4. ANALYSIS OF CONTINOUS TIME SIGNALS USING FOURIER TRANSFORM 4.1 Fourier …old.staff.neu.edu.tr/~fahri/signals_6.pdf · 2011. 10. 17. · 4. Fourier Analysis of Continuous-Time

4. Fourier Analysis of Continuous-Time Signals

56

4. ANALYSIS OF CONTINOUS –TIME SIGNALS USING FOURIER TRANSFORM

4.1 Fourier Series

A periodic signal x(t) with period T>0 can be expressed in terms of set of periodic

basis approximation functions { k(t)}, which also called a set of basis function

k

k

kk )t(a)t(x (4.1)

where, ak are the coefficients.

The representation of the periodic signal in the form of Eq.(4.1) is referred to as Fourier

series. The term for k=0 is a dc component. The two terms for k=±1 have fundamental period

equal to T0 and are collectively referred to as the fundamental components or first harmonic

components. The two terms for k=±2 are periodic with half the period (or equivalently twice

the frequency) of the fundamental component and are referred to as the second harmonic

components. More generally the components for k=±n are referred to as n-th harmonic

components.

Example 4.1

Consider a periodic signal x(t) which is expressed in term of complex exponential basic

function

Suppose 0=2or

Collecting together each pair of harmonic components, we obtain

3

3k

tjk

k0ea)t(x

10 a4

111 aa

2

1aa 22

3

133 aa

t6cos3

2t4cost2cos

2

11)t(x

)ee(3

1)ee(

2

1)ee(

4

11ea)t(x 6jk6jk4jk4jk2jk2jk

3

3k

t2jk

k

Equivalently using Euler’s relations

j2

eetsin;

2

eetcos

tjtj

0

tjtj

0

0000

We can write x(t) in the form

(4.2)


57

In Figure 4.1 we illustrate graphically for this example how the signal x(t) is built up from its

harmonic components.

One of commonly encountered form for the Fourier series is:

where Bk and Ck are both real coefficients

The existence of a Fourier series of a periodic signal x(t) with period T is determined by

Dirichlet conditions:

1k

0k0k0 tksinCtkcosB2a)t(x

1

x(t)

cos4t

(2/3)cos6t

t

t

t

t

t

» syms t

x1=(1/2)*cos(2*pi*t);

» ezplot(x1)

» hold on

» x2=cos(4*pi*t);

» ezplot(x2)

» hold on

x3=(2/3)*cos(6*pi*t);

» ezplot(x3)

» hold on

» x=1+x1+x2+x3;

» ezplot(x)

Figure 4.1


58

1. x(t) must be absolutely integral able over any period

A periodic signal that violates the first Dirichlet condition is

Where x(t) is periodic function with period 1. This signal is illustrated in Figure 4.2(a).

2. In any finite interval of time, x(t) is of bounded variation that is, there are no more then a

finite number of maxim and minima during any single period of the signal. An example of a

time function, which violates the second Dirichlet condition, is shown in Figure 4.2 (b)

For this function (periodic with T0 =1)

It has, however, an infinite number of maxim and minima in the interval.

3. In any finite interval of time there are only a finite number of discontinuities. Furthermore,

each of these discontinuities must be finite. An example of a time function that violates

condition 3 is illustrated in Figure 4.2(c). The signal s(t) (of period T =8 ) is composed of an

infinite number of sections each of which is half the height and half the width of the previous

section. For 0the value of x(t) decreases by a factor of 2 whenever the distance from t

to 8 decreases by a factor of 2; that is x(t)= 1, 0t


59

)/4t2(j)/4t2(jtjtjtjtj 000000 ee2

1eeee

j2

11)t(x

1/8, 7t


60

Thus, the Fourier series coefficients for this example are :

a0=1;

We have plotted the magnitude of ak on bar graph in which each line represents the magnitude

of the corresponding harmonic component of x(t).

