Section 3.8
PROPERTIES OF FOURIER REPRESENTATIONS
Periodic
(t,n)
Time
property
Nonperiodic
(t,n)
• Fourier Series (FS)
C.T.
(t)
x(t ) 
(3.19)
X [k ] 
(3.20)

jk 0t
X
[
k
]
e
, 0 

k  
1
T

T
0
2
T
x(t )e  jk 0t dt ,
(T: period)
• Four Transform (FT)
1 
jt
x(t ) 
X
(
j

)
e
d



2
(3.35)

X ( j )   x(t )e jt dt
(3.36)
• Discrete-Time Fourier
Transform (DTFT)
jk 0 n
N 1
1 
2
j
jn
x[n] 
X
(
e
)
e
d
x[n]   X [k ]e
, 0 



2
N
k 0
(3.10)
(3.31)
1
X [k ] 
N
(3.11)
N 1
 x[n]e
n 0
Discrete
[k]
 jk 0 n
(N: period)
X (e j ) 
(3.32)
(k,)

• Discrete-Time Fourier
Series (DTFS)
D.T.
[n]
Nonperiodic

 x[n]e
Periodic
(k,)
 jn
n  
Continuous
(, )
Freq.
property
• Linearity and symmetry
FT
z (t )  ax(t )  by (t ) 
Z  j )  aX  j )  bY  j )
FS ; 0
z (t )  ax(t )  by (t ) 
 Z [k ]  aX [k ]  bY [k ]
DTFT
z[n]  ax[n]  by[n] 
 Z e j )  aX e j )  bY e j )
DTFS ; 0
z[n]  ax[n]  by[n] 
 Z [k ]  aX [k ]  bY [k ]
E Example 3.30, p255:
z (t ) 
3
1
x(t )  y (t )
2
2
Find the frequency-domain
representation of z(t).
Which type of freq.-domain representation?
• FT, FS, DTFT, DTFS ?
Periodic signals, continuous time. Thus, FS.
3
1
X [k ]  Y [k ]
2
2
1 Tx
X [k ]   x(t )e  jk 0 xt dt
Tx 0
Z [k ] 
1

Tx

1/ 8
Tx / 2
Tx / 2
 e
1 / 8
x(t )e  jk 0 xt dt
 jk 2t
1
dt 
e  jk 2t
 jk 2
1
8
 18
1
 j 2 sin 4 k ))

 jk 2
1

sin 4 k )
k
1
Y [k ] 
sin 2 k )
k
3
1
FS ; 2
Z [k ] 

sin 4 k ) 
sin 2 k )
2k
2k
Symmetry:
We will develop using continuous, non-periodic signals. Results for
other 2 cases may be obtained in a similar way.
*
a) Assume x(t ) real  x (t )  x(t )
X
*
 j )    x(t )e


 j t
dt 

*

  x * (t )e jt dt


  x(t )e  j (  ) t dt

 X  j )
If x(t ) is real  X ( j ) is conjugate symmetric
 Further assume
x(t) even (and real)
x(-t)  x(t), x*(t)  x(t)  x*(t)  x(t )
X
*

 j )   x(t )e  j ( t ) dt
replace  with  t

  x( )e  j d

 X  j )
(why?)

b
a
a
x(t )dt    x(t )dt
b
  t  d  dt
If x(t ) is real and even  X ( j ) is real.
 Further assume x(t) odd (and real)
x(-t)   x(t), x*(t)  x(t)  x*(t)   x(t )
X
*

 j )    x(t ))e j ( t ) dt
  X  j )
If x(t ) is real and odd  X ( j ) is purely imaginary.
b) Assume x(t ) imaginary  x* (t )   x(t )
X
*

 j )   x(t )e j (  )t dt
  X  j )
If x(t ) is purely imaginary 
 Real part of X ( j ) has odd symmetry
 Imaginary part of X ( j ) has even symmetry
• Convolution:
Applied to non-periodic signals.
Let
y (t )  h(t )  x(t )

  h( ) x(t  )d

If
FT
x(t ) 
X ( j ) 
1
x(t   ) 
2



X ( j )e j (t  ) d


1
y (t )   h( ) 2  X ( j )e jt e  j d d
 





1
 2   h( )e  j d  X ( j )e jt d

 




H ( j )

