Transcript Discrete-Time Fourier Transform

Signals & systems
Ch.3 Fourier Transform of
Signals and LTI System
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Signals and systems in the Frequency domain
Fourier
transform
Time [sec]
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Frequency
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[sec-1, Hz]
2
3.1 Introduction
• Orthogonal vector => orthonomal vector


vo  (1/ 2,1/ 2 ) v1  (1/ 2,1/ 2 )
 
 
 
 
vo  v1  0 vo  v0  1 v1  v1  1 vi  v j   [i  j]
for i, j  0,1
Any vector in the 2- dimensional space can be represented by weighted sum of
2 orthonomal vectors


Fourier
h  (h0 , h1 )  H  (H0 , H1 )



h  H 0vo  H1v1
• What is meaning of magnitude of H?
Fourier Transform(FT)
Inverse FT
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


h  H 0vo  H1v1
 
 
H 0  h  vo H1  h  v1
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3.1 Introduction cont’
• CDMA?
1 1 1 1 v  (1 , 1 ,  1 ,  1) v  (1 ,  1 ,  1 , 1) v  (1 ,  1 , 1 ,  1)
vo  ( , , , ) 1
2
3
2 2
2
2
2
2
2 2
2
2 2
2
2 2 2 2
Orthogonal?
vo  v1  0 vo  v0  1 v1  v1  1
vi  v j   [i  j] for i, j  0,1, 2,3
Any vector in the 4- dimensional space can be represented by weighted sum of
4 orthonomal vectors
Orthonormal function?
vk (t )  A cos( k0t ) where 0 
vk (t )  e jk0t
where 0 
2
T
2
T

T 
vi (t )v j (t )dt   [i  j ]
 vm (t )vk* (t ) 
1
*
v
(
t
)
v
(t )dt   [m  k ]
m
k
T T 
1
1 j ( m  k ) 0t
*
v
(
t
)
v
(
t
)
dt

e
dt
m
k
T T 
T T 
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3.1 Introduction cont’
1
1 j ( m  k ) 0t
1
*
v
(
t
)
v
(
t
)
dt

e
dt

1  dt  1
m
k
T T 
T T 
T T 
T
 j 2 (Tm  k ) t 
1 j ( m  k ) 0t
1
 e
dt 
e
 0
T T 
j (m  k )2 
0
i) If m  k ,
ii) If m  k ,
• Fourier Series (FS)
 Any periodic function(signal) with period T can be represented by weighted
sum of orthonormal functions.
f (t )  f (t  T ) 

 F v (t )
k 
k k
where Fk  
T 
f (t )vk (t )dt
What is meaning of magnitude of Fk?
Think about equalizer in audio system.
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3.2 Complex Sinusoids and
Frequency Response of LTI Systems
cf) impulse response

 h[k ]x[n  k ]
y[n] 

 h[k ]e

k  
y[n]  e
jn

 h[k ]e
j ( n  k )
.
k  
 jk
j
 H ( e )e
jn
j
H (e ) 
,

y(t )   h( )e

j ( t  )
d  e
jt



h( )e
 j
 h[k ]e
 jk
.
k  
k  
How about x[n]  z n for complex z?

(3.1)
d  H ( j)e
How about x(t )  e st for complex s?
jt

,
H ( j )   h( )e  j d .

(3.3)
y(t )  H ( j) e j (t arg{H ( j )}) .
Magnitude to kill or not?
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Phase  delay
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3.3 Fourier Representations for Four
Classes of signals
• 3.3.1 Periodic Signals: Fourier Series Representations
“Any periodic function can be represented by weighted sum of
basic periodic function.”
Fourier said
(periodic) - (discrete)
(discrete) - (periodic)
Time Property
Continuous
(t)
Discrete [n]
DTFS
xˆ[n]   A[k ]e jk0n ,
k
CTFS
xˆ (t )   A[k ]e jk0t ,
(3.5)
Periodic?
k
Periodic
Fourier Series
(FS)
Discrete-Time
Fourier Series
(DTFS)
2
N
2
0 
T
0 
Non-periodic
Fourier Transform
(CTFT)
Discrete-Time
Fourier Transform
(DTFT)
(3.4)
e j ( N k )0n  e jN 0n e jk0n  e j 2 ne jk0n  e jk0n .
yes! With the period of N
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Fourier transform
Time domain
N 1
Periodic discrete
2
2
0 
0 
T
N
Periodic continuous
x[n]   X [k ]e
jk0 n
frequency domain
1 N 1
X [k ]   x[n]e  jk0n
N n 0
,.
k 0
x(t ) 


X [k ]e jk0t
X [k ] 
k 
1
j
jn
x[n] 
X
(
e
)
e
d
a-periodic discrete


2


2
1 
X ( j )e jt d 
a-Periodic continuous x(t ) 

2 
z-transform
x[n] 
1
 X ( z) z
n 1
2 j
1
st
X
(
s
)
e
ds
Laplace transform x(t ) 

2 j
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dz
j
1
T

X (e ) 
T 
x(t )e  jk0t dt

 x[n]e
 jn
n 

X ( j )   x(t )e jt dt


X ( z) 
 x[n]z
n
n 

X (s)   x(t )e st dt

8
3.3 Fourier Representations for Four
Classes of signals cont’
• 3.3.2 Non-periodic Signals: Fourier-Transform
Representations
(aperiodic) - (continuous)
(continuous) - (aperiodic)
Inverse continuous time Fourier Transform (CTFT)
(aperiodic)
(discrete)



X ( j )e jt d.
- (continuous)
- (periodic)
Inverse discrete time Fourier Transform (DTFT)
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1
xˆ (t ) 
2
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1
xˆ[n] 
2

  X (e

j
)e jn d.
9
3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series
DTFS ;0
x[n] 
 X [k ]
(periodic)
(discrete)
- (discrete)
- (periodic)
N 1
Inverse DFT
x[n]   X [k ]e jk0n ,.
DFT
1 N 1
X [k ]   x[n]e  jk0n
N n 0
k 0
(3.10)
Example 3.2 Determining DTFS Coefficients
Figure 3.5 Time-domain signal for Example 3.2.
Find the frequency-domain representation of the signal depicted in Fig. 3.5
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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
1 2
1
X [k ]   x[n]e  jk 2n / 5   x[2]e jk 4 / 5  x[1]e jk 2 / 5  x[0]e j 0  x[1]e  jk 2 / 5  x[2]e  jk 4 / 5
5 n2
5

