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Chapter 4 The Continue-Time Fourier
Transform
Instructor: Hongkai Xiong (熊红凯)
Distinguished Professor (特聘教授)
http://min.sjtu.edu.cn
TAs: Yuhui Xu，Qi Wang
Department of Electronic Engineering
Shanghai Jiao Tong University
2019-04
Foreword of the Chapter
• By exploiting the properties of superposition and time invariance, if we know the response of an LTI system to some inputs, we actually know the response to many inputs
• If we can find sets of “basic” signals so that
▫ We can represent rich classes of signals as linear combinations of these building block signals.
▫ The response of LTI Systems to these basic signals are both simple and insightful.
• Candidate sets of basic signal
▫ Unit impulse function and its delays
▫ Complex exponential signals (Eigenfunctions of all LTI systems)
• In this Chapter, we will focus on: why, how, what
▫ Can we represent aperiodic signals as “sums or integrals” of complex exponentials
▫ How to represent aperiodic signals as “sums or integrals” of complex exponentials
▫ What kinds of aperiodic signals can we represent as “sums or integrals” of complex exponentials? (how large types of such signals can benefit from the Fourier Transform?)
][/)( nt
sttj ee / nnj ze /
Topic
4.0 Introduction
4.1 The Continuous-Time Fourier Transform
4.2 The Fourier Transform for Periodic Signals
4.3 Properties of the Continuous-Time Fourier Transform
4.4 The Convolution Property
4.5 The multiplication Property
4.6 System Characterized by Linear Constant-Coefficient
Differential Equations
Topic
4.0 Introduction
4.1 The Continuous-Time Fourier Transform
4.2 The Fourier Transform for Periodic Signals
4.3 Properties of the Continuous-Time Fourier Transform
4.4 The Convolution Property
4.5 The multiplication Property
4.6 System Characterized by Linear Constant-Coefficient
Differential Equations
4.0 Introduction
• Fourier Series Representation
▫ It decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex exponentials). The discrete-time Fourier transform is a periodic
• Fourier Transform
▫ A representation of aperiodic signals as linear combinations of complex exponentials
T1 kept fixed T increases
Motivating Example
Discrete
frequency
points
become
denser in ω
as T
increases
• Then for periodic square wave, the spectrum of x(t), i.e. {ak},
are , the spectrum space is
• Then for square pulse, the spectrum X(jω) are , the
spectrum space is , i.e. the complex exponentials
occur at a continuum of frequencies
Tk
Tkak
0
10 )sin(2
T
20
)sin(2 1T
02
0 T
Topic
4.0 Introduction
4.1 The Continuous-Time Fourier Transform
4.2 The Fourier Transform for Periodic Signals
4.3 Properties of the Continuous-Time Fourier Transform
4.4 The Convolution Property
4.5 The multiplication Property
4.6 System Characterized by Linear Constant-Coefficient
Differential Equations
4.1.1 Development
• To derive the spectrum for aperiodic signals x(t), we can
approximate it by a periodic signal with infinite period
T
)(~ tx
-T1 T1
…… ……
-T1 0 T1 T
)(tx
)(~ tx
)()(~lim txtxT
Assuming (1) is converged, we define
• Thus
𝑥 𝑡 =
𝑘
𝑎𝑘𝑒𝑗𝑘𝜔0𝑡 =
𝑘
1
𝑇𝑋(𝑗𝑘𝜔0)𝑒
𝑗𝑘𝜔0𝑡
=1
2𝜋
𝑘=−∞
∞
𝑋(𝑗𝑘𝜔0)𝑒𝑗𝑘𝜔0𝑡𝜔0
• When 𝑇 → ∞
𝑥 𝑡 =1
2𝜋න−∞
∞
𝑋(𝑗𝜔)𝑒𝑗𝜔𝑡𝑑𝜔
