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数字信号处理 Sanjit K.Mitra 第四版 2-7章答案.pdf

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数字信号处理 Sanjit K.Mitra 第四版 2-7章答案.pdf

Not for sale. 2 Chapter 2 2.1 Using Eq. 2.9, we get a b    2.2 To show this, we start with the definitions from Eq. 2.9 and square them 2 2  [ ] 2  [ ]        2  12. The middle inequality is a generalization of the triangle inequality. We can take square roots of both sides of this result, because everything within the equations is positive, getting 2.3 a [ ] [ 3] {20 16 320}, 6 0. b [ ] [ 2] {82 7 3011}, 3 3. c [ ] [ ] {02 36 102}, 3 3. d [ ] [ 3] [ 3] {82 5 3 17 220}, 2 6. e [ ] [ 2] [ ] {3 6}, 2 1. f [ ] [ 4] [ 3] { 3 6122 4 8 3011}, 5 5. g [ ] 4.1 [ ] {32.8 8.2 28.7 12.304.1 4.1}, 5 1. 2.7 a Let [] be the signal in the middle of the structure. Then  We can see that . Putting these together, we get [ ] 0 [ ] 1 [ 1] 2 [ 2] 1 [ 1] 1 [ 2] 2 [ 3] 2 [ 2] 1 [ 3] 2 [ 4] 0 [ ] 1 1 [ 1] 2 1 2 [ 2] 1 2 [ 3] 2 [ 4]. b Again, if we let [] be the signal in the middle of the structure, then Not for sale. 3 We can see that Combining the above two equations we get c If we let [] and [] represent the signals past each feed-forward component, then Substituting the top two equations into the last equation, we get which leads to [ ] [0] [ ] [0] 11 12 13 [ 1] [0] 21 12 11 22 11 13 12 13 [ 2] [0] 12 21 11 22 13 21 11 12 13 13 22 11 23 12 23 [ 3] [0] 22 21 12 13 21 11 13 22 11 12 23 22 23 [ 4] [0] 21 13 22 21 12 23 22 23 11 [ 5] [0] 21 22 23 [ 6] d Let 0[ ], 1[ ], and denote the outputs of the three adders Not for sale. 4 Combining the above, we get 2.5 Since is of length and defined for the convolution sum reduces to Thus, will be nonzero for all those values of and for which satisfies The minimum value of is 0, and occurs for the lowest at and The maximum value of and occurs for maximum value of at Thus and Hence the total number of nonzero samples 2.6 Let  be the length of the convolution sum of [] and [], which are of length  and , respectively. From Problem 2.5, we know that  The length of the convolution of [n] with [], is thus or 2.7 To show that the two convolutions are equal, we simply evaluate both convolution sums 2.8 a 1[ ] 1[ ] 1 23 21{ }, 2 2. b 2[ ] 2[ ] 121 2 4 2121{ }, 0 8. c 3[ ] 3[ ] 40 120170 1204{ }, 4 4. 2.9 a Given [] and [] from Problem 2.3, their convolution sum [] is given by [ ] [ ] [ ] 164 2240 5 279 6 13 120{ }, 8 4. b Given [] and [] from Problem 2.3, their convolution sum [] is given by [ ] [ ] [ ] 6 12 51640 82322210920{ }, 5 7. c Given [] and [] from Problem 2.3, their convolution sum g[] is given by Not for sale. 5 [ ] [ ] [ ] 2454 17 374152 19 53 2451271{ }, 7 5. 2.10 First, express [] as a convolution and then rewrite the convolution of [] as a function of  Now Let Then 2.11 Using the same steps as in the solution of Problem 2.10, we first express [] as a convolution and then reevaluate [] in terms of these convolution sums where Now Define Then from the solution of Problem 2.10, Hence Therefore, making use of the solution of Problem 2.10 again we get 2.12 Two results are needed for this problem Length of Sequence Max Min 1 Length of Convolution a The sequence , formed from the convolution of [] with itself, will have a length , or . To find the range of indices over which the convolution will have nonzero values, it is necessary to think about the nature of the convolved signals. Both of them, since they are the same, have nonzero values on either side of the origin, which will still be true after one of them is time reversed within the convolution sum formula. Thus, if the original has values in the range , then the time reversed version will have values in the range The first left-shift at which point these two overlap will occur at 2, since the time-reversed version must be shifted left  points to have the negative portions overlap, and then again by  points to be outside of the range of overlap. A similar principle applies to the right-most region Not for sale. 6 of overlap. The time-reversed signal must be shifted by  twice to the right to be just outside of the region of overlap, so that the right-most boundary of the resulting convolution is 2. Therefore the convolution is nonzero in the range [2, 2]. Note that these boundaries also give the correct length of the resulting signal 2 2 1 2 2 1. b The sequence 2[], formed from the convolution of [] with itself, will have length