Errata for First Printing of Discrete-Time Signal Processing by ...

Errata for First Printing of Discrete-Time Signal Processing by Oppenheim and Schafer with Buck Page xxv

Where 2nd paragraph, 4th line

55

2 lines after Eq. (2.154c)

87

Prob. 2.67(a), Last sentence

91

Prob. 2.82, 1st line

91


91


137 148

Fig. P3.46-2 Eq. (4.21)

160

Ex. 4.6, 3rd Eq. from end

164 165 171

Last line before Example 4.9 Last line before Example 4.10 Fig. 4.22(c)

171

Fig. 4.22(e)

178

Fig. 4.29(e)

182

Fig. 4.34

187

1st paragraph, 5th line

295

Line after Eq. (5.135)

419

Fig. P6.2-1

432

Fig. 6.34-1

513

Prob. 7.8, second sentence

Correction Delete “e” from “Kelley” to it reads “thanks W. Kelly Mosley for his . . . ” Insert minus sign so it reads xo [n] = −xo [−n] Change ω in exponent to π so it reads (−1)n = ejπn Change 2nd x[n] to s[n] so it reads x[n] = s[n] + e[n] Insert “zero-mean” so it reads . . . are independent zero-mean stationary. . . Insert |a| < 1 so it reads an u[n] with |a| < 1. Label on left-most pole should be − 34 . Insert k so the RHS reads ∞ 1 Xc (j (Ω − kΩs )) T k=−∞

Remove ] at end. Insert π (2x) so RHS reads jΩ0 πδ(Ω − Ω0 ) − jΩ0 πδ(Ω + Ω0 ) Change Example 4.5 to Example 4.10 Change Example 4.5 to Example 4.10 Slope for ω > 2π should be the same as slope for 0 ≤ ω ≤ 3π/2. See attached artwork correction. Slopes for ω > 2π and ω < −2π are wrong. Slope for ω > 2π should match slope for 0 ≤ ω ≤ π. Slope for ω < −2π should match slope for −π ≤ ω ≤ 0. See attached artwork correction. Heights should be TL , not T1 . See attached artwork correction. Output should be labeled y[n], not w[n]. See attached artwork correction. Insert 2 so it reads “. . . e.g. at 2M ΩN .” Change 1st ω to α so it reads . . . where α and β are constants. . . Upper arrow of left side should point upwards. See attached artwork correction. All three arrows along the bottom should point to the right See attached artwork correction. Change 9 to 11, 4 to 5, and -4 to -5 so it reads The impulse response of the desired filter has length 11; i.e., h[n] = 0 for n

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Page 566

Where 3 lines from bottom

577 577 606

Last sentence before 8.7.1 First sentence of 8.7.2 Prob. 8.20

614 626

Table P8.37-2 Prob 8.65(d)

627

Fig. P8.67-1

671

Prob. 9.4(c), 1st sentence

697

2nd paragraph of Example 10.2

737

Eq. (10.70)

758

Prob. 10.16, 2nd sentence

760

Prob. 10.24(a), 2nd sentence

830

Ans. to 2.2(a)

833

Ans. to 3.3(c)

834

Ans. to 3.12(a)

834

Ans. to 3.12(b)

835

Ans. to 4.4(b)

839

Ans. to 6.9(b)

843

Ans. to 7.6(a)

843

Ans. to 7.8(a)

843

Ans. to 7.8(b)

846

Ans. to 8.19

859 865 865

Index Index Index

Correction Insert − sign so equation reads −(N −m)k WNmk = WN Change Example 8.15 to Example 8.11 Change Example 8.15 to Example 8.11 Change sign of exponent so it reads X1 [k] = X[k]ej2πk2/N . Change X3 [k] to H3 [k]. Change last term on RHS of Eq. (P8.65-7) to read XH [((−k))N ]SN [n0 k] 3rd Box should be H[k] Output of 3rd Box should be Y [k], no subscript. Output of 4th Box should be y[n], no subscript. See attached artwork corrections. Insert “for n ≥ 0 ” so it reads h[n] should be periodic for n ≥ 0. Insert comma so equation reads V [k] = V ∗ [((−k))N ], k = 0, 1, . . . , N − 1 Insert { so LHS reads ¯ E I(ω) Insert “continuous-time” so it reads Determine the continuous-time frequency spacing . . . Insert “continuous-time” so it reads What is the effective continuous-time frequency spacing. . . Change equation for N5 to read N5 = N1 + N3 . Insert z −1 so it reads −N 2 −1 1 − z Xc (z) = z 2 (1 − z −1 ) Leave ROC as is. Change art so pole is at z = −2, not z = 2. See attached art correction. Change art so poles are at z = − 12 and z = 23 . See attached art correction. Change answer to read Not unique. T = 11/100. Add x[n − 2] − 8x[n − 3] to RHS so equation is y[n] + y[n − 1] − 8y[n − 2] = x[n] + 3x[n − 1] + x[n − 2] − 8x[n − 3] Change to read δ = 0.03, β = 2.181. Change answer to read Six alternations. L = 5, so this does not satisfy the alternation theorem Change answer to read Seven alternations, which satisfies the alternation theorem for L = 5. Change “-2” to “2” (2x) so it reads m = 2. This is not unique. Any m = 2 + 6 works Delete “k” of “Bankdpass filters.” “Group delay and attenuation, effects of” should be 243-245. “Group delay defined” should be 243.

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