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Symmetry Properties of the Fourier Series

Module by: Justin Romberg

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

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 =1T0Tft ω 0 ntdt=t,ft=ft¯:1T0Tft¯- ω 0 ntdt¯=1T0Tft- 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)
m10838ae.png
Subfigure 1.1
m10838ce.png
Subfigure 1.2
Figure 1: c n = c - n c n c - n , and c n =- c - n c n c - n .
m10838be.png
Subfigure 2.1
m10838de.png
Subfigure 2.2
Figure 2: | c n |=| c - n | c n c - n , and c n =- c - n c n c - n .

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-T2T2ft- ω 0 ntdt=1T-T20ft- ω 0 ntdt+1T0T2ft- ω 0 ntdt=1T0T2f-t ω 0 ntdt+1T0T2ft- ω 0 ntdt=2T0T2ftcos 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=2n=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=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 
triwave.png
Figure 3: T=1 T 1 and ω 0 =2π ω 0 2 .
ft f t is real and odd. c n =4Aπ2n2ifn=-11-7-3159-4Aπ2n2ifn=-9-5-137110ifn=-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 ?
m10838ee.png
Figure 4: The Fourier series of a triangle wave.
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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