Just inner product to orthonormal vectors in the 5 dimensional space!!
X[k] 
1  * 1
x  c k  ( x[2], x[1], x[0], x[1], x[2])  (e jk 4 / 5 , e jk 2 / 5 , e j 0 , e  jk 2 / 5 , e  jk 4 / 5 )
5
5

c2  (e  j 8 / 5 , e  j 4 / 5 , e j 0 , e j 4 / 5 , e j 8 / 5 )

c3  (e j12 / 5 , e j 6 / 5 , e j 0 , e j 6 / 5 , e j12 / 5 )
1   * 1  j 8 / 5  j 4 / 5 j 0 j 4 / 5 j 8 / 5
c 2  c 3  (e
,e
,e ,e
,e
)  (e j12 / 5 , e j 6 / 5 , e j 0 , e  j 6 / 5 , e  j12 / 5 )
5
5
1  *
ci  c j   [i  j ]
5
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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
1 1
1
 1
X [k ]  1  e jk 2 / 5  e  jk 2 / 5   1  j sin(k 2 / 5).
5 2
2
 5
1
sin( 4 / 5)
j
 0.232 e  j 0.531
5
5
1
X [ 0]   0.2e j 0
DC!!
5
X [2] 
X [1] 
sin( 2 / 5)
1
 j
 0.276 e j 0.760
5
5
X [1] 
X [2] 
1
sin( 2 / 5)
j
 0.276 e  j 0.760
5
5
1
sin(4 / 5)
j
 0.232e j 0.531.
5
5
Figure 3.6 Magnitude and phase of the DTFS coefficients for the signal in Fig. 3.5.
Recall “weighted sum of orthonormal vectors in the 5 dimensional space”
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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
Example 3.3 Computation of DTFS Coefficients by Inspection





v  a 0 c0  a1c1 orthonormal c0 and c1
DFT
(v0 , v1 ) 
(a0 , a1 )
Determine the DTFS coefficients of , using the method of inspection.
x[n] 
e


j  n  
3

e
2


 j  n  
3



j n
j n
1
1
 e  j e 3  e j e 3
2
2
Figure 3.8 Magnitude and phase of DTFS coefficients for Example 3.3.
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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
Example 3.4 DTFS Representation of an Impulse Train
Find the DTFS Coefficients of the N-periodic impulse train
x[n] 

 [n  lN ],
l  
Figure 3.9 A discrete-time impulse train with period N.
1 N 1
X [k ]   [n]e  jkn2 / N
N n 0

1
.
N
How about in other period?
1
X [k ] 
N
3N / 2
  [n  N ]e
n N / 2
 jkn2 / N

1
.`
N
Draw X[k] in the k-axis.
(impulse train) - (impulse train)
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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
Example 3.5 The Inverse DTFS of periodic X[k]
Use Eq. (3.10) to determine the time-domain signal x[n] from the DTFS
coefficients depicted in Fig. 3.10
Figure 3.10 Magnitude and phase of DTFS coefficients for Example 3.5.
x[n] 
4
 X [k ]e
jk 2n / 9
k  4
 e j 2 / 3e  j 6n / 9  2e j / 3e  j 4n / 9  1  2e  j / 3e j 4n / 9  e  j 2 / 3e j 6n / 9
 2 cos(6n / 9  2 / 3)  4 cos(4n / 9   / 3)  1.
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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
Example 3.6 DTFS Representation of a Square Wave
Find the DTFS coefficients for the N-periodic square wave
-M n M
.
M n N M
1,
x[n]  
0,
Figure 3.11 Square wave for Example 3.6.
1
X [k ] 
N
N  M 1
 x[n]e
 jk0 n
n M

1
N
N  M 1
e
n M
 jk0 n
1

N
1 2 M  jk0 ( mM ) 1  jk0M 2 M  jk0m
X [k ]   e
 e
e
.

N m 0
N
m 0
1
X [k ] 
N
2M  1
1 
,

N
m0
2M
e jk0 M
X [k ] 
N
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 1  e  jk0 ( 2 M 1) 

,
 jk 0
1

e


k  0, N ,2 N , 
k  0, N ,2 N ,  ,
M
e
 jk0 n
.
n M
(3.15) 등비수열의 합
DC=mean=평균
1  e jk0 M
X [k ]  
N 1
 1  e  jk0 ( 2 M 1) 


 jk 0
1

e


1  e  jk0 / 2  e jk0 ( 2 M 1) / 2  e  jk0 ( 2 M 1) / 2 

   jk0 / 2 
jk 0 / 2
 jk 0 / 2
N e
e
e


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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
1 sin(k 0 (2M  1) / 2)
,
N
sin(k 0 / 2)
k  0, N ,2 N ,.
 1 sin(k (2M  1) / N )
,

sin(k / N )
X [k ]   N
(2M  1) / N ,

k  0, N ,2 N ,
.
k  0, N ,2 N ,
X [k ] 
DC, also
 1 sin(k (2M  1) / N )  2M  1

 
.
k 0,  N , 2 N , N
sin(
k

/
N
)
N


Then X [k ] 
1 sin(k (2M  1) / N )
. for all k.
N
sin(k / N )
lim
M=0 (impulse train)
2M+1 = N?
X [k ] 
1 sin(k )
?
N sin(k / N )
(square) - (sinc)
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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
Example 3.7 Building a Square Wave form DTFS Coefficients
1 sin(k (2M  1) / N )
.
N
sin(k / N )
X [k ] 

x[n] 
 X [k ]e
k  
jk0 n
symmetric

X [0]  2 X [k ] cos(k 0 n)
k 1
Evaluate one period of the Jth term in Eq. (3.18) and the 2J+1 term approximation
xˆ J [n]
for J=1, 3, 5, 23, and 25.
J
xˆ J [n]   B[k ] cos(k 0 n), (3.18)
k 0
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3.4 Discrete-Time Periodic Signals:
The Discrete-Time Fourier Series cont’
Example 3.8 Numerical Analysis of the ECG
Evaluate the DTFS representations of the two electrocardiogram (ECG) waveforms
depicted in Figs. 3.15(a) and (b).
(a) Normal heartbeat.
(b) Ventricular tachycardia.
(c) Magnitude spectrum for the normal
heartbeat.
(d) Magnitude spectrum for ventricular
tachycardia.
Figure 3.15 Electrocardiograms for two
different heartbeats and the first 60
coefficients of their magnitude spectra.
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3.5 Continuous-Time Periodic Signals:
The Fourier Series
Any periodic function can be represented by weighted sum of basic periodic functions.
(periodic)  (discrete)
(continuous)  (aperiodic)

 X [k ]e
jk0t
Inverse FT
x(t ) 
FT
1 T
X [k ]   x(t )e  jk0t dt
T 0
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k  
,.
where
(3.19) Recall “orthonormal”!!
(3.20)
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3.5 Continuous-Time Periodic Signals:
The Fourier Series cont’
Example 3.9 Direct Calculation of FS Coefficients
Determine the FS coefficients for the signal
x(t ) depicted in Fig. 3. 16.
Solution :
1 2 2t  jkt
1 2 ( 2 jk )t
1
X [k ]   e e dt   e
dt 
e ( 2 jk )t
2 0
2 0
2(2  jk )
2
0
1  e4

