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barcode reader code in asp.net R + v (t) R in Software
R + v (t) R QR Code Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Print QR Code In None Using Barcode creator for Software Control to generate, create QR image in Software applications. L + v (t) L + i(t) vC (t) Decode QR Code ISO/IEC18004 In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. Generate QRCode In Visual C#.NET Using Barcode encoder for .NET framework Control to generate, create QR image in .NET framework applications. i(t ) dt vS (t) = 0 QR Code Maker In .NET Using Barcode printer for ASP.NET Control to generate, create QRCode image in ASP.NET applications. Generating QR Code In VS .NET Using Barcode creation for Visual Studio .NET Control to generate, create QR image in .NET framework applications. (58) QR Code 2d Barcode Generator In VB.NET Using Barcode generation for .NET Control to generate, create QR Code image in .NET applications. Creating Bar Code In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. vS (t) + _ European Article Number 13 Maker In None Using Barcode creation for Software Control to generate, create EAN / UCC  13 image in Software applications. Encode UPC A In None Using Barcode creation for Software Control to generate, create UPCA image in Software applications. Equation 58 is called an integrodifferential equation, because it contains both an integral and a derivative This equation can be converted into a differential equation by differentiating both sides, to obtain: R 1 d 2 i(t) dvS (t) di(t) + i(t) = +L 2 dt dt C dt (59) Drawing GS1128 In None Using Barcode generation for Software Control to generate, create UCC.EAN  128 image in Software applications. Making USS Code 39 In None Using Barcode creation for Software Control to generate, create ANSI/AIM Code 39 image in Software applications. Figure 57 Secondorder circuit
Encoding USD  8 In None Using Barcode creator for Software Control to generate, create USD  8 image in Software applications. EAN / UCC  14 Reader In VB.NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in VS .NET applications. or, equivalently, by observing that the current owing in the series circuit is related to the capacitor voltage by i(t) = CdvC /dt, and that equation 58 can be rewritten as: RC d 2 vC (t) dvC + LC + vC (t) = vS (t) dt dt 2 (510) USS Code 128 Generator In None Using Barcode encoder for Font Control to generate, create Code 128B image in Font applications. Bar Code Printer In .NET Using Barcode creation for ASP.NET Control to generate, create barcode image in ASP.NET applications. Note that, although different variables appear in the preceding differential equations, both equations 59 and 510 can be rearranged to appear in the same general form, as follows: a2 d 2 x(t) dx(t) + a1 + a0 x(t) = F (t) dt 2 dt (511) Barcode Maker In Java Using Barcode maker for Eclipse BIRT Control to generate, create barcode image in BIRT reports applications. UPC  13 Creator In C# Using Barcode generator for VS .NET Control to generate, create GS1  13 image in Visual Studio .NET applications. where the general variable x(t) represents either the series current of the circuit of Figure 57 or the capacitor voltage By analogy with equation 57, we call equation 511 a secondorder ordinary differential equation with constant coef cients As the number of energystorage elements in a circuit increases, one can therefore Creating Bar Code In VS .NET Using Barcode creation for ASP.NET Control to generate, create barcode image in ASP.NET applications. UPCA Supplement 2 Drawer In None Using Barcode generation for Font Control to generate, create Universal Product Code version A image in Font applications. 5
Transient Analysis
expect that higherorder differential equations will result Computer aids are often employed to solve differential equations of higher order; some of these software packages are speci cally targeted at the solution of the equations that result from the analysis of electrical circuits (eg, Electronics WorkbenchTM ) EXAMPLE 52 Writing the Differential Equation of an RLC Circuit
Problem
R1 vC (t) L iL(t) vS (t) + _ C R2
Derive the differential equation of the circuit shown in Figure 58
Solution
Known Quantities: R1 = 10 k ; R2 = 50 Find: The differential equation in iL (t) ; L = 10 mH; C = 01 F
Figure 58 Secondorder circuit of Example 52
Assumptions: None Analysis: Apply KCL at the top node (nodal analysis) to write the rst circuit equation Note that the top node voltage is the capacitor voltage, vC dvC vS vC iL = 0 C R1 dt Now, we need a second equation to complete the description of the circuit, since the circuit contains two energy storage elements (secondorder circuit) We can obtain a second equation in the capacitor voltage, vC , by applying KVL to the mesh on the righthand side: diL R 2 iL = 0 dt diL + R 2 iL vC = L dt Next, we can substitute the above expression for vC into the rst equation, to obtain a secondorder differential equation, shown below vC L R2 L diL d vs iL C R1 R1 dt R1 dt L diL + R2 iL iL = 0 dt Rearranging the equation we can obtain the standard form similar to equation 511: R1 CL d 2 iL diL + (R1 + R2 ) iL = vS + (R1 R2 C + L) dt 2 dt Comments: Note that we could have derived an analogous equation using the capacitor
voltage as an independent variable; either energy storage variable is an acceptable choice You might wish to try obtaining a secondorder equation in vC as an exercise In this case, you would want to substitute an expression for iL in the rst equation into the second equation in vC TRANSIENT RESPONSE OF FIRSTORDER CIRCUITS
Firstorder systems occur very frequently in nature: any system that has the ability to store energy in one form and to dissipate the energy stored is a rstorder Part I
Circuits
system In electrical circuits, we recognize that any circuit containing a single energy storage element (inductor or capacitor) and a combination of sources and resistors (and possibly switches) is a rstorder system In other domains, we also encounter rstorder systems For example, a mechanical system that has mass and damping (eg, friction), but not elasticity, will be a rstorder system A uid system with uid resistance and uid capacitance ( uid storage) will also be of rst order; an example of a rstorder uid system is a storage tank with a valve In thermal systems, we also encounter rstorder systems quite frequently: The ability to store heat (heat capacity) and to dissipate it leads to a rstorder thermal system; heating and cooling of bodies is, at its simplest level, described by rstorder behavior In the present section we analyze the transient response of rstorder circuits In what follows, we shall explain that the initial condition, the steadystate solution, and the time constant of the rstorder system are the three quantities that uniquely determine its response Natural Response of FirstOrder Circuits Figure 59 compares an RL circuit with the general form of the series RC circuit, showing the corresponding differential equation From Figure 59, it is clear that equation 512 is in the general form of the equation for any rstorder circuit: dx(t) a1 (512) + a2 x(t) = f (t) dt where f is the forcing function and x(t) represents either vC (t) or iL (t) The constant a = a2 /a1 is the inverse of the parameter , called the time constant of the system: a = 1/ To gain some insight into the solution of this equation, consider rst the natural solution, or natural response, of the equation,1 which is obtained by setting the forcing function equal to zero This solution, in effect, describes the response of the circuit in the absence of a source and is therefore characteristic of all RL and RC circuits, regardless of the nature of the excitation Thus, we are interested in the solution of the equation dxN (t) 1 + xN (t) = 0 dt or 1 dxN (t) (514) = xN (t) dt where the subscript N has been chosen to denote the natural solution One can easily verify by substitution that the general form of the solution of the homogeneous equation for a rstorder circuit must be exponential in nature, that is, that xN (t) = Ke at = Ke t/ (515) (513) + _ vS (t)