1
2
j t


H
(
j

)
X
(
j

)
e
d


FT
y(t )  x(t )  h(t ) 
Y ( j )  X ( j ) H ( j )
Conclusion:
Convolution in time domain  Multiplication in freq. domain.
E Example 3.31:
Let
x(t )  1t sin( t ) be input to a system with impulse response
h(t )  1t sin( 2t ). Find the system output y(t )
y (t )  x(t )  h(t )
Y ( j )  X ( j ) H ( j )
From results of example 3.26, p264.
1, |  | 
x(t )  X ( j )  
0, |  | 
FT
1, |  | 2
h(t )  H ( j )  
0, |  | 2
FT
1, |  |  FT
Y ( j )  
 y (t )  1t sin( t )
0, |  | 
E Example 3.32:
FT
x(t ) 
X ( j )  42 sin 2 ( ).
Recall that (Example 3.25, p244)
1, | t | 1 FT 2
z (t )  
  sin(  )
0, | t | 1
Find x(t ).
Let
Z(jω)  2 sin(  ). Then, X (j )  Z (j )Z (j ).
Thus,
x(t )  z (t )  z (t ).
The same convolution properties hold for discrete-time, non-periodic signals.
DTFT
y[n]  x[n]  h[n] 
Y (e j )  X (e j ) H (e j )
Convolution properties for periodic (DT or CT) and periodic with non-periodic
signals will be discussed in Chapter 4.
• Differentiation and integration:
(Section 3.11)
– Applicable to continuous functions: time (t) or frequency ( or )
– FT (t, ) and DFTF ()
Differentiation in time:
1
x(t ) 
2
d
1
x(t ) 
dt
2



X ( j )e jt d
jt


j

X
(
j

)
e
d


d
FT
x(t ) 
jX ( j )
dt
E Find FT of

)
d  at
e u (t ) , a  0
dt
it follows
1
e u (t ) 
(in - class example 3.24, p243)
a  j
d  at
j
FT
e u (t ) 
dt
a  j
 at
FT

)
E Find x(t) if
 j , |  | 1
X ( j )  
 0, |  | 1
1, |  | 1
Let Z ( j )  
then,
0, |  | 1
X ( j  )  j Z ( j  )
1
FT
Z ( j ) 
z (t ) 
sin( t ) (example 3.26, p246)
t
d
Thus, x(t )  z (t )
dt
1
1

cos(t )  2 sin( t )
t
t
If x(t) is periodic, frequency-domain representation is Fourier Series (FS):
x(t ) 

jk 0t
X
[
k
]
e

k  
d
FS ; 0
x(t ) 
 jk 0 X [k ]
dt
Differentiation in frequency:

X ( j )   x(t )e  jt dt


d
X ( j )   ( jt) x(t ) e  jt dt ,

d
d
FT
 jtx(t ) 
X ( j )
d
thus
E Example 3.40, p275
1  t22
A Gaussian pulse is given as : g (t ) 
e . Find its FT.
2
d
g (t ) 
 tg (t )
()
dt


1 d
j  G ( j )
G ( j )
Thus
j d
d
1 d
G ( j )   G ( j )
j  G ( j ) 
G ( j ) 
d
j d
This differential equation (it has the same mathematical form as
(*), and thus the functional form of G(j) is the same as that of
g(t) ) has a solution given as:
G( j )  ce
 2 / 2
Constant c can be determined as:




g (t )dt  1  G( j 0)  g (t )e jt dt | 0  c  1

Thus,
1  t22 FT
 2 / 2
e  e
2
Integration:
– In time: applicable to FT and FS
– In frequency: applicable to FT and DTFT
t
Let y (t )   x( )d

d
y (t )  x(t ) 
dt
1
j Y ( j )  X ( j )  Y ( j ) 
X ( j )
j
if   0
For =0, this relationship is indeterminate. In general,
1
 x( )d  j X ( j )  X ( j0) ( )
t
FT
E Determine the Fourier transform of u(t).
t
u (t )    ( )d

FT
 (t ) 1
1
U ( j ) 
  ( j )
j
FT
E Problem 3.29, p279: Fund x(t), given
1
X ( j ) 
  ( j )
j ( j  1)
1
1
1
1
X ( j ) 

  ( j ) 
  ( j ) 
j j  1
j
1  j

x(t )  u (t )  e t u (t )
Review Table 3.6. Commonly used properties.
d
 2t

x
(
t
)

2
te
u (t ) ).
X ( j )  ?
Problem
3.22(b),
p271.
E
dt
1
FT
 2t
 e u (t ) 
d
2  j
FT
 jtx(t ) 
X ( j )
d
1
FT
 2t
 te u (t ) 
2
2  j )
d
FT
x(t ) 
jX ( j )
d
2 j
FT
 2t
 2te u (t ) )
dt
dt
2  j )2
• Time and frequency shift
Time shift:
Let z (t )  x(t  t0 )

