Chapter 4 Frequency Analysis of Signalscwlin/courses/dsp/notes/DSP-chap4.pdf · 1 Frequency Analysis of Signals 清大電機系林嘉文 [email protected] 03-5731152 Chapter 4

1

Frequency Analysis of Signals

清大電機系林嘉文[email protected]

Chapter 4

The Fourier transform is a mathematical tool that is useful in the analysis and design of LTI systems

These signal representations basically involve the decomposition of the signals in terms of sinusoidal (or complex exponential) components

With such a decomposition, a signal is said to be represented in the frequency domain

Most signals of practical interest can be decomposed into a sum of sinusoidal signal components For the class of periodic signals, such a decomposition is called a

Fourier series For the class of finite energy signals, the decomposition is called

the Fourier transform

2010/4/14 Introduction to Digital Signal Processing 2

Frequency Analysis of Signals

2


The Father of Fourier Transform


A prism can be used to break up white light (sunlight) into the colors of the rainbow.

Analysis

Synthesis

Frequency Analysis of Continuous-Time Signals

3


Frequency analysis of a signal involves the resolution of the signal into its frequency (sinusoidal) components

Signal waveforms are basically functions of time The role of the prism is played by the Fourier analysis

tools that we will develop: the Fourier series and the Fourier transform

Examples of periodic signals encountered in practice are square waves, rectangular waves, triangular waves, sinusoids and complex exponentials

Frequency Analysis of Continuous-Time Signals

The Fourier Series for Continuous – Time Periodic Signals


The basic mathematical representation of periodic signals is the Fourier series , which is a linear weighted sum of harmonically related sinusoids or complex exponentials

Fundamental period:

determines the fundamental period of and the coefficients specify the shape of the waveform

02( ) j kF tk

k

x t c e π∞

=−∞

= ∑

0

1pT

F=

A periodic signal

0F ( )x t

kc

4



Suppose that we are given a periodic signal with period Tp. We can represent the periodic signal, called a Fourier series

The fundamental frequency F0 is selected to be the reciprocal of the given period Tp

To determine the expression for the coefficients ck, we first multiply both sides of

by the complex exponentialwhere l is an integer

( )x t

02( ) j kF tk

k

x t c e π∞

=−∞

= ∑02j F lte π−



Signal period, from . Where is an arbitrary but mathematically convenient starting value. Thus we obtain

We interchange the order of the summation and integration and combine the two exponentials. Hence

0 00 0 0

0 0

2 2 2( )p pt T t Tj lF t j lF t j kF t

kt tk

x t e dt e c e dtπ π π∞+ +− −

=−∞

=

∑∫ ∫

00

00

0

0

2 ( )2 ( )

02 ( )

( )

p

p

t Tj F k l tt T j F k l t

k ktk k t

k p l pk

ec e dt c

j F k l

c T k l c T

ππ

π

δ

+−∞ ∞+ −

=−∞ =−∞

∞

=−∞

= −

= − =

∑ ∑∫

∑

0 pt T+ 0t

5



We can get

Therefore

The integral for the Fourier series coefficients will be written as

That is, the signal x(t) and its Fourier series representation

are equal at every value of t

00

0

2( )pt T j lF t

l ptx t e dt c Tπ+ − =∫

021( )

p

j lF tl T

p

c x t e dtT

π−= ∫

00

0

21( )

pt T j lF tl t

p

c x t e dtT

π+ −= ∫

02( ) j kF tk

k

x t c e π∞

=−∞

= ∑



Dirichlet conditions: The signal has a finite number of discontinuities in any

period

The signal contains a finite number of maxima and minima during any period

The signal is absolutely integrable in any period, that is

All periodic signals of practical interest satisfy these conditions

( )x t

( )x t

( )x t

( )pT

x t dt < ∞∫

6



The weaker condition: signal has finite energy in one period,

guarantees that the energy in the difference signal

is zero, although x(t) and its Fourier series may not be equal for all value of t

Frequency Analysis of Continuous-Time Periodic Signals

02( ) ( ) j kF tk

k

e t x t c e π∞

=−∞

= − ∑

2( )

pTx t dt < ∞∫

Synthesis equation

Analysis equation

02( ) j kF tk

k

x t c e π∞

=−∞

= ∑021

( )p

j kF tk T

p

c x t e dtT

π−= ∫



In general, the Fourier coefficients ck are complex valued.

If the periodic signal is real, ck and c−k are complex conjugates.

