X0 = −1 4 All other coefficients are zero. (d) The Fourier coefficients for this case are X−3 = X3 = 1 8 ; X−1 = X1 = 3 8 All other coefficients are zero. Problem 2.16 The expansion interval is T0 = 4 and the Fourier coefficients are Xn = 1 4 Z 2 −2 2t 2e−jn(π/2)t dt = 2 4 Z 2 0 2t 2 cos µnπt 2 ¶ dt which follows by the eveness of the integrand. Let u = nπt/2 to obtain the form Xn = 2 µ 2 nπ ¶3 Z nπ 0 u2 cos u du = 16 (nπ) 2 (−1)n If n is odd, the Fourier coefficients are zero as is evident from the eveness of the function being represented. If n = 0, the integral for the coefficients is X0 = 1 4 Z 2 −2 2t 2dt = 8 3 The Fourier series is therefore x (t) = 8 3 + X∞ n=−∞, n6=0 (−1)n 16 nπ ejn(π/2)t 10 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Problem 2.17 Parts (a) through (c) were discussed in the text. For (d), break the integral for x (t) up into a part for t < 0 and a part for t > 0. Then use the odd half-wave symmetry contition. Problem 2.18 This is a matter of integration. Only the solution for part (b) will be given here. The integral for the Fourier coefficients is (note that the period really is T0/2) Xn = A T0 Z T0/2 0 sin (ω0t) e−jnω0t dt = − Ae−jnω0t ω0T0 (1 − n2) [jn sin (ω0t) + cos (ω0t)] ¯ ¯ ¯ ¯ T0/2 0 = A ¡ 1 + e−jnπ¢ ω0T0 (1 − n2) , n 6= ±1 For n = 1, the integral is X1 = A T0 Z T0/2 0 sin (ω0t) [cos (jnω0t) − j sin (jnω0t)] dt = −jA 4 = −X∗ −1 This is the same result as given in Table 2.1. Problem 2.19 (a) Use Parseval’s theorem to get P|nf0| ≤ 1/τ = X N n=−N |Xn| 2 = X N n=−N µAτ T0 ¶2 sinc2 (nf0τ) where N is an appropriately chosen limit on the sum. We are given that only frequences for which |nf0| ≤ 1/τ are to be included. This is the same as requiring that |n| ≤ 1/τf0 = T0/τ = 2. Also, for a pulse train, Ptotal = A2τ/T0 and, in this case, Ptotal = A2/2. Thus P|nf0| ≤ 1/τ Ptotal = 2 A2 X 2 n=−2 µA 2 ¶2 sinc2 (nf0τ) = 1 2 X 2 n=−2 sinc2 (nf0τ) = 1 2 £ 1+2 ¡ sinc2 (1/2) + sinc2 (1)¢¤ = 1 2 " 1+2 µ2 π ¶2 # = 0.91 2.1. PROBLEM SOLUTIONS 11 (b) In this case, |n| ≤ 5, Ptotal = A2/5, and P|nf0| ≤ 1/τ Ptotal = 1 5 X 5 n=−5 sinc2 (n/5) = 1 5 n 1+2 h (0.9355)2 + (0.7568)2 + (0.5046)2 + (0.2339)2 io = 0.90 Problem 2.20 (a) The integral for Yn is Yn = 1 T0 Z T0 y (t) e−jnω0t dt = 1 T0 Z T0 0 x (t − t0) e−jnω0t dt Let t0 = t − t0, which results in Yn = · 1 T0 Z T0−t0 −t0 x ¡ t 0 ¢ e−jnω0t0 dt0 ¸ e−jnω0t0 = Xne−jnω0t0 (b) Note that y (t) = A cos ω0t = A sin (ω0t + π/2) = A sin [ω0 (t + π/2ω0)] Thus, t0 in the theorem proved in part (a) here is −π/2ω0. By Euler’s theorem, a sine wave can be expressed as sin (ω0t) = 1 2j ejω0t − 1 2j e−jω0t Its Fourier coefficients are therefore X1 = 1 2j and X−1 = − 1 2j . According to the theorem proved in part (a), we multiply these by the factor e−jnω0t0 = e−jnω0(−π/2ω0) = ejnπ/2 For n = 1, we obtain Y1 = 1 2j ejπ/2 = 1 2 For n = −1, we obtain Y−1 = − 1 2j e−jπ/2 = 1 2 which gives the Fourier series representation of a cosine wave as y (t) = 1 2 ejω0t + 1 2 e−jω0t = cos ω0t 12 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY We could have written down this Fourier representation directly by using Euler’s theorem. Problem 2.21 (a) Use the Fourier series of a triangular wave as given in Table 2.1 with A = 1 and t = 0 to obtain the series 1 = ··· + 4 25π2 + 4 9π2 + 4 π2 + 4 π2 + 4 9π2 + 4 25π2 + ··· Multiply both sides by π2 8 to get the series in given in the problem. Therefore, its sum is π2 8 . (b) Use the Fourier series of a square wave (specialize the Fourier series of a pulse train) with A = 1 and t = 0 to obtain the series 1 = 4 π µ 1 − 1 3 + 1 5 − ···¶ Multiply both sides by π 4 to get the series in the problem statement. Hence, the sum is π 4 . Problem 2.22 (a) In the expression for the Fourier series of a pulse train (Table 2.1), let t0 = −T0/8 and τ = T0/4 to get Xn = A 4 sinc ³n 4 ´ exp µ j πnf0 4 ¶ (b) The amplitude spectrum is the same as for part (a) except that X0 = 3A 4 . Note that this can be viewed as having a sinc-function envelope with zeros at multiples of 4 3T0 . The phase spectrum can be obtained from that of part (a) by adding a phase shift of π for negative frequencies and subtracting π for postitive frequencies (or vice versa). Problem 2.23 (a) There is no line at dc; otherwise it looks like a squarewave spectrum. (b) Note that xA (t) = K dxB (t) dt where K is a suitably chosen constant. The relationship between spectral components is therefore XA n = K (jnω0) XB n where the superscript A refers to xA (t) and B refers to xB (t). 2.1. PROBLEM SOLUTIONS 13 Problem 2.24 (a) This is the right half of a triangle waveform of width τ and height A, or A (1 − t/τ). Therefore, the Fourier transform is X1 (f) = A Z τ 0 (1 − t/τ) e−j2πf tdt = A j2πf · 1 − 1 j2πfτ ³ 1 − e−j2πfτ ´¸ where a table of integrals has been used. (b) Since x2 (t) = x1 (−t) we have, by the time reversal theorem, that X2 (f) = X∗ 1 (f) = X1 (−f) = A −j2πf · 1 + 1 j2πfτ ³ 1 − ej2πfτ ´¸ (c) Since x3 (t) = x1 (t) − x2 (t) we have, after some simplification, that X3 (f) = X1 (f) − X2 (f) = jA πf sinc (2fτ) (d) Since x4 (t) = x1 (t) + x2 (t) we have, after some simplification, that X4 (f) = X1 (f) + X2 (f) = Aτ sin2 (πfτ) (πfτ) 2 = Aτsinc2 (fτ) This is the expected result, since x4 (t) is really a triangle function. Problem 2.25 (a) Using a table of Fourier transforms and the time reversal theorem, the Fourierr transform of the given signal is X (f) = 1 α + j2πf − 1 α − j2πf Note that x (t) → sgn(t) in the limit as α → 0. Taking the limit of the above Fourier transform as α → 0, we deduce that F [sgn (t)] = 1 j2πf − 1 −j2πf = 1 jπf 14 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (b) Using the given relationship between the unit step and the signum function and the linearity property of the Fourier transform, we obtain F [u (t)] = 1 2 F [sgn (t)] + 1 2 F [1] = 1 j2πf + 1 2 δ (f) (c) The same result as obtained in part (b) is obtained. Problem 2.26 (a) Two differentiations give d2x1 (t) dt2 = dδ (t) dt − δ (t − 2) + δ (t − 3) Application of the differentiation theorem of Fourierr transforms gives (j2πf) 2 X1 (f)=(j2πf) (1) − 1 · e−j4πf + 1 · e−j6πf where the time delay theorem and the Fourier transform of a unit impulse have been used. Dividing both sides by (j2πf) 2 , we obtain X1 (f) = 1 j2πf − e−j4πf − e−j6πf (j2πf) 2 = 1 j2πf − e−j5πf j2πf sinc (2f) (b) Two differentiations give d2x2 (t) dt2 = δ (t) − 2δ (t − 1) + δ (t − 2) Application of the differentiation theorem gives (j2πf) 2 X2 (f)=1 − 2e−j2πf + e−j4πf Dividing both sides by (j2πf) 2 , we obtain X2 (f) = 1 − 2e−j2πf + e−j4πf (j2πf) 2 = sinc2 (f) e−j2πf (c) Two differentiations give d2x3 (t) dt2 = δ (t) − δ (t − 1) − δ (t − 2) + δ (t − 3) 2.1. PROBLEM SOLUTIONS 15 Application of the differentiation theorem gives (j2πf) 2 X3 (f)=1 − e−j2πf − e−j4πf + e−j6πf Dividing both sides by (j2πf) 2 , we obtain X3 (f) = 1 − e−j2πf − e−j4πf + e−j6πf (j2πf) 2 (d) Two differentiations give d2x4 (t) dt2 = 2Π (t − 1/2) − 2δ (t − 1) − 2 dδ (t − 2) dt Application of the differentiation theorem gives (j2πf) 2 X4 (f)=2sinc (f) e−jπf − 2e−j2πf − 2 (j2πf) e−j4πf Dividing both sides by (j2πf) 3 , we obtain X4 (f) = 2e−j2πf + (j2πf) e−j4πf − sinc (f) e−jπf 2 (πf) 2 Problem 2.27 (a) This is an odd signal, so its Fourier transform is odd and purely imaginary. (b) This is an even signal, so its Fourier transform is even and purely real. (c) This is an odd signal, so its Fourier transform is odd and purely imaginary. (d) This signal is neither even nor odd signal, so its Fourier transform is complex. (e) This is an even signal, so its Fourier transform is even and purely real. (f) This signal is even, so its Fourier transform is real and even. Problem 2.28 (a) Using superposition, time delay, and the Fourier transform of an impulse, we obtain X1 (f) = ej16πt +2+ e−j16πt = 4 cos2 (6πt) The Fourier transform is even and real because the signal is even. (b) Using superposition, time delay, and the Fourierr transform of an impulse, we obtain X2 (f) = ej12πt − e−j12πt = 2j sin (12πf) The Fourier transform is odd and imaginary because the signal is odd. 16 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (c) The Fourier transform is X3 (f) = X 4 n=0 ¡ n2 + 1¢ e−j4πnf It is complex because the signal is neither even nor odd. Problem 2.29 (a) The Fourier transform of this signal is X1 (f) = 2 (1/3) 1 + (2πf/3)2 = 2/3 1+[f/ (3/2π)]2 Thus, the energy spectral density is G1 (f) = ½ 2/3 1+[f / (3/2π)]2 ¾2 (b) The Fourier transform of this signal is X2 (f) = 2 3 Π µ f 30¶ Thus, the energy spectral density is X2 (f) = 4 9 Π2 µ f 30¶ = 4 9 Π µ f 30¶ (c) The Fourier transform of this signal is X3 (f) = 4 5 sinc µf 5 ¶ so the energy spectral density is G3 (f) = 16 25sinc2 µf 5 ¶ (d) The Fourier transform of this signal is X4 (f) = 2 5 · sinc µf − 20 5 ¶ + sinc µf + 20 5 ¶¸ 2.1. PROBLEM SOLUTIONS 17 so the energy spectral density is G4 (f) = 4 25 · sinc µf − 20 5 ¶ + sinc µf + 20 5 ¶¸2 Problem 2.30 (a) Use the transform pair x1 (t) = e−αt u (t) ←→ 1 α + j2πf Using Rayleigh’s energy theorem, we obtain the integral relationship Z ∞ −∞ |X1 (f)| 2 df = Z ∞ −∞ df α2 + (2πf) 2 df = Z ∞ −∞ |x1 (t)| 2 dt = Z ∞ 0 e−2αt dt = 1 2α (b) Use the transform pair x2 (t) = 1 τ Π µ t τ ¶ ←→ sinc (τf) = X2 (f) Rayleigh’s energy theorem gives Z ∞ −∞ |X2 (f)| 2 df = Z ∞ −∞ sinc2 (τf) df = Z ∞ −∞ |x2 (t)| 2 dt = Z ∞ −∞ 1 τ 2 Π2 µ t τ ¶ dt = Z τ/2 −τ/2 dt τ 2 = 1 τ (c) Use the transform pair x3 (t) = e−α|t| ←→ 2α α2 + (2πf) 2 The desired integral, by Rayleigh’s energy theorem, is I3 = Z ∞ −∞ |X3 (f)| 2 df = Z ∞ −∞ · 1 α2 + (2πf) 2 ¸2 df = 1 (2α) 2 Z ∞ −∞ e−2α|t| dt = 1 2α2 Z ∞ 0 e−2αt dt = 1 4α3 (d) Use the transform pair 1 τ Λ µ t τ ¶ ←→ sinc2 (τf) 18 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY The desired integral, by Rayleigh’s energy theorem, is I4 = Z ∞ −∞ |X4 (f)| 2 df = Z ∞ −∞ sinc4 (τf) df = 1 τ 2 Z ∞ −∞ Λ2 (t/τ) dt = 2 τ 2 Z τ 0 [1 − (t/τ)]2 dt = 2 τ Z 1 0 [1 − u] 2 du = 2 3τ Problem 2.31 (a) The convolution operation gives y1 (t) =    0, t ≤ τ − 1/2 1 α £ 1 − e−α(t−τ+1/2)¤ , τ − 1/2 < t ≤ τ + 1/2 1 α £ e−α(t−τ−1/2) − e−α(t−τ+1/2)¤ , t> τ + 1/2 (b) The convolution of these two signals gives y2 (t) = Λ(t) + tr(t) where tr(t) is a trapezoidal function given by tr(t) =    0, t< −3/2 or t > 3/2 1, −1/2 ≤ t ≤ 1/2 3/2 + t, −3/2 ≤ t < −1/2 3/2 − t, 1/2 ≤ t < 3/2 (c) The convolution results in y3 (t) = Z ∞ −∞ e−α|λ| Π (λ − t) dλ = Z t+1/2 t−1/2 e−α|λ| dλ Sketches of the integrand for various values of t gives the following cases: y3 (t) =    R t+1/2 t−1/2 eαλdλ, t ≤ −1/2 R 0 t−1/2 eαλdλ + R t+1/2 0 e−αλdλ, −1/2 < t ≤ 1/2 R t+1/2 t−1/2 e−αλdλ, t> 1/2 Integration of these three cases gives y3 (t) =    1 α £ eα(t+1/2) − eα(t−1/2)¤ , t ≤ −1/2 1 α £ e−α(t−1/2) − e−α(t+1/2)¤ , −1/2 < t ≤ 1