An Instructor’s Solutions Manual to Accompany Fundamentals of Digital Signal Processing using MATLAB, 2nd Edition Robert J. Harris http://www.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS. © 2012, 2005 Cengage Learning ISBN-13: 978-1-1114-2603-3 ISBN-10: 1-111-42603-1 ALL RIGHTS RESERVED. 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HARRIS Contents Chapter 1 1 Chapter 2 57 Chapter 3 155 Chapter 4 274 Chapter 5 384 Chapter 6 467 Chapter 7 572 Chapter 8 675 Chapter 9 765 Chapter 1 1.1 Suppose the input to an amplifier is xa (t) = sin(2πF0 t) and the steady-state output is ya (t) = 100 sin(2πF0 t + φ1 ) − 2 sin(4πF0 t + φ2 ) + cos(6πF0 t + φ3 ) (a) Is the amplifier a linear system or is it a nonlinear system? (b) What is the gain of the amplifier? (c) Find the average power of the output signal. (d) What is the total harmonic distortion of the amplifier? Solution (a) The amplifier is nonlinear because the steady-state output contains harmonics.2), the amplifier gain is K = 100.4), the output power is d20 1 2 d1 + d+ 22 + d23 Py = + 4 2 = . All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.2 Consider the following signum function that returns the sign of its argument. 1 , t>0 ∆ sgn(t) = 0 , t=0 −1 , t < 0 (a) Using Appendix 1, find the magnitude spectrum (b) Find the phase spectrum Solution (a) From Table A2 in Appendix 1 1 Xa (f ) = jπf Thus the magnitude spectrum is Aa (f ) = |Xa(f )| 1 = |jπf | 1 = π|f | (b) The phase spectrum is φa (f ) = 6 Xa (f ) = −6 jπf π = −sgn(f ) 2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.3 Parseval’s identity states that a signal and its spectrum are related in the following way. Z ∞ Z ∞ 2 |xa (t)| dt = |Xa(f )|2 df −∞ −∞ Use Parseval’s identity to compute the following integral. Z ∞ J = sinc2 (2Bt)dt −∞ Solution From Table A2 in Appendix 1 if xa (t) = sinc(2Bt) then µa (f + B) − µa (f − B) Xa (f ) = 2B Thus by Parseval’s identity Z ∞ J = sin2 (2Bt)dt Z−∞ ∞ = |xa (t)|2 dt Z−∞ ∞ = |Xa(f )|2 df −∞ Z B 1 = df 2B −B = 1 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.4 Consider the causal exponential signal xa (t) = exp(−ct)µa (t) (a) Using Appendix 1, find the magnitude spectrum. (b) Find the phase spectrum (c) Sketch the magnitude and phase spectra when c = 1. Solution (a) From Table A2 in Appendix 1 1 Xa (f ) = c + j2πf Thus the magnitude spectrum is Aa(f ) = |Xa(f )| 1 = |c + j2πf | 1 = p c + (2πf )2 2 (b) The phase spectrum is Aa (f ) = |Xa (f )| = 1 − 6 (c + j2πf ) 6 2πf = − tan−1 c © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.4 (c) Magnitude and Phase Spectra, c = 1 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.5 If a real analog signal xa (t) is square integrable, then the energy that the signal contains within the frequency band [F0 , F1 ] where F0 ≥ 0 can be computed as follows. Z F1 E(F0 , F1 ) = 2 |Xa(f )|2 df F0 Consider the following double exponential signal with c > 0. Solution (a) From Table A2 in Appendix 1 2c Xa (f ) = c2 + 4π 2 f 2 Thus the total energy of xa (t) is Z ∞ E(0, ∞) = 2 |Xa(f )|2 df Z0 ∞ 2c = 2 df 0 c2 + 4π 2 f 2 ∞ 4c 2πf = tan−1 2πc c 0 2 π = π 2 = 1 (b) Using part (a), the percentage of the total energy that lies in the frequency range [0, 2] Hz is 100E(0, 2) p = E(0, ∞) = 100E(0, 2) 2πf 2 200 = tan−1 π c 0 200 4π = tan−1 % π c © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.6 Let xa (t) be a periodic signal with period T0 . The average power of xa(t) can be defined as follows. Z T0 1 Px = |xa (t)|2 dt T0 0 Find the average power of the following periodic continuous-time signals. (a) xa(t) = cos(2πF0 t) (b) xa(t) = c (c) A periodic train of pulses of amplitude a, duration T , and period T0 . Solution (a) Using Appendix 2, Z 1/F0 Px = F0 cos2 (2πF0 t)dt 0 Z 1/F0 F0 = [1 + cos(4πF0 t)]dt 2 0 1 = 2 (b) Z T0 1 Px = c2 dt T0 0 = c2 (c) Z T 1 Px = a2 dt T0 0 a2 T = T0 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.7 Consider the following discrete-time signal where the samples are represented using N bits. x(k) = exp(−ckT )µ(k) (a) How many bits are needed to ensure that the quantization level is less than . What is the average power of the quantization noise? Solution (a) For k ≥ 0, the signal ranges over 0 ≤ x(k) ≤ 1. Thus xmin = 0 and xmax = 1 and from (1.3) the quantization level is 1 q = 2N Setting q = .001 yields 1 1 = 2N 1000 Taking the log of both sides, −N ln(2) = − ln(1000) or ln(1000) N = ceil ln(2) = ceil(9.8) the average power of the quantization noise using N = 8 bits is q2 E[e2] = 12 1 = 12(2N )2 = 1. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.8 Show that the spectrum of a causal signal xa(t) can be obtained from the Laplace transform Xa(s) be replacing s by j2πf . Is this also true for noncausal signals? Solution For a causal signal xa (t), the one-sided Laplace transform can be extended to a two-sided transform without changing the result. Thus the spectrum of a causal signal can be obtained from the Laplace transform as follows. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.9 Consider the following periodic signal. (b) Suppose xa(t) is sampled with a sampling frequency of fs = 8 Hz. Sketch the magnitude spectrum of xa (t) and the sampled signal, x̂a (t). (c) Does aliasing occur when xa (t) is sampled at the rate fs = 8 Hz? What is the folding frequency in this case? (d) Find a range of values for the sampling interval T which ensures that aliasing will not occur. The folding frequency is fs fd = 2 = 4 Hz (d) The signal xa (t) is bandlimited to 5 Hz.1, to avoid aliasing, the sampling rate must satisfy fs > 10.
