discrete -time signals and systems continuous-time signals &...
TRANSCRIPT
-
Yogananda Isukapalli
Discrete - Time Signals and Systems
Continuous-Time signals & systemsFourier Transform
-
Fourier Transform
Motivation:We need to define the frequency spectrum for
a more general class of continuous time signals.Fourier Transform is at the heart of moderncommunication systems
We have defined the spectrum for a limited class of signals such as sinusoids and periodic signals. These kind of spectrum are part ofFourier series representation of signals
-
( ) ( ) (
1( ) ( ) 2
1)j t
j t
Inverse continuous ti
Forward continuous time Fourier Transform
X j x
me Fourier Transform
x t
t e
X j e dt
dt
w
ww
wp
¥
¥-
-
-¥
¥
=
=
-
-
ò
ò
(
(2)
() )FFrequency domTime doma aii n
X j
n
x t w¬¾®
:Fourier Transform Definition
-
Example
7
7
( ) ( )
; ( ) ( )
( )
j t
t
t
Forward continuous
Determine the Fourier Transform of the one - sided
expontime Fourier Transform
X j x t
ential signal
e d
x
t
t
e u
t
t
e u
ww¥
-
-
¥
-
-
=
-
=
=
ò
7
0
j t
t j t
e dt
e e d (notice the change in limits)t
w
w
¥-
-¥¥
- -=
ò
ò
-
( )( )
( )( ) ( )( )
( )
( )( )( )
( )
77
0 0
7 7 0
7
7
= 7
7
1
7
1
7
0
j t
j t
F
j j
t
j
ee dt
j
e e
j
j
Ti Frequency domainme dom i
j
e
a n
e
ww
w w
w
w
w
w
w
¥- +¥- +
- + ¥ -
+ ¥
-
-
+
=- +
-=
- +
=+
+
=
¬¾®
ò
!
Example contd…
-
Fourier Series & Fourier Transform
periodic signal, with period
Time ‘t’0 20T20T-0T- 0T23 0T-25 0T- 02T-
0T
Interpretation of Fourier Transform from Fourier Series
Note that the shape of the signal is drawn arbitrarily
0( )Tx t
0( )Tx t
Fig.18.1
-
Fourier Series & Fourier Transform contd…
0
0
0
0
2
0 20
0
;
1 ( ) ,
2
,
( )
k
kj tT k
T k
T
T
k
j tk
k
Fourier Analysis equation
a
And the Fourier synthesis equation
x t a e
x e tT
kT
t d
w
w pw w
¥
=-¥
-
-
=
=
= =ò
å
The periodic signal can be approximated by summingup harmonically related periodic exponential signals
ka Spectrum
Frequency, 0ww k=0 0w 03w …..0w-03w-………..
-
Now, assume that the period of the signal x(t) is infinite
Time ‘t’
x(t)
0 20T20T-
00
( )
li
)
(
'
m
'
TT
Note that the above equation also implies thatall non periodicsignals can be consider
x
ed periodicwith a peri
x t
od f
t
o
®¥=
-¥
Fourier Series & Fourier Transform contd…
Fig.18.3
-
0
0
0
0
0
0
2
0
2
02
2
0
0
0
;
( ) ( ) (
1 ( )
)
( ) ,) (
k
k
Tj t
kk
k T kT
k k
Tj t
k TT
Now defi
Consider the Fourier Analysis e
ne
