Post on 31-Dec-2015
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The Fourier Series for Discrete-Time Signals
Suppose that we are given a periodic sequence Suppose that we are given a periodic sequence with period N. The Fourier series with period N. The Fourier series representation for x[n] consists of N representation for x[n] consists of N harmonically related exponential functionsharmonically related exponential functionseej2j2kn/Nkn/N, k = 0, 1,2,…….,N-1, k = 0, 1,2,…….,N-1and is expressed as and is expressed as
1N
0k
N/kn2jkec]n[x
where the coefficients ck can be computed as:
0n
N/kn2jk e]n[x
N
1c
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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]Solution: (a) x[n] = [1, 1, 0, 0]
3
0n
N/kn2j1N
0n
N/kn2jk e]n[x
4
1e]n[x
N
1c
Now 2
10011
4
1]3[x]2[x]1[x]0[x
4
1]n[x
4
1c
3
0n0
00e]1[x]0[x4
1e]n[x
4
1e]n[x
4
1c 2/j
3
0n
2/nj3
0n
4/n2j1
j14
1j01
4
1sinjcos11
4
122
3
j3
0n
nj3
0n
4/n22j2 e.11
4
1e]n[x
4
1e]n[x
4
1c
0sinjcos14
1
j14
1j01
4
1)2/3sin(j2/3cos1
4
1e]n[x
4
1c
3
0n
4/3n2j3
The magnitude spectra are:
2
1c0
4
2c1 0c2
4
2c3
and the phase spectra are:
00 41
undefined2 43
4
(b) x[n] = cosn/3
Solution: In this case, fSolution: In this case, f00 = 1/6 and hence x[n] is = 1/6 and hence x[n] is periodic with fundamental period N = 6.periodic with fundamental period N = 6.
NowNow
5
0n
3/knj6/kn2j5
0n
5
0n
N/kn2jk e
3
ncos
6
1e
3
ncos
6
1e]n[x
6
1c
5
0n
k1jk1j3/knj3/nj3/nj5
0n
3n
3n
ee12
1eee
2
1
6
1
0coscoscoscoscos0cos6
1
3
ncos
6
1
3
ncos2
12
1c
35
34
33
32
3
5
0n
5
0n0
Similarly, c2 = c3 = c4 = 0, c1 = c5 = ½.
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(c) Cos(2)n
Solution: The frequency fSolution: The frequency f00 of the signal is of the signal is
1/1/2 Hz. Since f2 Hz. Since f00 is not a rational number, is not a rational number,
the signal is not periodic. Cosequently, this the signal is not periodic. Cosequently, this signal cannot be expanded in a Fourier signal cannot be expanded in a Fourier series.series.
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Power density Spectrum of Periodic Signals
The average power of a discrete time periodic signal with period N is The average power of a discrete time periodic signal with period N is
1N
0n
2
x )n(xN
1P
The above relation may also be written as
1N
0n
1N
0n
N/kn2*k
1N
0n
*x ec]n[x
N
1]n[x]n[x
N
1P
or
1N
0n
221N
0kk
1N
0n
N/kn2j1N
0n
*kx
]n[xN
1c
e]n[xN
1cP
This is Parseval’s Theorem for Discrete-Time Power Signals.
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Similarly, for discrete time energy signals, the Parseval’s Theorem may be stated as Similarly, for discrete time energy signals, the Parseval’s Theorem may be stated as follows:follows:
21N
0kk
21N
0nx cN]n[xE
If the signal x[n] is real, [i.e. x*[n] = x[n]], then we can easily show that
|c-k| = |ck| (even symmetry)-c-k = ck (odd symmetry)|ck| = |cN-k| (Periodicity)ck = cN-k (periodicity)
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More specifically, we haveMore specifically, we have
|c|c00| = |c| = |cNN| | cc00 = - = - ccNN
|c|c11| = |c| = |cN-1N-1|| cc11 = - = - ccN-1N-1
|c|cN/2N/2| = |c| = |cN/2N/2| | ccN/2N/2 = 0 if N is even = 0 if N is even
|c|c(N-1)/2(N-1)/2| = |c| = |c(N+1)/2(N+1)/2| | cc(N-1)/2(N-1)/2 = = (N+1)/2(N+1)/2 if N if N
is is odd odd
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Example: Determine the Fourier Series Coefficients and the Power Density Spectrum of the following periodic signal.
