Chapter 3: Fourier Representation of Signals and LTI Systems

Chapter 3: Fourier Representation of Signals

and LTI Systems

Chih-Wei Liu

Outline Introduction Complex Sinusoids and Frequency Response Fourier Representations for Four Classes of Signals Discrete-time Periodic Signals Fourier Series Continuous-time Periodic Signals Discrete-time Nonperiodic Signals Fourier Transform Continuous-time Nonperiodic Signals Properties of Fourier representations Linearity and Symmetry Properties

Convolution Property

Lec 3 - [email protected]

Discrete-Time Fourier Transform (DTFT)


The DTFT-pair of a discrete-time nonperiodic signal x[n] and X(ej)

DTFT represents x[n] as a superposition of complex sinusoids Since x[n] is not periodic, there are no restrictions on the periods (or

frequencies) of the sinusoids to represent x[n]. The DTFT would involve a continuous of frequencies on (discrete-time sinusoids are unique only over a 2 interval of frequency)

X(ej) is termed as the frequency-domain representation of x[n] If x[n] has finite duration and is finite valued, the infinite sum converges definitely The infinite sum converges uniformly only if x[n] is absolutely summable If x[n] is not absolutely summable, but squarely summable (i.e. has finite energy),

the infinite sum converges in a MMSE sense, but does not converge pointwise.

DTFT jx n X e

1[ ] ( )2

j j nx n X e e d

( ) [ ]j j n

n

X e x n e

Example 3.17


Find the DTFT of the sequence x[n] = nu[n].<Sol.>1. DTFT of x[n]:

nj

n

nnj

n

nj eenueX

0

2. For < 1, we have

0

1( ) , 11

nj jj

n

X e ee

3. If is real valued, then

sincos11

jeX j

4. Magnitude and phase spectra if is real valued :

2/122/1222 cos21

1

sincos1

1

jeX

cos1sinarctanarg

jeX

Example 3.17 (conti.)


4

4 4

42

2 2

22

2 2

24

4 4

4

Magnitude spectrum(even)

Phase spectrum(odd)

Figure 3.29The DTFT of an exponential signal x[n] = ()nu[n]. (a) Magnitude spectrum for = 0.5. (b) Phase spectrum for = 0.5. (c) Magnitude spectrum for = 0.9. (d) Phase spectrum for = 0.9.

Example 3.18 DTFT of a Rectangular Pulse


Find the DTFT of x[n].

2 2

22 1M

2

2 1M

1,0,

n Mx n

n M

<Sol.>1. DTFT of x[n]:

4 ,2 ,0,12

4 ,2 ,0,-1

)1(1)(

)12(

Meee

eeX j

MjMjM

Mn

njj

Example 3.19 Inverse DTFT of a Rectangular Spectrum


Find the inverse DTFT of

1,,

0,j W

X eW n

W

W

W

Note that X(e j ) is specified only for < .

12

Wj n

W

x n e d

<Sol.>

1 , 02

j n WWe n

nj

1 sin , 0.Wn nn

Example 3.20, 3.21


2 2

2 2 3 4

12

1j j n

n

X e n e

1.DTFTn

1 ,2

DTFT

.

21)(

21][

dnxWe can define X(e j ) over all by writing it as an infinite sum of delta functions shifted by integer multiples of 2.

2 .j

k

X e k

Remarks


x[n] is not absolutely summable (or not square summable), but it is still a valid DTFT-pair.

Strictly speaking, the DTFT of the impulse train x[n] does not exist; however, we can still identify their DTFT-pair. That is, we still can utilize the DTFT as a problem-solving tool.

2 2 3 4

12

k

knnx )(21][

k

j keX )2()( DTFT

Example 3.22 Simple Low-Pass and High-Pass Filters


Consider two different moving-average systems described by the input-output equations

11 12

y n x n x n 21 12

y n x n x n and

Find the frequency response of each system and plot the magnitude responses.

<Sol.>1. The frequency response is the DTFT of the impulse response.2. For system # 1

The corresponding frequency response:

11 1 12 2

h n n n

2 22

1( )2

j jjj e eH e e

2 cos .

2j

e

For system # 2

2 22 2

21 1( ) sin .2 2 2 2

j jj jj j e eH e e je je

j

21 1 12 2

h n n n



Low-pass filter High-pass filter

Fourier Transform (FT)


The FT-pair of a continuous-time nonperiodic signal x(t) and X(j)

FT represents x(t) as a superposition of complex sinusoids Since x(t) is not periodic, there are no restrictions on the periods (or

frequencies) of the sinusoids to represent x(t). The FT would involve a continuous of frequencies ranging from - to .

X(j) is termed as the frequency-domain representation of x(t) The integral in FT-pair may not converge for all functions of x(t) and X(j) if x(t) is square integral, the integral converges in a MMSE sense, but not

converge poitwise. If x(t) satisfies Dirichlet’s conditions, the integral converges pointwise.

1( ) ( )2

j tx t X j e d

( ) ( ) j tX j x t e dt

.FTx t X j

arg{X(j)}

/4/2

/4

/4

X(j)

/2

Example 3.24


Find the FT of x(t) = e a t u(t).

For a 0, x(t) is not absolutely integrable.<Sol.>

Therefore, we consider a > 0. The FT of x(t) is

0

01 1

a j tat j t

a j t

X j e u t e dt e dt

ea j a j

Magnitude and phase spectra of X(j) are:

12 2 2

1X ja

arg arctan ,X j a

( )X j

0

3T

0

2T

0T

0

2T

0

3T

0T

Example 3.25


Find the FT of 0 0

0

1,0,

T t Tx t

t T

<Sol.>1. The rectangular pulse x(t) is absolutely integrable, provided that To < .

