Continuous-Time Signal Analysis: The Fourier Transform...Continuous-Time Signal Analysis: The...

Continuous-Time Signal

Analysis: The Fourier Transform

Chapter 7

Mohamed Bingabr

Chapter Outline

• Aperiodic Signal Representation by Fourier Integral

• Fourier Transform of Useful Functions

• Properties of Fourier Transform

• Signal Transmission Through LTIC Systems

• Ideal and Practical Filters

• Signal Energy

• Applications to Communications

• Data Truncation: Window Functions

Link between FT and FS

Fourier series (FS) allows us to represent periodic

signal in term of sinusoidal or exponentials ejnω ot.

Fourier transform (FT) allows us to represent

aperiodic (not periodic) signal in term of exponentials

ejω t.

xTo(t) ( ) ∑∞

−∞=

=n

tjn

nT eDtx 0

0

ω

tjn

T

T

Tn etxT

D 0

0

0

0

2/

2/0

)(1 ω−

−∫=

Link between FT and FS

( ) ( )txtx TT 00

lim∞→

=0

00→⇒⇒∞→ ωT

xTo(t)xT(t)

nD )(ωX

As T0 gets larger and larger the fundamental frequency ω0 gets

smaller and smaller so the spectrum becomes continuous.

ω0 ω

)(1

0

0

ωnXT

Dn =

The Fourier Transform Spectrum

The Inverse Fourier transform:

∫∞

∞−

= ωωπ

ω deXtx tj)(2

1)(

The Fourier transform:

)()()(

)()(

ω

ω

ωω

ω

X

tj

eXX

dtetxX

∠

−∞

∞−

=

= ∫

The Amplitude (Magnitude) Spectrum The Phase Spectrum

The amplitude spectrum is an even function and the phase is an

odd function.

ExampleFind the Fourier transform of x(t) = e-atu(t), the

magnitude, and the spectrum

Solution:

)/(tan)(1

)(

0a if 1

)(

1

22

0

aXa

X

jadteeX tjat

ωωω

ω

ωω ω

−

∞−−

−=∠+

=

>+

== ∫

How does X(ω) relates to X(s)?

-aRe(s) if 1

)(

1

)(0

)(

0

>+

=

+−==

∞+−

∞−−∫

sasX

esa

dteesX tasstat

S-plane

s = σ + jω

Re(s)

σ

jω

ROC

-a

Since the jω-axis is in the region of convergence then FT exist.

Useful Functions

Unit Gate Function

<

=

>

=

2/|| 1

2/|| 5.0

2/|| 0

τττ

τx

x

xx

rect

Unit Triangle Function

<−

≥=

∆2/|| /21

2/|| 0

τττ

τ xx

xx

τ/2-τ/2

τ/2-τ/2

1

1

x

x

Useful Functions

Interpolation Function

0for x 1)(sinc

for x 0)(sinc

sin)(sinc

==

±==

=

x

kx

x

xx

π

sinc(x)

x

Example

Find the FT, the magnitude, and the phase spectrum

of x(t) = rect(t/τ).

Answer

)2/sinc()/()(2/

2/

ωτττωτ

τ

ω∫−

− == dtetrectX tj

The spectrum of a pulse extend from 0 to ∞. However, much of

the spectrum is concentrated within the first lobe (ω=0 to 2π/τ)

What is the bandwidth of the above pulse?

Examples

Find the FT of the unit impulse δ(t).Answer

1)()( ∫∞

∞−

− == dtetX tjωδω

Find the inverse FT of δ(ω).Answer

)(21

impulsean isconstant a of spectrum theso

2

1)(

2

1)(

ωπδ

πωωδ

πω

↔

== ∫∞

∞−

detx tj

Examples

Find the inverse FT of δ(ω- ω0).

Answer

)(2 and )(2

impulse shifted a isexponent complex a of spectrum theso

2

1)(

2

1)(

00

0

00

0

ωωπδωωπδ

πωωωδ

π

ωω

ωω

+↔−↔

=−=

−

∞

∞−∫

tjtj

tjtj

ee

edetx

Find the FT of the everlasting sinusoid cos(ω0t).

Answer

( )

( ) [ ])()(2

1

2

1cos

00

0

00

00

ωωδωωδπ

ω

ωω

ωω

−++↔+

+=

−

−

tjtj

tjtj

ee

eet

Examples

Find the FT of a periodic signal.

