Digital Signal Processing
Frequency Analysis of Signals (1)
Moslem Amiri, V´aclav Pˇrenosil
Embedded Systems Laboratory
Faculty of Informatics, Masaryk University
Brno, Czech Republic
amiri@mail.muni.cz
prenosil@fi.muni.cz
April, 2012
Frequency Analysis of Continuous-Time Signals
Frequency analysis of a signal is resolution of signal into its frequency
(sinusoidal) components
For class of periodic signals, such a decomposition is called a Fourier
series
For class of finite energy signals, the decomposition is called Fourier
transform
The term spectrum is used when referring to frequency content of a
signal
Different signal waveforms have different spectra
Thus spectrum provides an identity or a signature for a signal (no other
signal has the same spectrum)
Process of obtaining spectrum of a signal using basic mathematical
tools is frequency or spectral analysis
In contrast, process of determining spectrum of a signal in practice,
based on actual measurements of signal, is called spectrum estimation
In a practical problem, the signal is some information-bearing signal
which does not lend itself to an exact mathematical description
Recombination of sinusoidal components to reconstruct original signal
is a Fourier synthesis problem
2 / 52
The Fourier Series for Continuous-Time Periodic Signals
Examples of periodic signals are square waves, rectangular waves,
triangular waves, sinusoids and complex exponentials
Basic mathematical representation of periodic signals is Fourier series
Fourier series is a linear weighted sum of harmonically related sinusoids
or complex exponentials
A linear combination of harmonically related complex exponentials of
form
x(t) =
∞
k=−∞
ckej2πkF0t
(1)
is a periodic signal with fundamental period Tp = 1/F0
This is called a Fourier series
3 / 52
The Fourier Series for Continuous-Time Periodic Signals
Given a periodic signal x(t) with period Tp, it can be represented by
Fourier series
Fundamental frequency F0 is reciprocal of given period Tp
To determine expression for {ck }, first multiply both sides of (1) by
e−j2πF0lt
(l is an integer)
and then integrate both sides of resulting equation from t0 to t0 + Tp
t0+Tp
t0
x(t)e−j2πlF0t
dt =
t0+Tp
t0
e−j2πlF0t
∞
k=−∞
ck ej2πkF0t
dt
=
∞
k=−∞
ck
t0+Tp
t0
ej2πF0(k−l)t
dt (2)
=
∞
k=−∞
ck
ej2πF0(k−l)t
j2πF0(k − l)
t0+Tp
t0
(3)
For k = l, (3) yields zero
4 / 52
The Fourier Series for Continuous-Time Periodic Signals
If k = l, from equation (2)
t0+Tp
t0
dt = t
t0+Tp
t0
= Tp
Consequently
t0+Tp
t0
x(t)e−j2πlF0t
dt = cl Tp
cl =
1
Tp
t0+Tp
t0
x(t)e−j2πlF0t
dt =
1
Tp Tp
x(t)e−j2πlF0t
dt
Table 1: Frequency analysis of continuous-time periodic signals
Synthesis equation x(t) = ∞
k=−∞ ckej2πkF0t
Analysis equation ck = 1
Tp Tp
x(t)e−j2πkF0tdt
5 / 52
The Fourier Series for Continuous-Time Periodic Signals
Fourier coefficients ck are complex valued
If periodic signal is real, ck and c−k are complex conjugates
ck = |ck |ejθk
−→ c−k = |ck |e−jθk
Consequently, Fourier series may also be represented as
x(t) = c0 + 2
∞
k=1
|ck | cos(2πkF0t + θk )
where c0 is real when x(t) is real
Expanding cosine function in equation above
cos(2πkF0t + θk ) = cos 2πkF0t cos θk − sin 2πkF0t sin θk
x(t) = a0 +
∞
k=1
(ak cos 2πkF0t − bk sin 2πkF0t)
where
a0 = c0
ak = 2|ck | cos θk
bk = 2|ck | sin θk
6 / 52
Power Density Spectrum of Periodic Signals
A periodic signal has infinite energy and a finite average power, given
as
Px =
1
Tp Tp
|x(t)|2
dt
x(t) =
∞
k=−∞
ckej2πkF0t
−→ Px =
1
Tp Tp
x(t)
∞
k=−∞
c∗
k e−j2πkF0t
dt
=
∞
k=−∞
c∗
k
1
Tp Tp
x(t)e−j2πkF0t
dt
=
∞
k=−∞
|ck|2
The established relation is called Parseval’s relation for power signals
Px =
1
Tp Tp
|x(t)|2
d(t) =
∞
k=−∞
|ck|2
7 / 52
Power Density Spectrum of Periodic Signals
Example
Suppose x(t) consists of a single complex exponential
x(t) = ckej2πkF0t
In this case, all Fourier series coefficients except ck are zero
Px =
∞
k=−∞
|ck|2
−→ Px = |ck|2
|ck |2
represents power in kth harmonic component of signal
Hence total average power in periodic signal is simply sum of average
powers in all harmonics
8 / 52
Power Density Spectrum of Periodic Signals
Plotting |ck|2 as a function of frequencies kF0, k = 0, ±1, ±2, . . .,
obtained diagram is called power density spectrum or power
spectrum of periodic signal x(t)
Figure 1: Power density spectrum of a continuous-time periodic signal.
