Digital Communications: A Discrete-Time Approach M. Rice Errata Foreword Page xiii, first paragraph, bare witness should be bear witness Page xxi, last paragraph, You know who you. should be You know who you are. Chapter 1 Page 3, second new paragraph, Pittsburg should be Pittsburgh Page 9, The end of the second lines reads, signal sideband AM. This should be single sideband AM. Page 1, the second line of Section 1., information baring should be information bearing Page 1, Equation (1.1) should read ( ω θ ) = ( ) ( θ( )) ( ω ) ( ) ( θ( )) ( ωt ) ( ) ( ) A t cos t t A t cos t cos t A t sin t sin Page 14, Second new paragraph, The same power/bandwidth exists with digitally modulated carriers. should read, The same power/bandwidth trade off exists with digitally modulated carriers. Chapter Page 7, The sentence after Equation (.14) should read An energy signal is a signal with finite nonzero energy whereas a power signal is a signal with finite nonzero power. Page 8, Equation (.19) should read n ( n) = ( k) ( n) = u( n) u( n 1) u δ δ k= Page 8, Equation (.) should read N n= N 1 x ( n) δ ( n n ) ( ) x n N n N = otherwise 1
Page 33, Equation (.3) should read s s H ( s) = = s s 5 ( s 1 j)( s 1 j) Page 36, Equation (.36) should read H ( s) Page 37: Equation (.37) should read H j j s = = 4 4 ( s 1 j)( s 1 j) s 1 j s 1 j jω jω = = 5 ω j ω [1 j( ω )][1 j( ω )] ( jω ) Page 4, the sixth row in Table.4.4 should be k= a e k jkω t ( ω = π f ) k k k = k = ( k ) a ( f kf ) π a δ ω ω δ Page 41, second line of text, complex plain should be complex plane Page 5, Equations (.57) and (.58) should read X! " k # N %1 $ = x ( n )e % j! N & nk k =,1,, N %1 (.57) ( ) = 1 N x n n= N %1! X! " k # $ e j N & nk n =,1,, N %1 (.58) k = Page 56, Equation (.7) should be 1 π = Xc ω k T k = T Page 65, the third paragraph of Section.6.. The two occurrences of Hc ( ) ( jω ) Hc jω should be
Page 75, Exercise.4, the equation for X ( f ) should be X j4π f 1 1 e = jπ f ( f ) δ( f ) δ( f ) Page 76, Exercise.7, the line after the equation should read where the integral on the left is the power contained in the interval B9% f B 9% Page 83, Exercise.46, the plot for G( f ) should be G( f ) 45 4 395 395 45 4 f Page 83, Exercise.47, the plot for G( f ) should be G( f ) 45 4 395 395 45 4 f Page 84, the second figure of Exercise.48 should be r ( t) H( f ) x( t) ( f t) cos π H( f ) 71 69 69 71 f (MHz) 7 7 Page 85: The first figure of Exercise.5 should be
R( f ) 117 116 115 115 116 117 f (khz) Page 86: The figure at the top of the page (Exercise.5) should be R( f ) 117 116 115 5 5 115 116 117 f (khz) Page 86: The first Figure in Exercise.51 should be R( f ) 117 116 115 115 116 117 f (khz) Page 87, The figure at the top of the page (Exercise.51) should be R( f ) 7 117 116 115 115 116 117 7 f (khz) Page 95, The figure of Exercise.63 should be
A cos c ( π f t θ( t) ) c d/dt envelope detector y( t) Ac π fc Page 97, Exercise.67. The Fibonacci sequence is, 1, 1,, 3, 5, 8, 13, 1, 34, 55, 89, 144, 377, 61, 987, Page 98, Exercise.67. The initial conditions should be x() =, x(1) = 1. Page 98, Exercise.69 should begin Consider an LTI system with input Page 99, Exercise.7 should begin Consider an LTI system with input Page 99, the figure of Exercise.71 should be x( n) H 1 ( z) y 1 ( n) ( z) H y ( n) H 1 ( z) 1 z = 11 1 1 z z z 6 6 1 3 H 7 3 1 ( z) = 6 8z z Page 1, the figure of Exercise.77 should be ( n) 1 1 Kz x 1 y( n) 1 z 1.5z 1 Page 13, Exercise.81 (c) should read x n 1 π π 8 4 ( n) = cos n u( n) Page 13, Exercise.81 (f) should read 5π 7π x( n) = cos n sin n 3 3
Page 13, Exercise.8 (a) should read X ( e jω ) Page 16, Exercise.9. The plot for x ( ) 3π π Ω< 8 3π 3π = 1 Ω 8 8 3π <Ω π 8 n should be x ( n) 1 1 3 4 5 6 7 8 9 n 1 and the statement after the figures should read Show that the length-4 DFT of both series have the same magnitude. (Hint: do the analysis in the time domain.) Chapter 3 Page 15, Forth line after equation (3.16): z 1 should be Page 15, Fifth line after equation (3.16): z 1 should be Page 13, Equation (3.3) should read H d jω ( e ) Page 133, Equation (3.4) should read H d z 1 z 1 Ω kπ = Hc j k = T Ω e = H j π Ω π T jω ( ) c for Page 138, second line Figure 3.3.8 requires three multipliers should read Figure 3.3.8 requires four multipliers
Page 154, Equation (3.66) should read, T T y( nt ) = y ( n ) T x nt x n Page 165, the figure for Exercise 3.15 (b) should be ( 1 ) ( ) (( 1) T ) x( n) 4 y( m) πn j e Page 174, Exercise 3.33 (b) should read Sample the impulse response at instants t = nt and scale by T to produce a discrete-time h n = Th nt. impulse response ( ) ( ) d c Page 175, Exercise 3.37, The last sentence should read Show that for small Ω, the DTFT is approximately jω / T.
