152 TIME-DEPENDENT CIRCUIT ANALYSIS

Substitution ofI(s) and solving forVBgives

VB=

10 ( 3 s+ 1 )(s+ 1 )

3 s

(

s^2 + 3 s+ 2

)

VA=

1 / 2

1 / 2 +s/ 2

VB=

1

s+ 1

VB=

10 ( 3 s+ 1 )

3 s(s+ 1 )(s+ 2 )

=

K 1

s

+

K 2

s+ 1

+

K 3

s+ 2

where

K 1 =

10 ( 0 + 1 )

3 ( 0 + 1 )( 0 + 2 )

=

5

3

K 2 =

10 (− 3 + 1 )

(^3) (− (^1) )(− 1 + (^2) )

20
3
K 3 =
10 (− 6 + 1 )
3 (− 2 )(− 2 + 1 )
=−
25
3
Therefore,
v(t)=
(
5
3

• 20
  3
  e−t−
  25
  3
  e−^2 t
  )
  u(t)V
  Note that Thévenin’s and Norton’s network theorems are applicable in the frequency domain.
  Network functions(also known assystem functions) are defined as the ratio of the response to
  the excitation in the frequency domain, as illustrated in Figure 3.3.3. Driving-point impedances
  and admittances are network functions, as shown in Figure 3.3.4.
  The concept oftransfer functionsis illustrated in Figure 3.3.5, in which the response measured
  at one pair of terminals is related to an excitation applied to another pair of terminals.
  x(t)
  x(t) y(t)
  (a)
  Linear system
  Excitation Differential
  equation
  y(t)
  Response
  X(s)
  H(s) =Y(s)
  X(s)
  (b)
  Linear system
  H(s)
  Y(s)
  Figure 3.3.3Linear system with no initial energy storage.(a)Time domain.(b)Frequency domain.
  Driving-point impedance Z(s) = V(s) / I(s)
  Driving-point admittance Y(s) = I(s) / V(s)


• 
  

  −
  V(s)
  I(s)
  Network
  Figure 3.3.4Network functions.