Time-domain analysis of LTIC systems - Continuous and Discrete Time Signals and Systems - page 121

h ( t )

x ( t )

1 1

e − t u ( t ) (1 − t )

t t

0 0 1

(a) (b)

h (− t ) h ( t − t )

1 1

(1 + t ) (1 − t + t )

t t

−1 0 ( t − 1) t 0

(c) (d)

x ( t ), h ( t − t ) x ( t ), h ( t − t )

case 1: t < 0 case 2: 0 < t ≤ 1

1 1

t t

( t − 1) t ( t − 1) t

0 0

(e) (f)

x ( t ), h ( t − t )

case 3: t > 1

1

Fig. 3.8. Convolution of the
input signal x ( t ) with the
impulse response h ( t )in
Example 3.8. Parts (a)-(g) are
discussed in the text.

t

( t − 1) t

0

(g)

Case 1 For t < 0, we see from Fig. 3.8(e) that the non-zero parts of h ( t − τ )
and x ( τ ) do not overlap. In other words, output y ( t ) = 0 for t < 0.

Case 2 For 0 ≤ t ≤ 1, we see from Fig. 3.8(f) that the non-zero parts of h ( t − τ )
and x ( τ ) do overlap over the duration τ = [0 , t ]. Therefore,

t t

−τ (1 − t + τ )d τ

y ( t ) = x ( τ ) h ( t − τ )d τ = e

0 t − 1

t t

−τ d τ −τ d τ

= (1 − t ) e + τ e .

0 0

integral I integral II