T-61.140 Signal Processing Systems
Spring 2003
The Discrete-Time Fourier Transform
The DiscreteDiscrete-Time
Fourier Transform
• The Fourier series representation of a
discrete-time periodic signal is a finite series,
as opposed to the infinite series
representation required for the continuoustime periodic signals
• The discrete-time Fourier analysis is discussed
• The differences between continuous-time and
discrete-time Fourier transforms are considered
(similar to those between CT and DT Fourier
series)
Tik-61.140 / Chapter 5
 Olli Simula
Development of the Discrete-Time
Fourier Transform
Representation of Aperiodic Signals
• An aperiodic signal x(t) was earlier (Chapter 4)
represented by first constructing a periodic signal
xp (t) that was equal to x(t) over one period
• The Fourier series representation for xp(t) converged
to the Fourier transform representation for x(t)
• The similar procedure is applied to discrete-time
signals in order to develop the Fourier transform
representation for discrete-time aperiodic sequences
• Consider a general sequence x[n] that is of finite duration, i.e., for
some integers N1 and N2, x[n] = 0 outside the range -N1 < n < N2
• A periodic signal xp[n] is constructed for which x[n] is one period
• As N approaches infinity, xp[n] = x[n] for any finite value n
x[n ]
3
Representation of Aperiodic Signals
∑
ak =
k= N
1
N
N2
k = − N1
 Olli Simula
Chapter 5 / O. Simula
− jk ( 2π / N ) n
0
N2
N
n
4
• Defining the function
X (e jω ) =
+∞
∑ x[ n]e− jωn
n =−∞
we see that the coefficients a k are proportional to samples of X(ejω)
1
∑ x p[ n]e− jk ( 2π / N) n
N k= N
∑ x[ n]e
-N1
Tik-61.140 / Chapter 5
 Olli Simula
k
ak =
• Since x[n]=xp[n] over a period that includes the interval -N1<n<N2 ,
it is convenient to choose the interval of summation to include this
interval, so that xp[n] can be replaced by x[n] in the summation
ak =
n
N2
x p [n]
Representation of Aperiodic Signals
• Let us examine the effect on the Fourier series representation o f xp[n]
• Fourier series: x [n] =
a e jk ( 2π / N )n
p
0
-N1
-N
Tik-61.140 / Chapter 5
 Olli Simula
2
=
x p [n] =
5
where ω 0 = 2π / N
• We can now express xp[n] in terms of X(ejω) as
1 +∞
∑ x[ n]e− jk (2π / N )n
N k =−∞
Tik-61.140 / Chapter 5
1
X (e jk ω 0 )
N
 Olli Simula
∑
k= N
1
X (e jkω 0 )e jkω 0n
N
Tik-61.140 / Chapter 5
6
1
T-61.140 Signal Processing Systems
Spring 2003
Representation of Aperiodic Signals
The Discrete
Discrete--Time Fourier
Transform
• Equivalently, since 2π/N=ω0
x p [n ] =
1
X (e jkω0 )e jkω 0nω 0
2π k =∑N
x[ n] =
• As N increases ω0 decreases, and as N approaches infinity, and the
summation passes to an integral
• As N approaches infinity, xp[n] -> x[n] and the above equation
becomes
x p [n ] =
X (e jω ) =
7
The Discrete-Time Fourier Transform
• The discrete -time Fourier transform is
periodic and there is finite interval of
integration in the synthesis equation
• Discrete -time complex exponentials that
differ in frequency by a multiple of 2π are
identical
• Periodicity of ejωn :
ω = 0 and ω = 2π yield the same signal
Tik-61.140 / Chapter 5
 Olli Simula
− jω n
 Olli Simula
The Discrete-Time Fourier Transform
• Signals at frequencies
near ω=0 and any
even multiple of of
2π are slowly varying
• High frequencies
in discrete-time
are the values of ω
near odd multiples
of π
9
 Olli Simula
Tik-61.140 / Chapter 5
10
Example 5.1: Lowpass Filter, a > 0
Example 5.1: Exponential Sequences
• Consider the signal
+∞
∑ x[n]e
n= −∞
1
X (e jω )e jωn d ω
2π ∫2π
Tik-61.140 / Chapter 5
 Olli Simula
1
X (e jω )e jωn dω