𝑥 𝑡 =1
2𝜋∞−∞𝑋(𝑗𝜔)𝑒𝑗𝜔𝑡𝑑𝜔
𝑋 𝑗𝜔 = ∞−∞𝑥(𝑡)𝑒−𝑗𝜔𝑡𝑑𝑡
Synthesis equation
Analysis equation
4.1.2 Convergence
• What kinds of signals can be represented in Fourier Transform (satisfies one of the following 2 conditions)
▫ 1、Finite energy
Then we are guaranteed that:
𝑋(𝑗𝜔) is finite
∞−∞
𝑒(𝑡) 2𝑑𝑡 = 0
(𝑒 𝑡 = ො𝑥 𝑡 − 𝑥(𝑡) ො𝑥 𝑡 =1
2𝜋∞−∞𝑋(𝑗𝜔)𝑒𝑗𝜔𝑡𝑑𝜔)
▫ 2、Dirichlet conditions, require that
𝑥(𝑡) be absolutely integrable
𝑥(𝑡) have a finite number of maxima and minima within any finite interval
𝑥(𝑡) have a finite number of discontinuities within any finite interval. Furthermore, each of these discontinuities must be finite
Then we guarantee that
ො𝑥 𝑡 is equal to 𝑥(𝑡) for any 𝑡 except at a discontinuity, where it is equal to the average of the values on either side of the discontinuity
𝑋(𝑗𝜔) is finite
Examples • Exponential function
Magnitude Spectrum Phase Spectrum
Even symmetry Odd symmetry
If α is complex, x(t)
is absolutelty
integrable as long
as Re{α}>0
22
2)(0，)(
jXetx
t
Examples
• Unit impulse 𝑥 𝑡 = 𝛿(𝑡)
𝑋 𝑗𝜔 = න−∞
+∞
𝛿(𝑡)𝑒−𝑗𝜔𝑡𝑑𝑡 = 1
• DC Signal
)(2)(1)( jXtx
)(21
)(2
1
2
1)(
2
1
de tj
Example
• Rectangle Pulse Signal
1
1
,0
,1)(
Tt
Tttx 1
1 1
sin( ) 2 =2 ( )a
TX j T S T
𝑆𝑎 𝑥 =𝑠𝑖𝑛𝑥
𝑥
𝑠𝑖𝑛𝑥 𝑥 =𝑠𝑖𝑛𝜋𝑥
𝜋𝑥
Topic
4.0 Introduction
4.1 The Continuous-Time Fourier Transform
4.2 The Fourier Transform for Periodic Signals
4.3 Properties of the Continuous-Time Fourier Transform
4.4 The Convolution Property
4.5 The multiplication Property
4.6 System Characterized by Linear Constant-Coefficient
Differential Equations
• For a periodic signal x(t) with fundamental frequency , what’s its FT?
T
20
k
tjk
keatx 0)(
k
tjk
k
k
tjk
k eaeatx ][][)]([ 00
?0 tjk
ethe question becomes:
• Thanks to the impulse function, suppose𝑋 𝑗𝜔 = 𝛿 𝜔 − 𝜔0
𝑥 𝑡 =1
2𝜋න−∞
∞
𝛿 𝜔 − 𝜔0 𝑒𝑗𝜔𝑡𝑑𝜔 =1
2𝜋𝑒𝑗𝜔0𝑡
• That is 𝑒𝑗𝜔0𝑡 ↔ 2𝜋𝛿 𝜔 − 𝜔0
• So
𝑥 𝑡 =
𝑘
𝑎𝑘𝑒𝑗𝑘𝜔0𝑡 ↔ 𝑋 𝑗𝜔 =
𝑘
2𝜋𝑎𝑘 𝛿 𝜔 − 𝑘𝜔0
— All the energy is
concentrated in one
frequency — ωo,
• So for a periodic signal x(t) with fundamental frequency , its FT is:
▫ Fourier Series Coefficient
▫ Fourier Transform
• The FT 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 kω0 is 2π times the kth F.S. coefficient ak
T
20
2
2
0
0
)(1
)(
T
T
tjk
k
tjk
k
dtetxT
a
eatx
0 0
2( ) ( 2 ( )k
k
x t X j a kT
） ，
• Example: )cos()( 0ttx
0 0
1 1
1 1( )
2 2
1
2
0 1
j t j t
k
x t e e
a a
a k
，
)]()([)( 00 jX
ka
k1-1
1/2
00
)( jX
Similarly:
)]()([sin 000 jt
• Example:
k
kTttx )()(
……
ka
0 0( ) ( )k k
t kT k ……
( )X j
k
Same function in the frequency-domain!
Note: (period in t) T ⇔ (period in ω) 2π/T
Topic
4.0 Introduction
4.1 The Continuous-Time Fourier Transform
4.2 The Fourier Transform for Periodic Signals
4.3 Properties of the Continuous-Time Fourier Transform
4.4 The Convolution Property
4.5 The multiplication Property
4.6 System Characterized by Linear Constant-Coefficient
Differential Equations
• Linearity
• Time Shifting
)()( jXtx )()( jYty
)()()()( jbYjaXtbytax
)()( jXtx
)()( 0
0 jXettx
tj
• Time and Frequency Scaling
)()( jXtx
)(||
1)(
a
jX
aatx
1a )()( jXtx for
compressed in time ⇔ stretched in frequency
( ) ?x at b
a
bj
ea
jX
abatx
)(||
1)(
• Example: Determine the Fourier Transform of the following signals
2( ) ( )tx t e u t
2( 1)( ) ( )tx t e u t
2( ) ( 1)tx t e u t
1.
