0195136047.pdf

126 TIME-DEPENDENT CIRCUIT ANALYSIS

however, that the functionestused here in finding transient solutions is not the generalized phasor

function used earlier for exponential excitations. Thus let us try a solution of the form

xtr(t)=Aest (3.2.8)

in whichAandsare constants yet to be determined. Substituting Equation (3.2.8) into Equation

(3.2.6), we get

sAest+aAest=0or(s+a)Aest= 0 (3.2.9)

which implies

s+a=0ors=−a (3.2.10)

sinceAcannot be zero (orxtrwould be zero for alltcausing a trivial solution), andestcannot be

zero for alltregardless ofs. Thus the transient solution is given by

xtr(t)=Ae−at (3.2.11)

in whichAis a constant yet to be determined.

We already know how to find the steady-state solutionxss(t) for dc or ac sources, and the

principle of superposition can be used if we have more than one source. To find that part of

the steady-state solution due to a dc source, deactivate or zero all the other sources, and replace

inductors with short circuits, and capacitors with open circuits, and then solve the resulting circuit

for the voltage or current of interest. To find that part of the steady-state solution due to an ac

source, deactivate or zero all the other sources and use the phasor method of Section 3.1. The

solution to Equation (3.2.3) or Equation (3.2.4) is then given by Equation (3.2.5) as the sum of

the transient and steady-state solutions,

x(t)=xtr(t)+xss(t)=Ae−at+xss(t) (3.2.12)

In order to evaluateA, let us apply Equation (3.2.12) fort= 0 +which is immediately aftert=0,

x( 0 +)=A+xss( 0 +) or A=x( 0 +)−xss( 0 +) (3.2.13)

wherex( 0 +)is the value ofx(t) at the initial time and denotes theinitial conditiononx(t), and

xss( 0 +)would be found fromxss(t) corresponding tot=0. Thus the total solution forx(t) for all

t≥0 is given by

x(t)=︸[x( 0 +)−x︷︷ss( 0 +)]e−at︸

(transient response)

+ x︸ss︷︷(t)︸

(steady-state response)

=︸[x( 0 +︷︷)e−at︸]

(natural response)

−︸xss( 0 +)e︷︷−at+xss(t)︸

(forced response)

(3.2.14)

Note that the transient solution involvinge−ateventually goes to zero as time progresses (assuming

ato be positive). At a timet= 1 /a, the transient solution decays to 1/e, or 37% of its initial value

att= 0 +. The reciprocal ofais known as the time constant with units of seconds,

τ= 1 /a (3.2.15)

Thus Equation (3.2.14) can be rewritten as

x(t)=

[

x

(

0 +

)

−xss

(

0 +

)]

e−t/τ+xss(t) (3.2.16)

which is an important result as a solution of Equation (3.2.3). Then we can easily write the solution

of Equation (3.2.2) corresponding to theRLcircuit of Figure 3.2.1, and that will involvei( 0 +),

which is the same asiL( 0 +), which is the value of the inductor current immediately after the

switch is closed.