2π ∫2π
x[ n] = a n u[ n ] , | a | < 1
• The Fourier transform is given by
X ( e jω ) =
+∞
∑ an u[n ] e− jωn
n = −∞
=
∑ (ae− jω )
+∞
n
n= 0
=
1
1 − ae− jω
• For a > 0, the system corresponds to a lowpass filter
• For a < 0, the system corresponds to a highpass filter
 Olli Simula
Chapter 5 / O. Simula
Tik-61.140 / Chapter 5
11
 Olli Simula
Tik-61.140 / Chapter 5
12
2
T-61.140 Signal Processing Systems
Spring 2003
Example 5.1: Highpass Filter, a < 0
Example 5.2:
x[n] = a| n| , | a |< 1
• Consider the signal
+∞
∑ a| n|e − jωn
X ( e jω ) =
X ( e jω ) =
X ( e jω ) =
∑ (ae− jω )
+∞
n
n =0
1
1 − ae− jω
+
+
+∞
−1
n =0
n = −∞
∑ an e − jω n + ∑ a−n e − jωn
=
n = −∞
∑ (ae jω )
+∞
m
m=1
ae jω
1 − ae jω
=
1 −a 2
1− 2a cosω + a 2
• In this case, X(ejω) is real !
Tik-61.140 / Chapter 5
 Olli Simula
13
Example 5.2:
Signal
Tik-61.140 / Chapter 5
 Olli Simula
14
Example 5.3: Rectangular Pulse
x[n]=a |n| and
its Fourier transform
• Consider the rectangular pulse
1, | n |≤ N1
x[ n] = 
0, | n | > N1
X ( e jω ) =
+ N1
∑e − jωn =
n = − N1
• The function is the discrete-time counterpart of the sinc function
• This function, however, isperiodic with period 2π,
whereas the sinc function is aperiodic
0 < a <1
 Olli Simula
sin ω ( N1 + ½ )
sin(ω / 2)
Tik-61.140 / Chapter 5
15
Example 5.3:
Rectangular Pulse and Its Fourier Transform
Tik-61.140 / Chapter 5
 Olli Simula
16
Example 5.4: Unit Impulse
• Let x[n] be a unit impulse
x[ n] = δ [n]
• The analysis equation gives
X (e jω ) = 1
• The unit impulse has a Fourier transform representation consisting of
equal contributions at all frequencies
• We obtain now
xˆ [n ] =
1
2π
∫
W
−W
e jωn dω =
sin Wn
πn
• The frequency of the oscillations in the approximation increases
as W is increased
• The amplitude of these oscillations decreases relative to the magnitude
of x[0] as W is increased, and the oscillations disappear for W =π
 Olli Simula
Chapter 5 / O. Simula
Tik-61.140 / Chapter 5
17
 Olli Simula
Tik-61.140 / Chapter 5
18
3
T-61.140 Signal Processing Systems
Spring 2003
Example 5.4: Approximation of the Unit Sample
The Fourier Transform for Periodic Signals
x[n] = e jω 0n
• Consider the signal
• In continuos-time, the Fourier transform of ejω0t was interpreted
as an impulse at ω = ω0
• The discrete-time Fourier transform must be periodic in ω with
period 2π
• Thus, the Fourier transform of x[n] should have impulses at
ω0 , ω0 +2π, ω0 +4π , etc. , i.e.,
X(jω)
jω
X (e ) =
+∞
∑ 2π δ (ω − ω 0 − 2πl )
2π
l =−∞
ω0 -4π
Tik-61.140 / Chapter 5
 Olli Simula
19
ω0 -2π
ω0
ω0+2π
ω0+4π
Tik-61.140 / Chapter 5
 Olli Simula
The Fourier Transform for Periodic Signals
The Fourier Transform for Periodic Signals
• Substituting into the synthesis equation
• Now, consider a periodic sequence x[n] with period N and with
the Fourier series
x[ n] =
+∞
1
1
X (e jω )e jω n dω =
∑ 2π δ (ω − ω 0 − 2π l ) e jωn dω
2π 2∫π
2π 2∫π l = −∞
x[ n] =
∑ ak e
ω
20
jk (2π / N ) n
k= N
• The Fourier transform is
• Any intervalof length 2π includes exactly one impulse, then if
the interval of integration chosen includes the impulse at ω0+2 πr,
we have
x[ n] =
1
2π
∫ X (e
jω
)e
jω n
dω = e
j(ω 0 + 2π r )n
=e
Tik-61.140 / Chapter 5
21
The Fourier Transform for Periodic Signals
x[n] = a0 + a1e j( 2π / N ) n + a2 e j 2( 2π / N ) n + ... + aN − 1e j( N −1)( 2π / N ) n
+∞
∑ 2π
akδ (ω −
2πk
)
N
Tik-61.140 / Chapter 5
 Olli Simula
22
Example 5.5: Periodic (Sinusoidal) Signal
• Consider the periodic signal
x[ n] = cosω 0 n = 12 e jω 0n + 12 e − jω 0n , with ω 0 =
X ( e jω ) =
ω 0 = 0,
+∞
∑πδ ω −
l =−∞
2π
5
2π
2π