Since power in a periodic signal exists only at discrete values of
frequencies, the signal is said to have a line spectrum
9 / 52
Power Density Spectrum of Periodic Signals
Since Fourier series coefficients {ck} are complex valued, i.e.,
ck = |ck|ejθk where θk = ck
instead of plotting power density spectrum, we can plot magnitude
spectrum {|ck|} and phase spectrum {θk} as a function of frequency
If periodic signal is real valued, then
c−k = c∗
k −→ |c−k|2 = |c∗
k |2 = |ck|2
Power spectrum is an even function
Magnitude spectrum is an even function
Phase spectrum is an odd function
Hence it is sufficient to specify spectrum for positive frequencies only
Total average power
Px = c2
0 + 2
∞
k=1
|ck |2
= a2
0 +
1
2
∞
k=1
(a2
k + b2
k )
a0 = c0
ak = 2|ck | cos θk
bk = 2|ck | sin θk
10 / 52
Power Density Spectrum of Periodic Signals
Example
Determine Fourier series and power density spectrum of following
rectangular pulse train signal
Figure 2: Continuous-time periodic train of rectangular pulses.
Since x(t) is an even signal, it is convenient to select integration
interval from −Tp/2 to Tp/2
ck =
1
Tp Tp
x(t)e−j2πkF0t
dt → c0 =
1
Tp
Tp/2
−Tp/2
x(t)dt =
1
Tp
τ/2
−τ/2
Adt =
Aτ
Tp
11 / 52
Power Density Spectrum of Periodic Signals
Example (continued)
c0 represents average value (dc component) of x(t)
For k = 0
ck =
1
Tp
τ/2
−τ/2
Ae−j2πkF0t
dt =
A
Tp
e−j2πF0kt
−j2πkF0
τ/2
−τ/2
=
A
πF0kTp
ejπkF0τ − e−jπkF0τ
j2
=
Aτ
Tp
sin πkF0τ
πkF0τ
, k = ±1, ±2, . . . (4)
(4) has the from (sin φ)/φ, where φ = πkF0τ
φ takes on discrete values since F0 and τ are fixed and k varies
However, plot of (sin φ)/φ with φ as a continuous parameter is shown in
Fig. 3
12 / 52
Power Density Spectrum of Periodic Signals
Example (continued)
Figure 3: The function (sin φ)/φ.
Since x(t) is even, Fourier coefficients {ck} are real
Phase spectrum is either zero, when ck is positive, or π when ck is
negative
Instead of plotting magnitude and phase spectra separately, we may plot
{ck } on a single graph
13 / 52
Power Density Spectrum of Periodic Signals
Example (continued)
When Tp is fixed and pulse width τ is allowed to vary
Tp = 0.25 seconds −→ F0 = 1/Tp = 4 Hz
Spacing between adjacent spectral lines is F0 = 4 Hz, independent of τ
Figure 4: Fourier coefficients of the rectangular pulse train when Tp is
fixed and the pulse width τ varies.
14 / 52
Power Density Spectrum of Periodic Signals
Example (continued)
If τ is fixed and Tp varies when Tp > τ
Spacing between adjacent spectral lines decreases as Tp increases
Figure 5: Fourier coefficients of a rectangular pulse train with fixed
pulse width τ and varying period Tp.
15 / 52
Power Density Spectrum of Periodic Signals
Example (continued)
ck =
Aτ
Tp
sin πkF0τ
πkF0τ
, k = ±1, ±2, . . .