Although the signal is no longer on both sides of the origin, when the time-reversed version is generated, there will be one copy whose values are in the range [,] and another whose values are in the range [, ]. In order to get the first point of overlap, the time-reversed version must be shifted to the right by 2 points, giving the left-most boundary of the convolution sum. The right-most boundary is similarly calculated by noting that the signal must be shifted by 2 points. Thus, the convolution will be nonzero in the range [2, 2]. Note, again, that these boundaries also give the correct length 2 2 1 2 2 1. c The sequence 3[], formed from the convolution of [] with itself, will have length     1, or 2 2 1. This situation is similar to that in part b, except that we are dealing with the mirror image of the signal so that the time reversed part will be in the range [,]. Because of the symmetry of the convolution operation, we will end up with inverted boundaries but the same length computation the convolution will be nonzero in the range [ 2, 2]. Also, these boundaries will give the correct length 2 2 1 2 2 1. d The sequence y4[n], formed from the convolution of [] with [], will have length   1   1 1, or 2   1. To find the range of indexes, it is again instructive to think of the time-reversed version of [], and the points at which it stops and starts overlapping with []. The left-most boundary will occur at  , while the right-most boundary will occur at 2, so that the convolution will be nonzero in the range [  , 2]. Again, these boundaries can be used to confirm the length 2   1 2   1. e The sequence 5[], formed from the convolution of [] with [], will have length   1   1 1, or     1. The range of nonzero indexes can be found similarly to that in d, by checking the left-most and right-most points of overlap. The resulting convolution will be nonzero in the range [ ,  ]. Not for sale. 7 And again, this confirms the length calculation     1     1. 2.13 [ ] and [ ] Thus, 2.14 Tthe maximum value will occur with the largest overlap during the convolution operation. Since one sequence is 2 samples shorter than the other, there will be three possible points where all terms of the shorter sequence overlap with terms of the longer sequence. Thus, the maximum of [] will be at the locations and the maximum value is 2.15 The convolution of a sequence of length  and a sequence of length  will produce a sequence of length  1. Thus, the length of [] can be computed by rearranging the equation and evaluating for  1. Rearranging the terms of the convolution formula, we can recursively compute [] because successive samples of [] are based purely on successive coefficients of []. For example, since [0] [0][0], we can find [0] [0]/[0]. From here, we can use the following formula to compute all other terms within [] a Using the formula for the length of [], we get  8 4 1 4. Using the above recursive formula for deconvolution, we arrive at . b Using the formula for the length of [], we get  5 3 1 3. The length of y[n] in this problem is effectively 5, because the first term is 0. Using the above recursive formula for deconvolution, we arrive at . c Using the formula for the length of [], we get  9 5 1 5. Using the above recursive formula for deconvolution, we arrive at . The above results can be derived using the function deconv in MATLAB. 2.16 We make use of the circular shifting operation given by Eqn. 2.25. The length of {[]} is 9. a Thus,   Therefore, Therefore, y[3] 4 b Thus, Therefore, Therefore, [2] 5. Not for sale. 8 2.17 We make use of the circular shifting operation given by Eqn. 2.25. Now, { } The length of {[]} is 8. a { } b { } 2.18 Using the definition of average power Eqn. 2.37, the average power of the odd and even portions of [] are thus given by and Combining the two, we get The quantity inside the parentheses is given by . 2.19 The given signal is The formula for energy is given by Eqn. 2.34 12 1 cos4 / 0 1 2 12 cos4 / 0 1 . Not for sale. 9 Let and Then This implies  0 Hence 2.20 Using Eqn. 2.30, the odd and even parts are determined as follows    and [] a [ ] 11 26 211{ }, 3 3, [ ]  1 110 11 1 , 3 3. b       { }   [ ] 41 4 2 10124 1 4{ }, 5 5. c [ ] 1354 14531{ }, 4 4, [ ] 1 3 220 2231{ }, 4 4. 