2.
3.
• Differentiation
• The differentiation operation enhances high-frequency components in the effective frequency band of a signal
• Without any further information about the DC component of the original signal, we cannot completely recover it from its differentials
)()( jXtx
)()(
jXjdt
tdx
• Integration
dxtg
t
)()( )()( jXtx
)()()()(*)()( jUjXjGtutxtg
0
1
u(t)
where u(t) is the unit step function, defined as
• 𝑢 𝑡 =1+𝑠𝑔𝑛(𝑡)
2
• 𝑥 𝑡 = 𝑠𝑔𝑛 𝑡 ↔ 𝑋 𝑗𝜔 =2
𝑗𝜔
0
1
u(t)
0
1
-1
sgn(t)
0
1
DC
)(2)(1)( jXtx
)(1
)(
j
jU
Odd
function
• 𝑢 𝑡 =1+𝑠𝑔𝑛(𝑡)
2
• 𝑥 𝑡 = 𝑠𝑔𝑛 𝑡 ↔ 𝑋 𝑗𝜔 =2
𝑗𝜔
0
1
u(t)
0
1
-1
sgn(t)
0
1
DC
)(2)(1)( jXtx
)(1
)(
j
jU
Odd
function
)()()()(*)()( jUjXjGtutxtg
)()0()(1
)()()(
XjXj
jGdxtg
t
)(1
)(
j
jU
according to:
The integration operation diminishes high-frequency components in the effective frequency band of a signal
• Example: triangle pulse
1
2
2
' ( )x t
2
2
2
2
)(tx
)2
||0
2||
||21
)(
t
tt
tx
，
，
'
1
11
'
1
( ) )
)( ) (0) ( )
(0)= ( ) 0
x t X j
X jX j X
j
X x t dt
（
（
又Since
)4
(2
)4
(sin8
)( 2
2
2
SajX
• 2 approaches to calculate X(0) ：
01. (0) ( ) |
2. (0) ( )
X X j
X x t dt
• Example: sgn(t)
)0(1
)0(1)sgn()(
t
tttx
，
，
0
1
-1
sgn(t)
• By defining the sgn function as a special exponential function
• By representing the sgn function in terms of unit step functions
0)0(
)0()(1
te
tetx
t
t
，
，
1
0
-1
)(1 tx
)(lim)sgn( 10
txt
jjXt
2)(lim)sgn( 1
0＝
)()()sgn( tutut
jjjt
2]
1)([)
1)(()sgn(
• By exploiting integration property
)()(2)(sgn1)(2)sgn( 1
' txtttut
1 1
( )( ) ( )
dx tx t X j
dt 设
)()()(1
xtxdttx
t
t
xdttxtx )()()( 1
Suppose
)()(2)()0()(
)( 11
xX
j
jXjX
2)(1 ＝jX
jj
t2
)()1(2)(22
)sgn(
When x(-∞)≠0
• Duality
▫ Both time and frequency are continuous and in general aperiodic
▫ Suppose f() and g() are two functions related by
Same except for
these differences
)()( jXtx )(2)( jxtX
• Examplesin
( )Wt
x tt
＝
W
WjX
,0
,1)(
• Example
• Example
1( )x t
t
00
0/2)()sgn()(
jjXttx
)sgn()( jjX
21
1)(
ttx
22
2)(0}Re{)(
jXetx
t，
ejX )(
• Other duality properties
▫ （1） Frequency Shifting
)()( jXtx
))(()( 00
jXtxetj
• Example:
)]()([
)]()([
][2
1sin
)]()([
][2
1cos
00
00
0
00
0
00
00
j
j
eej
t
eet
tjtj
tjtj
▫ （2）Differentiation in frequency domain
▫ （3）Integration in frequency domain
d
jdXtjtx
)()(
d
jdXjttx
)()(
dXtxtxjt
)()()0()(1
t
dXjt
tx)(
)(when x(0)=0,
• Example: 2( ) ( ) ?tx t te u t
2
2
2
1( )
2
1 1( ) ( )
2 (2 )
t
t
e u tj
dte u t j
d j j
2( ) ( 1) ?tx t te u t
• Example :To determine x(t) according to X(jω)
1 1
)( jX
1
)(' jX
11
Hints: To exploit the
differentiation property in
frequency domain
'
1
11
'
1
) ( )
)( ) (0) ( )
(0)= ( ) 0
X j x t
x tx t x t
jt
x X j d
（
（
又
• Conjugation and Conjugate Symmetry
)()( jXtx
)()( jXtx
If x(t) is real valued
)()( jXjX —Conjugate Symmetry
)]([)](Re[)( jXjIjXjX m
)](Re[)](Re[ jXjX
)](Im[)](Im[ jXjX
①
)(|)(|)( jXjejXjX
|)(||)(| jXjX ＝
)()( jXjX ＝
②
evenandrealtx )( evenandrealjX )(
oddandrealtx )( oddandimaginarypurelyjX )(
)]([)]()([2
1)( jXRetxtxtxe
)]([)]()([2
1)(0 jXjItxtxtx m