4  jk 2
X[0]=?
FT
Figure 3.16 Time-domain signal
for Example 3.9.
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Figure 3.17 Magnitude and phase
for Ex. 3.9.
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3.5 Continuous-Time Periodic Signals:
The Fourier Series cont’
Example 3.10 FS Coefficients for an Impulse Train. Determine the FS coefficients

x(t ) 
for the signal defined by
 (t  4l )
t  
Solution :
X [k ] 
1
1 2
1 6
 jk ( / 2 ) t
 jk ( / 2 ) t
x
(
t
)
e
dt


(
t
)
e
dt

 (t  4)e  jk ( / 2)t dt




T


2
2
4
4
4
(impulse train)  (impulse train)
Example 3.11 Calculation of FS Coefficients by Inspection
Determine the FS representation of the signal x(t )  3 cos(t / 2   / 4) using the
method of inspection.
Solution :
x(t )  3
e
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 X [k ]e
x(t ) 
j (t / 2   / 4 )
 3  j / 4
,
2 e

3
X [ k ]   e j / 4 ,
2
0,



e
2
jkt / 2
(3.21)
k  
 j (t / 2   / 4 )
3
3
 e j / 4e jt / 2  e j / 4e jt / 2
2
2
k  1
k 1
(3.22)
otherwise
Figure 3.18
Magnitude and phase for Ex. 3.11.
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3.5 Continuous-Time Periodic Signals:
The Fourier Series cont’
Example 3.12 Inverse FS : Find the time-domain signal x(t) corresponding to the
k
FS coefficients
. Assume that the fundamental periodTis  2.
X [k ]  (1/ 2) e jk / 20
Solution : 0


x(t )   (1/ 2) e
k
k 0
jk / 20
e
jkt

  (1/ 2) k e jk / 20 e jkt
k  1


k 0
l 1
등비수열의 합
  (1/ 2) k e jk / 20 e jkt  (1/ 2) l e  jl / 20 e  jlt
x(t ) 
1
1

1
j (t  / 20 )
 j (t  / 20 )
1  (1 / 2)e
1  (1 / 2)e
x(t ) 
3
5  4 cos(t   / 20)
Figure 3.20 FS coefficients for Problem 3.9.
Figure 3.21
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Square wave for Example 3.13.
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3.5 Continuous-Time Periodic Signals:
The Fourier Series cont’
Example 3.13 FS for a Square Wave Determine the FS representation of the square
wave depicted in Fig. 3.21.
X [k ] 
1 T /2
x(t )e  jkwot dt


T
2
T

1 To  jkwot
 1  jkwot To
e
dt

e
,
 To
T To
Tjkwo
2 sin(kwoTo )
2 e jkwoTo  e  jkwoTo

(
) 
,
Tkwo
2j
Tkwo
2T
1 To
For k=0, X [0]   dt  o
T To
T
X [k ] 
2 sin(kwoTo ) 2T0
2T

sin c(k 0 )
Tkw0
T
T
k 0
k 0
lim
k 0
2 sin(kwoTo ) 2To

Tkw0
T
where sin c(u ) 
sin(u )
u
Figure 3.22 The FS coefficients, X[k], –50  k  50, for three square waves.
In Fig. 3.21 Ts/T = (a) 1/4 . (b) 1/16. (c) 1/64.
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3.5 Continuous-Time Periodic Signals:
The Fourier Series cont’
To/T=1/2? (DC)  (impulse)
To  0? (impulse train)  (impulse train)
T  ? (aperiodic)  (continuous)
(square)  (sinc)
(sinc)  (square)
Figure 3.23
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Sinc function sinc(u) = sin(u)/(u)
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3.5 Continuous-Time Periodic Signals:
The Fourier Series cont’
Example 3.14 Square-Wave Partial-Sum Approximation
Let the partial-sum approximation to the
FS in Eq.(3.29), be given by
J
^
x j (t )   B[k ] cos(kwo t )
k 0
This approximation involves the exponential FS Coefficients with indices
Consider a square wave T  1 and T0 / T  1 / 4. Depict one period of the
 J  k  J.
J
th term in
^
this sum, and find x j (t ) for J  1,3,7,9, and 99.
Solution :
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k 0
1 / 2,

B[k ]  (2 /(k ))(1) ( k 1) / 2 , k odd
0,
k even

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3.6 Discrete-Time Non-periodic Signals :
The Discrete-Time Fourier Transform
(discrete)  (periodic)
1
x[n] 
2
j
X (e ) 




 x[n]e
X (e jn )e jn d
(3.31)
(3.32)
 jn
n  
DTFT
x[n] 
 X (e j )

 x[n]  
n  
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

x[n]  
2
n  
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3.6 Discrete-Time Non-periodic Signals : The
Discrete-Time Fourier Transform cont’
Example 3.17 DTFT of an Exponential Sequence
Find the DTFT of the sequence x[n]   n u[n]
Solution :
j
X (e ) 


n  
n
u[n]e
 jn

  e
n
 jn
n 0

  (e  j ) n 
n 0
1
,
 j
1  e
 1
1
1
1
X (e j ) 

((1   cos) 2   2 sin 2 )1 / 2 ( 2  1  2 cos)1 / 2
1   cos  j sin 
 sin 
arg{ X (e j )}   arctan(
)
1   cos 
 = 0.5
 = 0.9
X (e j ) 
x[n] = nu[n].
magnitude
phase
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 = 0.5
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 = 0.9
28
3.6 Discrete-Time Non-periodic Signals : The
Discrete-Time Fourier Transform cont’
Example 3.18 DTFT of a Rectangular Pulse
Let

1,
x[n]  

0,
n M
n M
Find the DTFT of x[n].
Solution : (square)  (sinc)
Figure 3.30 Example 3.18. (a) Rectangular pulse in the time domain.
(b) DTFT in the frequency domain.
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3.6 Discrete-Time Non-periodic Signals : The
Discrete-Time Fourier Transform cont’
2M
X (e )   e
j
 j ( m  M )
e
jm
m0
2M
e
m 9
 jm 1  e  j( 2 M 1)
,
e

1  e  j

2M  1,

j
X (e )  e
jM
 jm
  0,  2 ,  4 ,
  0,  2 ,  4 ,
e  j( 2 M 1) / 2 (e j( 2 M 1) / 2  e  j( 2 M 1) / 2 )
e  j / 2 (e j / 2  e  j / 2) )
e j( 2 M 1) / 2  e  j( 2 M 1) / 2

e j / 2  e  j / 2
X (e j ) 
sin((2M  1) / 2)
sin( / 2)
sin((2M  1) / 2)
 2M  1
0,  2 ,  4 , ,
sin( / 2)
lim
X (e j ) 
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sin((2M  1) / 2)
sin( / 2)
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3.6 Discrete-Time Non-periodic Signals : The
Discrete-Time Fourier Transform cont’
Example 3.19 Inverse DTFT of a Rectangular Spectrum
Find the inverse DTFT of
Solution :
x[n] 
(sinc)  (square)
1
2
W

e jn d 
W
1
W
sin(Wn) 
n 0 n

x[0]  lim

1,   W
X (e j )  