Chapter 4 Page 195, Equation (4.45) should be M = E x! µ {( )( x! µ ) T } Page 198, Equation (4.5): µ should be boldface so that Equation (4.5) reads as follows: AE{ ( x! µ )( x! µ ) T }A T = AMA T Page 198, Equation (4.54): µ should be boldface so that Equation (4.54) reads as follows: µ ʹ = Aµ Page 198, Equation (4.55): AMA should be boldface so that Equation (4.55) reads as follows: M' = AMA Page, the last equation on the page. RY ( k ) should be RYY ( k ). Page 1, figure 4.4.1 should be T X ( n) XX ( k) j Page 1, the first line of the equation for Y ( ) S Y S e Ω should be Ω ( ) YY( ) e = R k e j jωk k = Page 1, Example 4.4.1: In the properties of the Gaussian sequence, RX ( k) should be RXX ( k ). Page, Figure 4.4. should be R S X LTI system h(n) Yn ( ) ( k) =? jω jω ( e ) SY ( e ) R YY =? X ( n) R S XX X ( k) jω ( e ) LTI system h(n) Y ( n) R YY Y ( k) = h( k) h( k) R ( k) XX jω jω ( ) ( ) jω = X ( e ) S e H e S
Page 11, Exercise 4.19: µ should be boldface. Page 11, Exercise 4.: µ should be boldface. Page 1, Exercise 4.3: part (b) should read (b) Compute the probability that X1 X X3 X 4 > 1. Page 1, Exercise 4.4: The autocorrelation function should be RXX ( m) = ( ) m Page 1, Exercise 4.4: part (c) should read (c) Compute the probability that X1 X X3 X 4 > 1. Page 1, Exercise 4.6: the impulse response should be written as follows: hn ( ) = sin ( Wn) Page 13, Exercise 4.8: the impulse response should be written as follows: π n 1 n N hn ( ) = n > N 1 1
Chapter 5 Page 5, Equation (5.17) should read T T K 1 K 1 T s $ = argmin r ( t) dt r( t) φk( t) dt am, kam, kʹ φk( t) φkʹ ( t) dt sm () t S T k k' 1 T = = 1 T1 Page 9, First line after equation (5.9): should be A. Page 34, Four lines below equation (5.4), ADC should be DAC. Page 39, Equation (5.53) should be sin Page 39, Equation (5.54) should be Page 39, Equation (5.56) should be ( ω( T T1) ) sin ω( T T1) ω 1 1 ρ ω ω ( ) = T p ( t)sin( ωt) dt T 1 Page 41, Equation (5.6) should be ( LT ) ( LT ) p T p T ρ ω ω 1 1
Page 49, Figure 5.3.11 should be I r ( t) ( ) cos ωt h matched filter ( t) = p( t) x( t) t = kt s x( kt s ) decision aˆ ( k) r( t) LO π/ ( ) sin ωt Qr ( t) h matched filter ( t) = p( t) y( t) t = kt s y( kt s ) aˆ 1 ( k) Page 53, Figure 5.3.13 should be I r ( t) ( ) cos ωt ( t) = p( t) h matched filter x( t) t = kt s x( kt s ) decision aˆ ( k) r( t) r( t) LO π/ ( ) sin ωt Q r ( t) ( t) = p( t) h matched filter y( t) t = kt s y( kt s ) aˆ 1 ( k) I r ( t) x( t) ( a ( 1 ), aˆ ( 1) ) = ( 1, 1) ˆ 1 ( a ( ), aˆ ( ) ) = ( 1, 1) ˆ 1 Q r ( t) y( t) ( a ˆ ( 3, ) a ˆ1( 3) ) = ( 1, 1) ( a ( ), aˆ ( ) ) = ( 1, 1) ˆ 1 Page 54, The table at the end of Example 5.3.9 should be k 1 3 xkt ( s) 1. 1..97.97 ykt ( s ).95 1..98 1. aˆ ( k ) 1 1 1 1 aˆ ( k ) 1 1 1 1 1
