数字信号处理_第四版课后答案.pdf-资料库

2.1 Using Eq. (2.9), we get: (a) = = (b)   Chapter 2 =  2.2 To show this, we start with the definitions from Eq. (2.9) and square them: 2  2  2 [ ]      [ ] 2      2 . 1 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 2.7 (a) [ ] = [ + 3] = {2 0 1 6 3 2 0}, (b) [ ] = [ 2] = {8 2 7 3 0 11}, 6 3 0. 3. (c) [ ] = [ ] = {0 2 3 6 1 0 2}, 3 3. (d) [ ] = [ 3] + [ + 3] = {8 2 5 3 1 7 2 2 0}, 2 6. (e) [ ] = [ 2] [ ] = {3 6}, 2 1. (f) [ ] = [ + 4] [ 3] = { 3 612 2 4 8 3 0 11}, 5 5. (g) [ ] = 4.1 [ ] = {32.8 8.2 28.7 12.3 0 4.1 4.1}, 5 1. (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] 1 [ 3] + 2 = 0 [ ] + ( 1 (b) Again, if we let [] be the signal in the middle of the structure, then: 2 [ 1] + ( 2 ( 1 + 2) [ 2) [ 1) [ 2] 1 ) 4] 3] 2 [ 4]. Not for sale. 2

+ + = Combining the above two equations we get: = + + (c) If we let [] and [] represent the signals past each feed-forward component, then: ( + + + + + + ) = We can see that: = = = ( = ( 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] ( + [0] 22 21 + 12 13 21 + 11 13 22 + 11 12 23 + 22 23 ) [ 4] 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. 3

Combining the above, we get = = = ( + = + ) + ( + + + ( ) + + + + ) + + Since is of length and defined for the convolution sum reduces to = = Thus, will be nonzero for and for which satisfies The minimum all those values of value of of = = + is 0, and occurs for the lowest at = and = The maximum value and occurs for maximum value of Hence the total number of nonzero samples = at Thus + = and Let  be the length of the convolution sum of [] and [], which are of length  and , respectively. From Problem 2.5, we know that  convolution of [n] with [], is thus The length of the or + + + To show that the two convolutions are equal, we simply evaluate both convolution sums: 2.5 2.6 2.7 2.8 (a) 1[ ] 1[ ] { = 1 2 3 2 1 }, 2 2. = = = (b) (c) 2[ ] 2[ ] = 1 2 1 { 2 4 212 1 }, 0 8. 3[ ] 3[ ] = 4 0 { 12017 0 120 4 }, 4 4. (a) Given [] and [] from Problem 2.3, their convolution sum [] is given by: [ ] = [ ] [ ] = { = 16 4 2240 5 27 9 6 1 3 12 0 }, 8 4. (b) Given [] and [] from Problem 2.3, their convolution sum [] is given by: [ ] = [ ] [ ] { = 6 12 51640 82322210 92 0 }, 5 7. = (c) Given [] and [] from Problem 2.3, their convolution sum g[ ] is given by: 2.9 Not for sale. 4

[ ] = [ ] [ ] = { = 2454 17 374152 19 53 245127 1 }, 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 + + + + + + + , formed from the convolution of [] with itself, will have a Two results are needed for this problem: Length of Sequence = Max — Min + 1 Length of Convolution = (a) The sequence length 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 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 + . To find the range of indices over The first left-shift at , or Not for sale. 5

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. 6

2.13 = [ Thus, = And again, this confirms the length calculation:  —  — (— — ) + 1 =  +  +  —  + 1. ] + and = [ + = + ] + = 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 = value is and the maximum 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 2.16 We make use of the circular shifting operation given by Eqn. (2.25). The length of { []} deconv in MATLAB. is 9. (a) Thus,   Therefore, = + = = Therefore, y[—3] = — 4 (b) Thus, = Therefore, = = = Therefore, [2] = — 5. Not for sale. 7

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 = (2.34): = = = = 1 ( 1 cos(4 =0 = 1 2 = The formula for energy is given by Eqn. / ) ) = 2 1 2 1 cos(4 =0 / ) . Not for sale. 8

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