Answer

∑

∑

∞=

−∞=

∞=

−∞=

−=

==

n

n

n

tjnn

n

n

nDX

TeDtx

)(2)(

FT ofproperty linearity use and sideboth of FT theTake

/2)(

0

000

ωωδπω

πωω

Examples

Find the FT of the unit impulse train

Answer

)(0tTδ

∑

∑∞=

−∞=

∞=

−∞=

−=

=

n

n

n

n

tjn

T

nT

X

eT

t

)(2

)(

1)(

0

0

0

0

0

ωωδπ

ω

δ ω

Properties of the Fourier Transform•• Linearity:Linearity:

•• Let and Let and

thenthen

( ) ( )ωXtx ⇔ ( ) ( )ωYty ⇔

( ) ( ) ( ) ( )ωβωαβα YXtytx +⇔+

•• Time Scaling:Time Scaling:

•• LetLet

thenthen

( ) ( )ωXtx ⇔

( )

⇔a

Xa

atxω1

Compression in the

time domain results in

expansion in the

frequency domain

Internet channel A can transmit 100k pulse/sec and channel B

can transmit 200k pulse/sec. Which channel does require higher

bandwidth?

Properties of the Fourier Transform•• Time Reversal:Time Reversal:

•• LetLet

thenthen ( ) ( )x t X ω− ↔ −( ) ( )ωXtx ⇔

Example: Find the FT of eatu(-t) and e-a|t|

•• Left or Right Shift in Time:Left or Right Shift in Time:

•• LetLet

thenthen

( ) ( )ωXtx ⇔

( ) ( ) 0

0

tjeXttx ωω −⇔−Example: if x(t) = sin(ωt) then what is the FT of x(t-t0)?

Time shift effects the

phase and not the

magnitude.

Example: Find the FT of and draw its magnitude and

spectrum

|| 0ttae −−

Properties of the Fourier Transform•• Multiplication by a Complex Exponential (Freq. Shift Multiplication by a Complex Exponential (Freq. Shift Property):Property):

•• LetLet

then then 0

0( ) ( )j t

x t e Xω ω ω↔ −

( ) ( )ωXtx ⇔

•• Multiplication by a Sinusoid (Amplitude Modulation):Multiplication by a Sinusoid (Amplitude Modulation):

LetLet

thenthen

( ) ( )ωXtx ⇔

( ) ( ) ( ) ( )[ ]0002

1cos ωωωωω −++⇔ XXttx

cosω0t is the carrier, x(t) is the modulating signal (message),

x(t) cosω0t is the modulated signal.

Example: Amplitude Modulation

Example: Find the FT for the signal

-2 2

A

x(t)

ttrecttx 10cos)4/()( =

HW10_Ch7: 7.1-1, 7.1-5, 7.1-6, 7.2-1, 7.2-2, 7.2-4, 7.3-2

Amplitude Modulation

ttmt cAM ωϕ cos)()( =

Modulation]2cos1)[(5.0 cos)( ttmtt

ccAMωωϕ +=

Demodulation

Then lowpass filtering

Amplitude Modulation: Envelope Detector

Applic. of Modulation: Frequency-Division Multiplexing

1- Transmission of

different signals over

different bands

2- Require smaller antenna

Properties of the Fourier Transform

•• Differentiation in the Time Domain:Differentiation in the Time Domain:

LetLet

thenthen ( ) ( ) ( )n

n

n

dx t j X

dtω ω↔

( ) ( )ωXtx ⇔

•• Differentiation in the Frequency Domain:Differentiation in the Frequency Domain:

•• LetLet

thenthen ( ) ( ) ( )n

n n

n

dt x t j X

dω

ω↔

( ) ( )ωXtx ⇔

Example: Use the time-differentiation property to find the

Fourier Transform of the triangle pulse x(t) = ∆(t/τ)

Properties of the Fourier Transform•• Integration in the Time Domain:Integration in the Time Domain:

LetLet

ThenThen1

( ) ( ) (0) ( )

t

x d X Xj

τ τ ω π δ ωω−∞

↔ +∫

( ) ( )ωXtx ⇔

•• Convolution and Multiplication in the Time Domain:Convolution and Multiplication in the Time Domain:

LetLet

ThenThen ( ) ( ) ( ) ( )x t y t X Yω ω∗ ↔

( ) ( )( ) ( )ω

ωYty

Xtx

⇔

⇔

)()(2

1)()( 2121 ωω

πXXtxtx ∗↔ Frequency convolution

ExampleFind the system response to the input x(t) = e-at u(t) if the

system impulse response is h(t) = e-bt u(t).

Properties of the Fourier Transform

•• Parseval’s TheoremParseval’s Theorem: : sincesince xx((tt)) is nonis non--periodicperiodicand has FTand has FT XX((ωω)),, then it is an energy signals:then it is an energy signals:

( ) ( )∫∫∞

∞−

∞

∞−

== ωωπ

dXdttxE22

2

1

Real signal has even spectrum XX((ωω))= = XX((--ωω)),, ( )∫∞

=0

21ωω

πdXE

Example

Find the energy of signal x(t) = e-at u(t). Determine the frequency

ω so that the energy contributed by the spectrum components of

all frequencies below ω is 95% of the signal energy EX.

