chapter 4 continuous-time fourier transform

ELG 3120 Signals and Systems Chapter 4

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Chapter 4 Continuous-Time Fourier Transform

4.0 Introduction • A periodic signal can be represented as linear combination of complex exponentials which

are harmonically related. • An aperiodic signal can be represented as linear combination of complex exponentials, which

are infinitesimally close in frequency. So the representation take the form of an integral rather than a sum

• In the Fourier series representation, as the period increases the fundamental frequency decreases and the harmonically related components become closer in frequency. As the period becomes infinite, the frequency components form a continuum and the Fourier series becomes an integral.

4.1 Representation of Aperiodic Signals: The Continuous-Time Fourier Transform

4.1.1 Development of the Fourier Transform Representation of an Aperiodic Signal Starting from the Fourier series representation for the continuous-time periodic square wave:

<<

<=

2/,0

,1)(

1

1

TtT

Tttx , (4.1)

1T1T−

2

T T

2

T−

T− T2T2−

)(tx

The Fourier coefficients ka for this square wave are

TkTk

ak0

10 )sin(2ω

ω= . (4.2)

or alternatively


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0

)sin(2 1

ωωωω

kk

TTa

=

= , (4.3)

where ωω /)sin(2 1T represent the envelope of kTa • When T increases or the fundamental frequency T/20 πω = decreases, the envelope is

sampled with a closer and closer spacing. As T becomes arbitrarily large, the original periodic square wave approaches a rectangular pulse.

• kTa becomes more and more closely spaced samples of the envelope, as ∞→T , the Fourier

series coefficients approaches the envelope function.

This example illustrates the basic idea behind Fourier’s development of a representation for aperiodic signals. Based on this idea, we can derive the Fourier transform for aperiodic signals. Suppose a signal )(tx with a finite duration, that is, 0)( =tx for 1Tt > , as illustrated in the figure below. • From this aperiodic signal, we construct a periodic signal )(~ tx , shown in the figure below.


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• As ∞→T , )()(~ txtx = , for any infinite value of t . • The Fourier series representation of )(~ tx is

∑+∞

−∞=

=k

tjkk eatx 0)(~ ω , (4.4)

dtetxT

aT

T

tjkk ∫−

−=2/

2/0)(~1 ω . (4.5)

• Since )()(~ txtx = for 2/Tt < , and also, since 0)( =tx outside this interval, so we have

dtetxT

dtetxT

a tjkT

T

tjkk ∫∫

∞

∞−

−

−

− == 00 )(1

)(1 2/

2/

ωω .

• Define the envelope )( ωjX of kTa as

∫∞

∞−

−= dtetxjX tjωω )()( . (4.6)

we have for the coefficients ka ,

)(1

0ωjkXT

ak =

Then )(~ tx can be expressed in terms of )( ωjX , that is

00000 )(

21

)(1

)(~ ωωπ

ω ωω ∑∑+∞

−∞=

+∞

−∞=

==k

tjk

k

tjk ejkXejkXT

tx . (4.7)


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• As ∞→T , )()(~ txtx = and consequently, Eq. (4.7) becomes a representation of )(tx . • In addition, 00 →ω as ∞→T , and the right-hand side of Eq. (4.7) becomes an integral. We have the following Fourier transform:

∫∞

∞−= ωω

πω dejXtx tj)(

21

)( TransformFourierInverse (4.8)

and

∫∞

∞−

−= dtetxjX tjωω )()( TransformFourier (4.9)

4.1.2 Convergence of Fourier Transform If the signal )(tx has finite energy, that is, it is square integrable,

∞<∫∞

∞−dttx

2

)( , (4.10)

Then we guaranteed that )( ωjX is finite or Eq. (4.9) converges. If )()(~)( txtxte −= , we have

0)(2

=∫∞

∞−dtte . (4.11)

An alterative set of conditions that are sufficient to ensure the convergence: Contition1: Over any period, )(tx must be absolutely integrable, that is

∞<∫∞

∞−dttx )( , (4.12)

Condition 2: In any finite interval of time, )(tx have a finite number of maxima and mi nima. Condition 3: In any finite interval of time, there are only a finite number of discontinuities. Furthermore, each of these discontinuities is finite.


