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Fourier Series Properties

Module by: Justin Romberg, Benjamin Fite

Summary: An introduction to the general properties of the Fourier series

We will begin by refreshing your memory of our basic Fourier series equations:

ft=∑n=-∞∞ c n ⅇⅈ ω 0 nt f t n c n ω 0 n t (1)
c n =1T∫0Tftⅇ-ⅈ ω 0 ntdt c n 1 T t 0 T f t ω 0 n t (2)
Let ℱ· ℱ · denote the transformation from ft f t to the Fourier coefficients ℱft=∀n,n∈ℤ: c n ℱ f t n n c n ℱ· ℱ · maps complex valued functions to sequences of complex numbers.

Linearity

ℱ· ℱ · is a linear transformation.

theorem 1

If ℱft= c n ℱ f t c n and ℱgt= d n ℱ g t d n . Then ∀α,α∈ℂ:ℱαft=α c n α α ℱ α f t α c n and ℱft+gt= c n + d n ℱ f t g t c n d n

Proof

Easy. Just linearity of integral.

ℱft+gt=∀n,n∈ℤ:∫0Tft+gtⅇ-ⅈ ω 0 ntdt=∀n,n∈ℤ:1T∫0Tftⅇ-ⅈ ω 0 ntdt+1T∫0Tgtⅇ-ⅈ ω 0 ntdt=∀n,n∈ℤ: c n + d n = c n + d n ℱ f t g t n n t 0 T f t g t ω 0 n t n n 1 T t 0 T f t ω 0 n t 1 T t 0 T g t ω 0 n t n n c n d n c n d n (3)

Shifting

Shifting in time equals a phase shift of Fourier coefficients

theorem 2

ℱft- t 0 =ⅇ-ⅈ ω 0 n t 0 c n ℱ f t t 0 ω 0 n t 0 c n if c n =| c n |ⅇⅈ∠ c n c n c n ∠ c n , then |ⅇ-ⅈ ω 0 n t 0 c n |=|ⅇ-ⅈ ω 0 n t 0 || c n |=| c n | ω 0 n t 0 c n ω 0 n t 0 c n c n ∠ⅇ-ⅈ ω 0 t 0 n=∠ c n - ω 0 t 0 n ∠ ω 0 t 0 n ∠ c n ω 0 t 0 n

Proof

ℱft- t 0 =∀n,n∈ℤ:1T∫0Tft- t 0 ⅇ-ⅈ ω 0 ntdt=∀n,n∈ℤ:1T∫- t 0 T- t 0 ft- t 0 ⅇ-ⅈ ω 0 nt- t 0 ⅇ-ⅈ ω 0 n t 0 dt=∀n,n∈ℤ:1T∫- t 0 T- t 0 f t ~ ⅇ-ⅈ ω 0 n t ~ ⅇ-ⅈ ω 0 n t 0 dt=∀n,n∈ℤ:ⅇ-ⅈ ω 0 n t ~ c n ℱ f t t 0 n n 1 T t 0 T f t t 0 ω 0 n t n n 1 T t t 0 T t 0 f t t 0 ω 0 n t t 0 ω 0 n t 0 n n 1 T t t 0 T t 0 f t ~ ω 0 n t ~ ω 0 n t 0 n n ω 0 n t ~ c n (4)

Parseval's Relation

∫0T|ft|2dt=T∑n=-∞∞| c n |2 t 0 T f t 2 T n c n 2 (5)
Parseval's relation allows us to calculate the energy of a signal from its Fourier series.

note:

Parseval tells us that the Fourier series maps L20T L 0 T 2 to l2ℤ l 2 .

Figure 1
Figure 1 (pars.png)

Exercise 1

For ft f t to have "finite energy," what do the c n c n do as n→∞ n ?

Solution 1

| c n |2<∞ c n 2 for ft f t to have finite energy.

Exercise 2

If ∀n,|n|>0: c n =1n n n 0 c n 1 n , is f∈L20T f L 0 T 2 ?

Solution 2

Yes, because | c n |2=1n2 c n 2 1 n 2 , which is summable.

Exercise 3

Now, if ∀n,|n|>0: c n =1n n n 0 c n 1 n , is f∈L20T f L 0 T 2 ?

Solution 3

No, because | c n |2=1n c n 2 1 n , which is not summable.

The rate of decay of the Fourier series determines if ft f t has finite energy.

Differentiation in Fourier Domain

ℱft= c n ⇒ℱddtft=ⅈn ω 0 c n ℱ f t c n ℱ t f t n ω 0 c n (6)

Since

ft=∑n=-∞∞ c n ⅇⅈ ω 0 nt f t n c n ω 0 n t (7)
then
ddtft=∑n=-∞∞ c n ddtⅇⅈ ω 0 nt=∑n=-∞∞ c n ⅈ ω 0 nⅇⅈ ω 0 nt t f t n c n t ω 0 n t n c n ω 0 n ω 0 n t (8)
A differentiator attenuates the low frequencies in ft f t and accentuates the high frequencies. It removes general trends and accentuates areas of sharp variation.

note:

A common way to mathematically measure the smoothness of a function ft f t is to see how many derivatives are finite energy.
This is done by looking at the Fourier coefficients of the signal, specifically how fast they decay as n→∞ n . If ℱft= c n ℱ f t c n and | c n | c n has the form 1nk 1 n k , then ℱdmdtmft=ⅈn ω 0 m c n ℱ t m f t n ω 0 m c n and has the form nmnk n m n k . So for the m th m th derivative to have finite energy, we need ∑|nmnk|2<∞ n n m n k 2 thus nmnk n m n k decays faster than 1n 1 n which implies that 2k-2m>1 2 k 2 m 1 or k>2m+12 k 2 m 1 2 Thus the decay rate of the Fourier series dictates smoothness.

Integration in the Fourier Domain

If

ℱft= c n ℱ f t c n (9)
then
ℱ∫-∞tfτdτ=1ⅈ ω 0 n c n ℱ τ t f τ 1 ω 0 n c n (10)

note:

If c 0 ≠0 c 0 0 , this expression doesn't make sense.

Integration accentuates low frequencies and attenuates high frequencies. Integrators bring out the general trends in signals and suppress short term variation (which is noise in many cases). Integrators are much nicer than differentiators.

Signal Multiplication

Given a signal ft f t with Fourier coefficients c n c n and a signal gt g t with Fourier coefficients d n d n , we can define a new signal, yt y t , where yt=ftgt y t f t g t . We find that the Fourier Series representation of yt y t , e n e n , is such that e n =∑k=-∞∞ c k d n - k e n k c k d n - k . This is to say that signal multiplication in the time domain is equivalent to discrete-time convolution in the frequency domain. The proof of this is as follows

e n =1T∫0Tftgtⅇ-ⅈ ω 0 ntdt=1T∫0T∑k=-∞∞ c k ⅇⅈ ω 0 ktgtⅇ-ⅈ ω 0 ntdt=∑k=-∞∞ c k 1T∫0Tgtⅇ-ⅈ ω 0 n-ktdt=∑k=-∞∞ c k d n - k e n 1 T t 0 T f t g t ω 0 n t 1 T t 0 T k c k ω 0 k t g t ω 0 n t k c k 1 T t 0 T g t ω 0 n k t k c k d n - k (11)

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