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An introduction to the general properties of the Fourier series

Introduction

In this module we will discuss the basic properties of the Continuous-Time Fourier Series. We will begin by refreshing your memory of our basic Fourier series equations:

f t n c n ω 0 n t
c n 1 T t 0 T f t ω 0 n t
Let · denote the transformation from f t to the Fourier coefficients f t n n c n · maps complex valued functions to sequences of complex numbers .

Linearity

· is a linear transformation .

If f t c n and g t d n . Then α α α f t α c n and f t g t c n d n

Easy. Just linearity of integral.

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

Shifting

Shifting in time equals a phase shift of Fourier coefficients

f t t 0 ω 0 n t 0 c n if c n c n c n , then ω 0 n t 0 c n ω 0 n t 0 c n c n ω 0 t 0 n c n ω 0 t 0 n

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

Parseval's relation

t 0 T f t 2 T n c n 2
Parseval's relation tells us that the energy of a signal is equal to the energy of its Fourier transform.
Parseval tells us that the Fourier series maps L 0 T 2 to l 2 .

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

c n 2 for f t to have finite energy.

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If n n 0 c n 1 n , is f L 0 T 2 ?

Yes, because c n 2 1 n 2 , which is summable.

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Now, if n n 0 c n 1 n , is f L 0 T 2 ?

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

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

Parsevals theorem demonstration

ParsevalsDemo
Interact (when online) with a Mathematica CDF demonstrating Parsevals Theorem. To download, right click and save file as .cdf.

Symmetry properties

Even signals

    Even signals

  • f ( t ) = f ( - t )
  • c n = c - n
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( - t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( - t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T f ( t ) exp ( ı ω 0 n t ) d t + exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T f ( t ) 2 cos ( ω 0 n t ) d t

Odd signals

    Odd signals

  • f ( t ) = -f ( -t )
  • c n = c - n *
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( t ) exp ( - ı ω 0 n t ) d t - 1 T T 2 T f ( - t ) exp ( ı ω 0 n t ) d t
  • = - 1 T 0 T f ( t ) exp ( ı ω 0 n t ) d t - exp ( - ı ω 0 n t ) d t
  • = - 1 T 0 T f ( t ) 2 ı sin ( ω 0 n t ) d t

Real signals

    Real signals

  • f ( t ) = f * ( t )
  • c n = c - n *
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( - t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( - t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T f ( t ) exp ( ı ω 0 n t ) d t + exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T f ( t ) 2 cos ( ω 0 n t ) d t

Differentiation in fourier domain

f t c n t f t n ω 0 c n

Since

f t n c n ω 0 n t
then
t f t n c n t ω 0 n t n c n ω 0 n ω 0 n t
A differentiator attenuates the low frequencies in f t and accentuates the high frequencies. It removes general trends and accentuates areas of sharpvariation.
A common way to mathematically measure the smoothness of a function f t is to see how many derivatives are finite energy.
This is done by looking at the Fourier coefficients of thesignal, specifically how fast they decay as n .If f t c n and c n has the form 1 n k , then t m f t n ω 0 m c n and has the form n m n k .So for the m th derivative to have finite energy, we need n n m n k 2 thus n m n k decays faster than 1 n which implies that 2 k 2 m 1 or k 2 m 1 2 Thus the decay rate of the Fourier series dictates smoothness.

Fourier differentiation demonstration

FourierDiffDemo
Interact (when online) with a Mathematica CDF demonstrating Differentiation in the Fourier Domain. To download, right click and save file as .cdf.

Integration in the fourier domain

If

f t c n
then
τ t f τ 1 ω 0 n c n
If 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.

Fourier integration demonstration

fourierIntDemo
Interact (when online) with a Mathematica CDF demonstrating Integration in the Fourier Domain. To download, right click and save file as .cdf.

Signal multiplication and convolution

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

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
for more details, see the section on Signal convolution and the CTFS

Conclusion

Like other Fourier transforms, the CTFS has many useful properties, including linearity, equal energy in the time and frequency domains, and analogs for shifting, differentation, and integration.

Properties of the ctfs
Property Signal CTFS
Linearity a x ( t ) + b y ( t ) a X ( f ) + b Y ( f )
Time Shifting x ( t - τ ) X ( f ) e - j 2 π f τ / T
Time Modulation x ( t ) e j 2 π f τ / T X ( f - k )
Multiplication x ( t ) y ( t ) X ( f ) * Y ( f )
Continuous Convolution x ( t ) * y ( t ) X ( f ) Y ( f )

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Source:  OpenStax, Signals and systems. OpenStax CNX. Aug 14, 2014 Download for free at http://legacy.cnx.org/content/col10064/1.15
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