barcode reader code in asp.net Part I in Software

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Part I
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Circuits
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to form a closed circuit2 The inductor current will now continue to ow through the resistor, which dissipates the energy stored in the inductor By the reasoning in the preceding discussion, the inductor current will decay exponentially: iL (t) = IB e tR/L (521)
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That is, the inductor current will decay exponentially from its initial condition, with a time constant = L/R Example 53 further illustrates the signi cance of the time constant in a rst-order circuit
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EXAMPLE 53 First-Order Systems and Time Constants
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Problem
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Create a table illustrating the exponential decay of a voltage or current in a rst-order circuit versus the number of time constants
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Solution
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Known Quantities: Exponential decay equation Find: Amplitude of voltage or current, x(t), at t = 0, , 2 , 3 , 4 , 5 Assumptions: The initial condition at t = 0 is x(0) = X0 Analysis: We know that the exponential decay of x(t) is governed by the equation:
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x(t) = X0 e t/ Thus, we can create the following table for the ratio x(t)/X0 = e n / , n = 0, 1, 2, , at each value of t:
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x(t) X0
n 0 1 2 3 4 5
1 08 x/X 0 06 04 02 0 0 1 2 3 4 Time constants 5
1 03679 01353 00498 00183 00067
Figure 512 depicts the ve points on the exponential decay curve
Comments: Note that after three time constants, x has decayed to approximately 5
percent of the initial value, and after ve time constants to less than 1 percent
Figure 512 First-order exponential decay and time constants
2 Note
that in theory an ideal current source cannot be connected in series with a switch For the purpose of this hypothetical illustration, imagine that upon opening the right-hand-side switch, the current source is instantaneously connected to another load, not shown
5
Transient Analysis
EXAMPLE 54 Charging a Camera Flash Time Constants
Problem
t=0 R C + vC
A capacitor is used to store energy in a camera ash light The camera operates on a 6-V battery Determine the time required for the energy stored to reach 90 percent of the maximum Compute this time in seconds, and as a multiple of the time constant The equivalent circuit is shown in Figure 513
+ VB i
Solution
Known Quantities: Battery voltage; capacitor and resistor values Find: Time required to reach 90 percent of the total energy storage Schematics, Diagrams, Circuits, and Given Data: VB = 6 V; C = 1,000 F;
Figure 513 Equivalent circuit of camera ash charging circuit
R=1k
Assumptions: Charging starts at t = 0, when the ash switch is turned on The capacitor is completely discharged at the start Analysis: First, we compute the total energy that can be stored in the capacitor:
2 2 Etotal = 1 CvC = 1 CVB = 18 10 3 2 2
Thus, 90 percent of the total energy will be reached when Etotal = 09 18 10 3 = 162 10 3 J This corresponds to a voltage calculated from
1 2 CvC 2
= 162 10 3 2 162 10 3 = 5692 C V
vC =
Next, we determine the time constant of the circuit: = RC = 10 3 103 = 1 s; and we observe that the capacitor will charge exponentially according to the expression vC = 6 1 e t/ = 6 1 e t To compute the time required to reach 90 percent of the energy, we must therefore solve for t in the equation vC -90% = 5692 = 6 1 e t 0949 = 1 e t 0051 = e t t = loge (0051) = 297 s
The result corresponds to a charging time of approximately 3 time constants
Comments: This example demonstrates the physical connection between the time
constant of a rst-order circuit and a practical device If you wish to practice some of the calculations related to time constants, you might calculate the number of time constants required to reach 95 percent and 99 percent of the total energy stored in a capacitor
Part I
Circuits
Forced and Complete Response of First-Order Circuits In the preceding section, the natural response of a rst-order circuit was found by setting the forcing function equal to zero and considering the energy initially stored in the circuit as the driving force The forced response, xF (t), of the inhomogeneous equation dxF (t) 1 (522) + xF (t) = f (t) dt is de ned as the response to a particular forcing function f (t), without regard for the initial conditions3 Thus, the forced response depends exclusively on the nature of the forcing function The distinction between natural and forced response is particularly useful because it clari es the nature of the transient response of a rstorder circuit: the voltages and currents in the circuit are due to the superposition of two effects, the presence of stored energy (which can either decay, or further accumulate if a source is present) and the action of external sources (forcing functions) The natural response considers only the former, while the forced response describes the latter The sum of these two responses forms the complete response of the circuit: x(t) = xN (t) + xF (t) (523)
The forced response depends, in general, on the form of the forcing function, f (t) For the purpose of the present discussion, it will be assumed that f (t) is a constant, applied at t = 0, that is, that f (t) = F t 0 (524)
(Note that this is equivalent to turning a switch on or off) In this case, the differential equation describing the circuit may be written as follows: dxF xF = +F t 0 (525) dt For the case of a DC forcing function, the form of the forced solution is also a constant Substituting xF (t) = XF = constant in the inhomogeneous differential equation, we obtain 0= or XF = F Thus, the complete solution of the original differential equation subject to initial condition x(t = 0) = x0 and to a DC forcing function F for t 0 is x(t) = xN (t) + xF (t) or x(t) = Ke t/ + F
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