Time-domain analysis of LTIC systems - Continuous and Discrete Time Signals and Systems - page 123

Fig. 3.9. Output y ( t ) computed
in Example 3.8.

0.4

0.3

0.2

0.1

t

0

−1 0 1 2 3 4 5 6 7 8 9

−2

Solution
Functions x ( τ ), h ( τ ), h ( −τ ), and h ( t − τ ) are plotted in Figs. 3.10(a)-(d).
Depending on the value of t , the convolution integral takes five different forms.
We consider these five cases below.

Case 1 ( t < − 3). As seen in Fig. 3.10(e), the non-zero parts of h ( t − τ ) and
x ( τ ) do not overlap. Therefore, the output signal y ( t ) = 0.

Case 2 ( − 3 ≤ t ≤ 0). As seen in Fig. 3.10(f), the non-zero part of h ( t − τ )
partially overlaps with x ( τ ) within the region τ = [ − 2 , t + 1]. The product
x ( τ ) h ( t − τ ) becomes a rectangular function in the region with an amplitude
of 1 . 5 2 = 3. Therefore, the output for − 3 ≤ t ≤ 0isgivenby

t + 1

y ( t ) = 3d τ = 3( t + 3) .

− 2

Case 3 (0 ≤ t ≤ 2). As seen in Fig. 3.10(g), the non-zero part of h ( t − τ )
overlaps completely with x ( τ ). The overlapping region is given by τ =
[ t − 2 , t + 1]. The product x ( τ ) h ( t − τ ) is a rectangular function with an ampli-
tude of 3 in the region τ

= [ t − 2 , t + 1]. The output for 0 ≤ t ≤ 2 is given by

t + 1

y ( t ) = 3d τ = 9 .

t − 2

Case 4 (2 ≤ t ≤ 5). The non-zero part of h ( t − τ ) overlaps partially with x ( τ )
within the region τ

= [ t − 2 , 3]. Therefore, the output for 2 ≤ t ≤ 5isgiven

by

3

y ( t ) = 3d τ

= 3(5 − t ) .

t − 2

Case 5 ( t ≥ 0). We see from Fig. 3.10(i) that the non-zero parts of h ( t − τ )
and x ( τ ) do not overlap. Therefore, the output y ( t ) = 0.