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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 f⁢t f t is real it implies that f⁢t=f⁢t¯ f t f t ( f⁢t¯ f t is the complex conjugate of f⁢t 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⁢∫0Tf⁢t⁢ei⁢ ω 0 ⁢n⁢td t =∀ t ,f⁢t=f⁢t¯:1T⁢∫0Tf⁢t¯⁢e−(i⁢ ω 0 ⁢n⁢t)d t ¯=1T⁢∫0Tf⁢t⁢e−(i⁢ ω 0 ⁢n⁢t)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)

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 f⁢t=f⁢t¯ f t f t and f⁢t=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⁢∫−T2T2f⁢t⁢e−(i⁢ ω 0 ⁢n⁢t)d t =1T⁢∫−T20f⁢t⁢e−(i⁢ ω 0 ⁢n⁢t)d t +1T⁢∫0T2f⁢t⁢e−(i⁢ ω 0 ⁢n⁢t)d t =1T⁢∫0T2f⁢−t⁢ei⁢ ω 0 ⁢n⁢td t +1T⁢∫0T2f⁢t⁢e−(i⁢ ω 0 ⁢n⁢t)d t =2T⁢∫0T2f⁢t⁢cos ω 0 ⁢n⁢td 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)
f⁢t f t and cos ω 0 ⁢n⁢t ω 0 n t are both real which implies that c n c n is real. Also cos ω 0 ⁢n⁢t=cos−( ω 0 ⁢n⁢t) ω 0 n t ω 0 n t so c n = c - n c n c - n . It is also easy to show that f⁢t=2⁢∑ n =0∞ c n ⁢cos ω 0 ⁢n⁢t f t 2 n 0 c n ω 0 n t since f⁢t f t , c n c n , and cos ω 0 ⁢n⁢t 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 f⁢t=−f⁢−t f t f t and f⁢t=f⁢t¯ 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 f⁢t f t is odd, then we can expand it in terms of sin ω 0 ⁢n⁢t ω 0 n t : f⁢t=∑ n =1∞2⁢ c n ⁢sin ω 0 ⁢n⁢t 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

f⁢t= f e ⁢t+ f o ⁢t f t f e t f o t
(3)
which implies that, for any f⁢t f t , we can find a n a n and b n b n such that
f⁢t=∑ n =0∞ a n ⁢cos ω 0 ⁢n⁢t+∑ n =1∞ b n ⁢sin ω 0 ⁢n⁢t 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 =2⁢π ω 0 2 .
Figure 3 (triwave.png)

f⁢t f t is real and odd. c n ={4⁢Ai⁢π2⁢n2  if  n=…-11-7-3159…−4⁢Ai⁢π2⁢n2  if  n=…-9-5-13711…0  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 ?

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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