③
• Example:
)}({2)()()( || tueEtuetueetx t
v
ttt
jtue t
1
)(
222
2}
2
1Re{2)(
jtx
• Parseval’s Relation
dffXdjXdttx
222 |)(||)(|2
1|)(|
2|)(| jX
d
T
jXdttx
T TT
T
2
2 |)(|lim
2
1|)(|
1lim
T
jX
T
2|)(|lim
——Energy-density Spectrum
——Power-density Spectrum
and：
2|)(| fX——Energy per unit frequency (Hz)
• Example1：
• Example2：
)(tx
)( jX
2/
-1 -0.5 10.5
sin 2( )
tx t
t＝
dttx 2|)(|
dttx 2|)(|
To determine
To determine
• Example:
-1-2 1 2
1
2
)(tx
t
To use the FT of typical signals and FT
properties to determine the FT of the following
signals
• Solution 1:
2( ) 2 ( ) ( 2) ( 2)x t g t u t u t
2
1
-2
1
1-1
2
+ +
gτ(t) is the rectangle pulse with width of τand unit magnitude
( )g t
1
2
-
2
t
• Solution 2: '
1
1 1
( ) ( )
( ) ( )
x t x t
x t X j
设
11
( )( ) (0) ( ) 2 ( ) ( )
X jX j X x
j
则
-1
1
2
-2
'( )x t
t1 2
Assuming :
Then
• Example: To determine the FC of the periodic signal by using FT
2
1T
2
1T1T
2
3 1T
)(tf
t
。。。 。。。
• Let be the basic signal
•
)()( 00 Ftf
)(0 tf1
0 2/1T
)('
0 tf
0 2/1T1
2
T
)("
0 tf
0
1
2
T
1
2
T
" '1 10
1 1
2 2( ) ( ) ( ) ( )
2 2
T Tf t t t t
T T
1|)(
10
1
nn FT
F
• Example :To determine x(t) according to )( jX
0 0
A
|)(| jX )( jX
0 0 w
A
|)(| jX )( jX
00
2/
2/
1.
2.
Notes: They
have different
phase
spectrum
Topic
4.0 Introduction
4.1 The Continuous-Time Fourier Transform
4.2 The Fourier Transform for Periodic Signals
4.3 Properties of the Continuous-Time Fourier Transform
4.4 The Convolution Property
4.5 The multiplication Property
4.6 System Characterized by Linear Constant-Coefficient
Differential Equations
4.4.1 Convolution Property
• Example: the Triangle Impulse Signal
4.4.2 Frequency Response
• Definition:
dtth )( －stable system
( )( ) ( ) ( ) / ( )
( )
Y jY j X j H j H j
X j
Conditioned on:
Then:
4.4.2 Frequency Response
• The frequency response H(jω) can completely represent a stable LTI system (NOT all LTI systems)
H1(jω) H2(jω) H1(jω) · H2(jω)
Series interconnection of LTI systems (Cascaded system)
H1(jω)
H2(jω)
H1(jω) + H2(jω)
Parallel interconnection of LTI systems
4.4.2 Frequency Response
• The frequency response is the F.T. of the impulse response, it captures the change in complex amplitude of the Fourier transform of the input at each frequency ω
▫ For a complex exponential input x(t), as a consequence of the eigenfunction property, the output y(t) can be expressed as:
▫ For a sinusoid input x(t), as a consequence of the eigenfunction property, the output y(t) can be expressed as:
jHjejHjH
Magnitude gain Phase shifting
tjtjejHtyetx 0
0
0 |)()()(
))(cos()()()cos()( 0000 jHtjHtyttx
• Example: Consider an LTI system with If the input x(t)=sin(t), determine the
output y(t)
• Solution:
1( )
1H j
j
2
1
1( )
1
1( )
1
( ) tan ( )
H jj
H j
H j
4sin
2
1
))1(sin()1()(
t
jHtjHty
• Example: Consider an LTI system with
for the input x(t)
Determine the output of the system
)()( tueth t
)()( 2 tuetx t
)(*)()( thtxty
• Example: for a system with Gaussian response, i.e. the unit impulse response is Gaussian, consider the output of the system with a Gaussian input