X j X jk x t e dt T a
quation
a x t e dtT
XX j
T
aj
a
T
w
w
w w
ww
-
-
-
-
=
= = =
\ =
= ò
ò
Fourier Series & Fourier Transform contd…
-
0
2 2
0
0
0
0
0
0
, ;
,
( )
1( ) (
li
( )
m
)
1jk t jk t
k
T Tk
k
j t
j
T
j
j tT k
k
T kk
tk
k
pay at
Then from the Fourier synthesis equation
x t a
tention to the term e e e
e
x t XT
X je
j
T
e
p p
w
w
w
w
w
w
æ ö æ öç ÷ ç ÷ç ÷ ç ÷è ø è ø
¥
¥
=-¥
¥
=-
¥
®
-
¥
¥
=
= ®
=
\ =
æ ö= ç ÷
è øå
å
å kt
Fourier Series & Fourier Transform contd…
-
Fourier Series & Fourier Transform contd…
0
0
0
10 0 0
0
;2 ( 1) 2 2
2
1( ) ( )
1( ) ( ) ( )2
,
2
k
k
k
j tT k
k
j tkT
k k
k
k k kk k
substitute the result in the sysnthesis equ
Def
x t X j e
inek kT T T
ation
T
x t X j X j e
T
w
w
w
w w w wp
p p pw w
w
p
w
p
¥
=-
+
¥
¥ ¥
=-¥ =-¥
+D = - = - =
=D
=
D= = D
å
å å
-
Fourier Series & Fourier Transform contd…
0
0
0 0
0
0
'
2lim 0
2
lim
' ' '
k kT
k k
kT
k
The above equation implies that the discrete variablecan now be replaced with a conti
T
nuous variable
T
kk
TT
Going back to the orig
pw w
pw w
w w
w w
®¥
®¥
æ öÞ D ® D =ç
-
÷è ø
= = D
Þ ®
!
00
; lim ( )( ) TT
inal equation connecting the periodicand non perio xdic signa s xtl t
®¥=-
-
00
0
0
0
0
0
1lim ( )2
1 ( )
lim ( ) ( ) lim
1( ) ( )2
; ( ) lim (
( )2
)
( ) ( )
k
k
k
k
j tk k
k
T TT
j tT k k
k
j t
uk
X j
Because x u du X k u
x t x
e
x
t x t
x t X j e
t X j
u
x
d
t
e
w
w
w
w
w
w wp
w wp
w wp
¥ ¥
D ®=-¥-¥
®¥
¥
¥
D ®
D ®=-¥
-
-¥
¥
¥
=
D
=
= =
D
Þ
=
D
D
=
å
ò
ò
å
å
Fourier Series & Fourier Transform contd…
-
Fourier Series & Fourier Transform contd…
1( ) ( ) :
(1) (2) ,
( ) ( ) :
lim ( ;
) 0
(2)
1
( )
)
2
(j t
t
t
jx t X j
Fourier Transform integrals
equati
X j x t e d An
ons and exis
a
t if and
e d Synthesis
l
only if
tha
x t
x tt
s
s
y
i
siw
ww w
w w
p
®¥
¥-
-¥
¥
-
-¥
¥
¥
=
=
= ò
ò
ò ( )Sufficient condition for the existance
dt
of X jw
< ¥
-
Example
The periodicity is increased to show theeffect of higher period.Notice that for fig.c, norepetition is visible inthe plot. Also the spectrum plot for thethree cases shows aninteresting phenomenon.
-
0
0
0
2
02
2 2 2
02
2
0 2
;
1 ( )
(
sin( 2)2
)
k
k k
k
k
T
Tj t
kT
Tj t j T j T
k
j tk T
kT
k
k
k
T
a T e dt
notice the change in limits for the present case
e
Fourier Analysis equation
a x
e ea T
j
t
T
j
t e dT
w
w w w
w
w
w
w
w
-
-
-
- -
-
-
=
é ù-= = ú-
=
=
ê -ë û
ò
ò
Example contd…
-
17
Example contd…
0
0
0
0
, ,
sin( 2)( ) lim2
.