− 2π l  + ∑ πδ  ω +
− 2πl 
5
5
 l =−∞ 

+∞

2π 

2π 
X (e j ω ) = πδ  ω −  + πδ  ω +
 , −π ≤ω < π
5 
5 


2π
,
N
4π
,...
N
( N − 1) 2π
N
Chapter 5 / O. Simula
)=
k = −∞
• x[n] is a linear
combination of
signals with
 Olli Simula
jω
• The Fourier transform of a periodic signal can be directly
constructed from its Fourier coefficients
• This can be verified by noting that x[n] is the linear combination of
complex exponentials and so is the Fourier transform
jω 0 n
2π
 Olli Simula
X (e
Tik-61.140 / Chapter 5
23
 Olli Simula
Tik-61.140 / Chapter 5
24
4
T-61.140 Signal Processing Systems
Spring 2003
Example 5.6: Periodic Impulse Train
Properties of the Discrete-Time Fourier Transform
Consider the discrete-time counterpart of the periodic impulse train
+∞
x[n] =
∑δ
(n − kN )
k = −∞
−Ν
Fourier series coefficients:
X (e jω ) =
2π
N
+∞
∑δ
k = −∞
ak =
Ν
0
n
F
F
Fourier transform pairs: x1[ n] ←→ X1(e jω ) , x2[n] ←→ X2 (e jω)
ax1[n] + bx2[n]←→aX1(e jω ) +bX2(e jω )
F
• Linearity:
X(e jω)
Tik-61.140 / Chapter 5
ω
2π
N
25
Tik-61.140 / Chapter 5
 Olli Simula
jω
X k (e ) =
jω
X k (e ) =
(k )
[ n]e
− j ωn
=
+∞
∑x
n= −∞
( k)
[ rk ]e
− j ωrk
+∞
∑ x[ r]e
− j ( kω ) r
jω
= X (e )
n=−∞
F
x(k )[n]←

→X (e jkω )
27
Tik-61.140 / Chapter 5
 Olli Simula
28
Convolution Property
• Note that bas the
signal is spread out
and slowed down in
time by taking k>1
its Fourier transform
is compressed
• The application of the
time scaling is in
increasing and
decreasing the
sampling rate
Chapter 5 / O. Simula
+∞
∑x
n=−∞
• Furthermore, since x(k)[rk]=x[r], the Fourier transform can be written
as
Time Expansion
Tik-61.140 / Chapter 5
26
Time Expansion
• For k=3, x(k)[n] is obtained
fromx[n] by placing k-1
zeros between successive
values of the original signal
• Intuitively, we can think of
x(k)[n] as a slowed-down
version of x[n]
Tik-61.140 / Chapter 5
F
e jω0 nx[n]←

