Time-domain analysis of LTIC systems - Continuous and Discrete Time Signals and Systems - page 122

The two integrals simplify as follows:

−τ ] t 0 − t );

integral I = (1 − t )[ − e = (1 − t )(1 − e

−τ −τ ] t 0 − t − t .

integral II = [ −τ e − e = 1 − e − t e

For 0 ≤ t ≤ 1, the output y ( t )isgivenby

− t − t ) + (1 − e − t − t ) = (2 − t − 2e − t ) .

y ( t ) = (1 − t − e + t e − t e

Case 3 For t > 1, we see from Fig. 3.8(g) that the non-zero part of h ( t − τ )
completely overlaps x ( τ ) over the region τ = [ t − 1 , t ]. The lower limit of
the overlapping region in case 3 is different from the lower limit of the over-
lapping region in case 2; therefore, case 3 results in a different convolution
integral and is considered separately from case 2. The output y ( t ) for case 3 is
given by

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 .

t − 1 t − 1

integral I integral II

The two integrals simplify as follows:

−τ ] t − 1 − ( t − 1) − e − t );

integral I = (1 − t )[ − e = (1 − t )(e

−τ −τ ] t − 1 − ( t − 1) + e − ( t − 1) − t e − t − e − t

integral II = [ −τ e − e = ( t − 1)e

− ( t − 1) − t e − t − e − t .

= t e

For t > 1, the output y ( t )isgivenby

− ( t − 1) − t e − ( t − 1) − e − t − 1 − ( t − 1) − t e − t − t

y ( t ) = e + t e + t e − e

− ( t − 1) − 2e − t

= e .

Combining the above three cases, we obtain

0

t < 0

y ( t ) =

− t )

(2 − t − 2e 0 ≤ t ≤ 1

− ( t − 1) − 2e − t )

(e t > 1 ,

which is plotted in Fig. 3.9.

Example 3.9
Calculate the output for the following input signal and impulse response:

1 . 5 − 2 ≤ t ≤ 3 2 − 1 ≤ t ≤ 2

x ( t ) = and h ( t ) =

0 otherwise 0 otherwise.