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# Symmetry Properties of the Fourier Series

Module by: Justin Romberg. E-mail the author

Summary: This module looks at the different symmetry properties of the fourier series and its fourier coefficients.

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## Symmetry Properties

### Real Signals

Real signals have a conjugate symmetric Fourier series.

#### Theorem 1

If ft f t is real it implies that ft=ft¯ f t f t ( ft¯ f t is the complex conjugate of ft f t ), then c n = c - n ¯ c n c - n which implies that c n = c - n c n c - n , i.e. the real part of c n c n is even, and c n = c - n c n c - n , i.e. the imaginary part of c n c n is odd. See Figure 1. It also implies that | c n |=| c - n | c n c - n , i.e. that magnitude is even, and that c n =( c - n ) c n c - n , i.e. the phase is odd.

##### Proof

c - n =1T0Tftei ω 0 ntd t = t ,ft=ft¯:1T0Tft¯e(i ω 0 nt)d t ¯=1T0Tfte(i ω 0 nt)d t ¯= c n ¯ c - n 1 T t 0 T f t ω 0 n t t f t f t 1 T t 0 T f t ω 0 n t 1 T t 0 T f t ω 0 n t c n
(1)

### Real and Even Signals

Real and even signals have real and even Fourier series.

#### Theorem 2

If ft=ft¯ f t f t and ft=ft f t f t , i.e. the signal is real and even, then c n = c - n c n c - n and c n = c n ¯ c n c n .

##### Proof

c n =1TT2T2fte(i ω 0 nt)d t =1TT20fte(i ω 0 nt)d t +1T0T2fte(i ω 0 nt)d t =1T0T2ftei ω 0 ntd t +1T0T2fte(i ω 0 nt)d t =2T0T2ftcos ω 0 ntd t c n 1 T t T 2 T 2 f t ω 0 n t 1 T t T 2 0 f t ω 0 n t 1 T t 0 T 2 f t ω 0 n t 1 T t 0 T 2 f t ω 0 n t 1 T t 0 T 2 f t ω 0 n t 2 T t 0 T 2 f t ω 0 n t
(2)
ft f t and cos ω 0 nt ω 0 n t are both real which implies that c n c n is real. Also cos ω 0 nt=cos( ω 0 nt) ω 0 n t ω 0 n t so c n = c - n c n c - n . It is also easy to show that ft=2 n =0 c n cos ω 0 nt f t 2 n 0 c n ω 0 n t since ft f t , c n c n , and cos ω 0 nt ω 0 n t are all real and even.

### Real and Odd Signals

Real and odd signals have Fourier Series that are odd and purely imaginary.

#### Theorem 3

If ft=ft f t f t and ft=ft¯ f t f t , i.e. the signal is real and odd, then c n = c - n c n c - n and c n = c n ¯ c n c n , i.e. c n c n is odd and purely imaginary.

##### Proof

Do it at home.

If ft f t is odd, then we can expand it in terms of sin ω 0 nt ω 0 n t : ft= n =12 c n sin ω 0 nt f t n 1 2 c n ω 0 n t

## Summary

In summary, we can find f e t f e t , an even function, and f o t f o t , an odd function, such that

ft= f e t+ f o t f t f e t f o t
(3)
which implies that, for any ft f t , we can find a n a n and b n b n such that
ft= n =0 a n cos ω 0 nt+ n =1 b n sin ω 0 nt f t n 0 a n ω 0 n t n 1 b n ω 0 n t
(4)

## Example 1: Triangle Wave

ft f t is real and odd. c n ={4Aiπ2n2  if  n=-11-7-31594Aiπ2n2  if  n=-9-5-137110  if  n=-4-2024 c n 4 A 2 n 2 n -11 -7 -3 1 5 9 4 A 2 n 2 n -9 -5 -1 3 7 11 0 n -4 -2 0 2 4 Does c n = c - n c n c - n ?

## Note:

We can often gather information about the smoothness of a signal by examining its Fourier coefficients.
Take a look at the above examples. The pulse and sawtooth waves are not continuous and there Fourier series' fall off like 1n 1 n . The triangle wave is continuous, but not differentiable and its Fourier series falls off like 1n2 1 n 2 .

The next 3 properties will give a better feel for this.

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