0, W    
1 jn W
1
e
 sin(Wn ),
 W n
2nj
x[n] 
1
sin(Wn)
n
x[n] 
n0
W

si nc (Wn /  )
Figure 3.31 (a) Rectangular pulse in the frequency domain.
(b) Inverse DTFT in the time domain.
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3.6 Discrete-Time Non-periodic Signals : The
Discrete-Time Fourier Transform cont’
Example 3.20 DTFT of the Unit Impulse
Find the DTFT of x[n]   [n]
Solution : (impulse) - (DC)
j
X (e ) 

 [n]e
 jn
1
n
DTFT
 [n] 
1
Example 3.21 Find the inverse DTFT of a Unit Impulse Spectrum.
Solution : (impulse train)  (impulse train)
1
x[n] 
2

   ()e

jn
d
j
X (e ) 

 (  k 2 )
k  
1
DTFT

  (),
2
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    
32
3.6 Discrete-Time Non-periodic Signals : The
Discrete-Time Fourier Transform cont’
Example 3.22 Two different moving-average systems
y1 [n] 
1
( x[n]  x[n  1])
2
y 2 [ n] 
1
( x[n]  x[n  1])
2
h1 [n] 
1
1
 [n]   [n  1]
2
2
h2 [n] 
1
1
 [n]   [n  1]
2
2
solution :
Figure 3.36
H 1 ( e j ) 
e
j

2
e
j

2
e
2
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1 1  j
 e
2 2
j

2
e
j

2
H 2 ( e j ) 

j

e
cos( ) e 2
2
j

2

H 1 (e j )  cos( )
2
1 1  j
 e
2 2
e
2
j

2
 je
j

2

sin( )
2
arg{ H 1 (e j )}  
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
2

H 2 (e j )  sin( )
2


 /2 ,


2
arg{H 2 (e j )}  

   / 2,

 2
0
0
33
3.6 Discrete-Time Non-periodic Signals : The
Discrete-Time Fourier Transform cont’
Example 3.23 Multipath Channel : Frequency Response y[n]  x[n]  ax[n  1]
Solution :
H (e j )  1  ae j  1  a e j (arg{a})  1  a cos(  arg{a})  j a sin(  arg{a})
H (e j )  ((1  a cos(   arg{ a})) 2  a sin 2 (  arg{ a})) 1/ 2  (1  a  2 a cos(   arg{ a})) 1/ 2
2
2
H inv (e j ) 
H inv (e j ) 
1
,
1  ae  j
1
1 a e
H inv (e j ) 
 j ( arg{a})

a 1
1
1  a cos(  arg{})  j a sin(  arg{})
1
(1  a  2 a cos(  arg{a}))1 / 2
2

1
H (e j )
| H (e j ) | (a) a = 0.5ej2/3. (b) a = 0.9ej2/3.
| H inv (e j ) | (a) a = 0.5ej2/3. (b) a = 0.9ej2/3.
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34
3.7 Continuous-Time Non-periodic
Signals : The Fourier Transform
(continuous aperiodic)  (continuous aperiodic)
Inverse CTFT
CTFT
x(t ) 
1
2



(3.35)
X ( jw)e jwt dw

X ( jw)   x(t )e  jwt dt
(3.26)

x(t )  X ( jw)
CTFT
Condition for existence of Fourier transform:



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x(t ) dt  
2
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35
3.7 Continuous-Time Non-periodic
Signals : The Fourier Transform cont’
Example 3.24 FT of a Real Decaying Exponential
 at
Find the FT of x(t )  e u(t )

Solution : 0 eat dt  , a  0 Therefore, FT not exists.
a  0,


X ( jw)   e  at u (t )e  jwt dt   e  ( a  jw) t dt


1
e  ( a  jw) t
a  jw
0


0
1
a  jw
x(t )  e  at u(t )
X ( jw) 
1
a 2  w2
arg{X ( jw)}   arctan(w / a)
LPF or HPF? Cut-off from 3dB point?
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36
3.7 Continuous-Time Non-periodic
Signals : The Fourier Transform cont’
Example 3.25 FT of a Rectangular Pulse
1,
x(t )  
0,
 T0  t  T0
Find the FT of x(t).
t  T0
Solution : (square)  (sinc)

X ( jw)   x(t )e
 jwt

For w  0,
T0
dt   e
T0
T0
 jwt
1  jwt
2
dt  
e
 sin(wT0 ),
jw
w
T0
w0
2
sin( wT0 )  2T0
w 0 w
lim
X ( jw) 
2
sin( wT0 ) X ( jw)  2T0 s inc (wT0 /  )
w
X ( jw)  2
Example 3.25. (a) Rectangular pulse. (b) FT.
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sin(wT0 )
w
0, sin(wT0 ) / w  0
arg{X ( jw)}  
 , sin(wT0 ) / w  0
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3.7 Continuous-Time Non-periodic
Signals : The Fourier Transform cont’
Example 3.26 Inverse FT of an Ideal Low Pass Filter!!
Fine the inverse FT of the rectangular spectrum depicted in Fig.3.42(a) and given by
1,
X ( jw)  
0,
W  w  W
w W
Solution : (sinc) -- (square)
1
x(t ) 
2
1 jwt
e
dw

e
W
2 jt
W
W

jwt
W
1
sin(Wt ),
t
t0
1
sin(Wt )  W /  ,
t  0 t
1
W
Wt
x(t )  sin(Wt )
or x(t )  si nc( )
t


when
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t  0, lim
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38
3.7 Continuous-Time Non-periodic
Signals : The Fourier Transform cont’
Example 3.27 FT of the Unit Impulse x(t )   (t )
Solution : (impulse) - (DC)

X ( jw)    (t )e  jwt dt  1 

FT
 (t ) 
1
Example 3.28 Inverse FT of an Impulse Spectrum
Find the inverse FT of X ( jw)  2 (w)
Solution : (DC)  (impulse)
1
x(t ) 
2



2 ( w)e jwt dw  1
FT
1
2 (w)
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3.7 Continuous-Time Non-periodic
Signals : The Fourier Transform cont’
Example 3.29 Digital Communication Signals
 Ar ,
Rectangular (wideband) xr (t )  
0,
t  T0 / 2
t  T0 / 2
Separation between KBS and SBS.
Narrow band