Answer: ω=12.7a rad/sec

Properties of the Fourier Transform

•• Duality ( Similarity) :Duality ( Similarity) :

•• LetLet

thenthen ( ) 2 ( )X t xπ ω↔ −

( ) ( )ωXtx ⇔

HW11_Ch7: 7.3-3(a,b), 7.3-6, 7.3-11, 7.4-1, 7.4-2, 7.4-3, 7.6-1, 7.6-6

Data Truncation: Window Functions

1- Truncate x(t) to reduce numerical computation

2- Truncate h(t) to make the system response finite and causal

3- Truncate X(ω) to prevent aliasing in sampling the signal x(t)

4- Truncate Dn to synthesis the signal x(t) from few harmonics.

What are the implications of data truncation?

)(*)(2

1)( and )()()( ωωπ

ω WXXtwtxtxww==

Implications of Data Truncation

1- Spectral spreading

2- Poor frequency resolution

3- Spectral leakage

What happened if x(t) has

two spectral components of

frequencies differing by less

than 4π/T rad/s (2/T Hz)?

The ideal window for truncation

is the one that has

1- Smaller mainlobe width

2- Sidelobe with high rolloff rate

Data Truncation: Window Functions

Using Windows in Filter Design

=××××

Using Windows in Filter Design

=××××

Sampling TheoremA real signal whose spectrum is bandlimited to B Hz [X(ω)=0 for |ω| >2πB ] can be reconstructed exactly from its samples taken

uniformly at a rate fs > 2B samples per second. When fs= 2B then

fs is the Nyquist rate.

∑

∑

∑

∞=

−∞=

∞=

−∞=

∞=

−∞=

−=

==

−==

n

n

s

n

n

tjn

n

n

nXT

X

eT

txnTxtx

nTttxnTxtx

s

)(1

)(

1)()()(

)()()()(

ωωω

δ

ω

Reconstructing the Signal from the Samples

∑

∑

∑

−=

−=

−=

=

=

n

n

n

nTtBnTxtx

nTthnTxtx

nTtnTxthtx

nTxthtx

XHX

)(2(sinc)()(

)()()(

)()(*)()(

)(*)()(

)()()(

π

δ

ωωω

LPF

Example

Determine the Nyquist sampling rate for the signal

x(t) = 3 + 2 cos(10π) + sin(30π).

Solution

The highest frequency is fmax = 30π/2π = 15 HzThe Nyquist rate = 2 fmax = 2*15 = 30 sample/sec

AliasingIf a continuous time signal is sampled below the Nyquist rate

then some of the high frequencies will appear as low

frequencies and the original signal can not be recovered from

the samples.

LPF

With cutoff

frequency

Fs/2

Frequency above Fs/2

will appear (aliased) as

frequency below Fs/2

Quantization & Binary Representation

111

110

101

100

011

010

001

000

111

110

101

100

011

010

001

000

4

3

2

1

0

-1

-2

-3

4

3

2

1

0

-1

-2

-3

nL 2=

L : number of levels

n : Number of bits

Quantization error = ∆x/2

∆x

x(t)

1

minmax

−

−=∆

L

xxx

Example

A 5 minutes segment of music sampled at 44000 samples per

second. The amplitudes of the samples are quantized to 1024

levels. Determine the size of the segment in bits.

Solution

# of bits per sample = ln(1024) { remember L=2n }n = 10 bits per sample

# of bits = 5 * 60 * 44000 * 10 = 13200000 = 13.2 Mbit

Problem 8.3-4

Five telemetry signals, each of bandwidth 1 KHz, are

quantized and binary coded, These signals are time-division

multiplexed (signal bits interleaved). Choose the number of

quantization levels so that the maximum error in the peak

signal amplitudes is no greater than 0.2% of the peak signal amplitude. The signal must be sampled at least 20% above

the Nyquist rate. Determine the data rate (bits per second) of

the multiplexed signal.

Discrete-Time Processing of

Continuous-Time Signals

Discrete Fourier Transform

∫∞

∞−

−= dtetxXtjωω )()(

∑∞

−∞=

−=n

njenx

TX

ωω )(1

)(

∑=

−=1-N

0n

/2)()( NknjenxkX π

k

Link between Continuous and Discrete

∫∞

∞−

−= dtetxXtjωω )()( ∑

=

−=

1-N

0n

2

)()(n

N

kj

enxkX

π

x(t) x(n)Sampling Theorem

x(t)Laplace Transform

X(s) x(n) X(z)z Transform

x(t) X(jω) x(n) X(k)Fourier Transform Discrete Fourier Transform

∫∞

∞−

−= dtetxsX st)()( ∑∞=

−∞=

−=n

n

nznxzX )()(

t

x(t)

Continuous Discrete

x(n)

n