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4.1.3 Examples of Continuous-Time Fourier Transform Example : consider signal )()( tuetx at−= , 0>a . From Eq. (4.9),

ωωω ωω

jae

jadteejX tjatjat

+=

+−==

∞

+−∞ −−∫11)(

0

)(

0, 0>a (4.12)

If a is complex rather then real, we get the same result if { } 0Re >a The Fourier transform can be plotted in terms of the magnitude and phase, as shown in the figure below.

22

1)(

ωω

+=

ajX ,

−=∠ −

ajX

ωω 1tan)( . (4.13)

Example : Let taetx −=)( , 0>a

220

0 211)(

ωωωω ωωω

+=

++

−=+== ∫∫∫

∞ −−

∞−

−∞

∞−

−−

aa

jajadteedteedteejX tjattjattjta

The signal and the Fourier transform are sketched in the figure below.


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Example : )()( ttx δ= . (4.14)

1)()( == ∫∞

∞−

− dtetjX tjωδω . (4.15)

That is, the impulse has a Fourier transform consisting of equal contributions at all frequencies. Example : Calculate the Fourier transform of the rectangular pulse signal

>

<=

1

1

,0

,1)(

Tt

Tttx . (4.16)

1T1T−

)(tx

1

ωω

ω ωω 1sin21)()( 1

1

TdtedtetxjX

T

T

tjtj === ∫∫ −

−∞

∞−

− .

(4.17) The Inverse Fourier transform is

∫∞

∞−= ω

ωω

πω de

Ttx tj1sin

221)(ˆ , (4.18)

Since the signal )(tx is square integrable,

0)(ˆ)()(2

=−= ∫∞

∞−dttxtxte . (4.19)

)(ˆ tx converges to )(tx everywhere except at the discontinuity, 1Tt ±= , where )(ˆ tx converges to

½, which is the average value of )(tx on both sides of the discontinuity. In addition, the convergence of )(ˆ tx to )(tx also exhibits Gibbs phenomenon. Specifically, the integral over a finite-length interval of frequencies

∫−

W

W

tj deT

ωωω

πω1sin

221

)()( ttx δ= 1)( =ωjX


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As ∞→W , this signal converges to )(tx everywhere, except at the discontinuities. More over, the signal exhibits ripples near the discontinuities. The peak values of these ripples do not decrease as W increases, although the ripples do become compressed toward the discontinuity, and the energy in the ripples converges to zero. Example : Consider the signal whose Fourier transform is

>

<=

W

WjX

ω

ωω

,0

,1)( .

The Inverse Fourier transform is

tWt

detxW

W

tj

πω

πω sin

21

)( == ∫−.

Comparing the results in the preceding example and this example, we have

functionSincwaveSquareFT

FT

←→

−1

This means a square wave in the time domain, its Fourier transform is a sinc function. However, if the signal in the time domain is a sinc function, then its Fourier transform is a square wave. This property is referred to as Duality Property. We also note that when the width of )( ωjX increases, its inverse Fourier transform )(tx will be compressed. When ∞→W , )( ωjX converges to an impulse. The transform pair with several different values of W is shown in the figure below.


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4.2 The Fourier Transform for Periodic Signals The Fourier series representation of the signal )(tx is

∑∞

−∞=

=k

tjkkeatx 0)( ω . (4.20)

It’s Fourier transform is

)(2)( 0ωωδπω kajXk

k∑∞

−∞=

−= .

(4.21) Example : If the Fourier series coefficients for the square wave below are given

1T1T−

2

T T

2

T−

T− T2T2−

)(tx

kTk

ak πω 10sin

= , (4.22)

The Fourier transform of this signal is

)(sin2

)( 010 ωωδ

ωω k

kTk

jXk∑

∞

−∞=

−= . (4.23)


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Example : The Fourier transforms for ttx 0sin)( ω= and ttx 0cos)( ω= are shown in the figure below.


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Example : Calculate the Fourier transform for signal ∑∞

−∞=

−=k

kTttx )()( δ .

The Fourier series of this signal is

∫+

−

− ==2/

2/

1)(

10

T

T

tjk T

etT

a ωδ .

The Fourier transform is

)2

(2

)(0T

kT

jXk

πωδ

πω ∑

∞

−∞=

−= .

The Fourier transform of a periodic impulse train in the time domain with period T is a periodic impulse train in the frequency domain with period T/2π , as sketched din the figure below.