→X(ej(ω −ω0 ))
• Since x(k)[n] equals 0 unless n is a multipleof k, i.e., unless n=rk,
the Fourier transformof x(k)[n] can be given as
 x[n / k ], if n is a multipleo f k
xk [n] = 
 0,
if n is not a multipleo f k
 Olli Simula
F
• Frequency Shifting:
Time Expansion
• Let k be a positive integer, and define the signal
 Olli Simula
x[n− n0]←→e− jωn0 X (e jω )
• Time Shifting:
2π
N
0
 Olli Simula
2Ν
1
1
∑ x[ n] e− jk (2π / N )n = N
N n= N
2πk 

ω −

N 

X (e j (ω +2π )) = X (e jω )
• Periodicity:
x[n]
1
29
x[n]
h[n]
y[n]= x[n]*h[n]
In frequency domain:
Y (e jω ) = X (e j ω )H(e jω )
Convolution in the time domain corresponds to
multiplication in the frequency domain
• The frequency response H(ejω) captures the change in complex
amplitude of the Fourier transform of the input at frequency ω
 Olli Simula
Tik-61.140 / Chapter 5
30
5
T-61.140 Signal Processing Systems
Spring 2003
Example 5.11: Delay on n 0 Samples
• Impulse response:
Example 5.12: Ideal Lowpass Filter
h[ n] = δ [ n − n0 ]
• Frequency response: H (e jω ) =
h[ n ] =
+∞
1
2π
∑δ [n − n0 ] e− jωn = e− jωn
+π
∫ H (e
jω
) e jω n dω =
−π
1
2π
+ωc
∫e
h[ n] ≠ 0, for n < 0
Y (e jω ) = e − j ωn0 X (e j ω )
• The impulse response
is not causal and
its oscillatory behavior
is not desired
y[ n ] = x[ n − n0 ]
Tik-61.140 / Chapter 5
31
Example 5.14:
Ideal Bandstop Filter Based on Ideal Lowpass
Filters and the Frequency Shifting Property
Tik-61.140 / Chapter 5
 Olli Simula
32
Example 5.14: Ideal bandstop filter based on ideal lowpass
filters and the frequency shifting property
w 1[ n ] = ( − 1) n x [ n] = e jπ n x[ n ]
Frequency shifting :
W1 ( e jω ) = X ( e j (ω − π ) )
Convolution :
W2 (e j ω ) = H lp (e j ω )X (e j ( ω − π ) )
Frequency shifting :
W3 (e jω ) = W2 (e j ( ω − π ) ) = H lp (e j( ω − π ) )X (e j( ω − 2 π ) )
Periodicity:
W3 (e j ω ) = Hlp (e j ( ω −π ) ) X (e jω )
Hlp (ejω)
−π
4
Tik-61.140 / Chapter 5
33
Linearity:
π
4
Y ( e jω ) = W 3 ( e jω ) + W 4 ( e jω )
[
π
34
• The corresponding Fourier transforms:
Y (e jω ), X 1 (e jω ), X 2 (e jω )
]
Y (e jω ) =
[
H ( e j ω ) = H lp ( e j ( ω − π ) ) + H lp ( e jω )
]
Y ( e jω ) =
Y ( e jω ) =
Bandstop filter
Chapter 5 / O. Simula
π
2
y[n ] = x [ n] x [ n ]
jω)jω
jω−π
HH(e
lpH(e
lp (e )
 Olli Simula
−π
2
• Consider a product of two sequences
= H lp (e j (ω − π ) ) + H lp (e j ω ) X (e jω )
−π
−π
−π
Tik-61.140 / Chapter 5
The Multiplication Property
W 4 ( e jω ) = H lp ( e jω ) X ( e jω )
Frequency response:
Hlp (ej(ω−π) )
 Olli Simula
Example 5.14: continued...
Convolution:
sin ω cn
πn
• The impulse response
is a sinc function
• Thus, for any input x[n] with the Fourier transform X(ejω),
the Fourier transform of the output is
 Olli Simula
dω =
0
n= −∞
 Olli Simula
jω n
−ωc
−π
−π −π
22 44
ππ
44
Tik-61.140 / Chapter 5
ππ
22
π
35
 Olli Simula
1
+∞
+∞
n =−∞
n= −∞
2
∑ y[n]e− jωn = ∑ x1[n ]x2 [n]e− jωn
+∞
∑
 1