Gaussian × Gaussian = Gaussian ⇒ Gaussian ∗ Gaussian = Gaussian
Why: Log-Magnitude and Phase to illustrate the frequency
response
𝑌(𝑗𝜔) = 𝐻 𝑗𝜔 × 𝑋 𝑗𝜔
𝑙𝑜𝑔 𝑌(𝑗𝜔) = 𝑙𝑜𝑔 𝐻 𝑗𝜔 + 𝑙𝑜𝑔 𝑋 𝑗𝜔
∠𝐻 𝑗𝜔 = ∠𝐻1 𝑗𝜔 + ∠𝐻2(𝑗𝜔)
𝐻1(𝑗𝜔) 𝐻2(𝑗𝜔)
𝑙𝑜𝑔 𝐻(𝑗𝜔) = 𝑙𝑜𝑔 𝐻1 𝑗𝜔 + 𝑙𝑜𝑔 𝐻2 𝑗𝜔
∠𝑌 𝑗𝜔 = ∠𝐻 𝑗𝜔 + ∠𝑋(𝑗𝜔)
Easy to add
Easy to add
Cascading:
How: Plotting Log-Magnitude and Phase
• a) For real-valued signals and systems
• b) For historical reasons, log-magnitude is usually plotted in units of decibels (dB):
Plot for ω ≥ 0, often with a
logarithmic scale for
frequency in CT
Why 20 log10(.)power magnitude
So… 20 dB or 2 bels:
= 10 amplitude gain
= 100 power gain
• A Typical Bode plot for a second-order CT system
20 log10|H(jω)| and ∠ H(jω) vs. log10ω
4.4.3 Filtering
－a process in which the relative complex magnitudes of the frequency components in a signal are changed or some frequency components are completely eliminated
• Frequency-Selective Filters—systems that are designed to pass some frequency components
undistorted, and diminish/eliminate others significantly
• Typical types of frequency-selective filters▫ LPF(Low-pass Filter)
▫ HPF(High-pass Filter)
▫ BPF(Band-pass Filter)
▫ BSF (Band-stop Filter
)( jH
cc
1
c
cjH
,0
1)(
，LPF
HPF
BPF
passbandstopband
c
cjH
,0
1)(
，
21
21
,0
1)(
cc
ccjH
，，
cutoff frequency
upper cutoff frequencylower cutoff frequency
• Example :
1/4-1/4
1
x(t)
1-1 t
。。。。。。
)( jH
3 3
To determine the response of the LPF to the signal x(t)
• Some typical systems
▫ ① Delay
▫ ② Differentiator
0
0( ) ( )
( ) ( )j t
y t x t t
Y j X j e
0
)(
)()(
tje
jX
jYjH
)()(
)()(
jXjjY
dt
tdxty
j
jX
jYjH
)(
)()(
▫ ③ Integrator
when
)()0()(
)(
)()(
Xj
jXjY
dxty
t
0)()0(
dttxX
jjX
jYjH
1
)(
)()(
• Example: to determine outputs of the system with H(jω) in the figure with the following input signals
jtetx )(
)6)((
1)(
jjjX
-1 1
2j
-2j
)( jH
1、
2、
• Example: for the following signal x(t) with period of 1
To determine the output of the system with frequency response H(jω) with the input x(t)
1)2
1(,0
)2
1(,2sin
)(
mtm
mtmttx
)33(3
)(
jjH
• Solution:
∴ x(t) contains the frequency components:
Only the DC and the first order harmonic components are within
the passband of the LPF
k
k katx )(2)( 0
22
0 T
,4,2,0
)( jH Differe
ntiator
3
1
33
)(1 ty )(ty
and
Filters
• Zero-phase shifting Ideal LPF
)( jH
cc
c
cjH
||,0
||,1)(
sin( )
sin
c
c c
c
th t
t
t
t
)(1
2
1)( tSits c
dxx
xySi
sin)(
Unit impulse response Unit step response
where
The unit impulse response of the HPF is t
ttth c
sin)()(
• Linear Phase Ideal LPF
c
tjejH
||,)( 0
|)(| jH
cc
)( jH
cc
c
cjH
||,0
||,1|)(| ctjH ||,)( 0
Result: Linear phase ⇔ simply a rigid shift in time, no distortion
Nonlinear phase ⇔ distortion as well as shift
𝑌 𝑗𝜔 = 𝑒−𝑗𝜔𝑡0𝑋 𝑗𝜔 𝑦 𝑡 = 𝑥(𝑡 − 𝑡0)time-shift
▫ Unit impulse response:
▫ Unit step response:
)(
)(sin)(
0
0
tt
ttth c
)]([1
2
1)( 0ttSits c
• How do we think about signal delay when the phase is nonlinear?
