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10 =T

66 22 55

Compute the power in x(n),

The output is:

If x(n) is input to a time-varying system defined by y(n) = nx(n), find the power in the output signal (i.e., evaluate the sum)

d. Trace of the selection sort:

However, we may derive a closed-form expression for this sum as follow^.^ Differentibut not for C:,n2an. ating both sides of Eq. (1.19) with respect to a , we have

{22, {22, {AB, {AB, {AB, {AB, 44, 66, 88} 44, 66, 88} CD, EF} CD, EF} XYZ, EF} CD, EF}

.r(n) =

a[0] 44 22 a[1] 88 33 44 55 77 88 a[2] 55 a[3] 99 a[4] 66 a[5] 33 88 55 99 88 99 a[6] 22 44 a[7] 88 a[8] 77

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In this problem, we would like to perform a signal decomposition, expressing x ( n ) as a sum of scaled and shifted unit steps. There are several ways to derive this decomposition. One way is to express x ( n ) as a sum of weighted and shifted unit samples, x ( n ) = S(n) 2S(n - I) 3S(n - 2)

The array a[] contains four int elements. The array b[] is a duplicate of a[]. Similarly, the array d[] is a duplicate of the array c[], each containing three String elements. In both cases, the duplication is obtained by invoking the clone() method. Since it returns a reference to an Object, it must be cast to the array type being duplicated, int[] or String[]. The last part of the example shows that the cloned array d[] is indeed a separate copy of c[]: Changing c[1] to "XYZ" has no effect upon the value "CD" of d[1].

and use the fact that a unit sample may be written as the difference of two steps as follows:

e. Trace of the insertion sort:

Therefore,

CHAP. 2]

x ( n ) = u(n) - u(n - I )

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a[0] 44 a[1] 88 55 44 33 a[2] 55 88 66 55 44 a[3] 99 88 66 55 a[4] 66 99 88 66 a[5] 33 a[6] 22 a[7] 88 a[8] 77

+ 2[u(n- I) - u(n - 2)] + 3[u(n- 2 ) - u(n - 3)]

public static void sort(int[] a) { for (int i = 0; i < a.length-1; i++) { int j=i; for (int k = i+1; k < a.length; k++) { if (a[k] < a[j]) { j = k; } } swap(a, i, j); } } public static void sort(int[] a) { sort(a, a.length); } private static void sort(int[] a, int n) { if (n < 2) { return; } int j = 0; for (int k = 1; k < n; k++) { if (a[k] > a[j]) { j = k; } } swap(a, n-1, j); sort(a, n-1); }

which gives the desired decomposition:

33 22

CHAP. I]

The last loop invariant proves correctness So, like the proof for the bubble sort, we need only verify the loop invariants On the first iteration of the main loop (step 1), a[i] is the last element in the array, so the index k of the inner loop runs through every element after a[0] The value of the index j begins at 0 and then changes each time k finds a larger element Since j is always reset to the index of the larger element, a[j] will be the largest element of the array when the inner loop finishes This verifies the first loop invariant On each successive iteration of the outer loop, the index k runs through the remaining unsorted segment of the array, so for the same reason, a[j] will be the largest element of that remaining segment when the inner loop finishes.

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