Tutorials - Digital Signal Processing: An Introduction with MATLAB and Applications - page 238

Case 4: 2 t 5.

≤ ≤

2

s ( t − λ )

r ( λ )

λ

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

t − 2 t + 2

Case 5: If the start of s(t - k) [i.e., k = t - 2] moves past the end of r(k) [i.e.,

k = 3], no overlap occurs. The condition here is written as t - 2 [ 3,
or equivalently t [ 5. Here y(t) = 0.

Case 5: t ≥ 5, no overlap.

2

r ( λ ) s ( t − λ )

λ

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7

t − 2 t + 2

Summary of Results

8
<

9
= t 2 = 4 þ t þ 1 ;

2 t 1

ð 3 = 2 Þ t þ 3 = 4 ; 1\t\2

y ð t Þ¼

ð 15 = 4 Þþ t = 2 t 2 = 4 ;

: 2 t 5 ;

0 elsewhere

y ð t Þ¼ s ð t Þ r ð t Þ¼ R 1 1 s ð k Þ r ð t k Þ dk.

Q: Solve the above problem as You

should get the same answer, since convolution is commutative.

y ( t ) = r ( t ) * s ( t )

2

t

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6