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barcode reader project in c#.net y=yYs in Software
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Barcode Maker In Visual C#.NET Using Barcode generation for .NET Control to generate, create barcode image in VS .NET applications. European Article Number 13 Drawer In None Using Barcode creator for Online Control to generate, create EAN13 Supplement 5 image in Online applications. OF FIRSTORDER SYSTEMS
Code 128A Reader In Visual C#.NET Using Barcode decoder for .NET Control to read, scan read, scan image in Visual Studio .NET applications. GS1  13 Printer In None Using Barcode maker for Font Control to generate, create EAN13 image in Font applications. Taking the Laplace transform of this expression and rearranging the result give 1 Y(s) =7s + 1 X(s) (6.23) Barcode Decoder In VB.NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. UPCA Supplement 5 Generation In None Using Barcode maker for Font Control to generate, create UPC A image in Font applications. where T = V/q. This mixing process is, therefore, another firstorder process, for which the dynamics are now well known. We next bring in an example from DC circuit theory. RC Circuit Consider the simple RC circuit shown in Fig. 6.5 in which a voltage source v(t) is applied to a series combination of a resistance R and a capacitance C. For t < 0, v(t) = v,. Determine the transfer function relating et(t) to v(t), where et(t) is the voltage across the capacitor. Applying Kirchhoff s law, which states that in any loop the sum of voltage rises [v(t) in this example] must equal the sum of the voltage drops, gives i dt I Recalling that the current is the rate of change of charge with respect to time (coulombs per second), we may replace i by dqldt in Eq. (6.24) to obtain v(t) = Ri(t) + $ MO 1 v(t) = R + p(O dt
Since the voltage across the capacitance is given by the relationship
(6.25) 4 e, = c
the initial charge on the capacitor is simply
(6.26) qs = Ce,, Initially, when the circuit is at steady state and the capacitor is fully charged, the voltage across the capacitor is equal to the source voltage v,; therefore, Eq. (6.25) can be written for these steadystate conditions as VS = $4. = e,, RC circuit.
LINEAR OPENLOOP SYSTEMS
Subtracting Eq. (6.27) from Eq. (6.25) and introducing the deviation variables v=vvvs e=44s
E, = e,  e,, = g (6.28) we obtain the result
V,Rdp,rs
(6.29) V =RC%+E, (6.30) Taking the transform of Eq. (6.30) and rearranging the result give 1 E,(s) =7s + 1 V(s) (6.31) where G = RC. Again we obtain a firstorder transfer function. The three examples that have been presented in this section are intended to show that the dynamic characteristics of many physical systems can be represented by a firstorder transfer function. In the remainder of the book, more examples of firstorder systems will appear as we discuss a variety of control systems. Summary In each example of a firstorder system, the time constant has been expressed in terms of system parameters; thus mC r=hA
7=AR
V QZ.T=RC
for thermometer, Eq. (5.5)  . for liquidlevel process, Eq. (6.8) for mixing process, Eq. (6.23) for RC circuit, Eq. (6.31) LINEARIZATION
Thus far, all the examples of physical systems, including the liquidlevel system of Fig. 6.1, have been linear. Actually, most physical systems of practical importance are nonlinear. Characterization of a dynamic system by a transfer function can be done only for linear systems (those described by linear differential equations). The conve PHYSICAL EXAMPLES OF FIRSTORDER SYSTEMS
nience of using transfer functions for dynamic analysis, which we have already seen in applications, provides significant motivation for approximating nonlinear systems by linear ones. A very important technique for such approximation is illustrated by the following discussion of the liquidlevel system of Fig. 6.1. We now assume that the resistance follows the squareroot relationship q. = Ch 12 (6.32) where C is a constant. For a liquid of constant density and a tank of uniform crosssectional area A, a material balance around the tank gives s(t)  qo(t) = A$
Combining Eqs. (6.32) and (6.33) gives the nonlinear differential equation
q  Ch Q = f$
(6.33) (6.34) At this point, we cannot proceed as before and take the Laplace transform. This is owing to the presence of the nonlinear term h *, for which them is no simple transform. This difficulty can be circumvented as follows.X By means of a Taylorseries expansion, the function q.(h) may be expanded around the steadystate value h,; thus q. = qO(h,) + q;(h,)(h  h,) + q: (h );j  hS)2 + ...
where qA(h,) is the first derivative of q. evaluated at h S, qi(h ,) the second derivative, etc. If we keep only the linear term, the result is qo = q&s) + q:(k)(h  h,) (6.35) Taking the derivative of q. with respect to h in Eq. (6.32) and evaluating the derivative at h = h, gives q;(h;) = (1/2)Ch; Introducing this into Eq. (6.35) gives qo = qo, + $P  hs) where qo, = q&s) (6.36)