As a result, if

Consequently, the Fourier series may also be represented in the form

where c0 is real valued when x(t) is real

k

k

jk k

jk k

c c e

c c e

θ

θ−−

=

=then

0 01

( ) 2 cos(2 )k kk

x t c c kF tπ θ∞

=

= + +∑

7



Finally, we should indicate that yet another form for the Fourier series can be obtained by expanding the cosine function above page as

Then, we can obtain

where

0 0 0cos(2 ) cos(2 ) cos( ) sin(2 )sin( )k k kkF t kF t kF tπ θ π θ π θ+ = −

0 0 01

( ) ( cos(2 ) sin(2 ))k kk

x t a a kF t b kF tπ π∞

=

= + −∑

0 0 , 2 cos , 2 sink k k k k ka c a c b cθ θ= = =

Power Density Spectrum of Periodic Signals


A periodic signal has infinite energy and a finite average power, which is given as

If we take the complex conjugate and substitute for , we obtain

21( )

px T

p

P x t dtT

= ∫

0 022 2* *1 1

( ) ( )p p

j kF t j kF tx k k kT T

k k kp p

P x t c e dt c x t e dt cT T

π π∞ ∞ ∞

− −

=−∞ =−∞ =−∞

= = =

∑ ∑ ∑∫ ∫

*( )x t

8



Therefore, we have established the relation

which is called Parseval’s relation for power signals.

Suppose that x(t) consists of a single complex exponential

In this case, all the Fourier series coefficients except ckare zero

The average power in the signal is

2 21( )

px kT

kp

P x t dt cT

∞

=−∞

= = ∑∫

02( ) j kF tkx t c e π=

2

x kP c=



|ck|2 represents the power in the k-th harmonic component of the signal

Hence the total average power in the periodic signal is simply the sum of the average powers in all the harmonics.

Power density spectrum of a continuous-time periodic signal

9



The Fourier series coefficients ck are complex valued, that is, they can be represented as

where

If the periodic signal is real valued, the Fourier series coefficients ck satisfy the condition

Consequently

kjk kc c e θ=

k kcθ =

k kc c∗− =

22

k kc c∗=



The total average power can be expressed as

Example : Determine the Fourier series and the power density spectrum of the rectangular pulse train signal.

22 2 2 20 0

1 1

12 ( )

2x k k kk k

P c c a a b∞ ∞

= =

= + = + +∑ ∑

10


Power Density Spectrum of Periodic Signals Solution : The signal is periodic with fundamental period

Tp and, clearly, satisfies the Dirichlet conditions. So the signal can be represented in the Fourier series

As x(t) is an even signal (i.e., x(t) = x(−t)), it is convenient to select the integration interval from −Tp/2 to Tp/2

The term c0 represents the average value (DC component) of the signal x(t). For k ≠ 0 we have

2 20

22

1 1( )

p

p

T

T

p p p

Ac x t dt Adt

T T T

τ

ττ

−−= = =∫ ∫

02 02

02

sin1, 1, 2,...j kF t

kp p

kFAc Ae dt k

T T kF

τπ

τ

π ττπ τ

−

−= = = ± ±∫



The function

The power density spectrum for the rectangular pulse train is

sinφφ

2

2

2 2

0

0

, 0

sin, 1, 2,...

p

k

p

Ak

Tc

kFAk

T kF

τ

π ττπ τ

= =

= ± ±

11



Consider an aperiodic signal x(t) with finite duration. From this aperiodic signal, we can create a periodic signal xp(t) with period Tp, as shown below


The Fourier Transform for Continuous-Time Aperiodic Signals

12



Clearly, xp(t) = x(t) in the limit as Tp → ∞, that is

This interpretation implies that we should be able to obtain the spectrum of x(t) from the spectrum of xp(t) simply by taking the limit as Tp → ∞

We begin with the Fourier series representation of xp(t)

Since xp(t) = x(t) for

( ) lim ( )p

pTx t x t

→∞=

02( ) j kF tp k

k

x t c e π∞

=−∞

= ∑ 022

2

1( )

p

p

Tj kF t

Tk pp

c x t e dtT

π−

−= ∫

/ 2 / 2p pT t T− ≤ ≤

022

2

1( )

p

p

Tj kF t

Tkp

c x t e dtT

π−

−= ∫

Consequently, the limits on the integral above can be replaced by and . Hence

Let us now define a function X(F), called the Fourier transform of x(t), as

X(F) is a function of the continuous variable F The Fourier coefficients ck can be expressed in terms of

X(F)



−∞ ∞021

( ) j kF tk

p

c x t e dtT

π∞ −

−∞= ∫

2( ) ( ) j FtX F x t e dtπ∞ −

−∞= ∫

0

1( )k

p

c X kFT

= 0( )p kp

kT c X kF X

T

= =

or equivalently

13

From above, we can obtain

We define , then

The limit as Tp approaches infinity, xp(t) reduces to x(t)

This integral relationship yields x(t) when X(F) is known, and it is called the inverse Fourier transform



021( ) j kF t

pkp p

kx t X e

T Tπ

∞

=−∞

=

∑

( ) 2( ) j k Ftp

k

x t X k F e Fπ∞

∆

=−∞

= ∆ ∆∑

( ) 2

0lim ( ) ( ) limp

j k Ftp

T Fk

x t x t X k F e Fπ∞

− ∆

→∞ ∆ →=−∞

= = ∆ ∆∑

1

p

FT

∆ =

( ) 2( ) j Ftx t X F e dFπ∞

−∞= ∫

Frequency Analysis of Continuous-Time Aperiodic Signals

Let , since



Synthesis equation(inverse transform)

Analysis equation(direct transform)

( ) 2( ) j Ftx t X F e dFπ∞

−∞= ∫

2( ) ( ) j FtX F x t e dtπ∞ −

−∞= ∫

( )

( )

1( )

2

( )

j t

j t

x t X e d

X x t e dt

π

∞ Ω

−∞

∞ − Ω

−∞

= Ω Ω

Ω =

∫

∫

2 FπΩ = / 2dF d π= Ω

14


Dirichlet conditions: The signal x(t) has a finite number of discontinuities.