2.21 Using the formula for conjugate symmetric and conjugate anti-symmetric parts from Eqn. 2.28, we determine the components as follows [ ] and [ ] a        { }   1, [ ] 1 1 5 1 1{ }, 2 2. b From the formulas for sine and cosine, we have c From the properties and formulas for sine and cosine Not for sale. 10 3, [ ] 1 2 cos 2 7 sin 2 4 cos 2 7 sin 2 4 12 cos 27 sin 24 cos 27 sin 24 cos 27 sin 24 . 2.22 Since [] is conjugate symmetric it satisfies the condition and since [] is conjugate antisymmetric it satisfies the condition a    Thus, [] is conjugate symmetric. b Thus, [] is conjugate antisymmetric. c Thus, [] is conjugate symmetric. 2.23 An absolutely summable sequence {[]} satisfies the condition 0. Therefore, Likewise, from the definition for the odd sequence, we have [0] 0 and [ ] 12 [ ], 0. Therefore, b Since is causal, Also, We can express [] in terms of its real and imaginary parts and , which are and causal sequences, as follows From the definition of an conjugate anti-symmetric sequence  Hence, and [ ] 12 [ ], 0. Since is not known, cannot be fully recovered from . Likewise, from the definition of a conjugate symmetric sequence Hence and [ ] 12 [ ], 0. Not for sale. 11 Since is not known, cannot be fully recovered from . 2.25 From Eqn. 2.44, an -periodic extension of a signal [] is obtained as follows Therefore Substituting we get Hence, is a periodic sequence with a period 2.26 a Consider the sequence defined by If  0, the amplitude of the sample values is less than 1. Hence, [] is a bounded sequence. c [ ] [ 1], 0, the amplitude of the sample values is less than 1. Hence, [] is a bounded sequence. d is a causal complex-valued sequence. Its amplitude for all n 0 is 4 and as , the amplitude approaches . Thus [] is not bounded. e  is a two-sided sequence. As is a sinusoidal sequence with values between and for all values of , [] is a bounded sequence f is a causal sequence starting with a value [0] 0, and for all  0, the amplitude of the sample values is less than 1. Hence, [] is a bounded sequence. Not for sale. 13 2.29 . Now hence [] is not absolutely summable. 2.30 a Now, 2.47 In this problem we make use of the identity a 2 [ 1] 2 [ 1] O* [ 2] 1.5 [ ] [ 3] Not for sale. 22 2 [ 1] O* [ 2] 3 [ 1]O* [ ]2 [ 1]O* [ 3] 2 [ 1]O* [ 2] 3 [ 1]O* [ ] 2 [ 1]O* [ 3] 2 [ 3] 3 [ 1] 2 [ 2] 2 [ 1] 3 [ 1] 2 [ 4] 2 [ 3] [ 1] 3 [ 1] 2 [ 2] 2 [ 4]. b 3 [ 2] [ ] O* 3 [ 3] 2 [ 1] [ 1] 9 [ 2]O* [ 3] 6 [ 2]O* [ 1] 3 [ 2]O* [ 1] 3 [ ]O* [ 3] 2 [ ]O* [ 1] [ ]O* [ 1] 9 [ 5] 6 [ 3] 3 [ 1] 3 [ 3] 2 [ 1] [ 1] 9 [ 5] 3 [ 3] 5 [ 1] [ 1]. c 2 [ 1] 2 [ 1] O* 3 [ 3] 2 [ 1] [ 1] 6 [ 1]O* [ 3] 4 [ 1]O* [ 1] 2 [ 1]O* [ 1] 6 [ 1]O* [ 3] 4 [ 1]O* [ 1] 2 [ 1]O* [ 1] 6 [ 4] 4 [ 2] 2 [ ] 6 [ 2] 4 [ ] 2 [ 2] 6 [ 4] 2 [ 2] 6 [ ] 2 [ 2]. d   3 [ 2] [ ] O* [ 2] 1.5 [ ] [ 3]   3 [ 2]O* [ 2] 4.5 [ 2]O* [ ]3 [ 2]O* [ 3]  [ ]O* [ 2]1.5 [ ]O* [ ] [ ]O* [ 3]  3 [ 4] 4.5 [ 2] 3 [ 1]  [ 2]1.5 [ ] [  3]  3 [ 4] 3.5 [ 2]1.5 [ ] 3 [ 1] [  3]. 2.48 The sequences [], [], [] and [] are all fine-length rectangular sequences with sample values 1 in the ranges of  shown below , then the maximum value will be  and there will be samples with the maximum value. In the answers below, the location of the first such maximal value is listed. a The maximum of will occur at with a maximum value . b The maximum of will occur at with a maximum value . c The maximum of will occur at with a maximum value d The maximum of will occur at with a value of e The maximum of will occur at with a maximum value f The maximum of will occur at with a value of g The maximum of will occur at with a maximum value 2.49 Each of these uses the definitions of conjugate symmetry and conjugate anti-symmetry along with the definition of convolution a We will show that if both of the sequences being convolved are conjugate symmetric, then the result will also be conjugate symmetric. First, define [] to be the convolution sum Next, check whether this satisfies conjugate symmetry Not for sale. 24      b We will show that if one of the sequences being convolved isconjugate symmetric, and the other is conjugate anti-symmetric, then the combination will be conjugate anti-symmetric First, define y[n] to be the convolution sum Next, check whether this satisfies conjugate symmetry      c If both of the sequences being convolved are conjugate anti-symmetric, then the result will also be conjugate anti-symmetric. First, define y[n] to be the convolution sum Next, check whether this satisfies conjugate symmetry 2.50 The three parameters and of the continuous-time signal can be determined from by setti

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