barcode reader using vb net source code Figure 541 Two second-order circuits in Software

Generator QR-Code in Software Figure 541 Two second-order circuits

Figure 541 Two second-order circuits
Reading QR Code ISO/IEC18004 In None
Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.
Printing Denso QR Bar Code In None
Using Barcode maker for Software Control to generate, create QR Code ISO/IEC18004 image in Software applications.
where the natural response is the solution of the homogeneous equation without regard for the forcing function (ie, with f (t) = 0) and the forced response is the solution of the forced equation with no consideration of the effects of the initial conditions Once the general form of the complete response is found, the unknown
QR Code 2d Barcode Recognizer In None
Using Barcode scanner for Software Control to read, scan read, scan image in Software applications.
Denso QR Bar Code Printer In Visual C#.NET
Using Barcode creation for .NET Control to generate, create QR Code JIS X 0510 image in .NET framework applications.
5
QR Code JIS X 0510 Drawer In VS .NET
Using Barcode generation for ASP.NET Control to generate, create QR-Code image in ASP.NET applications.
Draw QR-Code In .NET
Using Barcode creation for VS .NET Control to generate, create Quick Response Code image in .NET framework applications.
Transient Analysis
Encoding QR Code 2d Barcode In Visual Basic .NET
Using Barcode printer for .NET framework Control to generate, create QR Code 2d barcode image in VS .NET applications.
EAN-13 Generator In None
Using Barcode generator for Software Control to generate, create EAN 13 image in Software applications.
constants are evaluated subject to the initial conditions, and the solution can then be divided into transient and steady-state parts, with x(t) = xT (t) + xSS (t) Transient part Steady-state part (549)
Code-39 Creation In None
Using Barcode creator for Software Control to generate, create USS Code 39 image in Software applications.
Barcode Creator In None
Using Barcode encoder for Software Control to generate, create barcode image in Software applications.
The aim of this section is to determine the natural response, which satis es the homogeneous equation: d 2 xN (t) dxN (t) +b + cxN (t) = 0 dt 2 dt (550)
Drawing EAN128 In None
Using Barcode maker for Software Control to generate, create UCC-128 image in Software applications.
Paint Universal Product Code Version A In None
Using Barcode maker for Software Control to generate, create UPC Code image in Software applications.
where b = a1 /a2 and c = a0 /a2 Just as in the case of rst-order circuits, xN (t) takes on an exponential form: xN (t) = Kest This is easily veri able by direct substitution in the differential equation: s 2 Kest + bsKest + cKest = 0 (552) (551)
RoyalMail4SCC Encoder In None
Using Barcode creator for Software Control to generate, create British Royal Mail 4-State Customer Barcode image in Software applications.
Print Bar Code In None
Using Barcode maker for Microsoft Excel Control to generate, create bar code image in Microsoft Excel applications.
and since it is possible to divide both sides by Kest , the natural response of the differential equation is, in effect, determined by the solution of the quadratic equation s 2 + bs + c = 0 (553)
Creating Code-128 In Visual Studio .NET
Using Barcode encoder for .NET framework Control to generate, create USS Code 128 image in .NET applications.
Data Matrix Scanner In C#
Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in VS .NET applications.
This polynomial in the variable s is called the characteristic polynomial of the differential equation Thus, the natural response, xN (t), is of the form xN (t) = K1 es1 t + K2 es2 t (554)
Decoding Bar Code In .NET Framework
Using Barcode Control SDK for ASP.NET Control to generate, create, read, scan barcode image in ASP.NET applications.
Barcode Generation In None
Using Barcode creation for Office Word Control to generate, create bar code image in Word applications.
where the exponents s1 and s2 are found by applying the quadratic formula to the characteristic polynomial: b 1 2 b 4c s1,2 = 2 2 (555)
Encode Code39 In Java
Using Barcode creator for Eclipse BIRT Control to generate, create USS Code 39 image in BIRT applications.
Matrix Barcode Drawer In C#.NET
Using Barcode encoder for Visual Studio .NET Control to generate, create 2D Barcode image in Visual Studio .NET applications.
The exponential solution in terms of the exponents s1,2 can take different forms depending on whether the roots of the quadratic equation are real or complex As an example, consider the parallel circuit of Figure 540, and the governing differential equation, 542 The natural response for iL (t) in this case is the solution of the following equation: 1 diL (t) 1 d 2 iL (t) + + iL (t) = 0 2 dt RC dt LC (556)
where R = RT in Figure 540 The solution of equation 556 is determined by solving the quadratic equation s2 + 1 1 s+ =0 RC LC (557)
The roots are s1,2 = 1 1 2RC 2 1 RC
4 LC
(558)
Part I
Circuits
where s1 = 1 1 + 2RC 2 1 1 2RC 2 1 RC 1 RC
4 LC 4 LC
(559a)
s2 =
(559b)
The key to interpreting this solution is to analyze the term under the square root sign; we can readily identify three cases:
Case I: (560)
1 2 4 > RC LC s1 and s2 are real and distinct roots: s1 = 1 and s2 = 2 Case II: 1 2 4 = RC LC s1 and s2 are real, repeated roots: s1 = s2 = Case III: 1 2 4 < RC LC s1 , s2 are complex conjugate roots: s1 = s2 = + j
(561)
(562)
It should be remarked that a special case of the solution (562) arises when the value of R is identically zero This is known as the resonance condition; we shall return to it later in this section For each of these three cases, as we shall see, the solution of the differential equation takes a different form The remainder of this section will explore the three different cases that can arise
EXAMPLE 59 Natural Response of Second-Order Circuit
Problem
Find the natural response of iL (t) in the circuit of Figure 542
+ vS + _ R1 R2 C
vC iL
Solution
Known Quantities: Resistor, capacitor, inductor values Find: The inductor current as a function of time
L = 1 H
Schematics, Diagrams, Circuits, and Given Data: R1 = 8 k ; R2 = 8 k ; C = 10 F; Assumptions: None Analysis: To determine the natural response of the circuit, we set the arbitrary voltage
source equal to zero by replacing it with a short circuit Next, we observe that the two
5
Transient Analysis
resistors can be replaced by a single resistor, R = R1 ||R2 , and that we now are faced with a parallel RLC circuit Applying KCL at the top node, we write: dvC vC +C + iL = 0 R dt We recognize that the top node voltage is also equal to the inductor voltage, and that diL dt Next, we substitute the expression for vC in the rst equation to obtain vC = vL = L d 2 iL 1 diL 1 + + iL = 0 dt 2 RC dt LC The characteristic equation corresponding to this differential equation is: s2 + with roots s1,2 = 1 1 2RC 2 1 RC
Copyright © OnBarcode.com . All rights reserved.