The signal x(t) has a finite number of maxima and minima.

The signal x(t) is absolutely integrable, that is


( )x t dt∞

−∞< ∞∫

The third condition follows easily from the definition of the Fourier transform

Hence

A weaker condition for the existence of the Fourier transform is that x(t) has finite energy; that is,



( ) ( )2( ) j FtX F x t e dt x t dtπ∞ ∞−

−∞ −∞= ≤∫ ∫

( )X F < ∞

( )2

x t dt∞

−∞< ∞∫

15

Note that if a signal is absolutely integrable, it will also have finite energy. That is, if

then

For example, the signal is a square integrable but is not absolutely integrable. This signal has the Fourier transform



( )x t dt∞

−∞< ∞∫

( )2

xE x t dt∞

−∞= < ∞∫

( ) 0sin 2 F tx t

t

ππ

=

0

0

1,(F)=

0,

F FX

F F

≤

>

( )x t

Let x(t) be any finite energy signal with Fourier transform X(F). Its energy is

To express X(F) as follow


Energy Density Spectrum of AperiodicSignals

( )2

xE x t dt∞

−∞= ∫

( ) ( ) ( ) ( )

( ) ( )

( )

* * 2

* 2

2

j Ftx

j Ft

E x t x t dt x t dt X F e dF

X F dF x t e dt

X F dF

π

π

∞ ∞ ∞ −

−∞ −∞ −∞

∞ ∞ −

−∞ −∞

∞

−∞

= =

=

=

∫ ∫ ∫

∫ ∫

∫Parseval’s relation for aperiodic signal

16

Therefore, we conclude that

Polar form Magnitude |X(F)|, phase spectrum Energy density spectrum

Finally, as in the case of Fourier series, it is easily shown that if the signal x(t) is real, then

We obtain



( ) ( )2 2

xE x t dt X F dF∞ ∞

−∞ −∞= =∫ ∫( ) ( ) ( )j FX F X F e Θ=

( ) ( )F X FΘ =

2( ) ( )xxS F X F=

( ) ( ) , ( ) ( )X F X F X F X F− = − = −

( ) ( )xx xxS F S F− =

A periodic sequence x(n) with period N, that is,

The Fourier series representation for x(n) consists of Nharmonically related exponential functions

And is expressed as

The ck are the coefficients in the series representation.


The Fourier Series for Discrete-Time Periodic Signals

2

, 0,1,..., 1kn

jNe k Nπ

= −

21

0

( )j knN

Nk

k

x n c eπ−

=

=∑

( ) ( ),x n x n N n= + ∀

17

To derive the expression for the Fourier coefficients, we use the following formula:

The geometric summation formula



21

0

, 0, , 2 ,...

0,

knN jN

n

N k N Ne

otherwise

π−

=

= ± ±=

∑

1

0

, 1

1, 1

1

Nn N

n

N a

a aa

a

−

=

=

= −≠ −

∑

The expression for the Fourier coefficients ck can be obtained by multiplying both sides of

by the exponential and summing the product from n=0 to n = N – 1. Thus



2 2 ( )1 1 1

0 0 0

( )ln k l nN N Nj j

N Nk

n n k

x n e c eπ π −− − −−

= = =

=∑ ∑∑

21

0

( )j knN

Nk

k

x n c eπ−

=

=∑

2 lnj

Neπ

−

18

If we perform the summation over n first, we obtain

Hence

Frequency Analysis of Discrete-Time Periodic Signals



2 ( )1

0

, 1 0, , 2 ,...

0,

k l nN jN

n

N k N Ne

otherwise

π −−

=

− = ± ±=

∑21

0

1( ) , 0,1,..., 1

lnN jN

ln

c x n e l NN

π− −

=

= = −∑

Synthesis equation

Analysis equation

21

0

( )knN j

Nk

k

x n c eπ−

=

=∑21

0

1( )

knN jN

kn

c x n eN

π− −

=

= ∑

Discrete-time Fourier Series (DTFS)

The average power of a discrete-time periodic signal with period N was defined as

We shall now derive an expression for Px in terms of the Fourier coefficient cx. We have

Then

If we are interested in the energy of the sequence x(n) over a single period, it implies that



12

0

1( )