' '
k
k
k
k
k
The frequecnciesget closer and closer as
and eventuallybecome dense in the interval
The quatitiesapproach a continuous
envelope functiT
X
on
j
k
T
a T
w w
w w
w
www®
=
®¥
-¥ <
=
< ¥
sin( 2) 2Tw
w=
Fig.18.5
-
0
= ( )
(
) (
( ) ( )
)
at j t
a
j t
t t
at
j
e
Forward continuous time Fourier Transform
(notice the change in l
u t e dt
e e d
X j x t e
x t e
imi
t
)t
u
t
t
s
d ww
w
w¥
- -
-¥¥
- -
¥-
-
-¥
=
=-
= òò
ò
Fourier Transform Pair 1
Fig.18.6
-
( )( )
( )( ) ( )( )
( )
( )( )( )
( )
0 0
0
=
1
1
0
( ) ,
a j ta j t
a
a j a j
a
j
Ft Fourier transform
ee dt
a j
e e
a j
a j
Time domain
e
Fr
is u
e
n
quency doma
e
ique
in
a ju t
ww
w
w
w
w
w
w
w
¥- +¥- +
- +
- +
¥ - +
¥
-
=
=- +
-=
- +
+
=+
¬¾®
ò
!
Fourier Transform Pair 1 contd….
-
Fourier Transform Pair 1 contd….
( )
( )
( ) ( ){ } { }
2 2
1
2 2
2 2 2 2
1
( ) ta
1( )
1
n
( )
( ) ;
(
(
;
)
)
1
a jX jw
a j a
a
a a
X jw
a
X w
a
j
j
a ja j
e X jw m X
X jw
jw
ww
w
w
w
w
w
w
ww
w-
-=
+ +
-
+
+
Ð = -
=+
+
=
æ öç ÷
æ ö-=ç ÷-è ø
 = Á =
è ø
-
Fourier Transform Pair 1 contd….
{ } 2 2( )a
ae X jw
w+Â =
{ } 2 2( ) am X jw w
w
-
+Á =
-
Fourier Transform Pair 2
[ ]
[ ]
( ) ( 2) ( 2)
( ) ( )
= ( 2 ) ) 2 ( j t
j t
Forward continuous time Fourier Transform
x t u t T u t T
u t T u t T e
X j x t e dt
dtw
ww¥
-
-¥
¥-
-¥
+
= + - -
- -
-
= ò
ò
Fig.18.8
-
Fourier Transform Pair 2 contd….
( )( )
( )( ) ( )( )
( )
22
2 2
2 2
=
2 sin( 2) sin( 2)
sin( )
,=2
;
j tj t
j j
TT
T T
T T
Time domain Frequ
ee dt
j
e e
j
j T
enc o
T
j
y d
'sinc' function
ww
w w
pqpq
w
w
w ww w
--
- -
-
=-
-=
-
-=
-
ò
[ ] ( 2) ( 2) sin( 2)2
F
main
u t u t TT
Tww
+ - ¬¾®-
-
0
0
sin( 2)
sin( 2)
2 os )
c ( 2
: ( 0) lim ,
( )=
2
' ; ( 0
2
) lim1 2
T
T T
T
Note X j
using L Hospitals rule T
X j
X j
w
w
w
w
www
w®
®
=
= =
Fourier Transform Pair 2 contd….