Example 4.4

We will encounter this signal on several occasions. This signal is periodic with fundamental

period T0 and fundamental frequency 0= 2Because of the symmetry of x(t) about t=0 it

is most convenient to choose the interval over which the integration is performed as -(T0/2


61

(4.5)

(4.6)

0k;kπ

2sink

a k

The Eq (4.5) shows that for k even ak = 0. Therefore

a1 = a-1 =1/ ; a3 = a-3 = -1/3; a5 = a-5 =1/5; an = a-n1 =1/n (n is odd)

Figure 4.6 shows the Fourier serie coefficients for for the T0 = 4T1.

4.2.1 Gibbs Phenomenon

Since the Fourier series represents a continuous time signal as a linear combination of

continuous function, therefore it should be expected that the Fourier series is well suited for

modelling smooth signals. Recall that the Fourier series is the infinite sum of weighted

complex exponentials given by:

k

0k tjkexpa)t(x

In practice, it may be more partial to consider using only a finite sum to approximate )(tx .

Using 2N+1 coefficients, the reconstructed approximation of )(tx shall be defined to be

N

Nk

0kN tjkexpa)t(x

ak

0.5

1/

-1/3

1/5

-1/7

1/9 k

-2-1 0 1 23 4 5 67 8 9 10 14

. .

1/

-1/3

1/5 1/9

-1/7

Figure 4.6

0k;kπ

Tsinkωa

;kπ

Tsinkω

Tkω

1Tsinkω2a

10k

10

00

0

k


62

» syms t N

» y=(1/(2*N+1))*sin((2*N+1)*t)

» z1=symsum(y,N,0,3)

z1 =

sin(t)+1/3*sin(3*t)+1/5*sin(5*t)+

1/7*sin(7*t)

» ezplot(z1)

» hold on


» ezplot(z2)

» hold on


» ezplot(z3)

» hold on


» ezplot(z)

» hold on


» ezplot(z)

Any difference between xN(t) and x(t) is attribute to the use of a finite number of terms (i.e.,

harmonics)to reconstruct x(t). Defining the approximation error to be

N

Nk

0kNN tjkexpatxtxtxte

The number of harmonics required to produce an error that does not exceed a given mean

error is signal-dependent.

Example 4.5

In the Figure 4.7 are shown the convergence of the Fourier series representation of a square

wave. Here we have depicted the finite series approximation of a square wave by different N.

The behavior of partial sum in the vicinity of discontinuity exhibits ripples. As N increases,

the ripples in the partial sum becomes compressed toward the discontinuity, but any finite

value of N , the peak amplitude of these ripples remain constant. This behavior is called the

Gibbs phenomenon.

N

Nk

tjkw

kN eatx0)(

t

N=31

N=3

N=7

N=11

N=21

Figure 4.7

t

t

t

t


63

4. 3 Representation of Aperiodic Signals

Considers again the periodic square wave discussed in Example 4.4. The Fourier series

coefficients are given as,

Example 4.6

In Figure 4.8 we plotted T0ak for the different values of T1/T0 =1/4;1/8;1/16 (or T1=2/4;

2/8; 2/16) using Matlab. Multiplying ak by T0 we obtain

The function (2sinT1)/ represents the envelope of T0ak. As shown in Figure 4.8, T0ak

become more and more closely spaced with increasing T0 and approaches the envelope of

function 2sinT1/ as To This example illustrates the basic idea behind Fourier’s

development of a representation for aperiodic signals. Specifically, we think of an a periodic

signal as the limit of a periodic signal as the period becomes arbitrarily large, and we examine

the limiting behavior of the Fourier series representation for this signal .

4. 4 The Fourier Transform

The Fourier series can be used to analyse periodic signals in the frequency-domain.

Unfortunately, signals such as the unit step Us(t), unit impulse (t) would not posses a Fourier series representation, since they are aperiodic. Therefore the frequency-domain

representation of such signals must be produced by other means. Generally the frequency

spectra of such signals are defined in terms of the Fourier transform. The continuous –time

Fourier transform (CTFT) of complex signal x ( t ) is given by analysis equation

(4.7)

Where X ( j )is complex and is called the frequency spectrum of x( t ), or simply the spectrum of x ( t ).