N

xn

P x nN

−

=

= ∑

21 1 1* *

0 0 0

1 1( ) ( ) ( )

knN N N jN

x kn n k

P x n x n x n c eN N

π− − − −

= = =

= =

∑ ∑ ∑

21 1 1 12 2*

0 0 0 0

1 1( ) ( )

knN N N NjN

x k kk n k n

P c x n e c x nN N

π− − − −−

= = = =

= = =

∑ ∑ ∑ ∑

1 12 2

0 0

( )N N

N kn k

E x n N c− −

= =

= =∑ ∑

19

If the signal x(n) is real, then, we can easily show that

Or equivalently

We obtain

More specifically, we have



*k kc c−=

Even symmetry

Odd symmetry

k k

k k

c c

c c−

−

=

− =

k N k

k N k

c c

c c−

−

=

= −

0 0

1 1 1 1

/2 /2 2

( 1)/2 ( 1)/2 ( 1)/2 ( 1)/2

, 0

,

, 0,

, ,

N N

N N

N N N

N N N N

c c c c

c c c c

c c c

c c c c

− −

−

− + − +

= = − =

= = −

= =

= = −

if N is evenif N is odd

The Fourier series can also be expressed in the alternative forms

Where , ,



01

01

2( ) 2 cos

2 2cos sin

L

k kk

L

k kk

x n c c knN

a a kn b knN N

πθ

π π=

=

= + +

= + −

∑

∑

0 0a c= 2 cosk k ka c θ= 2 sink k kb c θ=

/ 2

( 1) / 2

NL

N

=

−

If N is even

If N is odd

20

The Fourier transform of a finite-energy discrete-time signal x(n) is defined as

X(ω) represents the frequency content of the signal x(n)

X(ω) is a decomposition of x(n) into its frequency components.


The Fourier Transform of Discrete-Time Aperiodic Signals

( ) ( ) j n

n

X x n e ωω∞

−

=−∞

= ∑

X(ω) is periodic with period 2π, that is

Hence X(ω) is periodic with period 2π Let us evaluate the sequence x(n) from X(ω). Thus we

have

This interchange can be made if the series

converges uniformly to X(ω) as N → ∞



( 2 ) 2( 2 ) ( ) ( )

( ) ( )

j k n j n j kn

n n

j n

n

X k x n e x n e e

x n e X

ω π ω π

ω

ω π

ω

∞ ∞− + − −

=−∞ =−∞

∞−

=−∞

+ = =

= =

∑ ∑

∑

( ) ( )j m j n j m

n

X e d x n e e dπ πω ω ω

π πω ω ω

∞−

− −=−∞

=

∑∫ ∫

( ) ( )N

j nN

n N

X x n e ωω −

=−

= ∑

21

We can interchange the order of summation and integration

Consequently,

The desired result that



( ) 2 ,

0,j m n m n

e dm n

π ω

π

πω−

−

==

≠∫

( ) 2 ( ),( )

0,j m n

n

x m m nx n e d

m n

π ω

π

πω

∞−

−=−∞

==

≠∑ ∫

1( ) ( )

2j nx n X e d

π ω

πω ω

π −= ∫

Synthesis equation(inverse transform)

Analysis equation(direct transform)

2

1( ) ( )

2j nx n X e dω

πω ω

π= ∫

( ) ( ) j n

n

X x n e ωω∞

−

=−∞

= ∑

We assume that the seriesconverges uniformly to. By uniform convergence we mean that for each ω

Uniform convergence is guaranteed if x(n) is absolutely summable. If

then

Some sequences are not absolutely summable, but they are square summable. That is, they have finite energy


Convergence of the Fourier Transform( ) ( )

Nj n

Nn N

X x n e ωω −

=−

= ∑

lim sup ( ) ( ) 0NN

X Xω ω ω→∞

− =

( )n

x n∞

=−∞

< ∞∑

( ) ( ) ( )j n

n n

X x n e x nωω∞ ∞

−

=−∞ =−∞

= ≤ < ∞∑ ∑

2( )x

n

E x n∞

=−∞

= < ∞∑

22

This leads to a mean-square convergence condition:

the energy in the error X(ω) − XN(ω) tends toward zero

Example - Suppose that

X(ω) is periodic with period 2π The inverse transform of X(ω) results in the sequence


Convergence of the Fourier Transform

2( )x

n

E x n∞

=−∞

= < ∞∑

2lim ( ) ( ) 0NN

X X dπ

πω ω ω

−→∞− =∫

1,( )

0,c

c

Xω ω

ωω ω π

≤=

≤ ≤

sin1 1( ) ( ) , 0

2 2c

c

j n j n cnx n X e d e d nn

π ωω ω

π ω

ωω ω ω

π π π− −= = = ≠∫ ∫

For n = 0, we have

Hence

The sequence x(n) is expressed as



1(0)

2c c

cx dω ω

ω ωπ π

= − =∫

, 0

( )sin

, 0

c

c c

c

n

x nn

nn

ωπ

ω ωπ ω

== ≠

sin( ) ,cnx n n

n

ωπ

= −∞ < < ∞

23

Note, the above he sequence x(n) is not absolutely summable. Hence the infinite series

does not converge uniformly for all ω To elaborate on this point, let us

consider the finite sum

Right figures shows the function XN(ω) for several values of N.



sin( )

Nj nc

Nn N

nX e

nωω

ωπ

−

=−

= ∑

sin( ) j n j nc

n n

nx n e e

nω ωω

π

∞ ∞− −

=−∞ =−∞

=∑ ∑

The energy of a discrete-time signal x(n) is defined as

The energy Ex in terms of the spectral characteristic X(ω), first we have

Then



2( )x

n

E x n∞

=−∞

= ∑

* *1( ) ( ) ( ) ( )

2j n

xn n

E x n x n x n X e dπ ω

πω ω

π

∞ ∞−

−=−∞ =−∞

= = ∑ ∑ ∫

2*1 1( ) ( ) ( )

2 2j n

xn

E X x n e d X dπ πω

π πω ω ω ω

π π

∞−

− −=−∞

= =

∑∫ ∫

24

The energy relation between x(n) and X(ω) is

this is Parseval’s relation for discrete-time aperiodicsignals with finite energy.

