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barcode reader project in c#.net Step Response in Software
Step Response Code128 Reader In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128C Generator In None Using Barcode encoder for Software Control to generate, create ANSI/AIM Code 128 image in Software applications. If the forcing function is a unitstep function, we have X(s) = ; In terms of the damped vibrator shown in Fig. 8.1 this is equivalent to suddenly applying a force of magnitude K directed toward the right at time t = 0. This follows from the fact that X is defined by the relationship X(t) = F(t)IK . Super Code 128 Code Set C Reader In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. ANSI/AIM Code 128 Generator In Visual C# Using Barcode maker for Visual Studio .NET Control to generate, create Code 128C image in .NET applications. HIGHERORDER SYSTEMS: SECONDORDER AND TRANSPORTATION LAG
Creating Code 128B In .NET Using Barcode drawer for ASP.NET Control to generate, create Code 128 Code Set B image in ASP.NET applications. USS Code 128 Drawer In VS .NET Using Barcode drawer for Visual Studio .NET Control to generate, create Code 128B image in .NET applications. position will enable us to determine easily the response to a step function of any other magnitude. Combining Eq. (8.12) with the transfer function of Eq. (8.11) gives 1 Y(s) = 1 (8.13) s T2S2 + 257s + 1 The quadratic term in this equation may be factored into two linear terms that contain the roots Creating Code128 In VB.NET Using Barcode generator for VS .NET Control to generate, create Code 128 Code Set C image in Visual Studio .NET applications. Draw GS1  12 In None Using Barcode creator for Software Control to generate, create UPC Symbol image in Software applications. s1 = r+ 7 7
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Bar Code Printer In None Using Barcode encoder for Software Control to generate, create bar code image in Software applications. Printing USS Code 128 In None Using Barcode creation for Software Control to generate, create USS Code 128 image in Software applications. (8.14) (8.15) Making GS1  12 In None Using Barcode maker for Software Control to generate, create UPCE Supplement 2 image in Software applications. Encoding Data Matrix 2d Barcode In None Using Barcode creation for Word Control to generate, create DataMatrix image in Microsoft Word applications. Equation (8.13) can now be written 1172 y(s) = (s)(s  Sl)(S  s2) (8.16) GS1 128 Recognizer In VB.NET Using Barcode decoder for VS .NET Control to read, scan read, scan image in .NET framework applications. Drawing Data Matrix In Java Using Barcode creation for Java Control to generate, create ECC200 image in Java applications. The response of the system Y(t) can be found by inverting Eq. (8.16). The roots sr and s2 will be real or complex depending on the parameter 6. The nature of the roots will, in turn, affect the form of Y(t). The problem may be divided into the three cases shown in Table 8.1. Each case will now be discussed. USS128 Printer In VS .NET Using Barcode printer for VS .NET Control to generate, create GS1128 image in VS .NET applications. Barcode Creation In None Using Barcode drawer for Font Control to generate, create bar code image in Font applications. CASE I STEP RESPONSE FOR 6 < 1. For this case, the inversion of Eq. (8.16) Encoding Bar Code In VS .NET Using Barcode maker for ASP.NET Control to generate, create barcode image in ASP.NET applications. Data Matrix Generator In C#.NET Using Barcode creation for .NET Control to generate, create Data Matrix ECC200 image in .NET framework applications. yields the result (8.17) i 1 To derive Eq. (8.17), use is made of the techniques of Chap. 3. Since 6 < 1, Eqs. (8.14) to (8.16) indicate a pair of complex conjugate roots in the lefthalf plane and a root at the origin. In terms of the symbols of Fig. 3.1, the complex roots correspond to s:! and S; and the root at the origin to Sg. By referring to Table 3.1, we see that Y(t) has the form Y(f) = Cl + e 5f 7 C2 cos \/1_52$ + C3 sin i Y(f) = 1  J&ect sin Jt + tan CF f 47;) (8.18) The constants C 1, C2, and C3 are found by partial fractions. The resulting equation is then put in the form of Eq. (8.17) by applying the trigonometric identity used TABLE 8.1
CaSe
I II III
<I = 1 >l
Nature of roots Complex Real and equal Real
Description of response Underdamped or oscillatory Critically damped Overdamped or nonoscillatory
LINEAR OPENLOOP SYSTEMS
in Chap. 5, Eq. (5.23). The details are left as an exercise for the reader. It is evident from Eq. (8.17) that Y(t) f 1 as t + ~0. The nature of the response can be understood most clearly by plotting Eq. (8.17) as shown in Fig. 8.2, where Y(t) is plotted against the dimensionless variable t/r for several values of {, including those above unity, which will be considered in the next section. Note that, for 4 < 1, all the response curves are oscillatory in nature and become less oscillatory as c is increased. The slope at the origin in Fig. 8.2 is zero for all values of t. The response of a secondorder system for 6 < 1 is said to be underdamped. CASE II STEP RESPONSE FOR 5 = 1. For this case, the response is given by
expression
Y(t) = 1  (1 + ;,,1, (8.19) This is derived as follows: Equations (8.14) and (8.15) show that the roots st and s2 are real and equal. Referring to Fig. 3.1 and Table 3.1, it is seen that FIGURE 82 Response of a secondorder system to a unitstep forcing function.
HIGHERORDER SYSTEMS: SECONDORDER AND TRANSPORTATION LAG
Eq. (8.19) is in the correct form. The constants are obtained, as usual, by partial fractions. The response, which is plotted in Fig. 8.2, is nonoscillatory. This condition, b = 1, is called critical dumping and allows most rapid approach of the response to Y = 1 without oscillation. CASE III STEP RESPONSE FOR 5 > 1. For this case, the inversion of Eq.
(8.16) gives the result Y(t) = 1  e@ i where the hyperbolic functions are defined as p  ea sinh a = 2 ea + eMa cash a = 2
sinh ,@ $ i
(8.20) The procedure for obtaining Eq. (8.20) is parallel to that used in the previous cases. The response has been plotted in Fig. 8.2 for several values of 5. Notice that the response is nonoscillatory and becomes more sluggish as 4 increases. This is known as an overdumped response. As in previous cases, all curves eventually approach the line Y = 1. Actually, the response for f > 1 is not new. We met it previously in the discussion of the step response of a system containing two firstorder systems in series, for which the transfer function is 1 Y(s) = (71s + l)(r*s + 1) X(s) (8.21) This is true for f > 1 because the roots sr and ~2 are real, and the denominator of Eq. (8.11) may be factored into two real linear factors. Therefore, Eq. (8.11) is equivalent to Eq. (8.2 1) in this case. By comparing the linear factors of the denominator of Eq. (8.11) with those of Eq. (8.21), it follows that 71 = (l + &T)r 52 = ({  JG)r (8.22) (8.23)