If k = 0 and sin(πkF0τ) = 0, then ck = 0
Harmonics with zero power occur at frequencies kF0 such that
π(kF0)τ = mπ, m = ±1, ±2, . . .
Power density spectrum for rectangular pulse train
|ck|2
=



Aτ
Tp
2
, k = 0
Aτ
Tp
2
sin πkF0τ
πkF0τ
2
, k = ±1, ±2, . . .
16 / 52
Fourier Transform for Continuous-Time Aperiodic Signals
Periodic signals possess line spectra with equidistant lines
Line spacing is equal to fundamental frequency
Fundamental period provides number of lines per unit of frequency (line
density), as shown in Fig. 5
Allowing period to increase without limit, line spacing tends toward
zero
When period becomes infinite, signal becomes aperiodic and its
spectrum becomes continuous
Spectrum of an aperiodic signal is envelope of line spectrum in
corresponding periodic signal obtained by repeating the aperiodic signal
with some period Tp
17 / 52
Fourier Transform for Continuous-Time Aperiodic Signals
Consider an aperiodic signal x(t) with finite duration
We can create a periodic signal xp(t) with period Tp
x(t) = lim
Tp→∞
xp(t)
Figure 6: (a) Aperiodic signal x(t) and (b) periodic signal xp(t)
constructed by repeating x(t) with a period Tp.
18 / 52
Fourier Transform for Continuous-Time Aperiodic Signals
Fourier series representation of xp(t)
xp(t) =
∞
k=−∞
ck ej2πkF0t
, F0 =
1
Tp
where
ck =
1
Tp
Tp/2
−Tp/2
xp(t)e−j2πkF0t
dt
Since xp(t) = x(t) for −Tp/2 ≤ t ≤ Tp/2 and x(t) = 0 for |t| > Tp/2
ck =
1
Tp
∞
−∞
x(t)e−j2πkF0t
dt
Defining a function X(F), called Fourier transform of x(t)
X(F) =
∞
−∞
x(t)e−j2πFt
dt
ck =
1
Tp
X(kF0) or Tpck = X(kF0) = X
k
Tp
Fourier coefficients are samples of X(F) taken at multiples of F0 and scaled
by F0 (multiplied by 1/Tp) 19 / 52
Fourier Transform for Continuous-Time Aperiodic Signals
If we substitute for ck in Fourier series representation of xp(t)
xp(t) =
∞
k=−∞
ck ej2πkF0t
, F0 =
1
Tp
Tpck = X(kF0) = X
k
Tp
we obtain
xp(t) =
1
Tp
∞
k=−∞
X
k
Tp
ej2πkF0t
Defining ∆F = 1
Tp
xp(t) =
∞
k=−∞
X(k∆F)ej2πk∆Ft
∆F
lim
Tp→∞
xp(t) = x(t) = lim
∆F→0
∞
k=−∞
X(k∆F)ej2πk∆Ft
∆F
∆F→dF
−−−−−→
k∆F→F
x(t) =
∞
−∞
X(F)ej2πFt
dF (inverse Fourier transform)
20 / 52
Fourier Transform for Continuous-Time Aperiodic Signals
Table 2: Frequency analysis of continuous-time aperiodic signals
Synthesis equation (inverse transform) x(t) =
∞
−∞ X(F)ej2πFtdF
Analysis equation (direct transform) X(F) =
∞
−∞ x(t)e−j2πFtdt
Above Fourier transform pair can be expressed in terms of Ω = 2πF
Since dF = dΩ/2π
x(t) =
1
2π
∞
−∞
X(Ω)ejΩt
dΩ
X(Ω) =
∞
−∞
x(t)e−jΩt
dt
21 / 52
Energy Density Spectrum of Aperiodic Signals
Let x(t) be any finite energy signal with Fourier transform X(F)
Ex =
∞
−∞
|x(t)|2
dt
=
∞
−∞
x(t)x∗
(t)dt
=
∞
−∞
x(t)dt
∞
−∞
X∗
(F)e−j2πFt
dF
=
∞
−∞
X∗
(F)dF
∞
−∞
x(t)e−j2πFt
dt
=
∞
−∞
|X(F)|2
dF
Parseval’s relation for aperiodic, finite energy signals
Ex =
∞
−∞
|x(t)|2
dt =
∞
−∞
|X(F)|2
dF
22 / 52
Energy Density Spectrum of Aperiodic Signals
Spectrum X(F) of a signal is complex valued
X(F) = |X(F)|ejΘ(F)
where |X(F)| is magnitude spectrum and Θ(F) is phase spectrum
Θ(F) = X(F)
Energy density spectrum of x(t)
Sxx (F) = |X(F)|2
Sxx (F) is real and nonnegative, and does not contain any phase
information
It is impossible to reconstruct signal given Sxx (F)
If signal x(t) is real, then
|X(−F)| = |X(F)|
X(−F) = − X(F)
Energy density spectrum of a real signal has even symmetry
Sxx (−F) = Sxx (F)
23 / 52
Energy Density Spectrum of Aperiodic Signals
Example
Determine Fourier transform and energy density spectrum of
x(t) =
A, |t| ≤ τ/2
0, |t| > τ/2
Figure 7: Rectangular pulse.
