Unformatted text preview: First—Order Differential Equations Review We consider ﬁrstorder differential equations of the form:
d x t 1
0 xtf)=f(t) (1) +
Whereﬂt) is the forcing function. In general, the differential equation has two solutions: dt ? 1. complementary (or natural or homogeneous) solution, xdt) (When/(t) = 0), and
2. particular (or forced or nonhomogeneous) solution, x;(t) (whenf(t‘) at D). In our problems, ﬂ!) is often a constant, and therefore, the overall solution to the differential
equation is J«7(t‘)=xc(f)+xp(f)=K1P7”r +K2 (2)
The time constant, 1:, for ﬁrstorder electrical circuits is either T = RC or ‘E = UK Complementary Solution
The complementary solution is found by considering the homogeneous equation: dx(t) l _
dr +rx(t)—0 (3) The complementary solution is the system’s natural response, which is
xc (t) = K1 2"”
This result can be verified by substituting this answer into the differential equation, Eq. (3). Particular Solution
The particular solution is found by considering the full (nonhomogeneous) differential equation,
that is, Eq. (1). If the forcing function is a constant, then xP(t) is a constant (K2) also, and hence “yd, = 0. Substituting xP into the original differential equation, Eq. (1), yields: dxp l l
— + —x = 0 + —x =
dt 2' P 1' P f
Reducing the above expression pr'OVides the particular solution, which is the forced response:
x; = K2 = t f Note that, alternatively, K3 could have been found using Eq. (2) while considering the fnral value
reached by the variable of interest, and hence, in this case is also the steadystate response: x(°0) =Kl e‘” +K2 =K2 Total Solution
The total solution is the sum of the complementary and particular solutions: xr(t)=xc(t)+xp =K1e_’” +1'f (4)
The coefficient term, K1, is found from the initial conditions (at t= 0). Initial Conditions
Typically, the initial conditions are determined by considering that for dc steadystate conditions: (a) a capacitor is like an open circuit (ic = C ”V4, = 0 ), and (b) an inductor appears as a short circuit ("1. = L (ii/‘1’ = 0 ). Further, the cruient through an inductor and the voltage across a capacitor cannot change instantaneously, for example, i,_ (0—) = iL (0+) and vc (0—) = vc (0+) . SecondOrder Differential Equations Review The secondorder differential equations of interest are of the form: d 2 d ‘
_d12y +2gw°7i+w§ y(t)=f(t) (5)
wheref(t) is the forcing function. Like the ﬁrstorder case, this differential equation has two
solutions: (1) complementary (or natural) solution, ydt) when ﬁt) = 0, and (2) particular (or
forced) solution, yp(t) whenﬂt) # 0. In our problems, ﬁt) is often a constant, and therefore, the
overall solution to the differential equation is typically y(t)=yc(t)+y,,(t)=K1e"'+K2 e‘='+K3 (6) Complementary Solution The complementary solution is found by considering the homogeneous equation:
’1 %+a:—f+by(t)=0 (7)
where a = 2 gwo and b = mg. The complementary solution is the system’s natural response:
J’c(t)=Kr er't‘iKz 95" (8)
The natural ﬁ'equencies, :1 and 52, are the roots of the characteristic equation:
52 + a s +b = 0 (9) These roots lead to three possible cases: Roots (51, 32) Damping Real and unequal Overdamped Real and equal Critically damped
Complex conjugates Underdonated Natural Response
Sum of two decaying exponentials
Two faster decaying exponentials
Exponentially damped sinusoid Particular Solution
The particular solution is found by considering the full (nonhomogeneous) differential equation,
Eq. (5). If the forcing function is a constant, then yp(t) is also a constant (K3), and hence 2
dy" d t = 0 = d y /dt 2 . Substituting yp into the differential equation yields:
d 2 y dy
—+a—+b 't =O+a0+bt =
W dr y( ) r p f
Reducing the above expression provides the particular solution, which is the forced response
f
= K = _
yr 3 b Alternatively, K3 could have been determined from Eq. (6) and the ﬁnal value y(oo)t Initial Conditions
Initial conditions for y and its derivative are necessary to find K1 and K2; speciﬁcally needed are y(0) and 33L). A numerical value for f? M is generally found by reapplying KV'L or KCL at t=0r Practically speaking, as the KVL or KCL is performed, the inductor voltage and capacitor dv
and C C
A, is then equated to an analytical derivative of y(!) determined from Eq. (6). The initial conditions
for y(0) are found in the same manner as described for ﬁrstorder differential equations. current, respectively, are expressed as L d” dt . The numerical value of ,, [=0
{=0 ...
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 Spring '07
 Fuchs
 Physics

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