Page 7, Figure 5.5. (b) should be r( t) cos( πf c t) LPF ADC real serial to parallel ~ 4m r 4 T M ~ 4m 1 r 4 ~ 4m r 4 T M T M 4-point FFT x ( m) x 1 ( m) x ( m) decisions â ( m) â 1 ( m) â ( m) parallel to serial LPF ADC imag ~ 4m 3 r 4 T M x 3 ( m) â 3 ( m) sin( πf c t) Page 73, the sentence just above Equation (5.17) should read, Using Bayes rule, Page 86, The basis functions for Exercise 5.17 should be φ ( t) ( t) φ 1.5 1 Page 86, The constellation plot in Exercise 5.17 should be A A φ Page 86, The constellation plot of Exercise 5.18 should be φ 1 (, A) ( A,) φ
Page 86, The constellation of Exercise 5.19 should be φ 1 φ radius = A Page 86, The constellation of Exercise 5. should be φ 1 φ radius = A Page 87, The constellation of Exercise 5.1 should be φ 1 1 φ radius = A Page 91, The constellation of Exercise 5.3 should be 1 A A φ
Page 91, The constellation of Exercise 5.31 should be 1 11 1 3A A A 3A φ Page 91, The constellation of Exercise 5.3 should be 1 11 1 11 111 11 1 7A 5A 3A A A 3A 5A 7A φ Page 9, Exercise 5.33 should read Consider a binary PAM system using the SRRC pulse shape. (a) Produce an eye diagram corresponding to randomly generated symbols for a pulse shape with 1% excess bandwidth and L p = 6. Use a sampling rate equivalent to 16 samples/symbol. (b) Repeat part (a) for a pulse shape with 5% excess bandwidth and L p = 1. (c) Repeat part (a) for a pulse shape with 5% excess bandwidth and L p =. (d) Compare and contrast the eye diagrams from parts (a) (c). Page 9, Exercise 5.34 should read Consider a 4-ary PAM system using the SRRC pulse shape. (a) Produce an eye diagram corresponding to randomly generated symbols for a pulse shape with 1% excess bandwidth and L p = 6. Use a sampling rate equivalent to 16 samples/symbol. (b) Repeat part (a) for a pulse shape with 5% excess bandwidth and L p = 1. (c) Repeat part (a) for a pulse shape with 5% excess bandwidth and L p =. (d) Compare and contrast the eye diagrams from parts (a) (c).
Page 3, The figure for Exercise 5.46 should be φ 1 1 11 1 1 11 111 11 11 111 1111 111 φ A 1 11 111 11 A Page 31, The first line after the figure should read the average energy is 1.15 J
Chapter 6 Page 36, Equation (6.) should be BW = B B B Rs R T = = s log ( M) Page 36, Equation (6.3) should be 1 1 1 BW α α = = Rs = α R T log ( M ) s b b Page 38, Figure 6.1. (b). Fix the vertical alignment of!!!!!!! =! Page 313, The legend of Figure 6.1.3 is incorrect. Figure 6.1.3 should be as follows: 1 1-1 M = M = 4 M = 8 1 - P b 1-3 1-4 1-5 1-6 5 1 15 E b /N (db) Page 314, Equation (6.3) should be BW = B B B Rs R T = = s log ( M) Page 314, Equation (6.31) should be Page 318, Equation (6.55) should be 1 α 1 α BW = = ( 1 α ) Rs = R T log ( M) s b b!! s! =!! s! =!! s!" =!! s!"
