Transcript File
The Fourier Series for DiscreteTime Signals
Suppose that we are given a periodic sequence
with
period
N.
The
Fourier
series
representation for x[n] consists of N
harmonically related exponential functions
ej2kn/N, k = 0, 1,2,…….,N-1
and is expressed as
N 1
x[ n ]
cke
j 2 kn / N
k 0
where the coefficients ck can be computed as:
ck
1
N
x [ n ]e
j 2 kn / N
n0
1
Example 2: Determine the spectra of the
following signals:
(a) x[n] = [1, 1, 0, 0], x[n] is periodic with period 4
(b) x[n] = cosn/3
(c) x[n] = cos(2)n
Solution: (a) x[n] = [1, 1, 0, 0]
ck
1
N
Now
c1
1
N 1
x[n ]e
c0
1
4
x [ n ]e
4
1 1 cos
x[ n ]
1
4
n0
j2 n / 4
3
x[n ]e
1
1
4
2
x[ 0 ] x[1 ] x[ 2 ] x[ 3 ] 1 1 0 0
1
3
4
x [ n ]e
j sin
2
j n / 2
1
4
n0
2
j 2 kn / N
4 n0
3
n0
1
n0
3
4
j 2 kn / N
1
x[ 0 ] x[1 ]e
j / 2
00
1 0 j 1 j
1
1
4
4
2
c2
1
4
3
x [ n ]e
j2 2 n / 4
1
4
n0
3
c3
3
x[ n ]e
4
j2 n 3 / 4
1
4
n0
1
1
4
n0
4
1
x [ n ]e
j n
1 cos
1 1 .e
j
j sin 0
1
1
4
4
1 cos 3 / 2 j sin( 3 / 2 ) 1 0 j 1 j
The magnitude spectra are:
c0
1
2
c1
2
4
c2 0
c3
2
4
and the phase spectra are:
0 0
1
4
2 undefined
3
4
3
(b) x[n] = cosn/3
Solution: In this case, f0 = 1/6 and hence x[n] is
periodic with fundamental period N = 6.
Now
ck
5
1
6
x [ n ]e
j 2 kn / N
n0
e
6
2
1
5
1
1
5
6
cos
n
e
j n / 3
e
5
1
6
cos
n
j kn / 3
e
12
1
5
e
j kn / 3
3
n0
n0
j 3n 1 k
e
j 3n 1 k
n0
c0
1
5
2 cos
12
n0
3
n0
j n / 3
e
j 2 kn / 6
1
6
cos 0 cos
3
n
3
cos
1
5
n
n0
3
cos
6
2
3
cos
3
3
cos
4
3
Similarly, c2 = c3 = c4 = 0, c1 = c5 = ½.
cos
5
3
0
4
(c) Cos(2)n
Solution: The frequency f0 of the signal is
1/2 Hz. Since f0 is not a rational number,
the signal is not periodic. Cosequently, this
signal cannot be expanded in a Fourier
series.
5
Power density Spectrum of Periodic Signals
The average power of a discrete time periodic
signal with period N is
1
Px
N
N 1
x(n )
2
n0
The above relation may also be written as
N 1
*
* 2 kn / N
Px
x
[
n
]
x
[
n
]
x
[
n
]
c
ke
N n0
N n0
n0
1
or P
N 1
1
N 1
x
c
n0
k 0
1
N
2
N 1
*
k
ck
1
N
N 1
N 1
x [ n ]e
j 2 kn / N
n0
N 1
x[ n ]
2
n0
This is Parseval’s Theorem for Discrete-Time Power
Signals.
6
Similarly, for discrete time energy signals, the
Parseval’s Theorem may be stated as follows:
2
N 1
Ex
n0
x[ n ]
N 1
2
N ck
k 0
If the signal x[n] is real, [i.e. x*[n] = x[n]], then we
can easily show that
|c-k| = |ck|
-c-k = ck
|ck| = |cN-k|
ck = cN-k
(even symmetry)
(odd symmetry)
(Periodicity)
(periodicity)
7
More specifically, we have
|c0| = |cN|
c0 = - cN
|c1| = |cN-1|
c1 = - cN-1
|cN/2| = |cN/2|
cN/2 = 0 if N is even
|c(N-1)/2| = |c(N+1)/2| c(N-1)/2 = (N+1)/2 if N
is odd
8
Example: Determine the Fourier Series
Coefficients and the Power Density Spectrum of
the following periodic signal.
