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This module includes a table of properties of the DTFS.

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:

f n k N 1 0 c k ω 0 k n
c k 1 N n N 1 0 f n 2 N k n
Let · denote the transformation from f n to the Fourier coefficients f n k k c k · maps complex valued functions to sequences of complex numbers .

Linearity

· is a linear transformation .

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

Easy. Just linearity of integral.

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

Shifting

Shifting in time equals a phase shift of Fourier coefficients

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

Parseval's relation

n 0 N f n 2 N k N 1 0 c k 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 N 2 to l 2 .

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

c k 2 for f n to have finite energy.

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If k k 0 c k 1 k , is f L 0 N 2 ?

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

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Now, if k k 0 c k 1 k , is f L 0 N 2 ?

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

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

Parsevalstheorem demonstration

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

Symmetry properties

Even signals

    Even signals

  • f [ n ] = f [ - n ]
  • c k = c - k
  • 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 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 ]

Odd signals

    Odd signals

  • f [ n ] = -f [ -n ]
  • c k = c - k *
  • 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 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 ]

Real signals

    Real signals

  • f [ n ] = f * [ n ]
  • c k = c - k *
  • 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 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 ]

Differentiation in fourier domain

f n c k n f n k ω 0 c k

Since

f n k 0 N c k ω 0 k n
then
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
A differentiator attenuates the low frequencies in f n 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 n 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 k .If f n c k and c k has the form 1 k l , then n m f n k ω 0 m c k and has the form k m k l .So for the m th derivative to have finite energy, we need k k m k l 2 thus k m k l decays faster than 1 k which implies that 2 l 2 m 1 or l 2 m 1 2 Thus the decay rate of the Fourier series dictates smoothness.

Fourier differentiation demo

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

Integration in the fourier domain

If

f n c k
then
η 0 n f η 1 ω 0 k c k
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 demo

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

Signal multiplication

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

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

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.

Properties of the Discrete Fourier Transform
Property Signal DTFS
Linearity a x ( n ) + b y ( n ) a X ( k ) + b Y ( k )
Time Shifting x ( n - m ) X ( k ) e - j 2 π m k / N
Time Modulation x ( n ) e j 2 π m n / N X ( k - m )
Multiplication x ( n ) y ( n ) X ( k ) * Y ( k )
Circular Convolution x ( n ) * y ( n ) X ( k ) Y ( K )

Questions & Answers

Preparation and Applications of Nanomaterial for Drug Delivery
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Professor
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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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