4.3 Properties of The Continuous-Time Fourier Transform

4.3.1 Linearity If )()( ωjXtx F→← and )()( ωjYty F→← Then


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)()()()( ωω jbYjaXtbytax F +→←+ . (4. 20)

4.3.2 Time Shifting If )()( ωjXtx F→← Then

)()( 00 ωω jXettx tjF −→←− . (4. 20)

Or

{ } [ ]00 )(0 )()()( tjXjtj ejXjXettxF ωωω ωω −∠− ==− . (4. 20)

Thus, the effect of a time shift on a signal is to introduce into its transform a phase shift, namely,

t0ω− . Example : To evaluate the Fourier transform of the signal )(tx shown in the figure below.

)(tx

t1 2 3 4

11.5

)(2 tx

t

1

2

3

2

3−

1

)(1 tx

t

2

1−

2

1

The signal )(tx can be expressed as the linear combination

)5.2()5.2(21

)( 21 −+−= txtxtx . (4. 20)

)(1 tx and )(2 tx are rectangular pulse signals and their Fourier transforms are


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ωω

ω)2/sin(2

)(1 =jX and ω

ωω

)2/3sin(2)(2 =jX

Using the linearity and time-shifting properties of the Fourier transform yields

+

= −

ωωω

ω ω )2/3sin(2)2/sin()( 2/5jejX

4.3.3 Conjugation and Conjugate Symmetry If )()( ωjXtx F→← Then

)(*)(* ωjXtx F −→← . (4. 20)

Since ∫∫+∞

∞−

∗+∞

∞−

− =

= dtetxdtetxjX tjtj ωωω )(*)()(* ,

Replacing ω by ω− , we see that

∫+∞

∞−

−=− dtetxjX tjωω )(*)(* , (4. 20)

The right-hand side is the Fourier transform of )(* tx . If )(tx is real, from Eq. (4.20) we can get

)(*)( ωω jXjX =− . (4. 20) We can also prove that if )(tx is both real and even, then )( ωjX will also be real and even. Similarly, if )(tx is both real and odd, then )( ωjX will also be purely imaginary and odd. A real function )(tx can be expressed in terms of the sum of an even function

{ })()( txEvtxe = and an odd function { })()( txOdtxo = . That is

)()()( txtxtx oe += Form the Linearity property,


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{ } { } { })()()( txFtxFtxF oe += , From the preceding discussion, { })(txF e is real function and { })(txF o is purely imaginary. Thus we conclude with )(tx real,

)()( ωjXtx F→←

{ } { })(Re)( ωjXtxEv F→←

{ } { })(Im)( ωjXjtxOd F→← Example : Using the symmetry properties of the Fourier transform and the result

ωjatue Fat

+→←− 1

)( to evaluate the Fourier transform of the signal taetx −=)( , where 0>a .

Since { })(22

)()(2)()()( tueEv

tuetuetuetueetx at

atatatatta −

−−− =

−+=−+== ,

So 22

21Re2)(

ωωω

+=

+

=a

aja

jX

4.3.4 Differentiation and Integration If )()( ωjXtx F→← Then

)()(

ωω jXjdt

tdx F→←. (4. 20)

)()0()(1

)( ωδπωω

ττ XjXj

dx Ft+→←∫ ∞− . (4. 20)

Example : Consider the Fourier transform of the unit step )()( tutx = . It is know that


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1)()( →←= Fttg δ Also note that

∫ ∞−=

tdgtx ττ )()(

The Fourier transform of this function is

)(1

)()0(1

)( ωπδω

ωδπω

ω +=+=j

Gj

jX .

where 1)0( =G . Example : Consider the Fourier transform of the function )(tx shown in the figure below.

t1

11−

x(t)

= t

1

1− 1 +

t

1

1− 1

1− 1−

dttdx

tg)(

)( =

From the above figure we can see that )(tg is the sum of a rectangular pulse and two impulses.

ωω

ωω

ω jj eejG −−−

=

sin2)(

Note that 0)0( =G , using the integration property, we obtain

ωω

ωω

ωδπωω

ωjj

GjjG

jXcos2sin2

)()0()(

)( 2 −=+= .

It can be found )( ωjX is purely imaginary and odd, which is consistent with the fact that )(tx is real and odd.

