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Let h ( n ) = a n u ( n )and x ( n ) = bnu(n). Find the convolution y ( n ) = x ( n ) * h ( n ) assuming that a # 6 . If x ( n ) = anu(n),find the convolution y ( n ) = x ( n ) * x(n). The input to a linear shift-invariant system is the unit step, x ( n ) = u ( n ) , and the response is y ( n ) = S(n). Find the unit sample response of this system. If h(n) = A6(n) +(f )"u(n) is the unit sample response of a linear shift-invariant system, and s ( n ) is the step response s(n) = 0. (the response of the system to a unit step), find the value of the constant A so that lim,,, The unit sample response of a linear shift-invariant system is

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System.out.println(p);

Find the response of the system to the complex exponential x ( n ) = exp(jnrr/4). Evaluate the convolution of the sequence x ( n ) = n(i)"cos(rrn) with the unit step, h ( n ) = u(n). Let

CHAP. 13]

and h ( n ) = S(n - 2 ) S(n - 3 ) 6(n - 4), at what value of n will the convolution y ( n ) = x ( n ) * h ( n ) attain its maximum value, and what is this maximum value A linear system has a response h k ( n ) = S(2n - k ) to the unit sample 6(n - k). Find the response of the system to the input x ( n ) = u(n). Consider the interconnection of three linear shift-invariant systems shown in the figure below.

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the system automatically invokes that object s toString() method. Unless it has been overridden, the version of the toString() method that is defined in the Object class will execute, as it did in the program in Example 3.9. The string returned by that version merely contains the object s type (Node) followed by the @ sign and the memory address of the object (7182c1). So the last five lines of output report that the five Node objects are stored at the (hexadecimal) memory addresses 0x7182c1, 0x3f5d07, 0xf4a24a, 0xcac268, and 0xa16869. These then are the actual values stored in the reference variables start, start.next, start.next.next, start.next.next.next, and start.next.next.next.next. You can see from Figure 3.17 why we usually draw linked lists using arrows to represent the Node references. Showing the actual memory address values instead requires more effort to see which node references which. Moreover, those memory address values are runtime dependent: They will be different on different computers, and maybe even on the same computer at different times. One final note: At line 6 we use the chained assignment

x(n)

7 8 9 10 11 12 13

hAn)

p = p.next = new Node(22+11*i);

h~(n)

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pq.add("IT"); pq.add("ES"); while (!pq.isEmpty()) { System.out.printf("%s } } }

~ ( n )

It is important to remember the order of operations in such a statement. Here, the first thing that happens is the evaluation of the expression 22 + 11*i. When i is 1, that evaluates to 33; when i is 4, it evaluates to 66. After the value is obtained, it is passed to the Node class constructor at line 5 of Example 3.3 on page 50. That constructs a node with that value in its data field and null in its next field. The constructor returns a reference to the Node object. It is that reference that is assigned first to p.next, and then to p. The key is that the assignments are made from right to left. So we know that p is not updated until after its next field is. So, first the next field is set to point to the new node, and then the loop control variable p is advanced to that next node.

h3(n)

", pq.remove());

Consider the linear shift-invariant system described by the LCCDE y(n) = -iy(n - I) +2x(n) Find the response of this system to the input x(n) = 2 n = 0 , 2 , 4 , 6 ,... otherwise

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