( Ac / 2)(1  cos(2t /T 0)),
xc (t )  
0,


t  T0 / 2
t  T0 / 2
Figure 3.44 Pulse shapes used in BPSK communications. (a) Rectangular pulse.
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(b) Raised cosine pulse.
40
3.7 Continuous-Time Non-periodic
Signals : The Fourier Transform cont’
Solution : Pr 
Pc 
1
T0

Ac2

4T0
Figure 3.45 BPSK (a) rectangular pulse shapes
(b) raised-cosine pulse shapes.
T0 / 2
3 Ac2

8
T0 / 2
T0 / 2

T0 / 2
Ar2 dt  Ar2
( Ac2 / 4)(1  cos(2t / T0 ))2 dt
T0 / 2

1
T0
T0 / 2
[1  2 cos(2t / T0  1 / 2  1 / 2 cos(4t / T0 )]dt

8
Ar  Ac
3
the same power constraints
sin(wT0 / 2 )
sin(fT0 )
1 8 T0 / 2
X r' ( jf ) 
X c ( jw) 
(1  cos(2t / T0 ))e  jwt dt

w
f
2 3 T0 / 2
2 T0 / 2  jwt
1 2 T0 / 2  j ( w2 / T0 )t
1 2 T0 / 2  j ( w 2 / T0 )t
X c ( jw) 
e
dt

e
dt

e
dt
3 T0 / 2
2 3 T0 / 2
2 3 T0 / 2
T /2
sin(T0 / 2)
 jt
e
dt

2
T / 2

X r ( jw)  2
0
0
X c ( jw)  2
X c' ( jf ) 
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2 sin(wT0 / 2)
2 sin((w  2T0 )T0 / 2)
2 sin((w  2 / T0 )T0 / 2)


3
w
3
w  2 / T0
3
w  2 / T0
2 sin(fT0 )
2 sin( ( f  1 / T0 )T0 )
2 sin( ( f  1 / T0 )T0 )
 0.5
 0.5
3
t
3
 ( f  1 / T0 )
3
 ( f  1 / T0 )
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3.7 Continuous-Time Non-periodic
Signals : The Fourier Transform cont’
rectangular pulse.
One sinc
Raised cosine pulse
3 sinc’s
The narrower main lobe, the narrower bandwidth.
But, the more error rate as shown in the time domain
Figure 3.47 sum of three frequency-shifted sinc functions.
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Fourier transform
Time domain
Periodic discrete
2
2
0 
0 
T
N
Periodic continuous
x(t ) 


X [k ]e jk0t
frequency domain
X [k ] 
k 
1
x[n] 
X (e j )e jn d 
a-periodic discrete

2  2 
1 
X ( j )e jt d 
a-Periodic continuous x(t ) 

2 
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j
1
T

X (e ) 
T 
x(t )e  jk0t dt

 x[n]e
 jn
n 

X ( j )   x(t )e jt dt

43
9.1 Linearity and Symmetry Properties
of FT
z (t )  ax(t )  by(t ) FT
 Z ( jw)  aX ( jw)  bY ( jw)
; w0
z (t )  ax(t )  by(t ) FS
 Z [k ]  aX[k ]  bY[k ]
z[n]  ax[n]  by[n] DTFT
 Z (e j )  aX (e j )  bY (e j )
DTFS ; 0
z[n]  ax[n]  by[n] 
 Z [k ]  aX[k ]  bY[k ]
z (t ) 
2
1
x(t )  y (t )
3
2
x(t )
y(t )
FS ; 2
x(t ) 
 X [k ]  (1 /(k )) sin(k / 4)
FS ; 2
y(t ) 
 Y [k ]  (1 /(k )) sin(k / 2)
3
1
FS ; 2
z (t ) 
 Z [k ] 
sin(k / 4) 
sin(k / 2)
2k
2k
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9.1 Linearity and Symmetry Properties
of FT cont’
• 3.9.1 Symmetry Properties : Real and Imaginary
Signals


X ( jw)   x(t )e  jwt dt 
 







(3.37)
x  (t )e jwt dt
(real x(t)=x*(t))  (conjugate symmetric)

X  ( jw)   x(t )e  j (  w)t dt

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X  ( jw)  X ( jw)
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(3.38)
45
9.1 Linearity and Symmetry Properties
of FT cont’
• 3.9.2 SYSMMEYRY PROPERTIES : EVEN/ODD
SIGNALS
(even)  (real)
(odd)  (pure imaginary)
xe (t )  Re{X ( j )}
xo (t )  j Im{X ( j )}
For even x(t),
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X
*

 jw   x t e
 jwt 

dt   x e jw d  X  jw.
KyungHee University

real
46
3.10 Convolution Property
• 3.10.1 CONVOLUTION OF NON-PERIODIC
SIGNALS
(convolution)  (multiplication)

yt   ht * xt    h xt   d

But
xt    
1
2

 1
yt    h 

 2



X  jwe jwt  dw.
given xt  
1
2



X  jwe jwt dw.

X  jwe jwt e jw dw d




change the order of integration

1
2
  h e jw d  X  jwe jwt dw.
 



jwt
 yt   2  H  jwX  jwe dw,
1
FT
yt   ht * xt 
Y  jw  X  jwH  jw.
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3.40
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47
3.10 Convolution Property cont’
Example 3.31 Convolution problem in the frequency domain
xt   1 t sint  be the input to a system with impulse response ht   1 t sin2t .
Find the output yt   xt  * ht .
Solution:

1, w  
FT
xt 
X  jw  

0, w  
1, w  2
FT
ht 
H  jw  
.
0, w  2
FT
yt   xt  * ht 
Y  jw  X  jwH  jw,

1, w  
Y  jw  
,
0
,
w




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
yt   1 t sint .
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3.10 Convolution Property cont’
Example 3.32 Find inverse FT’S by the convolution property
Use the convolution property to find xt  , where
4
xt  X  jw  2 sin 2 w.
w
FT
2
Z  jw  sin w.
w

1, t  1 FT
z t   

 Z  jw,
0, t  1
Ex 3.32 (p. 261). (a) Rectangular z(t). (b) z (t ) * z (t )  xt 
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49
3.10 Convolution Property cont’
• 3.10.2 FILTERING
Continuous time
Discrete time
LPF
HPF
BPF
Figure 3.53
(p. 263)
Frequency dependent gain (power spectrum) 20 log H  jw or 20 log H e j .
Y  jw  H  jw X  jw ,
2
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2
2
kill or not (magnitude)
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50
3.10 Convolution Property cont’
Example 3.34 Identifying h(t) from x(t) and y(t)
t
The output of an LTI system in response to an input xt   e2tut  is yt   e ut  .
Find frequency response and the impulse response of this system.
Solution:
X  jw 
H  jw 
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1
jw  2
Y  jw 
1
.
jw  1
jw  2  jw  1 
1
1
 
 
 1
.
jw  1  jw  1  jw  1
jw  1
Y  jw
But H  jw  X  jw
 ht    t   e t ut .
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51
3.10 Convolution Property cont’
EXAMPLE 3.35 Equalization of multipath channel
X  jw  H inv  jwY  jw or X e j   H inv e j Y e j ,
Consider again the problem addressed in Example 2.13. In this problem, a distorted
received signal y[n] is expressed in terms of a transmitted signal x[n] as
yn  xn axn 1,
a  1.
hinv n* hn   n.
Then H inv e j  


1,

hn  a,
0,

n0
n 1
.
otherwise
   
H inv e j H e j  1,
1
.
H e j
 
 
DTFT
hn
 H e j  1  ae j .