4.3.5 Time and Frequency Scaling


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)()( ωjXtx F→← , Then

)(1

)(aj

Xa

atx F ω→← . (4. 20)

From the equation we see that the signal is compressed in the time domain, the spectrum will be extended in the frequency domain. Conversely, if the signal is extended, the corresponding spectrum will be compressed. If 1−=a , we get from the above equation,

)()( ωjXtx F −→←− . (4. 20) That is, reversing a signal in time also reverses its Fourier transform.

4.3.6 Duality The duality of the Fourier transform can be demonstrated using the following example.

ωω

ω 11

1

11

sin2)(

,0,1

)(T

jXTtTt

tx F =

→←><

=

>

<=→←=

W

WjX

tWT

tx F

ω

ωω

π ,0

,1)(

sin)( 2

12


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The symmetry exhibited by these two examples extends to Fourier transform in general. For any transform pair, there is a dual pair with the time and frequency variables interchanged.

Example : Consider using duality and the result 212

)(ω

ω+

=→←− jXe Ft to find the Fourier

transform )( ωjG of the signal

212

)(t

tg+

= .

Since 212

)(ω

ω+

=→←− jXe Ft , that is,

ωωπ

ω dee tjt ∫∞

∞−

−

+= 21

221

,

Multiplying this equation by π2 and replacing t by t− , we have

ωω

π ω dee tjt −∞

∞−

− ∫

+= 21

22

Interchanging the names of the variables t and ω , we find that

ωπ ωω det

e tj−∞

∞−

− ∫

+= 21

22 ⇒ ωπ −− =

+e

tF 2

12

21 .

Based on the duality property we can get some other properties of Fourier transform:

ωω

djdX

tjtx F )()( →←−

))(()( 00 ωωω −→← jXtxe Ftj

∫ ∞−→←+−

ωηηδπ dxtxtx

jtF )()()0()(

1


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4.3.7 Parseval’s Relation If )()( ωjXtx F→← , We have

ωωπ

djXdttx ∫∫∞

∞−

∞

∞−= 22 )(

21

)(

Parseval’s relation states that the total energy may be determined either by computing the energy per unit time 2)(tx and integrating over all time or by computing the energy per unit frequency

πω 2/)( 2jX and integrating over all frequencies. For this reason, 2)( ωjX is often referred to as the energy-density spectrum.

4.4 The convolution properties

)()()()()()( ωωω jXjHjYtxthty F =→←∗= The equation shows that the Fourier transform maps the convolution of two signals into product of their Fourier transforms.

)( ωjH , the transform of the impulse response, is the frequency response of the LTI system, which also completely characterizes an LTI system. Example : The frequency response of a differentiator.

dttdx

ty)(

)( = .

From the differentiation property,

)()( ωωω jXjjY = , The frequency response of the differentiator is

ωωω

ω jjXjY

jH == ))(

()( .

Example : Consider an integrator specified by the equation:


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∫ ∞−=

tdxty ττ )()( .

The impulse response of an integrator is the unit step, and therefore the frequency response of the system:

)(1

)( ωπδω

ω +=j

jH .

So we have

)()0()(1

)()()( ωδπωω

ωωω XjXj

jXjHjY +== ,

which is consistent with the integration property. Example : Consider the response of an LTI system with impulse response

)()( tueth at−= , 0>a to the input signal

)()( tuetx bt−= , 0>b To calculate the Fourier transforms of the two functions:

ωω

jbjX

+=

1)( , and

ωω

jajH

+=

1)( .

Therefore,

( )( )ωωω

jbjajY

++=

1)( ,

using partial fraction expansion (assuming ba ≠ ), we have

+

−+−

=ωω

ωjbjaab

jY111

)(

The inverse transform for each of the two terms can be written directly. Using the linearity property, we have


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[ ])()(1

)( tuetueab

ty btat −− −−

= .

We should note that when ba = , the above partial fraction expansion is not valid. However, with ba = , we have

( )2

1)(

ωω

jajY

+= ,

Considering ( )

+

=+ ωωω jad

djja

112 , and

ωjatue Fat

+→←− 1

)( , and

+

→←−

ωω jadd

jtute Fat 1)( ,

so we have

)()( tutetY at−= .

