Periodic signals having halfwave symmetry have the property
x
(
t
)
=

x
(
t

T
/
2
)
x
(
t
)
=

x
(
t
+
T
/
2
)
x
(
t
)
=

x
(
t

T
/
2
)
x
(
t
)
=

x
(
t
+
T
/
2
)
(1)It turns out that signals with this type of symmetry only have oddnumbered harmonics, the even harmonics are zero. To see this, lets look at the formula for the coefficients anan:
a
n
=
2
T
∫
t
0
t
0
+
T
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
=
2
T
∫
t
0
t
0
+
T
/
2
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
+
∫
t
0
+
T
/
2
t
0
+
T
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
=
2
T
I
1
+
I
2
a
n
=
2
T
∫
t
0
t
0
+
T
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
=
2
T
∫
t
0
t
0
+
T
/
2
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
+
∫
t
0
+
T
/
2
t
0
+
T
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
=
2
T
I
1
+
I
2
(2)Making the substitution τ=tT/2τ=tT/2 in I2I2 gives
I
2
=
∫
t
0
t
0
+
T
/
2
x
(
τ
+
T
/
2
)
cos
(
n
Ω
0
(
τ
+
T
/
2
)
)
d
τ
=

∫
t
0
t
0
+
T
/
2
x
(
τ
)
cos
(
n
Ω
0
(
τ
+
T
/
2
)
)
d
τ
I
2
=
∫
t
0
t
0
+
T
/
2
x
(
τ
+
T
/
2
)
cos
(
n
Ω
0
(
τ
+
T
/
2
)
)
d
τ
=

∫
t
0
t
0
+
T
/
2
x
(
τ
)
cos
(
n
Ω
0
(
τ
+
T
/
2
)
)
d
τ
(3)The quantity cos(nΩ0(τ+T/2))=cos(nΩτ+nπ)cos(nΩ0(τ+T/2))=cos(nΩτ+nπ) can be simplified using the trigonometric identity
cos
(
u
±
v
)
=
cos
(
u
)
cos
(
v
)
∓
sin
(
u
)
sin
(
v
)
cos
(
u
±
v
)
=
cos
(
u
)
cos
(
v
)
∓
sin
(
u
)
sin
(
v
)
(4)We have
cos
(
n
Ω
τ
+
n
π
)
=
cos
(
n
Ω
τ
)
cos
(
n
π
)

sin
(
n
Ω
τ
)
sin
(
n
π
)
=
(

1
)
n
cos
(
n
Ω
τ
)

0
cos
(
n
Ω
τ
+
n
π
)
=
cos
(
n
Ω
τ
)
cos
(
n
π
)

sin
(
n
Ω
τ
)
sin
(
n
π
)
=
(

1
)
n
cos
(
n
Ω
τ
)

0
(5)Therefore
I
2
=

(

1
)
n
∫
t
0
t
0
+
T
/
2
x
(
τ
)
cos
(
n
Ω
0
τ
)
d
τ
I
2
=

(

1
)
n
∫
t
0
t
0
+
T
/
2
x
(
τ
)
cos
(
n
Ω
0
τ
)
d
τ
(6)and we can write:
a
n
=
2
T
(
1

(

1
)
n
)
∫
t
0
t
0
+
T
/
2
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
a
n
=
2
T
(
1

(

1
)
n
)
∫
t
0
t
0
+
T
/
2
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
(7)From this expression we find that an=0an=0 whenever nn is even. In fact, we have
a
n
=
4
T
∫
t
0
t
0
+
T
/
2
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
,
n
,
odd
0
,
n
,
even
a
n
=
4
T
∫
t
0
t
0
+
T
/
2
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
,
n
,
odd
0
,
n
,
even
(8)A similar derivation leads to
b
n
=
4
T
∫
t
0
t
0
+
T
/
2
x
(
t
)
sin
(
n
Ω
0
t
)
d
t
,
n
,
odd
0
,
n
,
even
b
n
=
4
T
∫
t
0
t
0
+
T
/
2
x
(
t
)
sin
(
n
Ω
0
t
)
d
t
,
n
,
odd
0
,
n
,
even
(9)A good choice of t0t0 can lead to a considerable savings in time when calculating the Fourier Series of halfwave symmetric signals.
Note that halfwave symmetric signals need not have odd or even symmetry for the above formulae to apply. If a signal has halfwave symmetry and in addition has odd or even symmetry, then some additional simplification is possible. Consider the case when a halfwave symmetric signal also has even symmetry. Then clearly bn=0bn=0, and Equation 8 applies. However since the integrand in Equation 8 is the product of two even signals, x(t)x(t) and cos(nΩ0t)cos(nΩ0t), it too has even symmetry. Therefore, instead of integrating from, say, T/4T/4 to T/4T/4, we need only integrate from 0 to T/4T/4 and multiply the result by 2. Therefore the formula for anan for an even, halfwave symmetric signal becomes:
a
n
=
8
T
∫
0
T
/
4
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
,
n
,
odd
0
,
n
,
even
a
n
=
8
T
∫
0
T
/
4
x
(
t
)
cos
(
n
Ω
0
t
)
d
t
,
n
,
odd
0
,
n
,
even
(10)For an odd halfwave symmetric signals, a similar argument leads to
b
n
=
8
T
∫
0
T
/
4
x
(
t
)
sin
(
n
Ω
0
t
)
d
t
,
n
,
odd
0
,
n
,
even
b
n
=
8
T
∫
0
T
/
4
x
(
t
)
sin
(
n
Ω
0
t
)
d
t
,
n
,
odd
0
,
n
,
even
(13)