The spectrum X(ω) is, in general, a complex-valued function of frequency. It may be expressed as



2 21( ) ( )

2xn

E x n X dπ

πω ω

π

∞

−=−∞

= =∑ ∫

( )( ) ( ) jX X e ωω ω Θ=

where phase spectrum( ) ( )Xω ωΘ =

As in the case of continuous-time signals, the quantity

represents the distribution of energy as a function of frequency, namely the energy density spectrum of x(n)

Clearly, Sxx(ω) does not contain any phase information If x(n) is real, then it easily follows that

or equivalently

It also follows that



2( ) ( )xxS Xω ω=

*( ) ( )X Xω ω= −

( ) ( )

( ) ( )

X X

X X

ω ω

ω ω

− =

− = −

even symmetryodd symmetry

( ) ( )xx xxS Sω ω− = even symmetry

25

Example - Determine and sketch the energy density spectrum of the signal

Solution : Since , the sequence is absolutely summable, as

can be verified by applying the geometric summation formula,

Hence the Fourier transform of exists and is obtained. Thus



( )xxS ω

( ) ( ), 1 1nx n a u n a= − < <

0

1( )

1n

n n

x n aa

∞ ∞

=−∞ =

= = < ∞−∑ ∑

0 0

( ) ( )n j n j n

n n

x a e aeω ωω∞ ∞

− −

= =

= =∑ ∑

1a < ( )x n

( )x n

Since , use of the geometric summation formula again yields

The energy density spectrum is given by

Equivalently

Note that



2 1( ) ( ) ( ) ( )

(1 )(1 )xx j jS X X X

ae aeω ωω ω ω ω−

= = =− −

1( )

1 jx

ae ωω−

=−

2

1( )

1 2 cosxxSa a

ωω

=− +

1jae aω− = <

( ) ( )xx xxS Sω ω− =

26



The z-transform of a sequence x(n) is defined as

where is the region of convergence of X(z)

Let us express the complex variable z in polar form as

where and .


Relationship of the Fourier Transform to the z-Transform

2 1( ) ( ) , :n

n

X z x n z ROC r z r∞

−

=−∞

= < <∑

jz re ω=

2 1r z r< <

r z= zω =

27

Then, within the region of convergence of X(z), we can substitute in above

If X(z) converges for |z| = 1 , then



( ) | ( )j

n j n

z ren

X z x n r eωω

∞− −

==−∞

= ∑

( ) | ( ) ( )j

j n

z en

X z X x n eωωω

∞−

==−∞

≡ = ∑

jz re ω=



( ) n

n

x n r∞

−

=−∞

< ∞∑

The existence of the z-transform requires that the sequence

be absolutely summable for some value of r, that is

( ) nx n r−

28

There are sequences, however, that do not satisfy the requirement in above equation, for example, the sequence

This sequence does not have a z-transform. Since it has a finite energy, its Fourier transform converges in the mean-square sense to the discontinuous function X(ω), defined as



sin( ) ,cnx n n

n

ωπ

= −∞ < < ∞

1,( )

0,c

c

Xω ω

ωω ω π

<=

< ≤

There are some aperiodic sequences that are neither absolutely summable nor square summable. Hence their Fourier transforms do not exist. One such sequence is the unit step sequence, which has the z-transform

Another such sequence is the causal sinusodial signal sequence . This sequence has the z-transform


The Fourier Transform of Signals with Poles on the Unit Circle

1

1( )

1X z

z−=

−

10

1 20

1 cos( )

1 2 cos

zX z

z z

ωω

−

− −

−=

− +

0( ) (cos ) ( )x n n u nω=

29


Frequency-Domain Classification of Signals: The concept of Bandwidth

Low-frequency

High-frequency

Medium-frequency


Some examples of bandlimited signals

Frequency-Domain Classification of Signals: The concept of Bandwidth

30


The Frequency Ranges of Some Natural Signals

Frequency ranges of some biological signals



Frequency ranges of some seismic signals

31


Frequency ranges of electromagnetic signals



Frequency-Domain and Time-Domain Signal Properties

To summarize, the following frequency analysis tools have been introduced:

The Fourier series for continuous-time periodic signals.

The Fourier transform for continuous-time aperiodic signals.

The Fourier series for discrete-time periodic signals.

The Fourier transform for discrete-time aperiodic signals.

32



Let us briefly summarize the results of the previous sections.

Continuous-time signals have aperiodic spectra.

Discrete-time signals have periodic spectra.

Periodic signals have discrete spectra.

Aperiodic finite energy signals have continuous spectra.


