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Fourier series of periodic discrete-time signals
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Discrete-time signal x(n):
Defined for integer time instants n:
{x(n)} = {. . . , x(−2), x(−1), x(0), x(1), x(2), . . .}
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In analogy with continuous-time signals, discrete-
time signals can be expanded in terms of sinusoidal
components of form Ak cos(ωkn + φk)
−2 0 2 4 6 8 10 12
−1
0
1
cos(2πfn + φ), f = 0.1(Hz), φ = −2πf · 2(rad)
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Consider a periodic discrete-time signal with period
N :
x(n) = x(n + N), all n
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In analogy with continuous-time case the discrete
sinusoidal signals
cos(2πk
Nn), sin(
2πk
Nn), k = 0, 1, 2, . . .
have period N .
It follows that an N -periodic signal {x(n)} can be
expanded in terms of sinusoidal signals
cos(2πk
Nn), sin(
2πk
Nn), k = 0, 1, 2, . . .
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Important difference compared to continuous-time case:
We have
cos(nω) = cos(nω + 2πnl)
= cos (n(ω + 2πl)), all integers l
=⇒Frequencies ω and ω + 2πl give the same discrete-
time sequence
NOTE: works only because time n is an integer!
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Hence we need only include frequency components
cos(2π1N
n), cos(2π2N
n), . . . , cos(2πN − 1
Nn)
sin(2π1N
n), sin(2π2N
n), . . . , sin(2πN − 1
Nn)
and we obtain the Fourier series
x(n) = a0 +N−1∑
k=1
ak cos(k2π
Nn) +
N−1∑
k=1
bk sin(k2π
Nn)
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As before, we can express each frequency component
in terms of complex exponentials:
x(n) =N−1∑
k=−N+1
ckej2πkn/N
As time n is an integer we can eliminate the negative
exponentials:
ej2π(−m)n/N = ej2π(−m)n/N+j2πn = ej2π(N−m)n/N
=⇒Terms with k = −m < 0 can be replaced by a term
with k = N −m > 0
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NOTE:
As c−k = c∗k, we have
c−kej2π(−k)n/N = c∗ke
j2π(N−k)n/N
Fourier series of discrete-time periodic signal
An N -periodic discrete-time signal can be expanded
as
x(n) =N−1∑
k=0
dkej2πkn/N
where dN−k = d∗k
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Here the complex-valued frequency components are
given by
dk =1N
N−1∑n=0
x(n)e−j2πkn/N
Proof:
Multiply expression for x(n) with e−j2πmn/N and sum
over time n over one period:
1N
N−1∑n=0
[e−j2πmn/N ×
(x(n) =
N−1∑
k=0
dkej2πkn/N
)]
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Here the right-hand side becomes
1N
N−1∑
k=0
dk
N−1∑n=0
ej2π(k−m)n/N
which vanishes if m 6= k, because
N−1∑n=0
ej2π(k−m)n/N ={
N, if k −m = 0,±N,±2N, . . .
0, otherwise
Hence
1N
N−1∑
k=0
dk
N−1∑n=0
ej2π(k−m)n/N = dm
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and1N
N−1∑n=0
x(n)e−j2πmn/N = dm
giving the result
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EXAMPLE: Periodic discrete-time square wave with period
N = 4:
x(0) = 1x(1) = 1x(2) = 0x(3) = 1
-5 -4 -3 -2 -1 0 1 2 3 4 5
· · · · · ·
x(n)
n
and x(n) = x(n− 4), all n
Frequency components: 0, 2π4 , 2π·2
4 , 2π·34
(Next frequency 2π·44 = 2π is equivalent to frequency
component 0)
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Fourier series:
dk =14
3∑n=0
x(n)e−j2πkn/4
=14
[1 · e−j2πk·0/4 + 1 · e−j2πk·1/4 + 0 + 1 · e−j2πk·3/4
]
=14
[1 + e−jπk/2 + e−jπk·3/2
]
=14
[1 + 2 cos (πk/2)]
=⇒d0 =
34, d1 =
14, d2 = −1
4, d3 =
14
0 2ππ/2π
3π/2
dk
14
Hence
x(n) =3∑
k=0
dkej2πkn/4
=34
+14ej2π1·n/4 − 1
4ej2π2·n/4 +
14ej2π3·n/4
=34
+14
[cos (πn/2) + j sin (πn/2)]
−14
[cos (πn) + j sin (πn)] +14
[cos (3πn/2) + j sin (3πn/2)]
Note that as x(n) is real the imaginary components should
vanish
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Here we have:
- sin(πn) = 0
- cos(πn) = (−1)n
In addition, frequency components 3π/2 and π/2 are related,
because
- sin(3πn/2)=sin(3πn/2−2πn)=sin(−πn/2)=− sin(πn/2),and
- cos(3πn/2)=cos(3πn/2− 2πn)=cos(−πn/2)=cos(πn/2)
=⇒
x(n) =34
+12
cos(
2π
4n
)− 1
4(−1)n
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NOTE:
In above example the Fourier series coefficients dk were real.
