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barcode reader project in c#.net Sinusoidal Response in Software
Sinusoidal Response Recognize ANSI/AIM Code 128 In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128 Code Set A Creator In None Using Barcode printer for Software Control to generate, create ANSI/AIM Code 128 image in Software applications. To investigate the response of a firstorder system to a sinusoidal forcing function, the example of the mercury thermometer will be considered again. Consider a thermometer to be in equilibrium with a temperature bath at temperature x S. At some time t = 0, the bath temperature begins to vary according to the relationship x = xs +Asinot Recognizing Code128 In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Code 128C Drawer In C#.NET Using Barcode maker for Visual Studio .NET Control to generate, create Code128 image in Visual Studio .NET applications. (5.17) Code 128 Code Set C Creator In .NET Framework Using Barcode printer for ASP.NET Control to generate, create Code 128 Code Set B image in ASP.NET applications. Code 128 Code Set B Creation In .NET Framework Using Barcode generation for VS .NET Control to generate, create Code 128B image in VS .NET applications. where x = temperature of bath xs = temperature of bath before sinusoidal disturbance is applied A = amplitude of variation in temperature w = radian frequency, radkime In anticipation of a simple result we shall introduce a deviation variable X which is defined as x = x  x , (5,. 18) Using this new variable in Eq. (5.17) gives X = A sin ot By referring to a table of transforms, the transform of Eq. (5.19) is (5.19) Code 128A Creator In Visual Basic .NET Using Barcode generation for .NET framework Control to generate, create Code 128C image in Visual Studio .NET applications. European Article Number 13 Maker In None Using Barcode printer for Software Control to generate, create European Article Number 13 image in Software applications. X(s) = J$
Barcode Maker In None Using Barcode maker for Software Control to generate, create bar code image in Software applications. Generating UPC Symbol In None Using Barcode drawer for Software Control to generate, create GS1  12 image in Software applications. (5.20) Draw Bar Code In None Using Barcode printer for Software Control to generate, create barcode image in Software applications. Making USS Code 128 In None Using Barcode encoder for Software Control to generate, create Code 128 Code Set C image in Software applications. RESPONSE OF FIRSTORDER SYSTEMS
Painting Bookland EAN In None Using Barcode generation for Software Control to generate, create ISBN image in Software applications. Data Matrix Creator In VB.NET Using Barcode drawer for .NET Control to generate, create DataMatrix image in VS .NET applications. Combining Eqs. (5.7) and (5.20) to eliminate X(S) yields AW l/7 Y ( s ) =  s* + co* s + l/T
Painting Barcode In Java Using Barcode creator for Android Control to generate, create bar code image in Android applications. Printing UPC A In Visual Studio .NET Using Barcode printer for .NET Control to generate, create UPC Code image in Visual Studio .NET applications. (5.21) Printing Bar Code In None Using Barcode maker for Word Control to generate, create barcode image in Office Word applications. Drawing Bar Code In .NET Using Barcode encoder for Reporting Service Control to generate, create bar code image in Reporting Service applications. This equation can be solved for Y(t) by means of a partialfraction expansion, as described in Chap. 3. The result is Y(t) = Aore AWT A sin ot cos wt + r*w* + 1  7*&J* + 1 TW + 1 (5.22) Bar Code Creator In .NET Framework Using Barcode creator for ASP.NET Control to generate, create barcode image in ASP.NET applications. Bar Code Generator In .NET Framework Using Barcode generator for .NET Control to generate, create barcode image in .NET framework applications. Equation (5.22) can be written in another form by using the trigonometric identity p cos A + q sin A = r sin (A + 0) (5.23) where tane=~ 4 Applying the identity of Eq. (5.23) to (5.22) gives AWT t/7 + Y(t) = sin (wt + 4) r*W*+le where t$ = tan (on) As t + ~0, the first term on the right side of Eq. (5.24) vanishes and leaves only the ultimate periodic solution, which is sometimes called the steadystate solution Y ( t )s = J& sin W + 4) (5.25) r = Jpq (5.24) By comparing Eq. (5.19) for the input forcing function with Eq. (5.25) for the ultimate periodic response, we see that 1. The output is a sine wave with a frequency w equal to that of the input signal. 2. The ratio of output amplitude to input amplitude is l/ ,/T*w* + 1. This is always smaller than 1. We often state this by saying that the signal is attenuated. 3. The output lags behind the input by an angle 14 I. It is clear that lag occurs, for the sign of 4 is always negative.* *By convention, the output sinusoid lags the input sinusoid if C#J in Eq. (5.25) is negative. In terms of a recording of input and output, this means that the input peak occurs before the output peak. If C$ is positive in E$. (5.29, the system exhibits phase lead, or the output leads the input. In this book we shall always use the term phase angle (4) and interpret whether there is lag or lead by the convention phase lag phase lead LINEAR OPENLQOP
SYSTEMS
For a particular system for which the time constant r is a fixed quantity, it is seen from Eq. (5.25) that the attenuation of amplitude and the phase angle 4 depend only on the frequency o. The attenuation and phase lag increase with frequency, but the phase lag can never exceed 90 and approaches this value asymptotically. The sinusoidal response is interpreted in terms of the mercury thermometer by the following example. Example 5.2. A mercury thermometer having a time constant of 0.1 min is placed in a temperature bath at 100 F and allowed to come to equilibrium with the bath. At time t = 0, the temperature of the bath begins to vary sinusoidally about its average temperature of lOOoF with an amplitude of 2 F If the frequency of oscillation is 101~ cycles/min, plot the ultimate response of the thermometer reading as a function of time. What is the phase lag In terms of the symbols used in this chapter 7 = 0.1 x s = 100 F A = 2 F 10 f =  cycles/min 7iw = 27rf = 21r~ = 20 radlmin 7r From Eq. (5.25), the amplitude of the response and the phase angle am calculated; thus 2 = ___ = 0.896 F vGGTl JGT 4 = tan 2 = 63.5 or Phase lag = 63.5 The response of the thermometer is therefore Y(t) = 0.896 sin (20t  63.5 ) or Y(G) = 100 + 0.896 sin (20t  63.5 ) To obtain the lag in terms of time rather than angle, we proceed as follows: A frequency of lO/rr cycles/min means that a complete cycle (peak to peak) occurs in (10/r) min. Since one cycle is equivalent to 360 and the lag is 63.5 , the time corresponding to this lag is 63.5  X (time for 1 cycle) 360 A

