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Properties of the DTFS

By Stephen Kruzick

Introduction

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

fn= k =0N1 c k ei ω 0 kn f n k N 1 0 c k ω 0 k n
1
c k =1N n =0N1fne(i2πNkn) c k 1 N n N 1 0 f n 2 N k n
2
Let · · denote the transformation from fn f n to the Fourier coefficients fn= k ,kZ: c k f n k k c k · · maps complex valued functions to sequences of complex numbers.

Linearity

· · is a linear transformation.

Theorem 1

If fn= c k f n c k and gn= d k g n d k . Then α ,αC:αfn=α c k α α α f n α c k and fn+gn= c k + d k f n g n c k d k

Proof

Easy. Just linearity of integral.

fn+gn= k ,kZ: n =0N(fn+gn)e(i ω 0 kn)= k ,kZ:1N n =0Nfne(i ω 0 kn)+1N n =0Ngne(i ω 0 kn)= k ,kZ: c k + d k = c k + d k f n g n k k n 0 N f n g n ω 0 k n k k 1 N n 0 N f n ω 0 k n 1 N n 0 N g n ω 0 k n k k c k d k c k d k
3

Shifting

Shifting in time equals a phase shift of Fourier coefficients

Theorem 2

fn n 0 =e(i ω 0 k n 0 ) c k f n n 0 ω 0 k n 0 c k if c k =| c k |ei c k c k c k c k , then |e(i ω 0 k n 0 ) c k |=|e(i ω 0 k n 0 )|| c k |=| c k | ω 0 k n 0 c k ω 0 k n 0 c k c k e(i ω 0 n 0 k)= c k ω 0 n 0 k ω 0 n 0 k c k ω 0 n 0 k

Proof

fn n 0 = k ,kZ:1N n =0Nfn n 0 e(i ω 0 kn)= k ,kZ:1N n = n 0 N n 0 fn n 0 e(i ω 0 k(n n 0 ))e(i ω 0 k n 0 )= k ,kZ:1N n = n 0 N n 0 f n ~ e(i ω 0 k n ~ )e(i ω 0 k n 0 )= k ,kZ:e(i ω 0 k n ~ ) c k f n n 0 k k 1 N n 0 N f n n 0 ω 0 k n k k 1 N n n 0 N n 0 f n n 0 ω 0 k n n 0 ω 0 k n 0 k k 1 N n n 0 N n 0 f n ~ ω 0 k n ~ ω 0 k n 0 k k ω 0 k n ~ c k
4

Parseval's Relation

n =0N|fn|2=N k =0N1| c k |2 n 0 N f n 2 N k N 1 0 c k 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 L20N L 0 N 2 to l2Z l 2 .

Figure 1
Exercise 1

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

Solution

| c k |2< c k 2 for fn f n to have finite energy.

Exercise 2

If k ,|k|>0: c k =1k k k 0 c k 1 k , is fL2 0 N f L 0 N 2 ?

Solution

Yes, because | c k |2=1k2 c k 2 1 k 2 , which is summable.

Exercise 3

Now, if k ,|k|>0: c k =1k k k 0 c k 1 k , is fL2 0 N f L 0 N 2 ?

