Fourier series coefficients for a rectified sine wave

EE 341 Fall 2004
Consider the signal
|sin (ω 1 t)|
sin (ω 1 t)
A
0
−A
A
0
−A
EE 341
Fourier Series for Rectified Sine Wave
x(t) = A| sin(ω1 t)|
Rectified Sine and Sine
−2 T −T 0 T 2 T
−T1 0 T1
The period of the sinusoid (inside the absolute value symbols) is T1 = 2π/ω1. The period of the rectified
sinusoid is one half of this, or T = T1/2 = π/ω1. Therefore,
Thus we can write
The Fourier series coefficients are:
c x k
1
=
T
= 1
T
ω1 = 2π
T1
= π
T
x(t) = A
sin T
0
T
0
= ωo
2
ωo t
2
x(t) e −jkωot dt
A
sin
ωo t
e
2
−jkωot dt
From 0 to T , |sin(ωo t/2)| = sin(ωo t/2). Also, replace ωo by 2π/T :
c x k = 1
T
T
0
A sin
π t
e
T
−j2πkt/T dt
1

EE 341 Fall 2004
c x k = A
T
T
= A
j2T
= A
j2T
= − A
2T
0
T
ejπt/T − e−jπt/T e
j2
−j2πkt/T dt
e jπt(1−2k)/T
−jπt(1+2k)/T
− e dt
0
e jπt(1−2k)/T
jπ(1 − 2k)/T
ejπ(1−2k) π(1 − 2k)/T
T e−jπt(1+2k)/T
−
jπ(1 + 2k)/T
0
+ e−jπ(1+2k)
π(1 + 2k)/T −
1
π(1 − 2k)/T −
1
π(1 + 2k)/T
But e jπ(1−2k) = e jπ e −j2πk . e jπ = −1 and e −j2πk = 1, so e jπ(1−2k) = −1. Similarly, e −jπ(1+2k) = −1,
so:
For k = 0
c x k
A −1
= −
2T π(1 − 2k)/T +
−1
π(1 + 2k)/T −
1
π(1 − 2k)/T −
1
π(1 + 2k)/T
= − A
−2
2 π(1 − 2k) −
2
π(1 + 2k)
= A
1 1
+
π 1 − 2k 1 + 2k
= 2A
1
π 1 − 4k2
T
π t
A sin dt
0 T
= A
T
− cos(πt/T )
T π/T 0
= 2A
π
c x 0 = 1
T
Note: Because the signal is even, you could also find c x k by:
For an odd signal,
c x k
c x k = 2
T
= −j2
T
T/2
0
T/2
0
x(t) cos(kωot) dt
x(t) sin(kωot) dt
2