Frequency-Domain and Time-Domain Signal Properties Summary of analysis and synthesis formulas

33



We observe that there are dualities between the following analysis and synthesis equations: The analysis and synthesis equations of the continuous-time

Fourier transform.

The analysis and synthesis equations of the discrete-time Fourier series.

The analysis equation of the continuous-time Fourier series and the synthesis equation of the discrete-time Fourier transform.

The analysis equation of the discrete-time Fourier transform and the synthesis equation of the continuous-time Fourier series.


Properties of the Fourier Transform for Discrete-Time Signals

The Fourier transform for aperiodic finite-energy discrete-time signals described in the preceding section possesses a number of properties that are very useful in reducing the complexity of frequency analysis problems in many practical applications.

We develop the important properties of the Fourier transform. Similar properties hold for the Fourier transform of aperiodic finite-energy continuous-time signals.

34

For convenience, we adopt the notation

For the direct transform (analysis equation) and

For the inverse transform (synthesis equation). We also refer to x(n) and X(ω) as a Fourier transform pair and denote this relationship with the notation


Properties of the Fourier Transform for Discrete-Time Signals

( ) ( ) ( ) j n

n

X F x n x n e ωω∞

−

=−∞

≡ = ∑

1

2

1( ) ( ) ( )

2j nx n F X X e dω

πω ω ω

π−≡ = ∫

( ) ( )Fx n X ω←→

Suppose that both the signal x(n) and its transform X(ω)are complex-valued functions. Then they can be expressed in rectangular form as

By substituting above and andseparating the real and imaginary parts, we obtain


Symmetry Properties of the Fourier Transform

( ) ( ) ( )

( ) ( ) ( )R I

R I

x n x n jx n

X X jXω ω ω

= +

= +

cos sinje jω ω ω− = −

[ ]

[ ]

( ) ( ) cos ( ) sin

( ) ( ) sin ( ) cos

R R In

I R In

X x n n x n n

X x n n x n n

ω ω ω

ω ω ω

∞

=−∞

∞

=−∞

= +

= − −

∑

∑

35

In a similar manner, by substituting above and

we obtain



[ ]

[ ]

2

2

1( ) ( ) cos ( ) sin

21

( ) ( ) sin ( ) cos2

R R I

I R I

x n X n X n d

x n X n X n d

π

π

ω ω ω ω ωπ

ω ω ω ω ωπ

= −

= +

∫

∫

cos sinje jω ω ω= +

Real signals If is real, then and . Then

Since and , it follows from above



( ) ( ) cos

( ) ( )sin

Rn

In

X x n n

X x n n

ω ω

ω ω

∞

=−∞

∞

=−∞

=

= −

∑

∑

( )x n ( ) ( )Rx n x n= ( ) 0Ix n =

cos( ) cosn nω ω− = sin( ) sinn nω ω− = −

( ) ( )

( ) ( )R R

I I

X X

X X

ω ω

ω ω

− =

− = −

(even)

(odd)

36

If we combine above into a single equation, we have

In this case we say that the spectrum of a real signal has Hermitian symmetry.

We observe that the magnitude and phase spectra for real signals are



*( ) ( )X Xω ω= −

2 2

1

( ) ( ) ( )

( )tan

( )

R I

I

R

X X X

XX

X

ω ω ω

ωω

ω−

= +

=

The magnitude and phase spectra also possess the symmetry properties



( ) ( )

( ) ( )

X X

X X

ω ω

ω ω

= −

− = −

(even)

(odd)

37

In the case of the inverse transform of a real-valued signal [i.e., ], which implies that

Since both products and are even functions of , we have



[ ]2

1( ) ( ) cos ( )sin

2 R Ix n X n X n dπ

ω ω ω ω ωπ

= −∫

[ ]0

1( ) ( ) cos ( )sinR Ix n X n X n d

πω ω ω ω ω

π= −∫

( ) ( )Rx n x n=

( ) cosRX nω ω ( )sinIX nω ωω

Real and even signals If is real and even [i.e., ], then

is even and is odd. Hence

Thus real and even signals possess real-valued spectra, which, in addition, are even functions of the frequency variable .



1

0

( ) (0) 2 ( ) cos

( ) 0

1( ) ( ) cos

Rn

I

R

X x x n n

X

x n X ndπ

ω ω

ω

ω ω ωπ

∞

=

= +

=

=

∑

∫

(even)

( )x n ( ) ( )x n x n− =( ) cosx n nω ( )sinx n nω

ω

38



Real and odd signals If is real and odd [i.e., ], then

is odd and is even. Hence

Thus real-valued odd signals possess purely imaginary-valued spectra characteristics, which, in addition, are odd functions of the frequency variable .

