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Once again, notice that the Or operator only needs one operand to evaluate to True before the conclusion is True. Both VB .NET and C# also have a short-circuit version of the Or/| logical operator. They work by not comparing the second operand if the computer already knows that the first operand is True. This, again, saves the computer time and can make your program run a bit faster.

+ vout(t)

It is also known that dvout (t) iF (t) = CF (1265) dt from the fundamental de nition of the capacitor The source voltage can then be expressed as a function of the derivative of the output voltage: 1 dvout (t) vS (t) = RS C F dt 1 RS CF

t

(1266)

By integrating both sides of equation 1266, we obtain the following result: vout (t) = vS (t ) dt (1267)

The aim of this figure is to provide a snapshot of comparable groups by processes and related activities, marked with shades Please note that categories that cannot be mapped (for example, Reuse) are shown with a white background

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You can use substitution parameters to conditionally change the contents of a string like this: Console.WriteLine("result is {0}", Method1() | Method2()); Both VB .NET and C# allow you to use the {} operator to un-conditionally change a string value as well. When the program encounters this operator, it will replace each substitution parameter with the values listed after the comma. For example, if you typed in this command: Console.WriteLine("This is a {0} {1}", "test", "message"); the Console application would print out "This is a test message". The string "test" would be substituted for the {0}, while "message" would be substituted for the {1}. This takes place because of the order that the strings "test" and "message" are defined in the comma-separated list after the string. By creating two methods, one that returns true and one the returns false, you can create some test code and see an example of the short-circuit operators in action: VB .NET Function Method1() As Boolean Console.WriteLine("Method1 called") Return True End Function Function Method2() As Boolean Console.WriteLine("Method2 called") Return False End Function Public Sub Main()

This equation states that the output voltage is the integral of the input voltage There are numerous applications of the op-amp integrator, most notably the analog computer, which will be discussed in Section 125 The following example illustrates the operation of the op-amp integrator

Determine the output voltage for the integrator circuit of Figure 1231 if the input is a square wave of amplitude A and period T

12

t

Assumptions: Assume ideal op-amp The square wave starts at t = 0 and therefore Analysis: Following equation 1267, we write the expression for the output of the

'Both Method1 and Method2 will be called Console.WriteLine("Regular OR result is {0}", Method1() Or Method2()) 'Only Method1 will be called since it returns True Console.WriteLine("Short-circuit OR result is {0}", Method1() OrElse Method2()) End Sub C# static bool Method1() { Console.WriteLine("Method1 called"); return true; } //End of Method1 static bool Method2() { Console.WriteLine("Method2 called"); return false; }//End of Method2 public static void Main() { //Both Method1 and Method2 will be called Console.WriteLine("Regular OR result is {0}", Method1() | Method2()); //Only Method1 will be called since it returns true Console.WriteLine("Short-circuit OR result is {0}", Method1() || Method2()); } //End of Main

10:

0

Next, we note that we can integrate the square wave in a piecewise fashion by observing that vS (t) = A for 0 t < T /2 and vS (t) = A for T /2 t < T We consider the rst half of the waveform: vout (t) = 1 RF CS

= 100 0 +

T /2

T 100 2

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