Figure 1.

Riω=∫−∞∞rte−(iωt)d
t
=∫−TT1e−(iωt)d
t
=e−(iωt)−(iω)|−TT=(−1iω)(e−(iωT)−eiωT)
R
ω
t
r
t
ω
t
t
T
T
1
ω
t
T
T
ω
t
ω
1
ω
ω
T
ω
T

(1)
since

e−(iωT)−eiωT=−(2isinωT)
ω
T
ω
T
2
ω
T
we get:

Riω=2sinωTω=2TsinωTωT
R
ω
2
ω
T
ω
2
T
ω
T
ω
T

(2)
which is a sinc!

Moving average impulse responce,
p. B560.

Figure 2.