Page 35, the sentence just above Equation (6.87) should read, The union bound for the conditional probabilities are Page 35, Equation (6.88) should be!! s!!! avg! avg 3!!!! Page 39, the caption for Figure 6.3.4 should read, Probability of bit error versus E b /N for MQAM for M = 4, 16, 64, and 56. Page 343, Footnote 5. The last line should read utility of these added features. Page 349, Exercise 6.1 should read as follows: Use the union bound to upper bound the probability of bit error for the 4116 APSK constellation shown in Figure 5.3.5 using r 1 = A, r = A, r 3 = 3A, θ1 = π /4, θ =, and θ 3 =. Page 357-358, Exercise 6.38. The paragraph (on page 357) that begins The other kind of repeater is the regenerative repeater should be part (c). Part (d) is Suppose a 1 Mbit/s QPSK link has ( C/ N ) = 7dB W/Hz and 1 ( C/ N ) = 7 db W/Hz. Which type of repeater provides the lowest composite bit-error rate: the bent-pipe repeater or the regenerative repeater?
Chapter 7 Page 375, starting with the sentence just before equations (7.35) and (7.36), the rest of the section should read as follows: Using! =1/, B n T s =., N = 1, K = 16, and K p = in (C.61), the loop filter constants are K 1 = 1.6 1-3 (7.35) K = 4.3 1-5 (7.36) Page 47, Equation (7.83) should read ˆb k = ˆ! k! ˆ! k"1 ˆb k1 = ˆ! k1! ˆ! k Page 464, Equation (8.53) should read ( ) = x nt x kt I " Page 44, Exercise 7.7: (7.45) should be (7.44). n ( )h I kt I! nt ( )
Chapter 8 Page 445, Equation (8.) should be as follows: ek ( ) = xkt ( ˆ τ( k) ΔT) xkt ( ˆ τ( k) ΔT) s s s s ykt ( ˆ τ( k) ΔT) ykt ( ˆ τ( k) ΔT) s s s s Page 488, In the caption of Figure 8.4.6, Figure 8.4.6 should be Figure 8.4.5. Page 471, The lower diagram of Figure 8.4.16 should be x( nt) (( m( k) ) T) x (( m( k) ) T) x 1 z -1 1/6 z -1 z -1 1/6 z -1 ( m( k) T) x z -1 1/ z -1 1/ z -1 z -1 (( m( k) ) T) x 1 z -1 1/ z -1 z -1 1/ 1/6 1/ 1/3 v ( 3) v ( ) ( 1) v v( ) (( m( k) ( k ) T) x µ µ ( k) Page 57, Equation (8.13) should be sn ; k a k p nt kt n a k p nt kt ( n) ( T a ( ), τ) = ( ) ( s τ) cos( Ω ) 1( ) ( s τ) sin Ω Page 57, The exponential term of the first line of equation (8.131) should be N( k Lp ) 1 exp ( ) ( ) r nt p nt kts τ σ n= N( k L ) p Pages 517-518, Exercises 8.1-8.15 and 8.18 should be deleted. The remaining exercises should be renumbered in the obvious way.
Chapter 9 Page 56, the first line under Figure 9.3.3. Equation (9.6) should be (9.59). Page 56, the first new paragraph the text a filter s transfer frequency response should be a filter s transfer function Page 56: Equation (9.61) should be Page 579: Figure 9.4.1 should be H ( e j Ω) C H ( j( Ωπ = e ) ) θ ( ) x 4, y 4 ( ) x, y ( ) x, y θ 1 θ θ3 θ θ 1 ( x 4,) θ θ3 θ = θ θ θ θ 1 3 x cosθ sinθ x 4 y = 4 sinθ cosθ y (a) Page 58: Table 9.4.1 should be tan 1 x = x y 4 y = x ( θ θ θ θ ) (b) 1 3 k θ k k δθ e n n k n= (degrees) (degrees) (degrees) 45. 45. 5. 1 1.. 1 6.57 71.57 5. 1 1. 1. 14.4 57.53-1.57-1.5 1.5 3 7.13 5.4-7.53-1.88 1.38 4 3.58 46.83 -.4-1 1.5 1.7 5 1.79 48.6 3.17 1 1.13 1. 6.9 49.51 1.38 1 1.9 1.4 7.45 49.96.49 1 1.7 1.5 δ k ' x k ' y k
Page 581: Figure 9.4. should be 9 angles (degrees) 8 7 6 5 4 3 k δ n θ n n= θ 1 1 3 4 5 6 7 iteration index (k) Page 58: The first line of equation (9.17) should be δ = sgn{ y } Page 587: The first line of equation (9.117) should be k δ = sgn{ y } Page 591: The last word of the last sentence should be high. k Page 596: The first word of the second sentence of the third new paragraph on the page (1 lines from the bottom of the page) should be CoRDiC. Page 6: The first equation of Exercise 9.5 should be 1 c ln ( ) = 1 log e R Page 6: The first equation of Exercise 9.6 should be k k Ω xn ( ) = I( nt ) e I ( nt ) e c j n jωn c Page 63: The equation of Exercise 9.6 (c) should be 1 j Ar( nt) = Ic( nt) e Ω n