 
H inv e j 
1
.
1  ae  j
hinv n   a un.
n
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52
3.10 Convolution Property cont’
• 3.10.3 Convolution of periodic signals : Cyclic
convolution
Convolution in just one period Or better to derive in the frequency domain
yt   xt   zt    x zt   d
T
0 2
FS ;
T
yt   xt   zt Y k   T X k Z k .
3.44
EXAMPLE 3.36 Convolution of 2 periodic signals
Evaluate the periodic convolution of the sinusoidal signal
zt   2 cos2t   sin4t 


1,
1 2 j ,

Z k   
 1 2 j ,
0,
X k  
k  1
k 2
.
k  2
otherwize
2 sin k 2 
.
k 2
Figure 3.56 (p. 268) Square wave for Example 3.36.
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53
3.10 Convolution Property cont’
1  ,
Y k   X k Z k   
0,
k  1
, 
otherwise
yt   2  cos2t .
Xˆ j k   X k Z k ,
1,  J  k  J
0, otherwise



Z k 
N 1

z t  
sin t 2 J  1
. 3.45
sin t 
2
N
yn  xn  zn   xk zn  k  Y k   N X k Z k 
DTFS ;
k 0
Figure 3.57
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(p. 270) z(t) in Eq. (3.45) when J = 10.
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54
3.11 Differentiation and Integration Properties
• 3.11.1 DIFFERENTIATION IN TIME
1
xt  
2



X  jwe jwt dw.
d
1
xt  
dt
2



FT
X  jw jwe jwt dw 
jwX  jw.
EXAMPLE 3.37 The differentiation property implies that


d  at
jw
FT
e u t  
.
dt
a  jw


d at
e u t   ae at u t   e at  t   ae at u t     .
dt
d  at
a
jw
FT
e ut  
1 
dt
a  jw
a  jw

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
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55
3.11 Differentiation and Integration
Properties cont’
Example 3.38 Resonance in MEMS accelerometer
The MEMS accelerometer introduced in Section 1.10 is described by the
differential equation
d2
w d
yt   n
yt   wn2 yt   xt .
2
dt
Q dt

H  jw 
1
.
wn
2
2
 jw   jw  wn
Q
Fig 3.58
(p. 273) 20log10 | H  jw |
n = 10,000 rad/s
Q = 2/5, Q = 1, and Q = 200.
Resonance in 10,000 rad/s
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56
3.11 Differentiation and Integration
Properties cont’
Example 3.39 Use the differentiation property to find the FS of the triangular
wave depicted in Fig. 3.59(a)
0,

FS ;w0
z t   Z k    4 sin  k
 2


k
k 0


,
.

Y [k ]  ???
k0
Signals for (a) Triangular wave y(t). (b)
T 2 ,


k
FS ; w0
y t Y k    2T sin 
 2

2 2

 jk 
dy (t )
 z (t )
dt
k 0

.

, k  0
where
j 
2k
j,
T
If we differentiate z(t) once more???
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57
3.11 Differentiation and Integration
Properties cont’
• 3.11.2 DIFFERENTIATION IN FREQUENCY

X  jw   xt e

 jwt
dt, Differentiate w.r.t. ω,

d
X  jw    jtxt e  jwt dt ,

dw
d
FT
Then,  jtxt 
X  jw.
dw
Example 3.40 FT of a Gaussian pulse
Use the differentiation-in-time and differentiation-in-frequency properties for the
FT of the Gaussian pulse, defined by gt   1 2 et 2 and depicted in Fig. 3.60.


d
g t    t
dt
2 e
t 2 2

2
 tg t .
3.48
d
FT
g t   tg (t ) 
 jwG  jw,
dt
1 d
FT
G jw
  tg t 
j dw
FT
and  jtg t 
 dw G  jw   wG  jw.
d
Then G jw  ce w 2 . (But, c=?) G j 0   1

2
Figure 3.60 (p. 275) Gaussian pulse g(t).
4/13/2015
1

2 et
2
2
d
G  jw
dw

2 e t 2 dt  1.
2
FT

e w 2 .
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2
58
3.11 Differentiation and Integration
Properties cont’
• 3.11.3 Integration
yt    x d ;
t


t

FT
x d 

Y  jw 
1
X  jw.
jw
3.52
1
X  jw  X  j 0 w.
jw
3.53
FT
Ex) ut      d . Prove ut U  jw  jw   w.
1
t
u t  
1 1
 sgn t .
2 2
 1,

sgn t   0,
1,

3.54 
Note u(t )  e at u(t ) where a=0
t0
FT
2 w.
We know 1
t  0.
t0
d
sgn t   2 t .  jwSgn jw  2.
dt
Fig. a step fn. as the sum of a constant and a signum fn.
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 2
 , 0
S ( j )   j
since linear
0,
0

FT
 u t 
1
   
j
59
3.11 Differentiation and Integration
Properties cont’
Common Differentiation and Integration Properties.
d
FT
xt  
jX  j 
dt
d
FS ;0
xt  
 jk0 X k 
dt
d
FT
 jtxt  
X  j 
d
d
DTFT
 jnxn

X e j
d
t
1
FT
 x d  j X  j   X  j 0  
 
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60
3.12 Time-and Frequency-Shift
Properties
• 3.12.1 Time-Shift Property


 jt
Z  j    zt e
  xt  t0 e  jt dt



  x e

 j  t0 
d  e
 jt0



x e  j d  e  jt0 X  j 
arg[e j t0 X  j ]   t0  arg[X ( j)]
Table 3.7 Time-Shift Properties of Fourier Representations
FT
xt  t0 
e  jt0 X  j 
FS ;0
xt  t0 
 e  jk0t0 X k 
 
DTFT
xn  n0 
 e  jn0 X e j
DTFS ;0
xn  n0 
 e  jk0n0 X k 
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3.12 Time-and Frequency-Shift
Properties cont’
Example)
Figure 3.62
X  j  
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2

z t   xt  T1   Z  j   e jT X  j 
1
sin T0   Z  j   e jT
1
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2

sin T 
62
3.12 Time-and Frequency-Shift
Properties cont’
• 3.12.2 Frequency-Shift Property
1 
j t


z t  
Z
j

e
d



2
1 

X  j    e jt d

2 
Recall
z t  
1
2
 e jt



X  j e j   t d
1
2



X  j e jt d  e jt xt 
FT
xt  t0 
e jt0 X  j 
Table 3.8 Frequency-Shift Properties
FT
e jt xt  
X  j    
FS ;0
e jk00t xt 
 X k  k0 