2. FT of x(t):

0

0

0

0 01 2 sin , 0

Tj t j t

T

Tj tT

X j x t e dt e dt

e Tj

0 00

2lim sin 2 .T T

02 sin ,X j T

0 02 sinc .X j T T

Remarks


As T0 increases, the nonzero time extent of x(t) increases, while the X(j) becomes more concentrated about the frequency origin

Conversely, as T0 decreases, the nonzero duration of x(t) decreases, while the X(j) becomes less concentrated about the frequency origin

The duration of x(t) is inversely related to the bandwidth of X(j) The signal which is concentrated in one domain is spread out in the other

domain.

( )X j

0

3T

0

2T

0T

0

2T

0

3T

0T

0 0

0

1,0,

T t Tx t

t T

0 02 sinc .X j T T

Time-domain Frequency-domain

( )X j

W

3W

W

W 3

W 5

W

Example 3.26


Find the inverse FT of the rectangular spectrum 1,0,

W WX j

W

The inverse FT of x(t): 1 1 1 sin( ), 02

W j t j t WWW

x t e d e Wt tj t t

<Sol.>

0

1lim sin ,t

Wt Wt

1 sin ,x t Wtt

sinc ,W Wtx t

Example 3.27, 3.28


Find the FT of x(t) = (t).<Sol.>

x(t) does not satisfy the Dirichlet’s condition, since the discontinuity at the origin is infinite. However we attempt to process the FT of x(t) directly:

1j tX j t e dt

1,FTt

Find the inverse FT of X(j) = 2().<Sol.>Inverse FT of X(j) is 1 2 1

2j tx t e d

1 2FT

Duality between Example 3.27 and 3.28

Example 3.29 BPSK Modulation


In a simple digital communication system, BPSK (binary phase-shift keying) is a common scheme to assume that the signal representing ‘0’ is the negative of the signal and the signal representing “1” is the positive of the signal. Fig. 3.44 depicts two candidate signals for this approach: a rectangular pulse xr(t) and a raised-cosine pulse xc(t) (Note that each pulse is To seconds long)

0

0

, / 20, / 2r

A t Tx t

t T

0 0

0

/ 2 2cos 2 / , / 2

0, / 2c

c

A t T t Tx t

t T

For example for the sequence “1001011”, the transmitted BPSK signals for communicating are


1. With BPSK, the system has a transmission rate of 1/To bits per second. (Each user’s signal is transmitted within an assigned frequency band in order to prevent interference with others)

2. Suppose the frequency band assigned to each user is 20 kHz wide. Then, to prevent interference with adjacent channels, we assume that the peak value of the magnitude spectrum of the transmitted signal outside the 20-kHz band is required to be 30 dBbelow the peak in-band magnitude spectrum.

Problems: 1. Choose the constant Ar and Ac so that both BPSK signals have unit power. 2. Use FT to determine the maximum number of bits per second that can be transmitted

when the rectangular and raised-cosine pulse shapes are utilized.<Sol.>1. BPSK signal is a power signal. Powers in rectangular pulse and raised-cosine pulse are:

0

0

/ 2 2 2

/ 20

1 T

r r rTP A dt A

T and



0

0

0

0

2 22

20

2 2

0 0202

1 4 1 2cos 2 2

1 2cos 2 1 2 1 2cos 44

38

T

c cT

Tc

T

c

P A t dtT

A t T t T dtT

A

Hence, unity transmission power is obtained by choosing Ar = 1 and Ac = . 8/ 32. FT of the rectangular pulse xr(t): 0sin 2

2r

TX j

Rewrite it in terms of Hz: 0sin2r

fTX jf

f

rad/sec ( = 2f)

The normalized magnitude spectrum of the signal is given by

01002

10 )( log20)( log10 TjfXTjfX


The normalized by To removed the dependence of the magnitude on To.

Spectrum of rectangular pulse in dB, normalized by T0.


• From the figure, we find that the 10th sidelobe is the first one whose peak does not exceed -30 dB.

• This implies that we must choose T0 so that the 10th zero crossing is at 10 kHz in order to satisfy the constraint.

10K = 10/To Data transmission rate = 1K bits/sec.

3. FT of raised-cosine pulse xc(t): 0

0

2

02

1 8 1 2cos 22 3

T j tc T

X j t T e dt

0 0 00 0

0 0 0

2 2 22 2

2 2 2

2 1 2 1 23 2 3 2 3

T T Tj T t j T tj tc T T T

X j e dt e dt e dt

0 0 0 00

0 0

sin 2 2 sin 2 2sin 22 2 223 3 2 3 2c

T T T TTX j

T T

In terms of f (Hz), the above equation becomes

0 0 0 00

0 0

sin 1 sin 1sin2 2 20.5 0.53 3 1 3 1c

f T T f T TfTX jf

f f T f T

A superposition of three sinc spectra


for To = 1.

The normalized magnitude spectrum of the signal, i.e. , is 010 )( log20 TjfX

The sum of three spectra has lower sidelobs than that of the spectrum of the rectangular pulse

the peak value of the first sidelobe is below 30 dB, so we may satisfy the adjacent channel interference specifications by choosing the mainlobeto be 20 kHz wide:

10,000 = 2/To

To = 2 10 4 s

Data transmission rate = 5000 bits/sec.

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Chapter 3: Fourier Representation of Signals and LTI Systems