∫ X1 (e
n = −∞ 
 2π 2π
jθ

)e jθ n dθ  x2 [n ]e − jωn

 +∞

1
X 1 ( e jθ )  ∑ x 2[ n] e − j (ω − θ )n  dθ
2π 2∫π
 n = −∞

Tik-61.140 / Chapter 5
36
6
T-61.140 Signal Processing Systems
Spring 2003
The Multiplication Property
Summary of Fourier Series
and Transform Properties
• The bracketed summation is the discrete-time Fourier transform
 +∞

X 2 (e j (ω −θ ) ) =  ∑ x2 [n ]e − j (ω −θ ) n 
 n =−∞

and we have
Y ( e jω ) =
1
2π
∫ X 1 (e
jθ
) X 2 (e j(ω −θ ) )d θ
2π
This corresponds to a periodic convolution of X 1(ejω ) and X 2(ejω )
and the integral is evaluated over any interval of length 2π
Tik-61.140 / Chapter 5
 Olli Simula
37
Systems Characterized by Linear
Constant -Coefficient Difference Equations
∑ ak y[n − k] =
k =0
• Solving H(ejω)
M
∑ bk x[ n − k ]
H (e jω ) =
k= 0
• The frequency response of the system, H(ejω), can be determined
by applying the Fourier transform to both sides of the difference
equation and using the linearity and time-shifting operations
N
∑ ak e − jkωY (e j ω ) =
k =0
where
X (e
jω
F
) ←→ x [n ]
M
∑ bk e − jkω X (e jω )
k =0
and
F
Y (e jω ) ←→
y [n ]
Tik-61.140 / Chapter 5
 Olli Simula
39
• Consider the first order recursive or infinite impulse response
(IIR) filter
y[ n] − ay[n − 1] = x[ n] , with | a |< 1
 Olli Simula
1
1− ae − j ω
Chapter 5 / O. Simula
Tik-61.140 / Chapter 5
=
− jk ω
)
• The frequency response, H(ejω), is a ratio of polynomials in
the variable e-jω
• The coefficients ak and bk are the same as in the difference
equation
 Olli Simula
Tik-61.140 / Chapter 5
40
y[n] −
• The frequency response is
H (e j ω ) =
• Partial fraction expansion
parallel interconnection
of two 1 st order sections
h[ n] = a nu[ n]
• The impulse response is
41
3
1
y[n − 1] + y[n − 2] = 2x[n]
4
8
• The difference equation is
• Factoring the denominator
cascade of two 1 st order
sections
• The frequency response of this system is
• The impulse response is calculated earlier:
X (e
jω
∑k = 0 bk e
N
∑k = 0a k e − jkω
M
Y (e jω )
Example 5.19: Second Order IIR Filter
Example 5.18: First Order IIR Filter
H (e jω ) =
38
Systems Characterized by Linear
Constant -Coefficient Difference Equations
• An LTI system with input x[n] and output y[n] is described by
N
Tik-61.140 / Chapter 5
 Olli Simula
 Olli Simula
2
3 − j ω 1 − j 2ω
1− e
+ e
4
8
2
jω
H (e ) =
1 − 1 e − j ω 1 − 1 e − jω 



 2
 4

H (e
jω
)=
4
2
−
1 − jω
1 − jω
1− e
1− e
2
4
n
n
1
1
h[n] = 4  u[n] − 2  u[n]
2
4
Tik-61.140 / Chapter 5
42
7