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Symmetry Properties of the Fourier Series

Module by: Justin Romberg. E-mail the author

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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 ℜ c n =ℜ c - n c n c - n , i.e. the real part of c n c n is even, and ℑ c n =-ℑ 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 =1T∫0Tftⅇⅈ ω 0 ntdt=∀t,ft=ft¯:1T∫0Tft¯ⅇ-ⅈ ω 0 ntdt¯=1T∫0Tftⅇ-ⅈ ω 0 ntdt¯= 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)

Figure 1: ℜ c n =ℜ c - n c n c - n , and ℑ c n =-ℑ c - n c n c - n .
(a)
Figure 1(a) (m10838ae.png)
(b)
Figure 1(b) (m10838ce.png)
Figure 2: | c n |=| c - n | c n c - n , and ∠ c n =∠− c - n ∠ c n ∠ c - n .
(a)
Figure 2(a) (m10838be.png)
(b)
Figure 2(b) (m10838de.png)

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=f-t 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 =1T∫-T2T2ftⅇ-ⅈ ω 0 ntdt=1T∫-T20ftⅇ-ⅈ ω 0 ntdt+1T∫0T2ftⅇ-ⅈ ω 0 ntdt=1T∫0T2f-tⅇⅈ ω 0 ntdt+1T∫0T2ftⅇ-ⅈ ω 0 ntdt=2T∫0T2ftcos ω 0 ntdt 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=-f-t 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=1∞2 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

Figure 3: T=1 T 1 and ω 0 =2π ω 0 2 .
Figure 3 (triwave.png)

ft f t is real and odd. c n =4Aⅈπ2n2ifn=…-11-7-3159…-4Aⅈπ2n2ifn=…-9-5-13711…0ifn=…-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 ?

Figure 4: The Fourier series of a triangle wave.
Figure 4 (m10838ee.png)

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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