DTFT
e jn xn
 x e j  

DTFS ;0
e jk00n xn
 X k  k0 
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3.12 Time-and Frequency-Shift
Properties cont’
Example 3.42 FT by Using the Frequency-Shift Property
e j10t , t  
z t   
t 
0,
Solution: We may express zt  as the product of a complex sinusoid e j10t
and a rectangular pulse

1, t  
xt   

0, t  
FT
xt 
X  j  
2

zt   e j10t x(t )

sin  
FT
X  j   10  
 z t   e j10t x(t ) 
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sin   10  
  10
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3.12 Time-and Frequency-Shift
Properties cont’
Example 3.43 Using Multiple Properties to Find an FT
xt  



d 3t
e u t   e t u t  2
dt
Sol) Let wt   e3t ut  and vt   et ut  2
Then we may write xt  
d
wt   vt 
dt
By the convolution and differentiation properties X  j   jW  j V  j 
The transform pair
FT
e at u t 
 vt   e e
1
a  j
 W  j  
1
3  j
e j 2
ut  2  V  j   e
1  j
2 t 2 
2
je j 2
X  j   e
1  j 3  j 
2
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3.13 Inverse FT by Using PartialFraction Expansions
• 3.13.1 Inverse FT by using
FT
e at u t  
1
a  j
bM  j     b1  j   b0
B j 
X  j  

 j N  aN 1  j N 1    a1  j   a0 A j 
~
M N
B  j 
k
X  j    f k  j  
A j 
k 0
M
Let v  j, then v N  aN 1v N 1   a1v  a0  0
 N roots, d k
M
X  j  
for k  1,, N
 b  j 
k 0
N
k
N
k
  j  d 
 partial fraction  X  j   
k 1
k
Cx
j  d k
k 1
FT
e dt u t 
1
j  d
for d  0
N
N
xt    Ck e ut  X  j   
dk t
k 1
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FT
k 1
Ck
j  d k
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3.13 Inverse FT by Using PartialFraction Expansions cont’
• 3.13.2 Inverse Discrete-Time Fourier Transform
 M e jM    1e j  0
X e  
 N e jN   N 1e j( N 1)    1e j  1
j
Ne
 jN
  N 1e
 j  N 1
   1e
 j
N

 1   1  d k e  j

k 1
v N  1v N 1  2v N 2    N 1v   N  0
    1  dC e
X e
j
N
k
k 1
 j
where
k
N
DTFT
d k n un

1
1  d k e  j
Then xn   Ck d k  un
n
k 1
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3.13 Inverse FT by Using PartialFraction Expansions cont’
Example 3.45 Inversion by Partial-Fraction Expansion
 
X e j
5
 e  j  5
6

1
1
1  e  j  e  j 2 
6
6
Solution:
5
 e  j  5
C1
C2
6


1
1
1
1
1  e  j   e  j 2  1  e  j  1  e  j
6
6
2
3
Using the method of residues described in Appendix B, We obtain
And
5
 e  j  5
 1

6
C1  1  e  j 
 2
 1  1 e  j  1 e  j 2
6
6
5
 e  j  5
 1

6
C2  1  e j 
1
 3
 1  e j  1 e  j 2
6
6
5
 e  j  5
 6
4
1  j
1 e
3
e j 2
e j 2
5
 e j  5
 6
1
1  j
1 e
 j
3
e 3
e j 3
Hence, xn  41/ 2 un  1/ 3 un
n
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3.14 Multiplication (modulation) Property
Given xt  
1
2
yt   x(t ) z (t ) 



X  jv e jvt dv

1
2   
2

 
and z t  
1
2



Z  j e jt d
X  jv Z  j e j   v t ddv
Change of variable     to obtain
yt  
1
2
 1
  2





X  jv Z  j   v dve jt d

FT
y t   xt z t  
Y  j  
1
X  j    Z  j 
2
(3.56)

Where X  j  Z  j    X  jv Z  j   v dv

 
DTFT
yn  xnzn
 Y e j 

 
 
1
X e j  Z e j
2
(3.57)
denotes periodic convolution.
Here, X e j  and Z e j  are 2 -periodic.
 
 

 

X e j  Z e j   X e j Z e j   d
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3.14 Multiplication (modulation)
Property cont’
Example) Windowing in the time domain
FT
y t   x(t ) w(t ) 
Y  j  
W  j  
4/13/2015
2

1
X  j   W  j 
2
sin T0 
Figure 3.66 The effect of Truncating the
impulse response of a discrete-time system. (a)
Frequency response of ideal system. (b) F  
for  near zero. (c) F   for  slightly
greater than  / 2 (d) Frequency response of
system with truncated impulse response.
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3.14 Multiplication (modulation)
Property cont’
Example 3.46 Truncating the sinc function
Sol) hn 

1, n  M
1
 n 
sin  truncated by wn  
n  2 
0, otherwise
hn, n  M
ht n  
otherwise
0,
H t e j 
hn 
1
2


  H e W e

j  
j

d
1,    / 2
1
 n 
sin   H e j   
n  2 

0,  / 2    


W e j   
 
H t e j 
sin    2M  1 / 2
sin     / 2
1
2
 /2

 /2
  
F    H e W e
j
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DTFT
ht n
 Ht e j 
F  d
j  



j  

,
W e

0,


  /2
  /2
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3.14 Multiplication (modulation)
Property cont’
Example 3.47 Radar Range Measurement: RF Pulse Train
k  500
12 j ,

Solution) sin(2  500t )  S k    12 j , k  500

Pk   e  jkT00
X k  
sin k0T0 
k


0,
otherwise
x(t )  p(t ) sin(2  500t )  X [k ]  P[k ] * S[k ]
1  j k 500 T00 sin k  5000T0  1  j k 500 T00 sin k  5000T0 
e
 e
k  500
k  500
2j
2j
DTFT ; 2 / N
yn  xnzn

Y k   X k  Z k 
(3.59)
N 1
X k   Z k    X mZ k  m
m0
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3.14 Multiplication (modulation)
Property cont’
Figure 3.69 FS magnitude spectrum of FR
pulse train for 0  k  1000 The result is
depicted as a continuous curve, due to
the difficulty of displaying 1000 steps
Table 3.9 Multiplication Properties
FT
xt z t 
1
X  j   Z  j  
2
FS ;0
xt z t 
 X k  Z k 
1
DTFT
xnzn