Reason: Recall that
ak cos(2πkn/N)+bk sin(2πkn/N) = ckej2πkn/N+c∗ke−j2πkn/N
where
ck =12
(ak − jbk) , c∗k =12
(ak + jbk)
As x(n) is even, x(−n) = x(n), the sine-terms vanish so that
bk = 0 and hence ck and dk are real.
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Next example illustrates a signal which is not even.
Consider the periodic signal
x̃(0) = 1
x̃(1) = 1
x̃(2) = 1
x̃(3) = 0
and x̃(n) = x̃(n− 4), all n
We see that x̃(n) = x(n− 1), all n.
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Hence
x̃(n) = x(n− 1)
=3∑
k=0
dkej2πk(n−1)/4
=3∑
k=0
dke−j2πk/4ej2πkn/4
=3∑
k=0
d̃kej2πkn/4
where
d̃k = dke−j2πk/4
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or
d̃0 = d0ej·0 =34
d̃1 = d1e−jπ/2 =14· (−j)
d̃2 = d2e−jπ = −14· (−1)
d̃3 = d3e−j3π/2 =14· j
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Fourier series of non-periodic discrete-timesignals
In analogy with the continuous-time case a non-periodic
discrete-time signal consists of a continuum of frequencies
(rather than a discrete set of frequencies)
But recall that
cos(nω) = cos(nω + 2πnl)
= cos (n(ω + 2πl)), all integers l
=⇒Only frequencies up to 2π make sense
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Hence a discrete-time signal {x(n)} can be expanded as
x(n) =12π
∫ 2π
0
X(ω)ejωndω
where it can be shown that the Fourier transform X(ω) is given
by
X(ω) =∞∑
n=−∞x(n)e−jωn
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THE DISCRETE FOURIER TRANSFORM (DFT)
Next let’s introduce the standard definition of the discrete
Fourier transform of a sequence
{x(0), x(1), . . . , x(N − 1)}
of finite length N .
We do not make any assumptions about the signal (periodicity
etc) outside the range 0 ≤ n ≤ N − 1.
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Facts:
- from N signal values x(0), x(1), . . . , x(N − 1) at most N
independent frequency components can be computed
- frequency components ≥ 2π are redundant (identical with
frequencies < 2π
- lower bound on frequency resolution is 2π/N (= frequency
corresponding to period length equal to sequence length N)
Hence it makes sense to determine the frequency components
at the N equidistant frequencies
0,1N
2π,2N
2π, . . . ,N − 1
N2π
i.e., the same frequencies contained in an N -periodic signal
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DISCRETE FOURIER TRANSFORM (DFT):
The DFT {X(k)}N−1k=0 of a sequence {x(k)}N−1
k=0 is
defined as
X(k) =N−1∑n=0
x(n)e−j2π kNn, k = 0, 1, . . . , N−1 (DFT)
and the inverse DFT (IDFT) is given by
x(n) =1N
N−1∑
k=0
X(k)ej2π kNn, n = 0, 1, . . . , N − 1
Observe similarity with Fourier transform of periodic
signal
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EXAMPLE
Show that the Discrete Fourier Transform of the
sequence
{x(n)} = {1, 0, 0, 1}is given by
{X(k)} = {2, 1 + j, 0, 1− j}
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PROPERTIES OF DFT
• Symmetry property.