Z ( j )   z (t )e  jt dt


  x(t  t0 )e  jt dt
Replace t  t0 with 


   x( )e  j d  e  jt0
 

Z ( j )  e jt0 X ( j )
Note: Time shift  phase shift in frequency domain. Phase shift is a
linear function of . Magnitude spectrum does not change.
Table 3.7, p280:
FT
x(t  t0 ) 
e  jt 0 X ( j )
FS ; 0
x(t  t0 ) 
 e  jkt0 X [ k ]
DTFT
x[n  n0 ] 
 e  jn0 X (e j )
DTFS ; 0
x[n  n0 ] 
 e  jkn0 X [k ]
E Example 3.41:
Find Z(j)
z (t )  x(t  T1 )
x(t )  X ( j ) 
FT
2

z (t )  Z ( j )  e
FT
sin( T0 )
 jT1
2

sin( T0 )
(Example 3.25, p244)
E Problem 3.23(a), p282:
e j 4
X ( j ) 
.
Find x(t )
2
( 2  j )
1
FT
 2t
 Z ( j ) 
 z (t )  e u (t )
2  j
d
j
FT
  jtz(t ) 
Z ( j ) 

2
d
( 2  j )
1
FT
 2t
te u (t ) 
( 2  j ) 2
j 4
e
FT
2(t  4)
 (t  4)e
u (t  4) 
( 2  j ) 2
Frequency shift:
FT
x(t ) 
X ( j )
FT
Z ( j )  X ( j (   ))  ??
1 
j t
z (t ) 
Z
(
j

)
e
d



2
1 
j t

X
(
j
(



))
e
d

2 
 1 
 jt
jt
   X ( j )e d  e
 2 

jt
z (t )  e x(t )
(let      )
Note:
– Frequency shift  time signal multiplied by a complex sinusoid.
– Carrier modulation.
Table 3.8, p284:
FT
e jt x (t ) 
X ( j (   ))
FS ; 0
e jk0 0t x(t ) 
 X [k  k 0 ]
DTFT
e jn x[ n] 
 X (e j (   ) )
DTFS ; 0
e jk0 0 n x[n] 
 X [k  k 0 ]
E Example 3.42, p284:
Find Z(j).
e j10t , | t | 
z (t )  
 0, | t | 
1, | t |  FT 2
Let x(t )  
 sin(  )

0, | t | 
z (t )  x(t )e
j10t
2
 X ( j (  10)) 
sin((   10) )
  10
FT
E Example 3.43, p285:

)
)
d  3t
x(t ) 
e u (t )  e t u (t  2) . Find X ( j ).
dt
1
FT
 3t
Let w(t )  e u (t ) W ( j ) 
3  j
v(t )  e t u (t  2)
e
(t  2 )
 2  j 2
u (t  2)e V ( j )  e e
2
FT
 j 2
j

e
Thus, X ( j )  jW ( j )V ( j )  e  2
(3  j )(1  j )
1
1  j
• Multiplication
If z (t ), x(t ) are non - periodic, what is the FT of y (t )  x(t ) z (t ) ?
1
x(t ) 
2
1
z (t ) 
2






X ( jv)e jvt dv
READ derivation on p291!
Z ( j )e jt d
y (t )  x(t ) z (t ) 
1
(2 ) 2



X ( jv)Z ( j )e j ( v  )t dvd
Let   v   , fix v first  d  d . Then
1
y (t ) 
2
 1 
 jt
X
(
jv
)
Z
(
j
(


v
))
dv
  2 
e d



Inverse FT of
1
2
X ( j )Z ( j )
FT
y (t )  x(t ) z (t ) 
Y ( j ) 
1
2
X ( j )  Z ( j )
FT
Compare x(t )  z (t ) 
X ( j ) Z ( j )
For discrete - time signals x[n], z[n] :
DTFT
y[n]  x[n]z[n] 
 Y (e j ) 
Periodic convolution:
1
2
X (e j ) * Z (e j )

 )
j
- p296, CT, periodic signals:
FS ; 2 / T
y (t )  x(t ) z (t ) 
 Y [k ]  X [k ]  Z [k ]
X [k ]  Z [k ] 
)
X (e ) * Z (e )   X e j Z e j ( ) d

j

 X [m]Z [k  m]
m  
- p297, DT, periodic signals:
DTFS ; 2 / N
y[n]  x[n]z[n] 

 Y [k ]  X [k ] * Z [k ]
N 1
X [k ] * Z [k ]   X [m]Z [k  m]
m 0
• Scaling
Let z (t )  x(at )