Concept of Group Delay
When the signal is narrow-band and concentrated near ω0, H
(jw) ~ linear with ω near ω0, then the differential of H (jw) at ω0
reflects the time delay.
For frequencies “near” ω0
For w “near” ω0
Τ(ω0)Time delay of
the original signal
• Non-ideal LPF
R
C
1( ) ,H j
j RC
( )H j
2 2( )H j
1( ) tan ( )H j
)( jH
)()( tueth t )()1()( tuets t
1e
)(th
1 t
1
11 e
)(ts
t
1
Unit impulse response Unit step response
• causal h(t <0) = 0, decaying
• s(t) non-oscillation and non-overshoot
• Time domain and frequency domain aspects of non-ideal filter
▫ Trade-offs between time domain and frequency domain characteristics, i.e. the width of transition band ↔ the setting time of the step response
Definitions:
Passband ripple: δ1
Stopband ripple: δ2
Definitions:
Rise time: trSetting time: tsOvershoot: Δ
Ringing frequency ωr
Passband Transition Stopband
Rise time
Setting
time
Setting time: the time at which the step response settles to within δ (a specified tolerance) of its final value
Topic
4.0 Introduction
4.1 The Continuous-Time Fourier Transform
4.2 The Fourier Transform for Periodic Signals
4.3 Properties of the Continuous-Time Fourier Transform
4.4 The Convolution Property
4.5 The multiplication Property
4.6 System Characterized by Linear Constant-Coefficient
Differential Equations
4.5.1 Multiplication Property
)()()()( 2211 jXtxjXtx
)()()( 21 txtxtx
)]()([2
1)( 21
jXjXjX
• Example :
E
0 1
2
T
)]2
()([sin)( 11
TtututEtx
• Example :
t
t
sin
t
t
2
sin
]}2sin
[]sin
[{2
1)(
t
t
t
ttx
1
-1 1
1
-1/2 1/2
-3/2 -1/2 1/2 3/2
1/2
Then
2
sin sin2 ?
tt
dtt
• Example :
𝑟 𝑡 = 𝑠 𝑡 × 𝑃 𝑡 ↔ 𝑅 𝑗𝜔 =1
2𝜋[𝑆 𝑗𝜔 ∗ 𝑃 𝑗𝜔 ]
𝐹𝑜𝑟 𝑝 𝑡 = 𝑐𝑜𝑠𝜔0𝑡 ↔ 𝑃 𝑗𝜔 = 𝜋[𝛿 𝜔 − 𝜔0 + 𝛿(𝜔 + 𝜔0)]
𝑅 𝑗𝜔 =1
2[𝑆(𝑗 𝜔 − 𝜔0 ) + 𝑆(𝑗 𝜔 + 𝜔0 )]
4.5.2 Modulation
• Why?
▫ More efficient to transmit E&M signals at higher frequencies
▫ Transmitting multiple signals through the same medium using different carriers
▫ Transmitting through “channels” with limited passbands
▫ Others...
• How?
▫ Many methods
▫ Focus here for the most part on Amplitude Modulation (AM)
Amplitude modulation
(AM)
Drawn assuming:
Modulating
signal
Carrier
signal
Modulated
signal
• Synchronous Demodulation of Sinusoidal AM
𝑋 𝑗𝜔 =1
2𝜋𝑋𝑐 𝑗𝜔 ∗ 𝐶 𝑗𝜔
=1
2𝑋𝑐 𝑗 𝜔 − 𝜔0 + 𝑋𝑐 𝑗 𝜔 + 𝜔0
=1
2X jω +
1
4[𝑋𝑐(𝑗 𝜔 − 2𝜔0 ) +
𝑋𝑐(𝑗 𝜔 + 2𝜔0 )]
ො𝑥 𝑡 = 𝑥𝑐(𝑡) × 𝑐(𝑡)
If 𝜃 = 0
What if 𝜃 ≠ 0?
• Synchronous Demodulation (with phase error) in the Frequency Domain
𝑐𝑜𝑠(𝜔𝑐𝑡 + 𝜃) ↔ 𝜋𝑒𝑗𝜃𝛿 𝜔 − 𝜔0 + 𝜋𝑒−𝑗𝜃𝛿(𝜔 + 𝜔0)
• Asynchronous Demodulation
▫ Assume ωc>> ωM, so signal envelope looks like x(t)
▫ Add same carrier with amplitude A to signal
A = 0 ⇒ DSB/SC (Double Side Band, Suppressed Carrier)A > 0 ⇒ DSB/WC (Double Side Band, With Carrier)
In order for it to function properly, the envelope function mustbe
positive for all time, i.e.A+ x(t) > 0 for all t.
Demo: Envelope detection for asynchronous demodulation.
Advantages of asynchronous demodulation:
— Simpler in design and implementation.
