Tutorials - Digital Signal Processing: An Introduction with MATLAB and Applications - page 236

r ( λ )

1

λ

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

2

s ( − λ ) s ( λ )

λ

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

s ( t − λ )

2

λ

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

t − 2 t + 2

(start) (end)

Step 3: Integrate to find y(t). Consider the intervals of overlap between r(k) and

s(t - k). These intervals can be defined using the endpoints of r(k)
(which is fixed here) and s(t - k) (which is moving according to the
shift t, noting that positive t gives positive shift).

Case 1: End of s(t - k) \ Start of r(k) ) t +2 \ 0 ) t \ -2. No overlap,

hence, y(t) = 0.

Case 1: t < − 2, no overlap.

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 (start−1) (end−1)

(start−2) (end−2)

Case 2: As the end of s(t - k) [i.e., k = t + 2] moves right (when t increases),

s(t - k) moves right as well. The first overlap with r(k) occurs when the
end of s(t - k) [i.e., k = t + 2] exceeds the start of r(k) [i.e., k = 0].
Now consider that the end of s(t - k) is inside r(k), hence,
0 B t +2B 3, or equivalently -2 B t B 1. In this case we have:
y ¼ Z

t þ 2

dk ¼ t 2 = 4 þ t þ 1 ½ 1

|{z}
r ð k Þ

½ð t k Þ= 2 þ 1
|{z}
s ð t k Þ

0

(where -2 B t B 1, as shown above).