Z ( j )   z (t )e  jt dt


  x(at )e  jt dt

 1  x( )e  j ( / a ) d
a 
  
 j ( / a )
d
 1a  x( )e


1
| a|  

x( )e  j ( / a ) d
 |1a| X ( j ( / a ))
a0
a0
(let   at )
E Example 3.48, p300:
1 | t | 1
x(t )  
0 | t | 1
1 | t | 2
y (t )  
0 | t | 2
Find Y ( j )
y (t )  x(at ) |a  1
2
X ( j ) 
2

sin(  )
2
2
Y ( j )  2 X ( j 2 )  2
sin( 2 )  sin( 2 )
2

d  e j 2 
E Example 3.49, p301: Find x(t ) if X ( j )  j


d 1  j ( / 3) 
1
1  j
1
1
FT
z (t )  s (3t ) 
Z ( j ) 
3 1  j ( / 3)
FT
We know s(t )  e t u (t ) 
S ( j ) 
– Time scaling:
– Time shift:
e j 2
v(t )  3 z (t  2)  3s (3(t  2)) 
1  j ( / 3)
FT
d
FT
– Differentiation: tv(t )  j
d
Thus,
x(t )  tv(t )
 3tz(t  2)
 3ts(3(t  2))
 3te
 3( t  2 )
u (3(t  2))
 e j 2 


1  j ( / 3) 
FT
1
x(at ) 
|a| X ( j ( / a ))
FT
x(t  t0 ) 
e  j t 0 X ( j  )
FT
d
 jtx(t ) 
d X ( j )
Note : u(3(t  2))  u(t  2). Thus, x(t )  3te3(t  2)u(t  2)
• Parseval’s relationship:

2
Energy of CT, non - periodic signal x(t ) : Wx   | x(t ) | dt

1
x  (t ) 
2




  x  (t ) x(t )dt
X  ( j )e  jt d

 1  

 j t
Wx   x(t )   X ( j )e d dt

 2 


1  
 j t

X
(
j

)
x
(
t
)
e
dt d






2
1  

X ( j ) X ( j )d

2 

1 
2
2
Wx   | x(t ) | dt 
|
X
(
j

)
|
d




2



Note : a) | X ( j ) |2 : energy spectrum
b) Energy in time domain  energy in freq. domain
Table 3.10, p304:

1
 DTFT :  | x[n] | 
2
n  
2
1
 DTFS :
N
 FS :
1
T
N 1



| X ( e j ) | 2 d 
N 1
2
|
x
[
n
]
|

|
X
[
k
]
|


2
n 0

T
0
k 0
| x(t ) | dt 
2

 | X [k ] |
2
k  
E Example 3.50, p304:

sin( Wn)
sin 2 (Wn)
x[n] 
.
Determine E x  
n
(n) 2
n  
1,   W
sin( Wn) DTFT
j
x[n] 
 X e )  
n
0, W    

1 
2
j 2
E x   x [ n] 
X e ) d 



2
n  
1 W

1d  W / 


W
2
• Time-bandwidth product
Compression in time domain  expansion in frequency domain
Bandwidth: The extent of the signal’s significant contents. It is in
general a vague definition as “significant” is not mathematically defined.
In practice, definitions of bandwidth include
– absolute bandwidth
– x% bandwidth
– first-null bandwidth.
If we define

1/ 2

t
x
(
t
)
dt

 : RMS duration of an energy signal
Td   
2


x
(
t
)
dt
 

1/ 2
   2 X ( j ) 2 d 

 : RMS bandwidth, then
Bw   
2


X
(
j

)
d

 


2
2
Td Bw  1 / 2
• Duality
f (t )  F  j )
FT
F  jt )  2f ( )
FT
E Example 3.52, p308: Find X ( j ) if x(t ) 
1
1  jt
1
We know f (t )  e u (t ) 
 F ( j )
1  j
1
FT
Duality F ( jt ) 

2f ( ) 
1  jt
t
FT
X ( j )  2f ( )  2e u ( )
Check
1
x(t) 
2
1

2
0
E Problem 3.44, p309: Find x(t ) if X ( j )  u ( )

0

X ( j )e  jt d
2e e jt d

 e



 (1 jt )
1
d 
1  jt
X ( j )  u ( )
1
X ( jt )  u (t ) 
  ( )  2x( ) 
j
FT
 1
 1
x(t )  
  ((t )) 
 j (t )
 2
1 1

  (t )
2jt 2
Table 3.11, p311:
 DTFS :

t

FT
x( )d 
1
X ( j )  X ( j 0) ( )
j
DTFS ; 2 / N
 x[n] 

 X [k ]

DTFS ; 2 / N
X
[
n
]



 N1 x[k ]

 )
DTFT
 x[n] 
 X e j
 DTFT and FS : 
FS ;1
jt
X
e


 x[k ]

 DTFT and FS do not stay in their own class!
 )