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Unformatted text preview: = X + Y. Solution: We have fS = fX ∗ fY , and we will use Figure 4.15 to help work out the integral in
(4.16). As shown, the graph of the function fY (c − u) is a rectangle over the interval [c − 1, c].
f (u) f (c!u) fX (u) X Y f (c!u)
Y u
c !1 0 c u
c !1 1 0 1 c Figure 4.15: Calculating the convolution of two rectangle functions.
To check this, note that if u = c then fY (c − u) = fY (0), and as u decreases from c to c1, c − u
increases from zero to c, so fY (c − u) = 1 for this range of u. The product fX (u)fY (c − u) is equal
to one on [0, 1] ∪ [c − 1, c] and is equal to zero elsewhere. As shown in Figure 4.15, if 0 < c ≤ 1,
then the overlap is the interval [0, c], and therefore fX ∗ fY (c) = c. If 1 < c < 2, the overlap is the
interval [c − 1, 1] which has length 2 − c, and therefore fX ∗ fY (c) = 2 − c. Otherwise, the value of
the convolution is zero. Summarizing, 0<c≤1
c
2−c 1<c≤2
fS (c) = 0
else.
so the graph of fS has the triangular shape shown in Figure 4.16.
f (c)
S 1 c
0 1 2 Figure 4.16: The convoluti...
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 Spring '08
 Zahrn
 The Land

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