Fig.18.9
-
Fourier Transform Pair 2 -- Synthesis
( )X jw1
bw- bw w
[ ]1( )
(
( )
,
:
) ( ) ( )
2t
b b
j
The frequency domain information is given aboveFind the corresponding time domain sign
x t X j e d Synthesis equa
X j
ti
l
u
a
on
u
ww w
w w w
p
w w¥
-¥
-
= - -
=
+
ò
Fig.18.10
-
[ ]
( )
( )
( ) ( )( )( )
( ) ( )
1( ) ( ) ( )
2
1
= 2 2
2 sin( ) sin( )
2
( )
=
2
b bb
b
b
b
j t j t
j tb b
j t
j t
b b
notice the change in lim
e ee
jt j t
j t
x t u u
its
e d
e d
t
j t t
w
ww
w www
w
w
w w w w
p p
w wp
p
w
p
w
p
¥
-¥
-
-
-
= + - -
=
-
=
=
ò
ò
Fourier Transform Pair 2 – Synthesis contd…
-
[ ]( )
( ) ( )
sin
(
)
Fb
bb
Frequen
Note : finite - time signals are infinite in frequeny domain
finite - freq
cy domain Time doma
t
u
i
t
e
u
n
uw
w wp
w w+ - - ¬¾®
ncy signals are infinite in time domain
Fourier Transform Pair 2 – Synthesis contd…
Fig.18.11
-
Fourier Transform Pair 3
0
(
( ) ( ) =
)
( )
( )
)
(
j t j t
j
F
tt
X j x t
x t
A
e A t
t
dt t e d
A
A
e
A t
A
w w
w
d
w d
d¥ ¥
- -
-¥ -¥
-
=
=
=
=
¬¾®
=
ò ò
Fig.18.12 Fig.18.13
-
Fourier Transform Pair 3 -- Synthesis
(2 )p 1
( ) 2 ( )X jw pd w=
0
1( ) ( ) : 2
1 2 ( ) 2
= 1 j t
j t
j t
x t X j e d Synthesis equation
e d
e
w
w
w
w
w wp
pd w wp
¥
-¥¥
-¥
==
=
=
ò
ò
( )x t
w tFig.18.15Fig.18.14
-
Fourier Transform Pair 4
0
0
0
0
0
0
1( ) ( ) : 2
1 2 ( ) 2
2 ( )
( ) 2 ( )
=
2
j t
j t
j t
j tj
F
t
Fr Ti
Let X j
me do
x t X j e d Synthesis equation
e d
e quency domain main
e
e e
w
w
ww w
w
w
w pd w w
w wp
pd w w wp
pd w w
¥
-¥¥
-¥
=
= -
¬¾
=
=
®
= -
-
ò
ò
00 ( )
F j te wpd w w -¬¾®+
-
Fourier Transform Pair 5
0
0
0
0
0
0
0
( ) sin
( ) ( ),
(
2 ( )
) sin( )
( )2
F
Fj t
Fj
t
t
j t j
x t X j use the identities from previous exampl
Find the Fourier Transform f
e
e
e
or x t t
e ex t t
j
w
w
w ww
w
pd w
w
w-
-æ ö-= = ç ÷ç ÷
¬¾®
¬¾® -
¬¾®
ø
=
è
0
0 00
1 21 2
( ) ( )
- (
( ) ( )
) (
sin(
)
2 (
)
)
F
F
Time domain
ax t bx t
Frequency domain
aX j bX j
j
t j
pd
w w
pd w w pd w ww
w w
+
- +
¬¾®
¬¾® +
+
+
-
Fourier Transform Pair 5 contd….
-
Fourier Transform Pair 6
0
0
0
0
0
0
0
( ) ( ),
2 ( )
( ) cos( )2
(
) cos( )
F
Fj t
Fj
j t j t
t
x t X j use the identities from previous example
e
e
Find the Fourier Transfo
e ex
rm
t t
for x t tw w
w
w
w
pd w w
w
w
-
-æ ö+= = ç ÷
¬¾®
¬¾® -
=
¬¾®
ç ÷è ø
0
0 00
1 21 2
( ) ( )
( ) ( )
( )
2 (
()
)
)cos(
F
F
Frequency domain
aX