Solution

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

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

ParsevalsTheorem Demonstration

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

Symmetry Properties

Rule 1: Even Signals
Even Signals
  • f[n]=f[-n]f[n]=f[-n]
  • ck=c-kck=c-k
Proof
  • c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ] c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N 2 f [ - n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ - n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ - n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ - n ] exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] + exp [ - ı ω 0 k n ] = 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] + exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N f [ n ] 2 cos [ ω 0 k n ] = 1 N Σ 0 N f [ n ] 2 cos [ ω 0 k n ]
Rule 2: Odd Signals
Odd Signals
  • f [ n ] = -f [ -n ] f [ n ] = -f [ -n ]
  • ck=c-kck=c-k*
Proof
  • c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ] c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] - 1 N Σ N 2 N f [ - n ] exp [ ı ω 0 k n ] = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] - 1 N Σ N 2 N f [ - n ] exp [ ı ω 0 k n ]
  • = - 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] - exp [ - ı ω 0 k n ] = - 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] - exp [ - ı ω 0 k n ]
  • = - 1 N Σ 0 N f [ n ] 2 ı sin [ ω 0 k n ] = - 1 N Σ 0 N f [ n ] 2 ı sin [ ω 0 k n ]
Rule 3: Real Signals
Real Signals
  • f[n]=ff[n]=f*[n][n]
  • ck=c-kck=c-k*
Proof
  • c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ] c k = 1 N Σ 0 N f [ n ] exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ n ] exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N 2 f [ - n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ - n ] exp [ - ı ω 0 k n ] = 1 N Σ 0 N 2 f [ - n ] exp [ - ı ω 0 k n ] + 1 N Σ N 2 N f [ - n ] exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] + exp [ - ı ω 0 k n ] = 1 N Σ 0 N f [ n ] exp [ ı ω 0 k n ] + exp [ - ı ω 0 k n ]
  • = 1 N Σ 0 N f [ n ] 2 cos [ ω 0 k n ] = 1 N Σ 0 N f [ n ] 2 cos [ ω 0 k n ]

Differentiation in Fourier Domain

(fn= c k )(dfnd n =ik ω 0 c k ) f n c k n f n k ω 0 c k
6

Since

fn= k =0N c k ei ω 0 kn f n k 0 N c k ω 0 k n
7
then
d d n fn= k =0N c k dei ω 0 knd n = k =0N c k i ω 0 kei ω 0 kn n f n k 0 N c k n ω 0 k n k 0 N c k ω 0 k ω 0 k n
8
A differentiator attenuates the low frequencies in fn f n 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 fn f n 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 k k . If fn= c k f n c k and | c k | c k has the form 1kl 1 k l , then d m fnd n m =ik ω 0 m c k n m f n k ω 0 m c k and has the form kmkl k m k l . So for the m th m th derivative to have finite energy, we need k |kmkl|2< k k m k l 2 thus kmkl k m k l decays faster than 1k 1 k which implies that 2l2m>1 2 l 2 m 1 or l>2m+12 l 2 m 1 2 Thus the decay rate of the Fourier series dictates smoothness.

Fourier Differentiation Demo

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

Integration in the Fourier Domain

If

fn= c k f n c k
9
then
η =0nfη=1i ω 0 k c k η 0 n f η 1 ω 0 k c k
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 Demo

Figure 4: Interact (when online) with a Mathematica CDF demonstrating the effect of Integration in a Fourier Domain.

Signal Multiplication

Given a signal fn f n with Fourier coefficients c k c k and a signal gn g n with Fourier coefficients d k d k , we can define a new signal, yn y n , where yn=fngn y n f n g n . We find that the Fourier Series representation of yn y n , e k e k , is such that e k = l =0N c l d k - l e k l 0 N c l d k - l . This is to say that signal multiplication in the time domain is equivalent to discrete-time circular convolution in the frequency domain. The proof of this is as follows

e k =1N n =0Nfngne(i ω 0 kn)=1N n =0N l =0N c l ei ω 0 lngne(i ω 0 kn)= l =0N c l (1N n =0Ngne(i ω 0 (kl)n))= l =0N c l d k - l e k 1 N n 0 N f n g n ω 0 k n 1 N n 0 N l 0 N c l ω 0 l n g n ω 0 k n l 0 N c l 1 N n 0 N g n ω 0 k l n l 0 N c l d k - l
11

Conclusion

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

Property Signal DTFS
Linearity a x ( n ) + b y ( n ) a x ( n ) + b y ( n ) a X ( k ) + b Y ( k ) a X ( k ) + b Y ( k )
Time Shifting x ( n - m ) x ( n - m ) X ( k ) e - j 2 π m k / N X ( k ) e - j 2 π m k / N
Time Modulation x ( n ) e j 2 π m n / N x ( n ) e j 2 π m n / N X ( k - m ) X ( k - m )
Multiplication x ( n ) y ( n ) x ( n ) y ( n ) X ( k ) * Y ( k ) X ( k ) * Y ( k )
Circular Convolution x ( n ) * y ( n ) x ( n ) * y ( n ) X ( k ) Y ( K ) X ( k ) Y ( K )
Table 1