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L + v (t) L + i(t) vC (t)
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i(t ) dt vS (t) = 0
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Equation 58 is called an integro-differential 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)
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Figure 57 Second-order circuit
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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)
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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)
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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 second-order ordinary differential equation with constant coef cients As the number of energy-storage elements in a circuit increases, one can therefore
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5
Transient Analysis
expect that higher-order 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 Second-order 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 (second-order circuit) We can obtain a second equation in the capacitor voltage, vC , by applying KVL to the mesh on the right-hand 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 second-order 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 second-order 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 FIRST-ORDER CIRCUITS
First-order 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 rst-order
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 rst-order system In other domains, we also encounter rst-order systems For example, a mechanical system that has mass and damping (eg, friction), but not elasticity, will be a rst-order system A uid system with uid resistance and uid capacitance ( uid storage) will also be of rst order; an example of a rst-order uid system is a storage tank with a valve In thermal systems, we also encounter rst-order systems quite frequently: The ability to store heat (heat capacity) and to dissipate it leads to a rst-order thermal system; heating and cooling of bodies is, at its simplest level, described by rst-order behavior In the present section we analyze the transient response of rst-order circuits In what follows, we shall explain that the initial condition, the steady-state solution, and the time constant of the rst-order system are the three quantities that uniquely determine its response Natural Response of First-Order 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 rst-order 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 rst-order circuit must be exponential in nature, that is, that xN (t) = Ke at = Ke t/ (515) (513)
+ _ vS (t)
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