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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