For a real sequence {x(n)}, the DFT satisfies
X(N − k) = X(k)∗ (complex conjugate)
i.e.,
Re[X(N − k)] = Re[X(k)]
Im[X(N − k)] =−Im[X(k)], k = 1, . . . , N − 1
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Follows from the fact that in the definition of X(N − k),
x(n)e−j2πN−kN n = x(n)ej2π k
N n−j2π
= x(n)ej2π kN n
=(x(n)e−j2π k
N n)∗
• DC component.
For k = 0 we have
X(0) =N−1∑n=0
x(n)e−j2π 0Nn
= x(0) + x(1) + · · ·+ x(N − 1)
Hence X(0) is real if x(n) are real and the
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contribution of X(0) to x(n) is
1N
X(0)e0 =1N
X(0)
i.e., a constant component (usually called the DC
component).
• Oscillating component.For even values of N , we have for k = N/2
X(N/2) =N−1∑n=0
x(n)e−j2πN/2N n =
N−1∑n=0
x(n)e−jπn
= x(0)− x(1) + · · · − x(N − 1)
Hence X(N/2) is real if x(n) are real and the
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contribution of X(N/2) to x(n) is
1N
X(N/2)ej2πN/2N n =
1N
X(N/2) (−1)n
i.e., an oscillating component.
• Parseval’s relation.
The energy of the signal {x(n)} defined as
Px =N−1∑n=0
x(n)2
can be computed in the frequency domain using
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Parseval’s relation,
Px =N−1∑n=0
x(n)2 =1N
N−1∑
k=0
|X(k)|2
It follows that 1N |X(k)|2 can be interpreted as the
contribution to the total energy of the signal from
frequency component 2πk/N .
Note: important in signal compression!
• DFT of a δ function.
The δ function,
δ(n) ={
1, n = 00, n 6= 0
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has DFT X(k) = 1, k = 0, 1, . . . , N − 1.
Hence it consists of equal amounts of all
frequencies!
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DIRECT COMPUTATION OF DFT SUM ISNOT ADVISABLE IN PRACTICE:
- HIGH COMPUTATIONAL BURDEN:
Computation of sequence X(k) requires N 2
complex additions and multiplications (a lot for
long sequences)
- Numerical round-off errors add up when evaluation
long sum
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EFFICIENT ALGORITHM FOR SOLUTION:Fast Fourier Transform (FFT)
- FFT: N log2 N complex additions and
(N/2) log2 N complex multiplications (compare
N 2 using direct evaluation)
- Each element X(k) is obtained by only log2 N
complex additions (compare with N using direct
evaluation) ⇒ significantly smaller round-off errors
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FFT uses the fact that the discrete Fourier transformX(k) of a sequence {x(0), x(1), . . . , x(N − 1)} withN even can be determined as
X(k) = X11(k) + W kNX12(k), k = 0, 1, . . . , N/2− 1
X(k + N/2) = X11(k)−W kNX12(k), k = 0, 1, . . . , N/2− 1
where WN = e−j2π/N ,
{X11(k)} is the Fourier transform of the ’even
subsequence’ {x(0), x(2), . . . , x(N − 2)}, and
X12(k)} is the Fourier transform of the ’odd
subsequence’ {x(1), x(3), . . . , x(N − 1)}.
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INTERPRETATION OF FOURIER TRANSFORMCOMPONENTS
0
X(0)
X(1)
1
· · ·
X(k)
k
· · ·
X(N/2)
N/2
· · ·
X(N-k)=X(k)∗
N-k
· · ·X(N-1)=X(1)∗
N-1 N
Normalized frequency 0 1N
kN
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N−kN
N−1N 1 (Hz)
Actual frequency
Sampling frequency fs = 1/Ts0 1
Nfs
kN
fs12fs
N−kN
fsN−1
Nfs fs (Hz)
Matlab indexing X(1) X(2) X(k+1) X(N/2+1) X(N-k+1) X(N)
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