This signal is aperiodic
X(F) =
∞
−∞
x(t)e−j2πFt
dt → X(F) =
τ/2
−τ/2
Ae−j2πFt
dt = Aτ
sin πFτ
πFτ
24 / 52
Energy Density Spectrum of Aperiodic Signals
Example (continued)
X(F) is real and hence can be depicted using only one diagram
Figure 8: Fourier transform of rectangular pulse.
Zero crossings of X(F) occur at multiples of 1/τ
Width of main lobe, which contains most of signal energy, is 2/τ
As τ decreases (increases), main lobe becomes broader (narrower) and
more energy is moved to higher (lower) frequencies
25 / 52
Energy Density Spectrum of Aperiodic Signals
Example (continued)
Figure 9: Fourier transform of a rectangular pulse for various width values.
26 / 52
Energy Density Spectrum of Aperiodic Signals
Example (continued)
As shown in Fig. 9, as signal pulse is expanded (compressed) in time,
its transform is compressed (expanded) in frequency
Energy density spectrum of rectangular pulse
Sxx (F) = (Aτ)2 sin πFτ
πFτ
2
27 / 52
Frequency Analysis of Discrete-Time Signals
Fourier series representation of a continuous-time periodic signal can
consist of an infinite number of frequency components
Frequency spacing between two successive harmonically related
frequencies is 1/Tp (Tp is fundamental period)
Since frequency range for continuous-time signals extends from −∞ to
∞, it is possible to have signals that contain an infinite number of
frequency components
In contrast, frequency range for discrete-time signals is unique over
interval (−π, π) or (0, 2π)
A discrete-time signal of fundamental period N can consist of
frequency components separated by 2π/N radians or f = 1/N cycles
Consequently, Fourier series representation of discrete-time periodic
signal contains at most N frequency components
28 / 52
The Fourier Series for Discrete-Time Periodic Signals
Given a periodic sequence x(n) with period N (i.e., x(n) = x(n + N)
for all n), Fourier series representation for x(n) consists of N
harmonically related exponential functions
ej2πkn/N, k = 0, 1, . . . , N − 1
and is expressed as
x(n) =
N−1
k=0
ckej2πkn/N
(5)
where {ck} are coefficients in series representation
29 / 52
The Fourier Series for Discrete-Time Periodic Signals
Multiplying both sides of (5) by e−j2πln/N and summing from n = 0 to
n = N − 1
N−1
n=0
x(n)e−j2πln/N
=
N−1
n=0
N−1
k=0
ckej2π(k−l)n/N
(6)
N−1
n=0
ej2π(k−l)n/N
=
N, k − l = 0, ±N, ±2N, . . .
