Periodic signals having half-wave 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 odd-numbered 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=t-T/2τ=t-T/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 half-wave symmetric signals.
Note that half-wave symmetric signals need not have odd or even symmetry for the above formulae to apply. If a signal has half-wave symmetry and in addition has odd or even symmetry, then some additional simplification is possible. Consider the case when a half-wave 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/4-T/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, half-wave 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)
b
n
=
0
b
n
=
0
(11)
For an odd half-wave symmetric signals, a similar argument leads to
a
n
=
0
a
n
=
0
(12)
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)
