Signal Circular Convolution
Given a signal
ft
f
t
with Fourier coefficients
c
n
c
n
and a signal
gt
g
t
with Fourier coefficients
d
n
d
n
,
we can define a new signal,
vt
v
t
,
where
vt=ft⊛gt
v
t
⊛
f
t
g
t
We find that the
Fourier
Series representation of
yt
y
t
,
a
n
a
n
,
is such that
a
n
=
c
n
d
n
a
n
c
n
d
n
.
ft⊛gt
⊛
f
t
g
t
is the
circular convolution
of two periodic signals and is equivalent to the convolution
over one interval,
i.e.
ft⊛gt=∫0T∫0Tfτgt-τdτdt
⊛
f
t
g
t
t
0
T
τ
0
T
f
τ
g
t
τ
.
note:
Circular convolution in the time domain is equivalent to
multiplication of the Fourier coefficients.
This is proved as follows
a
n
=1T∫0Tvtⅇ-ⅈ
ω
0
ntdt=1T2∫0T∫0Tfτgt-τdτⅇ-ⅈ
ω
0
ntdt=1T∫0Tfτ1T∫0Tgt-τⅇ-ⅈ
ω
0
ntdtdτ=∀ν,ν=t-τ:1T∫0Tfτ1T∫-τT-τgνⅇ-ⅈ
ω
0
ν+τdνdτ=1T∫0Tfτ1T∫-τT-τgνⅇ-ⅈ
ω
0
nνdνⅇ-ⅈ
ω
0
nτdτ=1T∫0Tfτdnⅇ-ⅈ
ω
0
nτdτ=
d
n
1T∫0Tfτⅇ-ⅈ
ω
0
nτdτ=
c
n
d
n
a
n
1
T
t
0
T
v
t
ω
0
n
t
1
T
2
t
0
T
τ
0
T
f
τ
g
t
τ
ω
0
n
t
1
T
τ
0
T
f
τ
1
T
t
0
T
g
t
τ
ω
0
n
t
ν
ν
t
τ
1
T
τ
0
T
f
τ
1
T
ν
τ
T
τ
g
ν
ω
0
ν
τ
1
T
τ
0
T
f
τ
1
T
ν
τ
T
τ
g
ν
ω
0
n
ν
ω
0
n
τ
1
T
τ
0
T
f
τ
d
n
ω
0
n
τ
d
n
1
T
τ
0
T
f
τ
ω
0
n
τ
c
n
d
n
(1)
Example 1
Take a look at a square pulse with a period,
T
1
=T4
T
1
T
4
:
For this signal
c
n
=1Tifn=012sinπ2nπ2notherwise
c
n
1
T
n
0
1
2
2
n
2
n
Problem 1
What signal has Fourier coefficients
a
n
=
c
n
2=14sin2π2nπ2n2
a
n
c
n
2
1
4
2
n
2
2
n
2
?
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Solution 1
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