X e j  Z e j
2
DTFS ;0
xnzn
 X k   Z k 
 
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 
73
3.15 Scaling Properties

Z  j    zt e
 jt


  xate  jt dt

1 / a   x e j  / a  d , a  0


Z  j   

1 / a  x e j  / a  d , a  0



 Z  j   1 / a
 x e


 j  / a 
d
FT
1/ a X  j / a
zt   xat
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(3.60)
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3.15 Scaling Properties cont’
Example 3.48 SCALING A RECTANGULAR PULSE
Let the rectangular pulse
 1, t  1 

x(t )  

0
,
t

1



t  1, t  2 

y (t )  x( )  
2  0, t  2 
Solution :
X ( jw) 
2
sin( w).  y(t )  x(t / 2). a  1 / 2
w
Y ( jw)  2 X (i 2w)
2
 2( ) sin(2w)
2w
2
 sin(2w).
w
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3.15 Scaling Properties cont’
d
e j 2w
Example 3.49 Multiple FT Properties for x(t) when X ( jw)  j {
}.
dw 1  j (w / 3)
Solution)
s(t )  e t u (t ) 
S ( jw) 
FT
X ( jw)  j
1
1  jw
d
{e j 2 w S ( jw / 3)}.
dw
we define Y ( jw)  S ( jw / 3)
y(t )  3s(3t )  3e3t u(3t )  3e 3t u(t ).
Now we define W ( jw)  e j 2wY ( jw)
w(t )  y(t  2)  3e3(t  2)u(t  2).
Finally, since
X ( jw)  j
d
W ( jw),.
dw
x(t )  tw(t )  3te 3(t  2)u(t  2).
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3.16 Parseval’s Relationships
언제 꽝? | X ( jw) |2
| x(t ) |2
Wx  


Table 3.10 Parseval Relationships for the Four
Fourier Representations
Represent
ation
| x(t ) |2 dt,
Note that | x(t ) |2  x(t ) x * (t ).
1
x (t ) 
2
*




Wx   x(t )[

X * ( jw)e  jwt dw.
1
2



X * ( jw)e  jwt dw]dt.
1  *
Wx 
X ( jw) X ( jw) dw
2 
So concludethat

1 
2
2
|
x
(
t
)
|
dt

|
X
(
jw
)
|
dw.

2 
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Parseval Relation

1
| X ( jw) | dw 
2

FS

1 T
2
2
|
x
(
t
)
|
dt

|
X
[
k
]
|

T O
K  


DTFT
DTFS
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 | x[n] |2 
n  
1
N
N 1
1
2





FT
2
| X (e j ) | 2 d
N 1
 | x[n] |  | X [k ] |
2
K 0

| X ( jw) | 2 dw
2
k 0
77
3.16 Parseval’s Relationships cont’
Example 3.50 Calculate the energy in a signal
sin(Wn)
Let x[n] 
n

E∞=
| x[n]|
2
n  

sin 2 (Wn )

.
2 2
 n
n  
Use the Parseval’s theorem
Solution) E 
1
2
1
E 
2


W

| X (e j ) |2 d.
1d  W /  .
W
1,
|  | W
x[n]  X (e ) {
0, W |  | 
DTFT
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j
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3.17 Time –Bandwidth Product
1, | t | To FT
x(t )  {
 X ( jw)  2 sin(wTo ) / w.
0, | t | To
Figure 3.72 (p. 305) Rectangular pulse illustrating the inverse
relationship between the time and frequency extent of a signal.
Td

[

t 2 | x(t ) |2 dt




2
| x(t ) | dt
]1/ 2
Bw

[

w2 | X ( jw) |2 dw 1 / 2

] .

2
 | X ( jw) | dw

Td Bw  1 / 2.
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79
3.17 Time –Bandwidth Product cont’
Example 3.51 Bounding the Bandwidth of a Rectangular Pulse
1, | t |  T 
Let x(t )  

0
,
|
t
|

T
o 

Use the uncertainty principle to place a lower bound on the effective bandwidth of x(t).
Solution:
 To t 2 dt 
T
Td   Too 
 Todt 
1/ 2

 (1/(2To ))(1/ 3)t 3 |To To

1/ 2
 To / 3.
Bw  3 /(2To ).
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3.18.1 The Duality Property of the FT
1
x(t ) 
2



X ( jw)e
jwt

dw. 
 X ( jw)   x(t )e  jwt dt .

Fig. 3.73 Duality of
rectangular pulses and sinc
functions
1 
z ( )e jv dv.

2  
if we choose v  t and   w, then Eq. (3.66) im plies that
1 
y (t ) 
z ( w)e jwt dw.

2  
FT
y (t ) 
z ( w).
y (v ) 
1 
z (t )e  jwt dt,

2  
FT
z (t ) 
2y ( w),
y ( w) 
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FT
f (t ) 
F ( jw),
FT
F ( jt ) 
2f (  w).
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3.18.1 The Duality Property of the FT
cont’
Example 3.52 By using Duality, find the FT of x(t ) 
1
1  jt
Solution)
FT
FT
f (t ) 
F ( jw)  F ( jt ) 
2f (w).
FT
f (t )  et u (t ) 
F ( jw) 
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1
1  jw
 F ( jt ) 
1
1  jt
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3.18.2 & 3.18.3
• 3.18.2 The Duality Property of the DTFS
N 1
x[n]   X [k ]e
jk o n
K 0
1
X
[
k
]

. 
N
DTFS ; 2 / N
x[n] 

 X [k ]

N 1

x[n]e  jkon
n 0
DTFS ; 2 / N
X [k ] 


1
x[k ]
N
• 3.18.3 The Duality Property of the DTFT and FS
z (t ) 

Z [k ]e jkw0t  Z [k ]  1

2
k  


j
X (e ) 

  z (t )e

 jkt
dt
x[n]e  jn .
n  
Table 3.11 Duality Properties of Fourier Representations
FT
f (t ) 
F ( jw)
FT
DTFS
DTFS ; 2 / N
x[n] 

 X [k ]
DTFS
x[n] 
 X (e j )
DTFS
x[n] 
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1
2

  X (e

j
)e jn .
FT
F ( jt ) 
2f (w)
DTFS ; 2 / N
X [n] 

(1/ N ) x[k ]
FS ;1
X (e jt ) 
 x[k ]
DTFT
FS ;1
x[n] 
 X (e j )  X (e jt ) 
 x[k ].
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3.18.3 The Duality Property of the
DTFT and FS cont’
Example 3.53 FS-DTFT Duality
Use the duality property and the results of Example 3.39 to determine the inverse
DTFT of the triangular spectrum X (e j ) depicted in Fig 3.75(a).
Solution:
Z [k ]  e
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jw / 2
Y [k ]
, n0
 
 k 1
  4 j sin(k / 2) ,

k 2


n0 

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