1

0

( ) 0

( ) 2 ( )sin

1( ) ( )sin

R

In

I

X

X x n n

x n X ndπ

ω

ω ω

ω ω ωπ

∞

=

=

= −

= −

∑

∫

(odd)

( ) ( )x n x n− = −( ) cosx n nω ( )sinx n nω

( )x n

ω



Purely imaginary signals In this case and . Then

[ ]0

( ) ( ) sin

( ) ( ) cos

1( ) ( )sin ( ) cos

R In

I In

I R I

X x n n

X x n n

x n X n X n dπ

ω ω

ω ω

ω ω ω ω ωπ

∞

=−∞

∞

=−∞

=

=

= +

∑

∑

∫

(odd)

(even)

( ) 0Rx n = ( ) ( )Ix n jx n=

39



Purely imaginary signals If is odd [i.e., ], then

1

0

( ) 2 ( ) sin

( ) 0

1( ) ( )sin

R In

I

I R

X x n n

X

x n X ndπ

ω ω

ω

ω ω ωπ

∞

=

=

=

=

∑

∫

(odd)

( )Ix n ( ) ( )I Ix n x n− = −



Purely imaginary signals Similarly, If is even [i.e., ], then

An arbitrary, possibly complex-valued signal can be decomposed as

1

0

( ) 0

( ) (0) 2 ( ) cos

1( ) ( ) cos

R

I I In

I I

X

X x x n n

x n X ndπ

ω

ω ω

ω ω ωπ

∞

=

=

= +

=

∑

∫

(even)

( )Ix n ( ) ( )I Ix n x n− =

( )x n

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )e o e oR I R R I I e ox n x n jx n x n jx n j x n x n x n x n = + = + + + = +

40



Purely imaginary signals By definition

The superscripts e and o denote the even and odd signal components, respectively. We note that

and .

*

*

1( ) ( ) ( ) ( ) ( )

21

( ) ( ) ( ) ( ) ( )2

e ee R I

o oo R I

x n x n jx n x n x n

x n x n jx n x n x n

= + = + −

= + = − −

( ) ( )e ex n x n= − ( ) ( )o ox n x n− = −



Purely imaginary signals From above and the Fourier transform properties

established, we obtain the following relationships:

( ) ( ) ( ) ( ) ( ) ( ) ( )e e o oR I R I e ox n x n jx n x n jx n x n x n = + + + = +

( ) ( ) ( ) ( ) ( ) ( ) ( )e e o oR I R I e oX X jX X jX X Xω ω ω ω ω ω ω = + + − = +

41


Symmetry Properties of the Fourier Transform Symmetry properties of the discrete-time Fourier transform



Summary of symmetry properties of the Fourier transform

42


Fourier Transform Theorems and Properties

Linearity

1 1

2 2

( ) ( )

( ) ( )

F

F

x n X

x n X

ω

ω

←→

←→

1 1 2 2 1 1 2 2( ) ( ) ( ) ( )Fa x n a x n a X a Xω ω+ ←→ +

If

then



Example : Determine the Fourier transform of the signal

Solution : First, we observe that can be expressed as

( ) , 1 1nx n a a= − < <

( )x n

1 2( ) ( ) ( )x n x n x n= +

where 1

2

, 0( )

0, 0

, 0( )

0, 0

n

n

a nx n

n

a nx n

n

−

≥=

<

<=

≥and

43

Beginning with the definition of the Fourier transform, we have

The summation is a geometric series that converges to

Provided that



1 10 0

( ) ( ) ( )j n n j n j n

n n n

X x n e a e aeω ω ωω∞ ∞ ∞

− − −

=−∞ = =

= = =∑ ∑ ∑

1

1( )

1 jX

ae ωω−

=−

1j jae a e aω ω− −= ⋅ = <

Which is a condition that is satisfied in this problem. Similarly, the Fourier transform of is

By combining these two transforms, we obtain the Fourier transform of in the form



1 1

2 21

( ) ( ) ( ) ( )1

jj n n j n j n j k

jn n n k

aeX x n e a e ae ae

ae

ωω ω ω ω

ωω∞ − − ∞

− − − −

=−∞ =−∞ =−∞ =

= = = = =−∑ ∑ ∑ ∑

2

1 2 2

1( ) ( ) ( )

1 2 cos

aX X X

a aω ω ω

ω−

= + =− +

2 ( )x n

( )x n

44



The figure illustrates and for the case in which .

( )x n ( )X ω0.8a =

Time shifting

The proof of this property follows immediately from the Fourier transform of by making a change in the summation index. Thus



( ) ( )

( ) ( )

F

F j k

x n X

x n k e Xω

ω

ω−

←→

− ←→

[ ]( )( ) ( ) ( ) j X kj kF x n k X e X e ω ωωω ω −−− = =

If

then

( )x n k−

45

Time reversal

This property can be established by performing the Fourier transformation of by making a simple change in the summation index. Thus



( ) ( )

( ) ( )

F

F

x n X

x n X

ω

ω

←→

− ←→ −

( ) ( ) ( )j l

l

F x n x l e Xω ω∞

=−∞

− = = −∑

If

then

( )x n−

If is real, then we obtain



( ) ( )( ) ( ) ( ) ( )j X j XF x n X X e X eω ωω ω ω− −− = − = − =

( )x n

46

Convolution theorem

To prove above, we recall the convolution formula



1 1

2 2

( ) ( )

( ) ( )