0, otherwise
Right-hand side of (6) reduces to Ncl and hence
cl =
1
N
N−1
n=0
x(n)e−j2πln/N
, l = 0, 1, . . . , N − 1
30 / 52
The Fourier Series for Discrete-Time Periodic Signals
Table 3: Frequency analysis of discrete-time periodic signals
Synthesis equation x(n) = N−1
k=0 ckej2πkn/N
Analysis equation ck = 1
N
N−1
n=0 x(n)e−j2πkn/N
The synthesis equation above is often called discrete-time Fourier
series (DTFS)
From Analysis equation above, which holds for every value of k, we
have
ck+N =
1
N
N−1
n=0
x(n)e−j2π(k+N)n/N
=
1
N
N−1
n=0
x(n)e−j2πkn/N
= ck
Spectrum of a signal x(n), which is periodic with period N, is a periodic
sequence with period N
Any N consecutive samples of signal or its spectrum provide a complete
description of signal in time or frequency domains
31 / 52
The Fourier Series for Discrete-Time Periodic Signals
Example
Determine spectrum of signal
x(n) = cos
√
2πn
Since ω0 =
√
2π −→ f0 = 1/
√
2
f0 is not a rational number −→ signal is not periodic −→ signal cannot
be expanded in a Fourier series
Nevertheless, signal possesses a spectrum consisting of single frequency
component at ω = ω0 =
√
2π
32 / 52
The Fourier Series for Discrete-Time Periodic Signals
Example
Determine spectrum of signal
x(n) = cos πn/3
f0 = 1
6 −→ x(n) is periodic with fundamental period N = 6
ck =
1
N
N−1
n=0
x(n)e−j2πkn/N
→ ck =
1
6
5
n=0
x(n)e−j2πkn/6
, k = 0, 1, . . . , 5
x(n) = cos
πn
3
= cos
2πn
6
=
1
2
ej2πn/6
+
1
2
e−j2πn/6
Comparing x(n) with synthesis equation, x(n) = N−1
k=0 ckej2πkn/N
ej2πn/6 → k = 1 → c1 = 1
2
e−j2πn/6 (k = −1) → e−j2πn/6 = ej2π(5−6)n/6 = ej2π(5n)/6 → k = 5 →
c5 = 1
2
c0 = c2 = c3 = c4 = 0, c1 = 1
2, c5 = 1
2
33 / 52
The Fourier Series for Discrete-Time Periodic Signals
Example (continued)
Figure 10: Spectrum of the periodic signal discussed in Example.
34 / 52
The Fourier Series for Discrete-Time Periodic Signals
Example
Determine spectrum of signal
x(n) = {1
↑
, 1, 0, 0}
where x(n) is periodic with period N = 4
From analysis equation
ck =
1
N
N−1
n=0
x(n)e−j2πkn/N
→ ck =
1
4
3
n=0
x(n)e−j2πkn/4
, k = 0, 1, 2, 3
ck = 1
4(1 + e−jπk/2), k = 0, 1, 2, 3
c0 = 1
2, c1 = 1
4(1 − j), c2 = 0, c3 = 1
4(1 + j)
Magnitude and phase spectra are
|c0| = 1
2, |c1| =
√
2
4 , |c2| = 0, |c3| =
√
2
4
c0 = 0, c1 = −π
4 , c2 = undefined, c3 = π
4
35 / 52
The Fourier Series for Discrete-Time Periodic Signals
Example (continued)
Figure 11: Spectra of the periodic signal discussed in Example.
36 / 52
Power Density Spectrum of Periodic Signals
Average power of a discrete-time periodic signal with period N
Px =
1
N
N−1
n=0
|x(n)|2
=
1
N
N−1
n=0
x(n)x∗
(n)
x(n) =
N−1
k=0
ck ej2πkn/N
→ Px =
1
N
N−1
n=0
x(n)
N−1
k=0
c∗
k e−j2πkn/N
=
N−1
k=0
c∗
k
1
N
N−1
n=0
x(n)e−j2πkn/N
ck =
1
N
N−1
n=0
x(n)e−j2πkn/N
→ Px =
N−1
k=0
|ck |2
Average power in signal is sum of powers of individual frequency
components
|ck |2
for k = 0, 1, . . . , N − 1 is called power density spectrum
37 / 52
Power Density Spectrum of Periodic Signals
If x(n) is real (x∗(n) = x(n)), then
c∗
k = c−k
or
|c−k| = |ck| and − c−k = ck
Because
ck+N = ck
then
|ck | = |cN−k | and ck = − cN−k
Thus, for a real signal, the spectrum
ck , k = 0, 1, . . . , N/2 for N even,
or ck , k = 0, 1, . . . , (N − 1)/2 for N odd
completely specifies signal in frequency domain
This is consistent with the fact that the highest relative frequency that
can be represented by a discrete-time signal is π
0 ≤ ωk = 2πk/N ≤ π −→ 0 ≤ k ≤ N/2
38 / 52
Power Density Spectrum of Periodic Signals
Example
Determine Fourier series coefficients and power density spectrum of
periodic signal shown in Fig. 12
Figure 12: Discrete-time periodic square-wave signal.