The CTFT of x( t ) is denoted x( t ) F X( j ). The complex spectrum can be expressed in

Cartesian coordinates as X( j ) = Re )j(X + j lm )j(X , and in polar form as

X ( j ) = )j(X exp ( j ( j ) ). The term )j(X = 22 )j(XIm)j(XRe is called

the magnitude spectrum and ( j ) = tan 1 ( Im )j(X / Re )j(X ) is called the phase spectrum. The values of X ( j ) for > 0 are called the positive frequency spectrum and the values of X ( j ) for < 0 are called the negative frequency spectrum. The inverse Fourier transform is given by Synthesis equation

0k1

00

100k0 |

Tsin2

Tk

Tksin2TaT

dte)t(x)j(X tj

00

10k

Tk

Tksin2a


64

-20 -15 -10 -5 0 5 10 15 20 -1

-0.5

0

0.5

1

1.5

2

-20 -15 -10 -5 0 5 10 15 20 -0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

-20 -15 -10 -5 0 5 10 15 20 -0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

» T1=2*pi/4;

» omega=-16:1:16;

»T0ak=2*sin(omega*T1)./omega

» plot(omega,T0ak,’fill’,’k’)

» T1=2*pi/8;

» omega=-16:1:16;



» T1=2*pi/16;

» omega=-16:1:16;



Figure 4.8

T1/T0=1/4

T1/T0=1/8

T1/T0=1/16


65

(4.8)

The inverse Fourier transform is seen to have a form similar to the forward transform

(Eq. 4.7), with the principal difference being the sign of the complex exponentials exponent.

In both cases, the transform pair represents a mapping of a complex signal into another

complex signal. Substituting s= j, in to Eq.(4.7) from Fourier transform we obtain the Laplace transform

(4.9)

The conditions relating to the existence of a Fourier transform will require special attention.

The Fourier transform of x ( t ) will exist if x ( t ) has finite energy. That is

dt)t(x

2 (4.10)

However, the fact that x ( t ) does not satisfy Eq.(4.10) does not preclude the existence of

the Fourier transform of x ( t ). Another existence test for Fourier transforms is the Dirichlet

conditions (see Section 4.1).

Example 4.7 Find the Fourier transform of x(t) == e-at

Us(t)

For a=2 and this signal is shown in Figure 4.9 (a).

For =a=2,

22a

1)(X

0a;ja

1)(X

de)j(X2

1)t(x tj

dte)t(x)s(X st

;5.0a

1|)(X| 0max

Figure 4.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8 0.9

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

-8 –6 –4 –2 0 2 4 6

8 8

a)

b)

»w=-2*pi:.1:2*pi;

» a=2;

» x=1./sqrt(a^2+w.^2);

»plot(w,x)

35.02a

1)(X

a

2a

1

ja1

eja

1dtee)(X

0

t)ja(tj

0

at

(see Figure 4.9 (b)


66

Figure 4.9 (b) shows the Fourier transform of the signal.

Example 4.8 Let

22

tj

0

tatj0

tatjta

a

a2

ja

1

ja

1dteedteedtee)(X

The fragment of this signal for a=2 is sketched in figure 4.10 (a) .

For -2≤≥2, X() is shown in Figure 4.10 (b).

Example 4.9 Now let us determine the spectrum or the unit impulse

Substituting in to Eq.(4.7) we see that

|t|ae)t(x

)t()t(x

1dte)t()(X tj

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8 0.9

1

-2.0 -0.8 -0.6 -0.4 -0.2 0 1.2 1.4 1.6 1.8 2.0 0

0.1

0.2

0.3

0.4

0.5

-6 -4 -2 0 2 4 6

X()

Figure 4.10

» w=-2*pi:.1:2*pi;

» a=2;

» x=a./(a^2+w.^2);

» plot(w,x)

» axis([-6 6 0 0.5])

» t=-2:.1:2;

» a=2;

» t1=abs(t);

» x=exp(-a*t);

» plot(t,x)

a) b)


67

That is, the unit impulse has a Fourier transform representation consisting of equal

contributions at all frequencies.

