Time-domain analysis of LTIC systems - Continuous and Discrete Time Signals and Systems - page 115

0.4 0.4

0.35 0.35

0.3 0.3

0.25 0.25

0.2 0.2

0.15 0.15

0.1 0.1

0.05 0.05

t

t

0 0

−2 −1 0 1 2 3 4 5 6 7 8 9

−2 −1 0 1 2 3 4 5 6 7 8 9

(b)

(a)

Fig. 3.4. (a) Impulse response
h ( t ) of the LTIC system specified
in Example 3.5. (b) Output y ( t )
of the LTIC system for input
x ( t ) =

where constant c is determined from the zero initial condition. Substituting
h ( t ) = 0 for t

−

, in Eq. (3.30) gives c = 0. The impulse response of the
system in Eq. (3.28) is therefore given by h ( t ) = 2 exp( − 4 t ) u ( t ).

= 0

δ( t + 1) + 3δ( t − 2) +

2δ( t − 6) .

Example 3.5
The impulse response of an LTIC system is given by h ( t ) = exp( − 3 t ) u ( t ).
Determine the output of the system for the input signal x ( t ) = δ ( t + 1) +
3 δ ( t − 2) + 2 δ ( t − 6).

Solution
Because the system is LTIC, it satisfies the linearity and time-shifting properties.
Therefore,

δ ( t + 1) → h ( t + 1) ,
3 δ ( t − 2) → 3 h ( t − 2) ,

and

2 δ ( t − 6) → 2 h ( t − 6) .

Applying the superposition principle, we obtain

x ( t ) → y ( t ) = h ( t + 1) + 3 h ( t − 2) + 2 h ( t − 6) .

The impulse response h ( t ) is shown in Fig. 3.4(a) with the resulting output
shown in Fig. 3.4(b).