Disadvantages of asynchronous demodulation:
— Requires extra transmitting power [Acosωct]2to make sure A+
x(t) > 0 ⇒Maximum power efficiency = 1/3 (P8.27)
• Example:
)(tx )(ty
)(1 ty
t5cos
)( jH
t
ttx
2sin)(
)( jH
5 5 For
To determine )(tySignal processing in
frequency domain
• Double-Sideband (DSB) and Single-Sideband (SSB) AM
DSB, occupies 2ωMbandwidth in ω> 0.
Each sideband approach only occupies ωMbandwidth in ω> 0.
Since x(t) and y(t)
are real, from Conjugate symmetry both LSB and USB signals carry exactly the
same information.
• Single-Sideband (SSB) AM
Can also get SSB/SC
or SSB/WC
• An implementation of SSB modulation, p600,figure 8.21-22
Hilbert
Transform
tth
j
jjH
1)(
0
0)(
• Frequency-Division Multiplexing (FDM)(Examples: Radio-station signals and analog cell phones)
All the channels
can share the
same medium.
• FDM in the Frequency-Domain
• Demultiplexing and Demodulation
▫ Channels must not overlap ⇒Bandwidth Allocation
▫ It is difficult (and expensive) to design a highly selective bandpass filter with a tunable center frequency
▫ Solution –Superheterodyne Receivers
ωa needs to be tunable
• The Superheterodyne Receiver
▫ Operation principle:
— Down convert from ωc to ωIF, and use a coarse tunable BPF for the front end.
— Use a sharp-cutoff fixed BPF at ωIF to get rid of other signals.
4.5.3 Sampling
• Most of the signals we encounter are CT signals, e.g. x(t). How do we convert them into DT signals x[n] to take advantages of the rapid progress and tools of digital signal processing
▫ — Sampling, taking snap shots of x(t) every T seconds
• T –sampling period, x[n] ≡x(nT), n= ..., -1, 0, 1, 2, ... —Regularly spaced samples
• Applications and Examples
▫ —Digital Processing of Signals
▫ —Images in Newspapers
▫ —Sampling Oscilloscope
▫ —…
How do we perform sampling?
• Why/When Would a Set of Samples Be Adequate?▫ Observation: Lots of signals have the same samples
▫ By sampling we throw out lots of information –all values of x(t) between sampling points are lost.
▫ Key Question for Sampling:
Under what conditions can we reconstruct the
original CT signal x(t) from its samples?
• Impulse Sampling—Multiplying x(t) by the sampling function
• Analysis of Sampling in the Frequency Domain
Multiplication Property =>
=Sampling Frequency
• Illustration of sampling in the frequency-domain for a band-limited (X(jω)=0 for |ω| > ωM) signal
• Reconstruction of x(t) from sampled signals
If there is no overlap
between shifted
spectra, a LPF can
reproduce x(t) from xp(t)
Suppose x(t) is band-limited, so that
X(jω)=0 for |ω| > ωM
Then x(t) is uniquely determined by its
samples {x(nT)} if
where ωs = 2π/T
• Observations▫ (1) In practice, we obviously don’t
sample with
impulses or implement ideal lowpass filters
— One practical example: The Zero-Order Hold
▫ (2) Sampling is fundamentally a time varying operation, since we multiply x(t) with a time-varying function p(t). However, H(jω) is the identity system (which is TI) for band-limited x(t) satisfying the sampling theorem (ωs > 2ωM).
▫ (3) What if ωs <= 2ωM? Something different: more later.
).(
.
.