Example 4.10 Consider the rectangular pulse signal

(4.11)

The Fourier transform of this signal is

(4.12)

For T1=1 and T1=0.5 X() are sketched in figure 4.11 (b and d).

By examining figure 4.11 (a), (b), (c), (d) see that as T1 increases, X() becomes broader

while the main peak of x (t) at t = 0 becomes higher and the width of the first lobe of this

signal becomes narrower. In fact, in the limit as T1 X()=1 for all .

1T|t|0

1T|t|1)t(x

1

11

1T

2T

1tj

T

TsinT2

Tsin2dte)(X

x(t)

t

T1 -T1 0

1

x(t)

t

4

1T

4

1T

1

w=-6*pi:.1:6*pi;

» T1=1;

» x=sin(w*T1)./2;

» plot(w,x)

» hold on

» T1=0.5;

» x=sin(w*T1)./w;

» plot(w,x)

a)

b)

b)

-20 -10 0 10 20

-0.2

0.25

0.5

0

Figure 4.11

d)

/T1 -/T1

2T1

-20 -10 0 10 20

-0.5

0

1

2

/T1 -/T1

2T1

c)


68

Example 4.11. Consider the signal x(t) whose Fourier transform is given by Figure 4.12 (a))

(4.13)

Using the synthesis equation, we can determine

(4.14)

The transform is illustrated in figure 4.12 (b).

Comparing figures 4.11 and 4.12 or , equivalently, Eqs. (4.11) and (4.12) with Eqs. (4.13) and

(4.14), we see an interesting relationship. In each case the Fourier transform pair consists of a

(sin x)/x function and a rectangular pulses. However in Example 4.11 it is the signal x(t) that

is a pulse, while in Example 4.12 , it is the transform X(w), the special relationship that is

apparent here is direct done sequence of the duality property for Fourier transforms, which we

discuss in detail in the future.

4. 5 The Fourier Transform of Periodic Signal

Fourier transom for periodic signal is of the form of a linear combination of impulses equally

spaced in frequency that is,

(4.15)

Then the application of Eq. (4.8) yields

(4.16)

We see that Eq. (4.16) corresponds exactly to the Fourier series representation of a periodic

signal, as specified by Eq. (4.7). Thus the Fourier transform of a periodic signal with Fourier

series coefficients {ak} can be interpreted as a train of impulses occurring at the harmonically

related frequencies and for which the area of the impulse at the kth

harmonic frequency k0 is

2 times the kth

Fourier series coefficient ak.

W||0

W||1)(X

Wt

WtsinW

t

Wtsinde

2

1)t(x

W

W

tj

)k(a2)(X 0k

k

tjk

k

k0ea)t(x

X()

W -W 0

1 /W -/W

W/

x(t)

t

Figure 4.12

a) b)


69

Example 4.12

Consider again the square wave illustrated in Figure 4.13 (a). The Fourier series coefficients

for this signal are defined by Eq.(4.5)

:

Fourier transform is:

which is sketched in Figure 4.13 for T0 = 4T1.

Example 4.13 Let us

The Fourier series coefficients for this example are:

Thus, the Fourier transform is as shown in Figure 4.14 (a)

Example 4.14 Let

k

1Tksina 0k

)k(k

1Tksin2)(X 0

k

0

ja

2

11

ja

2

11 1or1k,0a k

11,0 orkak2

111 aaa

-T0 -T1 T1 T0

T0

x(t)

t

-0 0

X()

2 2

Figure 4.13 b)

a)

0 -0

/j

-/j

X()

a)

tsin)t(x 0

tcos)t(x 0

Figure 4.14

0 -0

X()

b)

Documents

4. ANALYSIS OF CONTINOUS TIME SIGNALS USING FOURIER TRANSFORM 4.1 Fourier …old.staff.neu.edu.tr/~fahri/signals_6.pdf · 2011. 10. 17. · 4. Fourier Analysis of Continuous-Time