4.5 The Multiplication Property

∫∞+

∞−−=→←= θθωθ

πω djPjSjRtptstr ))(()(

21

)()()()(

Multiplication of one signal by another can be thought of as one signal to scale or modulate the amplitude of the other, and consequently, the multiplication of two signals is often referred to as amplitude modulation. Example : Let )(ts be a signal whose spectrum )( ωjS is depicted in the figure below.


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Also consider the signal

ttp 0cos)( ω= , then

)()()( 00 ωωπδωωπδω ++−=jP . The spectrum of )()()( tptstr = is obtained by using the multiplication property,

)(21

)(21

))(()(21)(

00 ωωωω

θθωωπ

ω

++−=

−= ∫∞+

∞−

jSjS

djPjSjR

,

which is sketched in the figure below.

From the figure we can see that the signal is preserved although the information has been shifted to higher frequencies. This forms the basic for sinusoidal amplitude modulation systems for communications. Example : If we perform the following multiplication using the signal )(tr obtained in the preceding example and ttp 0cos)( ω= , that is,

)()()( tptrtg = The spectrum of )( ωjP , )( ωjR and )( ωjG are plotted in the figure below.


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If we use a lowpass filter with frequency response )( ωjH that is constant at low frequencies and zero at high frequencies, then the output will be a scaled replica of )( ωjS . Then the output will be scaled version of )(ts - the modulated signal is recovered.


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4.6 Summary of Fourier Transform Properties and Basic Fourier Transform Pairs


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4.7 System Characterized by Linear Constant-Coefficient Differential Equations An LTI system described by the following differential equation:

∑∑==

=M

kk

k

k

N

kk

k

k dttxd

bdt

tyda

00

)()(, (4. 67)

which is commonly referred to as an Nth-order differential equation. The frequency response of this LTI system

)()(

)(ωω

ωjXjY

jH = , (4. 68)

where )( ωjX , )( ωjY and )( ωjH are the Fourier transforms of the input )(tx , output )(ty and the impulse response )(th , respectively. Applying Fourier transform to both sides, we have

=

∑∑

==

M

kk

k

k

N

kk

k

k dttxd

bFdt

tydaF

00

)()( , (4. 69)

From the linearity property, the expression can be written as

∑∑==

=

M

kk

k

k

N

kk

k

k dttxd

Fbdt

tydFa

00

)()( . (4. 70)

From the differentiation property,

)()()()(00

ωωωω jXjbjYjaM

k

kk

N

k

kk ∑∑

==

= ⇒ ∑∑

=

=== N

kk

k

M

kk

k

ja

jb

jXjY

jH0

0

)(

)(

)()(

)(ω

ω

ωω

ω (4. 71)

)( ωjH is a rational function, that is, it is a ratio of polynomials in )( ωj .

Example : Consider a stable LTI system characterized by the differential equation

)()()(

txtaydt

tdy=+ , with 0>a .

The frequency response is


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ajjH

+=

ωω

1)( .

Te impulse response of this system is then recognized as

)()( tueth at−= . Example : Consider a stable LTI system that is characterized by the differential equation

)(2)(

)(3)(

4)(

2

2

txdt

tdxty

dttdy

dttyd

+=++ .

The frequency response of this system is

( )( )312

3)(4)(2)(

)( 2 +++

=++

+=

ωωω

ωωω

ωjj

jjj

jjH .

Then, using the method of partial-fraction expansion, we find that

32/1

12/1

)(+

++

=ωω

ωjj

jH .

The inverse Fourier transform of each term can be recognized as

)(21

)(21

)( 3 tuetueth tt −− += .

Example : Consider a system with frequency response of ( )( )312

)(++

+=

ωωω

ωjj

jjH and suppose

that the input to the system is

)()( tuetx t−= , find the output response. The output in the frequency domain is give as

( )( ) ( ) ( ))312

11

312)()()( 2 ++

+=

+

++

+==ωω

ωωωω

ωωωωjj

jjjj

jjXjHjY ,

Using partial-fraction expansion, we have


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( ) ( ))34/1

12/1

14/1

)( 2 ++

++

+=

ωωωω

jjjjY ,

By inspection, we get directly the inverse Fourier transform:

)(41

21

41

)( 3 tueteeth ttt

−+= −−− .

chapter 4 continuous-time fourier transform

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