Solution:Solution:
-N L N
AX[n]
n
1N
0n
N/kn2j1N
0n
N/kn2jk Ae
N
1e]n[x
N
1c
k = 0, 1, 2, …., N-1
1N,...,2,1k,
0k,e
N
Ac
N/k2j
N/kL2j
e1e1
NA
NALn1L
0n
N/k2jk
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But But
)N/ksin(
)N/kLsin(e
ee
ee
e
e
e1
e1
N/)1L(kj
N/kjN/kj
N/kLjN/kLj
N/kj
N/kLj
N/k2j
N/kL2j
Therefore,
otherwiseN/ksin
N/kLsin
0kc 2
2
NA
2
NAL
2
k
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The Fourier Transform of Discrete-Time Aperiodic Signals
The Fourier Transform of a finite energy discrete The Fourier Transform of a finite energy discrete time signal x[n] is defined astime signal x[n] is defined as
n
jwne]n[x)w(X
X(w) may be regarded as a decomposition of x[n] into its Frequency components. It is not difficult toVerify that X(w) is periodic with frequency 2.
The Inverse Fourier Transform of X(w) may be defined as
dwe)w(X2
1]n[x
2
jwn
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Energy Density Spectrum of Aperiodic Signals
Energy of a discrete time signal x[n] is defined asEnergy of a discrete time signal x[n] is defined as2
nx ]n[xE
Let us now express the energy Ex in terms of the spectral characteristic X(w). First we have
n n
jwn*x dwe)w(X
2
1]n[x]n[x]n[xE
If we interchange the order of integration and summation in the above equation, we obtain
dw)w(X2
1dwe]n[x)w(X
2
1E
2
n
jwnx
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Therefore, the energy relation between x[n] and Therefore, the energy relation between x[n] and X(w) isX(w) is
dw)w(X2
1]n[xE
22
nx
This is Parseval’s relation for discrete-time aperiodic signals.
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Example: Determine and sketch the energy density spectrum of the signal x[n] = anu[n], -1<a<1
Solution: Solution:
0njw
njwjwn
0n
n
n
jwn
ae1
1aeeae]n[x)w(X
The energy density spectrum (ESD) is given by
jwjw
2
xx ae1ae1
1)w(X)w(X)w(X)w(S
2awcosa21
1
0 w
X(w)
a = 0.5 a= -0.5
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Example: Determine the Fourier Transform and the energy density spectrum of the sequence
Solution:Solution:
otherwise,0
1Ln0,A]n[x
)2/wsin(
)2/wLsin(Ae
e1
e1AAee]n[x)w(X )1L)(2/w(j
jw
jwL1L
0
jwn
n
jwn
The magnitude of x[n] is
otherwise,A
0w,LA)w(X
)2/wsin()2/wLsin(
and the phase spectrum is
)2/wsin()2/wLsin(
2w )2L(A)w(X
The signal x[n] and its magnitude is plotted on the next slide. The Phase spectrum is left as an exercise.
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x[n]
|X(w)|
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Properties of Discrete Time Fourier Transform (DTFT)
Symmetry Properties:Symmetry Properties: Suppose that both the signal x[n] and its Suppose that both the signal x[n] and its transform X(w) are complex valued. Thentransform X(w) are complex valued. Then
x[n] = xx[n] = xRR[n] + jx[n] + jxII[n] (1)[n] (1)
X(w) = XX(w) = XRR(w) + jX(w) + jXII[w] (2)[w] (2)
n
jwne]n[x)w(X
Substitution of (1) and (2) gives
wnsinjwncos]n[x]n[xe]n[x]n[x)w(jX)w(Xn
IRjwn
nIRIR
Separating the real and imaginary parts, we have