X[n]
A
n
-N
N
L
Solution:
ck
1
N 1
N
x[ n ]e
j 2 kn / N
1
N 1
N
n0
Ae
j 2 kn / N
n0
k = 0, 1, 2, …., N-1
ck
L 1
e
N
A
n0
n
j2 k / N
A
N
AL
N
1 e
,
j 2 kL / N
1 e
j 2 k / N
k0
, k 1 , 2 ,..., N 1
9
But
1e
j 2 kL / N
1e
e
j2 k / N
j k ( L 1 ) / N
e j kL / N
j k / N
e
e j kL / N e j kL / N
j k / N
j k / N
e
e
sin( kL / N )
sin( k / N )
Therefore,
ck
2
AL 2
N
2
sin kL / N
A 2
N
sin k / N
k 0
otherwise
10
The Fourier Transform of Discrete-Time
Aperiodic Signals
The Fourier Transform of a finite energy discrete
time signal x[n] is defined as
X(w )
x[ n ]e
jwn
n
X(w) may be regarded as a decomposition of x[n]
into its Frequency components. It is not difficult to
Verify that X(w) is periodic with frequency 2.
The Inverse Fourier Transform of X(w) may be
defined as
x[ n ]
1
2
2
X ( w )e
jwn
dw
11
Energy Density Spectrum of Aperiodic
Signals
Energy of a discrete time signal x[n] is defined as
2
Ex
x[ n ]
n
Let us now express the energy Ex in terms of the
spectral characteristic X(w). First we have
1
E x x [ n ]x [ n ] x [ n ]
2
n
n
*
X ( w )e
jwn
dw
If we interchange the order of integration and
summation in the above equation, we obtain
jwn
Ex
X
(
w
)
x
[
n
]
e
2
n
1
1
dw
2
2
X ( w ) dw
12
Therefore, the energy relation between x[n] and
X(w) is
Ex
n
2
x[ n ]
1
2
2
X ( w ) dw
This is Parseval’s relation for discrete-time
aperiodic signals.
13
Example: Determine and sketch the energy
density spectrum of the signal x[n] = anu[n],
-1<a<1
Solution:
X(w )
x [ n ]e
jwn
n
n
a e
jwn
ae
n0
jw n
n0
1
1 ae
jw
The energy density spectrum (ESD) is given by
2
1
S xx ( w ) X ( w ) X ( w ) X ( w )
1 ae 1 ae
X(w)
1
jw
jw
1 2 a cos w a
a = 0.5
2
a= -0.5
w
0
14
Example: Determine the Fourier
Transform and the energy density
spectrum of the sequence
A,
x[ n ]
0,
0 n L 1
otherwise
Solution:
X(w )
x[ n ]e
jwn
L 1
n
Ae
jwn
A
0
1e
jwL
1e
jw
Ae
j ( w / 2 )( L 1 )
sin( wL / 2 )
sin( w / 2 )
The magnitude of x[n] is
X(w )
A
w 0
A L,
sin( wL / 2 )
sin( w / 2 )
,
otherwise
and the phase spectrum is
X (w ) A (L 2)
w
2
sin( wL / 2 )
sin( w / 2 )
The signal x[n] and its magnitude is plotted on the next slide. 15
The
Phase spectrum is left as an exercise.
x[n]
|X(w)|
16
Properties of Discrete Time Fourier
Transform (DTFT)
Symmetry Properties: Suppose that both the
signal x[n] and its transform X(w) are complex
valued. Then
x[n] = xR[n] + jxI[n]
(1)
X(w) = XR(w) + jXI[w]
(2)
X(w )
x [ n ]e
jwn
n
Substitution of (1) and (2) gives
X R ( w ) jX I ( w )
x
n
R
[ n ] x I [ n ]e
jwn
x
R
[ n ] x I [ n ]cos wn j sin wn
n
Separating the real and imaginary parts, we have
17