Basis Functions and Signal Representation - Signal Processing of Random Physiological Signals - page 25

Then from − M to M , the fractional error is given by (4.25):

M

1
E

2

η m = 1 − T | α n |

= 0

n

2

(4.25)

M

1
E

2
0

η = 1 − α + 2 T | α |

m n

n = 0

Example: Consider the periodic sequence, x ( t ) = A for 0 < t < t a , and x ( t ) = 0 for

< <

T : as in (4.26) and (4.27).

t a t

t a

1
T

Ae − jnw 0 t dt α = + 0 (for t a < t < T ) (4.26)

n

0

jnw 0 e − jnw 0 t

− 1

A
T

t a
0

α =

n

(1)

A
T

− 1 A

T

1
jnw 0

−

jnw 0 e − jnw 0 t a

α = − (4.27)

n

jnw 0 T 1

e − jnw 0 t

A

α n = −

By rewriting the terms in the bracket, you get (4.28):

2 e j 2

x x

2

x

e − j e − j

− , where x = nw 0 t (4.28)

and α

n becomes (4.29):

e j nw 0 t a e − j nw 0 t 2

e − j nw 0 t a

−

2 2

α = (4.29)

n

jnw 0

The [] (square bracket) term is a trig identity [(4.30) and (4.31)] with circular functions
where

e jx e − jx

2 j

−

sin( x ) = (4.30)

nw 0 t a
2

With x = (4.31)