Fourier Series: Trigonometric - Signal Processing of Random Physiological Signals - page 132

+V

Volts

T/4

0 3T/4 t

-T/2

T/2

T

-T

FIGURE 12.6 : Square waveform for example problem

By inspection, the average value over one period is zero so that a 0 = 0.

Value of a 1 with n = 1is

V

ω 0 tdt V V

T / 4 3 / 4 T

T

2
T

a n = cos n ω 0 tdt − cos n ω 0 tdt + cos n

T / 4 3 / 4

0 T

0 t 0 t 0 t

T / 4
0 −

3 / 4

T T

2 V
T

1 1 1

a n = sin n ω sin n ω + sin n ω

n ω n ω n ω

T / 4 3 / 4

0 0 0

T

sin ω sin 3 sin

2 V
ω

0 T
4

ω

0 T
4

sin ω

0 T
4

sin 3

ω

0 T
4

a n = − − + ω 0 T −

0 T

2 T

Since ω

0 T

= · T = 2 π , then

sin 2 sin 3 sin 2

2 V
2

2

sin 2 sin 3

2

for n = 1 , a 1 = − − + π −

π

V
π

V
π

4 V
π

a 1 = [1 − ( − 1 − 1) + (0 − ( − 1))] = [1 + 2 + 1] =

V

0 tdt V V

T / 4 3 / 4

T T

2
T

= cos n ω − cos n ω + cos n ω

a n 0 tdt 0 tdt

T 3 / 4

/ 4

0 T

0 t 0 t 0 t

T / 4
0 −

3 / 4

T T

2 V
T

1 1 1

a n = sin n ω sin n ω + sin n ω

n ω n ω n ω

T / 4 3 / 4 T

0 0 0