Half-Wave Symmetry
ECE 3100
John Stahl
Western Michigan University
Half-Wave Triangle
Solve for the Fourier Coefficients and plot the resulting sum. The period
(T) of the sequence is 2 sec
𝑓 𝑡 =
∞
6𝑡,
0≤𝑥<1
& 1−𝑡 ,
6
1≤𝑥<2
𝑓 𝑡 = & 𝑎0 +
𝑎𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝑏𝑛 sin 𝑛𝜔𝑜 𝑡
𝑛=1
𝑓 𝑡
-3
-2
-1
0
1
2
3
T
𝑎𝑜 = 0
4
𝑎𝑛 =
𝑇
𝑇
2
0
𝑓 𝑡 cos 𝑛𝑜𝑑𝑑 𝜔𝑜 𝑡 𝑑𝑡
4
𝑏𝑛 =
𝑇
𝑇
2
0
𝑓 𝑡 sin 𝑛𝑜𝑑𝑑 𝜔𝑜 𝑡 𝑑𝑡
By inspect the average value of the wave form is
zero, but we will complete the integral to
demonstrate this.
1
𝑎𝑜 =
𝑇
𝑇
𝑇=2
𝑓 𝑡 𝑑𝑡
0
1
=
2
1
0
1
6𝑡 ∙ 𝑑𝑡 +
2
2
6 1 − 𝑡 ∙ 𝑑𝑡
1
1
=3
2
𝑡 ∙ 𝑑𝑡 + 3
0
2
1 ∙ 𝑑𝑡 − 3
1
𝑡 ∙ 𝑑𝑡
1
3 21
3 22
2
= 𝑡 |0 + 3𝑡|1 − 𝑡 |1
2
2
=
3
3
+6−3−6+
2
2
𝑎𝑜 = 0
From the definition a function with Half-wave symmetry only
exhibits odd harmonics. We will derive this for the function.
2
𝑎𝑛 =
𝑇
𝑇
0
𝑓 𝑡 cos 𝑛𝜔𝑜 𝑡 𝑑𝑡
2
=
2
=
2𝜋
𝜔𝑜 =
=𝜋
𝑇
1
0
2
6𝑡 ∙ cos 𝑛𝜔𝑜 𝑡 𝑑𝑡 +
2
2
1
6 1 − 𝑡 ∙ cos 𝑛𝜔𝑜 𝑡 𝑑𝑡
6
6
cos 𝜋𝑛𝑡 2
1
cos
𝜋𝑛𝑡
+
𝜋𝑛𝑡
∙
sin
𝜋𝑛𝑡
|
−
−sin
𝜋𝑛𝑡
+
𝑡
sin
𝜋𝑛𝑡
+
|1
0
(𝜋𝑛)2
𝜋𝑛
𝜋𝑛
6
6
=
𝜋𝑛 ∙ sin 𝜋𝑛 + cos 𝜋𝑛 − 1 −
𝜋𝑛 ∙ sin 2𝜋𝑛 − cos 𝜋𝑛 + cos 2𝜋𝑛
(𝜋𝑛)2
(𝜋𝑛)2
6
=
2cos 𝜋𝑛 − 2
(𝜋𝑛)2
n
cos 𝜋𝑛 − 1
1
cos 𝜋𝟏 − 1
−2
2
cos 𝜋𝟐 − 1
0
3
cos 𝜋𝟑 − 1
−2
4
cos 𝜋𝟒 − 1
0
5
cos 𝜋𝟓 − 1
−2
⋮
⋮
⋮
𝑛𝑜𝑑𝑑
−2
−24
𝑎𝑛 =
(𝜋𝑛𝑜𝑑𝑑 )2
If we had used the short-cut equation for half-wave symmetry we would arrive at the
same result.
2
𝑏𝑛 =
𝑇
𝑇
0
𝑓 𝑡 sin 𝑛𝜔𝑜 𝑡 𝑑𝑡
2
=
2
=
=
1
0
2
6𝑡 ∙ sin 𝑛𝜔𝑜 𝑡 𝑑𝑡 +
2
2
1
6 1 − 𝑡 ∙ sin 𝑛𝜔𝑜 𝑡 𝑑𝑡
6
6
sin 𝜋𝑛𝑡 2
1
sin
𝜋𝑛𝑡
−
𝜋𝑛𝑡
∙
cos
𝜋𝑛𝑡
|
−
cos
𝜋𝑛
−
𝜋𝑛
∙
cos
𝜋𝑛
+
|1
0
(𝜋𝑛)2
𝜋𝑛
𝜋𝑛
6
6
sin
𝜋𝑛
−
𝜋𝑛
∙
cos
𝜋𝑛
+
sin 𝜋𝑛 − sin 2𝜋𝑛 + 𝜋𝑛 ∙ cos 2𝜋𝑛
(𝜋𝑛)2
(𝜋𝑛)2
0 for all n
0 for all n
0 for all n
=
6
cos 2𝜋𝑛 − cos 𝜋𝑛
𝜋𝑛
n
cos 2𝜋𝑛 − cos 𝜋𝑛
1
cos 2𝜋𝟏 − cos 𝜋𝟏
2
2
cos 2𝜋𝟐 − cos 𝜋𝟐
0
3
cos 2𝜋𝟑 − cos 𝜋𝟑
2
4
cos 2𝜋𝟒 − cos 𝜋𝟒
0
5
cos 2𝜋𝟓 − cos 𝜋𝟓
2
⋮
⋮
⋮
𝑛𝑜𝑑𝑑
2
12
𝑏𝑛 =
𝜋𝑛𝑜𝑑𝑑
If we had used the short-cut equation for half-wave symmetry we would arrive at the
same result.
∞
𝑓 𝑡 =0+
𝑛𝑜𝑑𝑑 =1
−24
12
cos
𝜋𝑛𝑡
+
sin 𝜋𝑛𝑡
(𝜋𝑛𝑜𝑑𝑑 )2
𝜋𝑛𝑜𝑑𝑑