F

F

x n X

x n X

ω

ω

←→

←→

1 2 1 2( ) ( ) ( ) ( ) ( ) ( )Fx n x n x n X X Xω ω ω= ∗ ←→ =

1 2 1 2( ) ( ) ( ) ( ) ( )k

x n x n x n x k x n k∞

=−∞

= ∗ = −∑

If

and

then

By multiplying both side of this equation by the exponential and summing over all n, we obtain



1 2( ) ( ) ( ) ( )j n j n

n n k

X x n e x n x n k eω ωω∞ ∞ ∞

− −

=−∞ =−∞ =−∞

= = −

∑ ∑ ∑

j ne ω−

47

The correlation theorem

In this case, we have



1 1

2 2

( ) ( )

( ) ( )

F

F

x n X

x n X

ω

ω

←→

←→

1 2 1 2 1 2( ) ( ) ( ) ( )Fx x x xr m S X Xω ω ω←→ = −

1 2 1 2( ) ( ) ( )x xk

r n x k x k n∞

=−∞

= −∑

If

and

then

By multiplying both sides of this equation by the exponential and summing over all n, we obtain

The function is called the cross-energy density spectrum of the signals and .



1 2 1 2 1 2( ) ( ) ( ) ( )j n j nx x x x

n n k

S r n e x k x k n eω ωω∞ ∞ ∞

− −

=−∞ =−∞ =−∞

= = −

∑ ∑ ∑

j ne ω−

1 2( )x xS ω

1( )x n 2 ( )x n

48

The Wiener-Khintchine theorem Let be a real signal. Then

That is, the energy spectral density of an energy signal is the Fourier transform of its autocorrelation sequence.



( )x n

( ) ( )Fxx xxr l S ω←→

Frequency shifting

Illustration of the frequency-shifting property of the Fourier transform .



( ) ( )Fx n X ω←→0

0( ) ( )j n Fe x n Xω ω ω←→ −

If

then

0 2 mω π ω≤ −

49

The modulation theorem



( ) ( )Fx n X ω←→

[ ]0 0 0

1( )cos ( ) ( )

2Fx n n X Xω ω ω ω ω←→ + + −

If

then

Parseval’s theorem



1 1

2 2

( ) ( )

( ) ( )

F

F

x n X

x n X

ω

ω

←→

←→

* *1 2 1 2

1( ) ( ) ( ) ( )

2n

x n x n X X dπ

πω ω ω

π

∞

−=−∞

=∑ ∫

If

and

then

50

To prove this theorem, we eliminate on the right-hand side of above. Thus we have



*1 22

*1 22

*1 2

1( ) ( )

2

1( ) ( )

2

( ) ( )

j n

n

j n

n

n

x n e X d

x n X e d

x n x n

ω

π

ω

π

ω ωπ

ω ωπ

∞−

=−∞

∞−

=−∞

∞

=−∞

=

=

∑∫

∑ ∫

∑

1( )X ω

In the special case where , Parseval’s relation reduces to

Therefore, we conclude that



2 2

2

1( ) ( )

2n

x n X dπ

ω ωπ

∞

=−∞

=∑ ∫

2 2

2

1 1(0) ( ) ( ) ( )

2 2x xx xxn

E r x n X d S dπ

π πω ω ω ω

π π

∞

−=−∞

= = = =∑ ∫ ∫

2 1( ) ( ) ( )x n x n x n= =

51

Multiplication of two sequences (Windows theorem)



1 1

2 2

( ) ( )

( ) ( )

F

F

x n X

x n X

ω

ω

←→

←→

3 1 2 3

1 2

( ) ( ) ( ) ( )

1( ) ( )

2

Fx n x n x n X

X X dπ

π

ω

λ ω λ λπ −

≡ ←→

= −∫

If

and

then

We begin with the Fourier transform of and use the formula for the inverse transform, namely,

Thus, we have



1 1

1( ) ( )

2j nx n X e d

π λ

πλ λ

π −= ∫

3 3 1 2

( )1 2 1 2

1 2

( ) ( ) ( ) ( )

1 1( ) ( ) ( ) ( )

2 2

1( ) ( )

2

j n j n

n n

j n j n j n

n n

X x n e x n x n e

X e d x n e X d x n e

X X d

ω ω

π πλ ω ω λ

π π

π

π

ω

λ λ λ λπ π

λ ω λ λπ

∞ ∞− −

=−∞ =−∞

∞ ∞− − −

− −=−∞ =−∞

−

= =

= =

= −

∑ ∑

∑ ∑∫ ∫

∫

3 1 2( ) ( ) ( )x n x n x n=

52

Differentiation in the frequency domain

To prove this property, we use the definition of the Fourier transform and differentiae the series term by term with respect to . Thus we obtain



( ) ( )Fx n X ω←→( )

( ) F dXnx n j

d

ωω

←→

( )( ) ( ) ( )j n j n j n

n n n

dX d dx n e x n e j nx n e

d d dω ω ωω

ω ω ω

∞ ∞ ∞− − −

=−∞ =−∞ =−∞

= = = −

∑ ∑ ∑

If

then

ω



Properties of the Fourier transform for discrete-time signals

53



Some useful Fourier transform pairs for discrete-time aperiodic signals

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