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A second-order system has two poles at z = 0.5 and a pair of complex zeros at z = e * ~ = / ~ . Geometrically find the gain, A, of the filter so that I H ( e J W ) is equal to unity at w = 0. J

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Because the length of the vectors from the two zeros at z = e'J"12 to the point z = 1 on the unit circle is equal to A,and because the distance from the two poles at z = 0.5 to z = 1 is equal to 0.5, the magnitude of the frequency response at o = 0 is

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Theorem 149 The merge sort runs in O(n lgn) time In general, the merge sort works by repeatedly dividing the array in half until the pieces are singletons, and then it merges the pieces pairwise until a single piece remains This is illustrated by the diagram in Figure 141 The number of iterations in the first part equals the number of times n can be halved: that is, lg n 1 In terms of the number and sizes of the pieces, the second part of the process reverses the first So the second part also has lg n 1 steps So the entire algorithm has O(lg n) steps Each step compares all n elements So the total number of comparisons is O(n lg n) Theorem 1410 The merge sort is correct The proof follows from the preconditions and postconditions given in the code In the Figure 14.

public static void main(String[] args) { for (int n=0; n<10; n++) { System.out.println("f("+n+") = "+f(n)); } }

A = !8

Therefore, the frequency response may be written as follows:

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1 The merge sort main sort() method, the array is already sorted if its length is 0 or 1 Otherwise, the postcondition of the three-parameter sort() method guarantees that the array will be sorted after that method returns because the entire array is passed to it That postcondition is the same as the postcondition of the merge() method, which is invoked last, so it remains to verify that the merge() method s postcondition will be true The merge() method s postcondition follows from its loop invariant, because when that loop has finished, the tmp[] array is sorted and that is copied back into a[] in the same order So it remains to verify the loop invariant for all k < q - p Suppose the invariant is false for some k, that is, tmp[0.k) is not sorted Then there must be some x and y in tmp[0..

The factorial function is also easy to implement iteratively:

Therefore, we may write the frequency response as follows:

where N isodd. Therefore, h ( n ) is symmetric about the half-integer, N j 2 , and the symmetry condition isequivalent to

k), where x was copied into tmp[] before y but x > y We may assume without loss of generality that x was copied from the left half of a[] and y from the right half, as Figure 142 The merge sort shown in Figure 142 Thus, x = a[r] and y = a[s] for some indexes r and s such that p r < i and m s < j Now the two halves of a[] are each already separately sorted Then for every element z in a[m.s], z a[s] But a[s] = y < x Therefore, every element z in a[m.s] must have been copied into tmp[] before x was, because this assignment.

1 2 3 4 5 6 7

Therefore,

tmp[k++] = ( a[i]<=a[j] a[i++] : a[j++] );

N+1 b(k)=2h(T-k)

public static int f(int n) { int f = 1; for (int i = 2; i <= n; i++) { f *= i; } return f; }

k = 1 . 2. . . . ,

always copies the smaller element first. But that means that a[s] was copied into tmp[] before x. But a[s] = y, which was assumed to have been copied into tmp[] after x. This contradiction proves that the invariant must be true.

How would the derivations in the previous two problems be modified to find the form of the frequency response for types 111 and IV linear phase filters

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