154 TIME-DEPENDENT CIRCUIT ANALYSIS

K 2 =

(^2) (− 3 + (^2) )
(− 3 + 1 )(− 3 )
=−
1
3
K 3 =
2 ( 0 + 2 )
( 0 + 1 )( 0 + 3 )

4
3
Thus,
y(t)=−e−t−
1
3
e−^3 t+
4
3
The first two terms on the right-hand side are the natural response arising fromH(s),
whereas the last term is the forced response arising fromX(s). Note thatK 3 isH(s)|s= 0 ,
whileX(s)= 1 /s.
(c) Forx(t)=e−^4 t, X(s)= 1 /(s+ 4 ). Hence,
Y(s)=H(s)X(s)=H(s)
1
s+ 4

2 (s+ 2 )
(s+^1 )(s+^3 )
1
(s+^4 )

K 1
s+ 1

• K 2
  s+ 3


• 
  

  K 3
  s+ 4
  where
  K 1 =
  2 (− 1 + 2 )
  (− 1 + 3 )(− 1 + 4 )

  
  

  1
  3
  K 2 =
  2 (− 3 + 2 )
  (− 3 + 1 )(− 3 + 4 )
  = 1
  K 3 =
  2 (− 4 + 2 )
  (− 4 + 1 )(− 4 + 3 )
  =−
  4
  3
  Thus,
  y(t)=
  1
  3
  e−t+e−^3 t−
  4
  3
  e−^4 t
  The first two terms on the right-hand side are the natural response arising fromH(s),
  whereas the last term is the forced response arising fromX(s). Note thatK 3 isH(s)|s=− 4 ,
  whereasX(s)= 1 /(s+ 4 ).
  (d)Y(s)=H(s)X(s)=
  2 (s+ 2 )
  (s+ 1 )(s+ 3 )
  X(s)
  or(s+ 1 )(s+ 3 )Y(s)= 2 (s+ 2 )X(s),or
  (
  s^2 + 4 s+ 3
  )
  Y(s)=( 2 s+ 4 )X(s). Rec-
  ognizing that multiplication bysin the frequency domain corresponds to differentiation
  in the time domain, the differential equation is expressed as
  d^2 y
  dt^2

  
  


• 4
  dy
  dt


• 3 y= 2
  dx
  dt


• 4 x

3.4 Frequency Response

Earlier we examined the circuit response to sinusoids that have a fixed frequency. Now let us

examine the response of a circuit to a sinusoidal source, called anoscillator, whose frequency can

be varied. Known as thefrequency response, it is often expressed as a network function, which is an

output–input ratio. In order to visualize the changes in phase shift and amplitude as the frequency of