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How to Easily Create a PDF Document in ASP.NET Core Web API
Jun 18, 2018 · Let's imagine that we have a .NET Core Web API project in which we need to generate a PDF report. Even though it shouldn't suppose to be ...

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page 50) does not initialize it. In Java, every class field that is an object reference (i.e., its type is either a class or an interface) is automatically initialized to null, unless it is initialized by its constructor to some existing object. In the figures that follow, each Node object is shown as a box with two parts: the left side contains the integer data, and the right side contains the next reference. This simply abbreviates the versions shown in Figure 3.9. Continuing the code in Example 3.4, at line 2, the start node s next field is assigned to a new Node object containing the data 33. Now the list has two nodes, as shown in Figure 3.12. Figure 3.12 Adding a node

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CHAP. 3]

h(n)= [-2(;)"

+3(a)"]u(n)

The next node is added to the end of the list at line 3. To do that, we have to assign it to the next field of the node that contains 33. But the only node to which we have external access (i.e., the only node that has a variable name) is the first node. Its name is start. So we have to use the expression start.next.next to refer to the next field of the node that contains 33. Similarly, the fourth node is added at line 4 using the expression start.next.next.next, and the fifth node is added at line 5 using the expression start.next.next.next.next. That finally gives us the five-node list shown in Figure 3.13.

5 6 7 8 9

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(b) To find the response of the system to x ( n ) = u ( n ) - u ( n - l o ) , we may proceed in one of two ways. First, we may perform the convolution of h ( n ) with x ( n ) :

Alternatively, noting that the input is a sum of two steps, we may find the step response of the system, s ( n ) , and then using linearity. write the response as

The code in Example 3.4 is clumsy and unsuited for generalization. Obviously, if we wanted to build a linked list of 50 nodes, this approach would be unworkable. The solution is to use a local reference variable that can walk through the list, locating one node after the Figure 3.13 The five-node list other and thereby giving local access to the nodes. Traditionally, the variable p (for pointer ) is used for this purpose. Since it will refer to individual nodes, it should be declared to be a Node reference, like this:

p.next = new Node(44); p = p.next; p.next = new Node(55); p = p.next; p.next = new Node(66);

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y ( n ) = s ( n ) - s ( n - 10) = [ 2 ( f ) " - ( : ) " ] u ( n ) - [ 2 ( f )

Node p;

11-10

This code may not seem much better than the other version in Example 3.4. But one big advange is that it is easily managed within a loop. For example, the same list can be built with the three lines of code in Example 3.6.

( i ) n - l o ] u ( n- 10)

And since our only access to the nodes is from the start node, we should initialize p like this:

(c) With x(n) = (f)"u(n), note that .r(n) has the same form as one of the terms in the homogeneous solution. Therefore, the particular solution will not be of the form y,(n) = C(:)" as indicated in Table 1-2. If we were to substitute this particular solution into the difference equation, we would find that no value of C would work. As is the case when a root of the characteristic equation is of second order, the particular solution has the form

1 2 3 4

We now must solve for the constants A1 and A> As we did in part (a),with zero initial conditions we find that y(0) = I and y ( l ) = $. Therefore, evaluating Eq. (1.27) at 11 = 0 and n = 1, we obtain the following two equations in the two unknowns A l and A2:

Node p=start;

A $100,000 mortgage is to be paid off in equal monthly payments of d dollars. Interest, compounded monthly, is charged at the rate of 10 percent per annum on the unpaid balance [e.g., after the first month . the total debt equals ($100,000 ~ $ l 0 0 , 0 0 0 ) ] Determine the amount of the payment, d, so that the mortgage is paid off in 30 years, and find the total amount of payments that are made over the 30-year period.

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