Let's find the Fourier series representation for the half-wave
rectified sinusoid.
st=sin2πtTif0≤t<T20ifT2≤t<T
s
t
2
t
T
0
t
T
2
0
T
2
t
T
(1)
Begin with the sine terms in the series; to find
b
k
b
k
we must calculate the integral
b
k
=2T∫0T2sin2πtTsin2πktTdt
b
k
2
T
t
0
T
2
2
t
T
2
k
t
T
(2)
The key to evaluating such integrals is the classic
trigonometric identities.
sinαsinβ=12cosα-β-cosα+β
α
β
1
2
α
β
α
β
(3)
cosαcosβ=12cosα+β+cosα-β
α
β
1
2
α
β
α
β
(4)
sinαcosβ=12sinα+β+sinα-β
α
β
1
2
α
β
α
β
(5)
Using these identities turns our integral of a product of
sinusoids into a sum of integrals of individual sinusoids,
which are much easier to evaluate.
2T∫0T2sin2πtTsin2πktTdt=1T∫0T2cos2πk-1tT-cos2πk+1tTdt=12ifk=10if{k|k≥2∧k∈ℕ}
2
T
t
0
T
2
2
t
T
2
k
t
T
1
T
t
0
T
2
2
k
1
t
T
2
k
1
t
T
1
2
k
1
0
k
k
2
k
(6)
Thus,
b
1
=12
b
1
1
2
(7)
b
2
=
b
3
=…=0
b
2
b
3
…
0
(8)
On to the cosine terms. The average value, which corresponds
to
a
0
a
0
, equals
1π
1
The remainder of the cosine coefficients are easy to find, but
yield the complicated result
a
k
=-2π1k2-1ifk∈24…0ifk is odd
a
k
2
1
k
2
1
k
2
4
…
0
k is odd
(9)
Thus, the Fourier series for the half-wave rectified sinusoid
has non-zero terms for the average, the fundamental, and the
even harmonics. Plotting the Fourier coefficients reveals at
what component frequencies the half-wave rectified sinusoid
has energy ( Figure 1 ).
Furthermore, this figure shows what the Fourier series sum
looks like with these coefficients as we add more and more
terms. Presumably, you now believe more in the Fourier series.