Fourier Series: Trigonometric - Signal Processing of Random Physiological Signals - page 133

Since 2 ω 0 T = 4 π , then:

sin 2 sin 2 sin 2

2 V T

4 −

3 T
4

T
4

3 T
4

For n = 2 , a 2 = ω ω − sin 2 ω ω

0 T

− sin 2 ω

0 0 0 0

2 ω

0 T

sin 4 sin 4

)

2 V
4

4 − π · 3

a 2 = − sin π +

( sin 4

π − sin 3 π

π 4

2 V
4

a 2 = [0 − (0 − 0) + (0 − 0)] = 0

π

Likewise, a n = 0 for all even values of a n .

Since ω 0 T = 2 π, then n ω 0 T for n = 3is3 ω 0 T = 6 π.

sin 6 sin 6 sin 6

2 V
6

4 − π · 3

sin 6

4

sin 6

π · 3

For n = 3 , a 3 = − + π −

π 4 4

V
3

a 3 = [ − 1 − ( + 1 − ( − 1)) + 0 − ( + 1)]

π

V
3

4 V
3

a 3 = [ − 1 − 2 − 1] =−

π
Applying the same procedure for all n , we find that

π

⎧
⎨

⎫
⎬

4 V
n

+

π ,

= 1 , 5 , 9 ,...

n

4 V
n

=

a n

−

π ,

n = 3 , 7 , 11 ,...

⎩ ⎭

0 , n = even

The imaginary term b n values are calculated as follows:

V

0 tdt V V

T / 4 3 / 4

T T

2
T

b n = sin n ω 0 tdt − sin n ω 0 tdt + sin n ω

T / 4 3 / 4

0 T

(

T
/ 4

3 / 4

T T

2 V
n

b n = − cos n ω 0 t ) − − cos n ω 0 t ) + − cos n ω 0 t )

( (

ω

0 T

T 3

/ 4 / 4

0 T

2 V
n

cos 2

4 +

cos 6

4 +

cos 2

4

cos 6

4

b 1 = − cos 0 − − + − cos 2 π +

ω

0 T

2 V
n

b 1 = [0 + 1 + 0 − 0 − 1 + 0] = 0

ω 0 T

2 V
2

b 1 =

( 0 )

= 0

π