Signal representation using Fourier series - Continuous and Discrete Time Signals and Systems - page 148

Case 2 By following the procedure outlined in case 1, it is straightforward to
show that

0

m = n

sin( m ω 0 t ) sin( n ω 0 t )d t = T 0

2

(4.16)

m = n ,

T 0

+ +

for m , n ∈ Z . Equation (4.16) proves that the set { sin( n ω 0 t ), n ∈ Z } contains

mutually orthogonal functions over interval t = [ t 0 , t 0 + T 0 ] with T 0 = 2 π/ω 0 .

Case 3 To verify that functions { cos( m ω 0 t ) } and { sin( n ω 0 t ) } are mutually
orthogonal, consider the following:

t 0 + T 0

cos( m ω 0 t ) sin( n ω 0 t )d t = cos( m ω 0 t ) sin( n ω 0 t )d t

t 0

T 0

t 0 + T 0

1
2

[sin(( m + n ) ω 0 t ) − sin(( m − n ) ω 0 t )]d tm = n

t 0

=

t 0 + T 0

1
2

[sin(2 m ω 0 t )d t m = n

t 0

t 0 + T 0 t 0 + T 0

cos(( m + n ) ω 0 t )
( m + n ) ω 0

cos(( m − n ) ω 0 t )
( m − n ) ω 0

− 1
2

+ 1
2

m = n

t 0 t 0

=

t 0 + T 0

− 1
2

cos(2 n ω 0 t )
2 m ω 0

m = n

t 0

0 m = n

=

0 m = n ,

(4.17)

+

for m , n ∈ Z , which proves that { cos( m ω 0 t ) } and { sin( n ω 0 t ) } are orthogonal

over interval t = [ t 0 , t 0 + T 0 ] with T 0 = 2 π/ω 0 .

Case 4 The following proof demonstrates that the function “1” is orthogonal
to cos( m ω 0 t ) } and { sin( n ω 0 t ) } :

sin( m ω 0 t )
m ω 0

t 0 + T 0

1 cos( m ω 0 t )d t =

t 0

T 0

sin( m ω 0 t 0 + 2 m π ) − sin( m ω 0 t 0 )

= = 0 (4.18)

m ω 0

and

− cos( m ω 0 t )
m ω 0

t 0 + T 0

1 sin( m ω 0 t )d t =

t 0

T 0

cos( m ω 0 t 0 + 2 m π ) − cos( m ω 0 t 0 )

m ω 0

=− = 0 (4.19)