Time domain
ax t j bX jbx
t
t w w
pd w w pd w w
pd w w
w
¬¾®
¬ - +¾® +
+
+
+
-
Periodic Signals
{ }
00
0
0
0
00
2
0 2
( ) ( )2
1
( )
( ) ,
(
( )
( )
)
:
k
F
jk t
k
Tjk
j tk
t
k
k
k
k
T
x t x t Tk
kT
a x t e dt
x
Whe
x t X j
F x t F a e
t
r
a
e
e
Tw
w
w
pw w
w
¥
=-¥
¥
-
=-¥
-
=
= +
=
¬¾®
ì ü=þ
=
ý
=
íî
å
å
ò
-
0
0
0
0
0
0 2 ( )
2 ( )
2 (
( ) 2 ( )
)
k F
k
Fj t
Fjk t
j tk
k
kkk
e
e k
Time domain Frequen
X
c
j a k
a e
y domain
a
k
w
w
w
w
p
p d
pd
w w
w w
pd w w
d w w
¥
=-
¥
=-
¥
=
¥
¥¥ -
¬¾® -
\ ¬¾
¬
= -
¾
®
-®
-
å
å å
Periodic Signals contd…
-
Example: Square Wave
0 0
0
02 2 2
0 0 0
0
02
2
2
1
1 ( ) ,
Tjk t
Tjk t
kT
jk T jk T
T
a x
e e eT jk k T
d
j
t e tT
w
w w w
w w
-
-
-
-
-æ ö æ ö-= =ç ÷ ç ÷ç ÷ ç ÷- -è
=
è ø ø
ò
Fig.18.17
-
0
02
00
0
0
0
2
0
0
000
sin( 2)
50% , ,
2
0
1
sin( 2
1 1 = 2
) sin( 2
0
2
2
2
;
2
)
T
T
k
Assume a duty cycle which means
and
k T
Ta dt
k kka
k
kk T k
T
T TT
T
T
TT pw pw p
w pw
p
p
p-
\
æ ö æ ö=ç
= =
ì ¹ïï= íï =
÷ ç ÷è øè
=
ø
= =
ïî
=
ò
Example: Square wave contd…
-
Example: Square wave contd…
0
02sin( 2)( ) ( ) (
2
)
( )kj tkk
k
k
k
F aa ke
kX j kk
w
pw pd w d w
p d w w
w
¥
=-¥
¥
¥
=-¥
=-¥¬
æ ö= + -ç
¾
÷
®
è ø
-åå
å
Fig.18.18
-
Example: Impulse Train
0
22
0 2 2
( ) ( )
1 1( ) ( )s
s
s
TTjk t jk t
k
n
sT
s
Ta x t e dt t e dt
T T
p t t nT
w w
d
d
¥
=
- -
-
-¥
-=
= -
=ò
å
ò
Fig.18.19
-
2
2
2(
1 1 ( )
) 2 ( ) ( )
ss
s
Tjk t
s sT
k s sk k s
P j a k k
t e tT T
T
dw
pw p d w w d w w
d
¥ ¥
=- -¥
-
=
-
¥= = -
=
-
=
å
ò
å
Example: Impulse Train contd…
Fig.18.20
-
Table 1: Fourier Transform Pairs
( )( )
( )( )
( )
1
1
: ( )
( ), 0
( ),
: ( )
sin( 2)
b 0
2
( ) ( )
( 2) ( 2)
sin( )
( )
( )
1
d
at
bt
b
d
b b
j t
e
be
a j
j
Time domain x Ft
u t a
u t
u t T u t T
tt
t
requency domain X
t
j
T
u u
t e w
w
w
w
wpd
ww
w
d
w w w
-
-
+
-
>
- >
-
+ - -
+ - -
-
Table 2: Fourier Transform Pairs
0
0
0
0 00
0
0
0 0
0 0
0
: ( )1
( )
2 ( )
2 ( )
( ) ( )( ) ( )
- ( ) ( )
2
: ( )
( )
1
cos( )cos( )sin(
( )
)
(
j j
j
kk
t
jk tk
k
FrequencTime domain x t
u t
e
A ttt
a e
t
y domain X j
j
Ae Ae
nT
j j
a k
f f
w
w
w
w
pd ww
pd w
pd w w
p d w w p d w wpd w w pd w wpd w w pd w w
p
fw
d w
w
d
w
-
¥
=-¥
¥
=-¥
+
-
- + +- + +-
+
+ +
-
-
åå
2( ))
n kk
Tp d w w
¥
=- ¥¥
¥
=--å å
-
James H. McClellan, Ronald W. Schafer and Mark A. Yoder, “ 11.1- 11.4 “SignalProcessing First”, Prentice Hall, 2003
Reference