Tutorials - Digital Signal Processing: An Introduction with MATLAB and Applications - page 237

Case 2: − 2 ≤ t ≤ 1.

s ( t − λ )

2

r ( λ )

1

λ

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

t − 2 t + 2

Case 3: As the end of s(t - k) [i.e., k = t + 2] exceeds the end of r(k) [i.e.,

k = 3], s(t - k) will fully overlap r(k) until the start of s(t - k) [i.e.,
k = t - 2] reaches the start of r(k) [i.e., k = 0]. Hence, we consider
two conditions: t +2[ 3 and t - 2 \ 0, which give t [ 1 and t \ 2, or
equivalently 1 \ t \ 2. In this case we have:

y ¼ Z

3

½ 1
|{z}
r ð k Þ

½ð t k Þ= 2 þ 1
| {z }
s ð t k Þ

dk ¼ð 3 = 2 Þ t þ 3 = 4

0

(where 1 \ t \ 2, as shown above).

Case 3: 1 < t < 2.

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 4: As the start of s(t - k) [i.e., k = t - 2] moves beyond the start of r(k)

[i.e., k = 0], overlap takes place only between k = t - 2[start of
s(t - k)] and k = 3[end of r(k)]. Hence, we consider the condition
0 B t - 2 B 3, or equivalently, 2 B t B 5. Here we have:

y ¼ Z

3

dk ¼ 15 = 4 þ t = 2 t 2 = 2

½ 1
|{z}
r ð k Þ

½ð t k Þ= 2 þ 1
|{z}
s ð t k Þ

t 2

where 2 B t B 5, as shown above).