)(,
2
2
,...,2,1,0),()(
.0)(
:
txequalexactlywillsignaloutputresultingthe
iffrequencycutoffandTgainwithfilterlowpassidealan
throughprocessedthenistrainimpulseThisvaluessamplesuccessive
arethatamplitudeshaveimpulsesuccessivewhichintrainimpulse
periodicageneratingbytxtreconstruccanwesamplestheseGiven
Twhere
ifnnTxsamplesitsbydetermineduniquelyistx
ThenforjXwithsignallimitedbandabeusLet
TheoremSampling
mcms
c
s
s
ms
s
m
• Example:
:
.0)()(
signalsfollowingtheforrateNyquisttheDetermine
forjXwithtxsignallimitedbandaConsider m
dt
tdx
tx
tx
)()3(
)()2(
1)(2)1(
2
• Time-Domain Interpretation of Reconstruction of Sampled Signals —Band-Limited Interpolation
The lowpass filter interpolates the samples assuming x(t) contains no
energy at frequencies >= ωc
• Graphic Illustration of Time-Domain Interpolation
▫ Original CT signal
▫ After Sampling
▫ After passing the LPF
• Interpolation Methods (1): Band-limited Interpolation: ideal LPF, i.e. sinc function in time domain
)]([)(
)()()()(
scs
n
c
cc
s
n
sr
nTtSanTx
SanTtnTxtx
• Interpolation Methods (2): Zero-Order Hold
)(txr
0T
)(0 th
])2/sin(2
[)( 2/
0
T
ejH Tj
)2/sin(2
)(
)(
)()(
2
0T
jHe
jH
jHjH
Tj
r
T
Reconstruct filter)(0 th)(0 tx
)()( jHth rr )(
)(
tx
tr)(tx p
)(tp
)(tx
H(jω) – Ideal interpolation filter (LPF)
• Interpolation Methods (3): First-Order Hold —Linear interpolation
2
1 ]2/
)2/sin([
1)(
T
TjH
-T
)(1 th
T
)(txr
• Under sampling and Aliasing
When ωs ≦ 2ωM => Under-sampling
Xr (jω)≠X(jω) Distortion due to aliasing
— Higher frequencies of x(t) are “folded back” and take on the “aliases”of lower frequencies
— Note that at the sample times, xr(nT) = x(nT)
• Example:
X(t) = cos(ω0t + Φ)
Sampling of cosω0t
Aliasing case:
Then with the ideal LPF with
cut off frequency of ωM<
ωc< ωs- ω0 , the
reconstructed signal is
cos((ωs-ω0)t)
Ref. Q7.38
• Example: AM with an Arbitrary Periodic Carrier
C(t) – periodic with period T, carrier frequency ωc = 2π/T
• Example: Modulating a (Periodic) Rectangular Pulse Train
In practice, we can use a (periodic) rectangular pulse train instead of impulses, since the later is impractical
• Discussions on modulating a (Periodic) Rectangular Pulse Train
▫ 1) We get a similar picture with any c(t) that is periodic with period T
▫ 2) As long as ωc= 2π/T > 2ωM, there is no overlap in the shifted and
scaled replicas of X(jω). Consequently, assuming a0≠0:
x(t) can be recovered by passing y(t) through a LPF
▫ 3) Pulse Train Modulation is the basis for Time-Division Multiplexing
▫ Assign time slots instead of frequency slots to different
channels, e.g. AT&T wireless phones
▫ 4) Really only need samples{x(nT)} when ωc> 2ωM⇒Pulse Amplitude
Modulation
Topic
4.0 Introduction
4.1 The Continuous-Time Fourier Transform
4.2 The Fourier Transform for Periodic Signals
4.3 Properties of the Continuous-Time Fourier Transform
4.4 The Convolution Property
4.5 The multiplication Property
4.6 System Characterized by Linear Constant-Coefficient
Differential Equations
LTI Systems Described by LCCDE’s
(Linear-constant-coefficient differential equations)
Using the Differentiation Property
Transform both sides of the equation
1) Rational, can usePFE to get h(t)
2) If X(jω) is rationale.g. x(t)=Σcie
-at u(t)
then Y(jω) is also rational
PFE: Partial-fraction expansion
• Example：
)(2)(
)(3)(
4)(
2
2
txdt
tdxty
dt
tdy
dt
tyd
3)(4)(
2)(
2
jj
jjH
• Zero-state response of LTI systems——Partial-fraction expansion method
• Example:
31)3)(1(
2
3)(4)(
2)( 21
2
j
A
j
A
jj
j
jj
jjH
vj
2
1|
3
2|)()1( 111
vv
v
vvHvA
2
1|
1
2|)()3( 332
vv
v
vvHvA
3
2
1
1
2
1
)(
jj
jH
)(1
tuej
t
)(2
1)(
2
1)( 3 tuetueth tt
Let
then
and
• Example: To calculate the zero-state response of the system discussed in previous example
3)1(1)3()1(
2)()()( 2
2
1211
2
j
A
j
A
j
A
jj
jjXjHjY
4
1|
)3(
1|]
3
2[|)]()1[(
)!12(
11211
2
11
vvv
vv
v
dv
dvYv
dv
dA
2
1|)()1( 1
2
12 vvYvA
jtue t
1
)(
4
1|
)1(
2|)()3( 3232
vv
v
vvYvA
2
1 1 1
4 2 4( )1 ( 1) 3
Y jj j j
2)(
1]
1[)(
jjd
djtute t
)(]4
1
2
1
4
1[)( 3 tueteety ttt
high-order pole point
• Homework BASIC PROBLEMS WITH ANSWER: 4.1, 4.4
BASIC PROBLEMS: 4.21, 4.22, 4.25, 4.32, 6.21, 6.22, 7.3, 7.4, 8.22, 8.30
Many Thanks
Q & A
134