Chapter 1 Page 65: Just below Equation (1.53),! = k "!#F / M should be! = k "!#F. Page 658: Figure 1..16 should be
Page 659: Figure 1..17 should be
Page 66: Figure 1..18 should be H ( ( z M / D ) r( nt) commutator (generalized serial-to-parallel converter) H ( 1( z M / D 1 ) e jπk / M e jπkd/ Mm u~ k ( mdt ) M / D 11 ( ) H M z j e πk ( M )/ M 1 e jπr / M u~ r ( mdt) e jπrd / Mm j e πr( M )/ M 1 Page 661: line 15, the parenthetical statement should read (ratio of RF power out to DC power in) Page 66: The matrix equation for Exercise 1.1 (a) should be!(!)!(!) = 1 1! sin(!) 1! cos(!)!(!)!(!)
Page 663: The matrix equation for Exercise 1.3 (a) should be!(!)!(!) = 1 1! sin(!) (1!) cos(!)!(!)!(!) Page 664: The second line of Exercise 1.6 should reference Figure 1.1. (d). Page 665: Exercise 1.8 should read The QAM modulator of Exercise 1.6 was derived from the QAM modulator of Figure 1.1. (d) by using polyphase partitions of pc ( nt ) and ps ( nt ). The QAM modulator of Exercise 1.7 was derived from the QAM modulator of Figure 1.1.4 by using a polyphase partition of gnt ( ). Show that the filterbanks in these two modulators are exactly identical. Page 668: The second line of Exercise 1. should begin consists of 14 channels Page 669: The second line of Exercise 1. should begin consists of 14 channels Page 669, Exercise 1.6: the last sentence above the figure should read Assume f1 < f and c f < ( fc f1). Page 669, Exercise 1.6 (b) should read Derive an expression for f in terms of f IF c, f, and 1 f. Page 67, Exercise 1.8 (a) should read Express the product of rtand () the LO in terms of baseband signals and double frequency signals (centered at ω c rads/s).
Appendix A Page 677: Three of the equation numbers in the paragraph after Equation (A.8) are wrong. The paragraph should read Assuming r p (t) satisfies the no-isi condition (A.6), the right-hand side of (A.8) reduces to the k = term, which proves that the right-hand side of (A.7) is a sufficient condition. Assuming the right-hand side of (A.7) is true, the Poisson sum formula is satisfied when r p (t) satisfies the no-isi condition (A.6). This proves that the right-hand side of (A.7) is necessary and the proof is complete. This proof relies on the Poisson sum formula. The derivation of the Poisson sum formulates explored in Exercise A.4. Page 677, Equation (A.9) should read Page 679, Equation (A.11) should read R p r p ( f ) Ts f < B = f > B πt sin T πt T s ( t) = φ ( t) s Page 68, the sentence on the last line of the page should read The absolute bandwidth of R p ( f ) is B abs = 1!. T s
Page 684, Figure A..4 should be R p ( f ) α = T s α =.5 T s α = 1 f 1 T s 1.5 T s 1 T s.5 T s.5 T s 1 T s 1.5 T s 1 T s r p ( t) α = 1 α = t α =.5 4T s 3T s T s T s Ts Ts 3T s 4Ts Page 685, Figure A..5 should be
Appendix B Page 715, the equation below the first line of Exercise B.18 should be rt () = I() t cos( ω t) Q ( t) sin( ω t) r r Page 715, the equation below the first line of Exercise B.18 (b) should be Bibliography Page 75. Ref. 3 should be jωt { [ Ir( t) jqr t ] e } r() t = Re () T. Cover and J. Thomas, Elements of Information Theory, John Wiley & Sons, New York, 1991. Page 758: Ref.18 should be C. Dick and f. harris, On the structure, performance, and applications of recursive all-pass filters with adjustable and linear group delay, Proceedings of the IEEE International Conference on Acoustics, Speech, and Signal Processing, Orlando, FL, May 14-17,, pp. 1517-15.