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Properties of the CTFS

By Justin Romberg, Benjamin Fite

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:

ft= n = c n ei ω 0 nt f t n c n ω 0 n t
1
c n =1T0Tfte(i ω 0 nt)d t c n 1 T t 0 T f t ω 0 n t
2
Let · · denote the transformation from ft f t to the Fourier coefficients ft= n ,nZ: c n f t n n c n · · maps complex valued functions to sequences of complex numbers.

Linearity

· · is a linear transformation.

Theorem 1

If ft= c n f t c n and gt= d n g t d n . Then α ,αC:αft=α c n α α α f t α c n and ft+gt= c n + d n f t g t c n d n

Proof

Easy. Just linearity of integral.

ft+gt= n ,nZ:0T(ft+gt)e(i ω 0 nt)d t = n ,nZ:1T0Tfte(i ω 0 nt)d t +1T0Tgte(i ω 0 nt)d t = n ,nZ: c n + d n = c n + d n 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

ft t 0 =e(i ω 0 n t 0 ) c n 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

ft t 0 = n ,nZ:1T0Tft t 0 e(i ω 0 nt)d t = n ,nZ:1T t 0 T t 0 ft t 0 e(i ω 0 n(t t 0 ))e(i ω 0 n t 0 )d t = n ,nZ:1T t 0 T t 0 f t ~ e(i ω 0 n t ~ )e(i ω 0 n t 0 )d t = n ,nZ:e(i ω 0 n t ~ ) c 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
4

Parseval's Relation

0T|ft|2d t =T n =| c n |2 t 0 T f t 2 T n c n 2
5
Parseval's relation tells us that the energy of a signal is equal to the energy of its Fourier transform.
Note:
Parseval tells us that the Fourier series maps L2 0 T L 0 T 2 to l2Z l 2 .

Figure 1
Exercise 1

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

Solution

| 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 fL2 0 T f L 0 T 2 ?

Solution

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 fL2 0 T f L 0 T 2 ?

Solution

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.

Parsevals Theorem Demonstration

Figure 2: Interact (when online) with a Mathematica CDF demonstrating Parsevals Theorem.

Symmetry Properties

Rule 1: Even Signals
Even Signals
  • f(t)=f(-t)f(t)=f(-t)
  • cn=c-ncn=c-n
Proof
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t 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 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 ) 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 = 1 T 0 T f ( t ) 2 cos ( ω 0 n t ) d t
Rule 2: Odd Signals
Odd Signals
  • f ( t ) = -f ( -t ) f ( t ) = -f ( -t )
  • cn=c-ncn=c-n*
Proof
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t 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 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 ) 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 = - 1 T 0 T f ( t ) 2 ı sin ( ω 0 n t ) d t
Rule 3: Real Signals
Real Signals
  • f(t)=ff(t)=f*(t)(t)
  • cn=c-ncn=c-n*
Proof
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t 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 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 ) 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 = 1 T 0 T f ( t ) 2 cos ( ω 0 n t ) d t

Differentiation in Fourier Domain

(ft= c n )(dftd t =in ω 0 c n ) f t c n 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
d d t ft= n = c n dei ω 0 ntd t = 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 ft= c n f t c n and | c n | c n has the form 1nk 1 n k , then d m ftd t m =in ω 0 m c n 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 n |nmnk|2< n n m n k 2 thus nmnk n m n k decays faster than 1n 1 n which implies that 2k2m>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.

Fourier Differentiation Demonstration

Figure 3: Interact (when online) with a Mathematica CDF demonstrating Differentiation in the Fourier Domain.

Integration in the Fourier Domain

If

ft= c n f t c n
9
then
tfτd τ =1i ω 0 n c n 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.

Fourier Integration Demonstration

Integration Demo
Figure 4: Click on the above thumbnail image (when online) to view an interactive Mathematica Player demonstrating the effect of Integration on a Fourier Domain.
Figure 5: Interact (when online) with a Mathematica CDF demonstrating Integration in the Fourier Domain.

Signal Multiplication and Convolution

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 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 =1T0Tftgte(i ω 0 nt)d t =1T0T k = c k ei ω 0 ktgte(i ω 0 nt)d t = k = c k (1T0Tgte(i ω 0 (nk)t)d t )= 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
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.

Property Signal CTFS
Linearity a x ( t ) + b y ( t ) a x ( t ) + b y ( t ) a X ( f ) + b Y ( f ) a X ( f ) + b Y ( f )
Time Shifting x ( t - τ ) x ( t - τ ) X ( f ) e - j 2 π f τ / T X ( f ) e - j 2 π f τ / T
Time Modulation x ( t ) e j 2 π f τ / T x ( t ) e j 2 π f τ / T X ( f - k ) X ( f - k )
Multiplication x ( t ) y ( t ) x ( t ) y ( t ) X ( f ) * Y ( f ) X ( f ) * Y ( f )
Continuous Convolution x ( t ) * y ( t ) x ( t ) * y ( t ) X ( f ) Y ( f ) X ( f ) Y ( f )
Table 1: Properties of the CTFS