Applying analysis equation
ck =
1
N
N−1
n=0
x(n)e−j2πkn/N
=
1
N
L−1
n=0
Ae−j2πkn/N
, k = 0, 1, . . . , N − 1
ck =
A
N
L−1
n=0
(e−j2πk/N
)n
=
AL
N , k = 0
A
N
1−e−j2πkL/N
1−e−j2πk/N , k = 1, 2, . . . , N − 1
39 / 52
Power Density Spectrum of Periodic Signals
Example (continued)
Simplifying last expression further
1 − e−j2πkL/N
1 − e−j2πk/N
=
e−jπkL/N
e−jπk/N
ejπkL/N
− e−jπkL/N
ejπk/N − e−jπk/N
= e−jπk(L−1)/N sin(πkL/N)
sin(πk/N)
Therefore
ck =
AL
N , k = 0, ±N, ±2N, . . .
A
N e−jπk(L−1)/N sin(πkL/N)
sin(πk/N) , otherwise
Power density spectrum
|ck |2
=



AL
N
2
, k = 0, ±N, ±2N, . . .
A
N
2 sin(πkL/N)
sin(πk/N)
2
, otherwise
40 / 52
Power Density Spectrum of Periodic Signals
Figure 13: Power density spectrum |ck |2
for L = 2, N = 10 and 40, and A = 1.
41 / 52
The Fourier Transform of Discrete-Time Aperiodic Signals
Fourier transform of a finite-energy discrete-time signal x(n)
X(ω) =
∞
n=−∞
x(n)e−jωn
(7)
X(ω) is a decomposition of x(n) into its frequency components
Frequency range for a discrete-time signal is unique over frequency
interval of (−π, π) or, equivalently, (0, 2π)
X(ω + 2πk) =
∞
n=−∞
x(n)e−j(ω+2πk)n
=
∞
n=−∞
x(n)e−jωn
e−j2πkn
=
∞
n=−∞
x(n)e−jωn
= X(ω)
Hence X(ω) is periodic with period 2π
42 / 52
The Fourier Transform of Discrete-Time Aperiodic Signals
Multiplying both sides of (7) by ejωm and integrating over (−π, π)
π
−π
X(ω)ejωm
dω =
π
−π
∞
n=−∞
x(n)e−jωn
ejωm
dω
On right-hand side, order interchange of and can be made if
XN(ω) =
N
n=−N
x(n)e−jωn
converges uniformly to X(ω) as N → ∞
I.e., for every ω, XN (ω) → X(ω), as N → ∞
∞
n=−∞
x(n)
π
−π
ejω(m−n)
dω =
2πx(m), m = n
0, m = n
x(n) =
1
2π
π
−π
X(ω)ejωn
dω
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The Fourier Transform of Discrete-Time Aperiodic Signals
Table 4: Frequency analysis of discrete-time aperiodic signals
Synthesis equation (inverse transform) x(n) = 1
2π 2π X(ω)ejωndω
Analysis equation (direct transform) X(ω) = ∞
n=−∞ x(n)e−jωn
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Energy Density Spectrum of Aperiodic Signals
Energy of a discrete-time signal x(n) is
Ex =
∞
n=−∞
|x(n)|2
=
∞
n=−∞
x(n)x∗
(n)
=
∞
n=−∞
x(n)
1
2π
π
−π
X∗
(ω)e−jωn
dω
=
1
2π
π
−π
X∗
(ω)
∞
n=−∞
x(n)e−jωn
dω
=
1
2π
π
−π
|X(ω)|2
dω
Parseval’s relation for discrete-time aperiodic signals with finite energy
Ex =
∞
n=−∞
|x(n)|2
=
1
2π
π
−π
|X(ω)|2
dω
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Energy Density Spectrum of Aperiodic Signals
X(ω) is a complex-valued function
X(ω) = |X(ω)|ejΘ(ω)
where
Θ(ω) = X(ω)
is phase spectrum and |X(ω)| is magnitude spectrum
Energy density spectrum of x(n)
Sxx (ω) = |X(ω)|2
if x(n) is real, then
X∗(ω) = X(−ω)
|X(−ω)| = |X(ω)|
X(−ω) = − X(ω)
Sxx (−ω) = Sxx (ω)
Similar to real discrete-time periodic signals, frequency range of real
discrete-time aperiodic signals can also be limited further to one-half of
period
0 ≤ ω ≤ π
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Energy Density Spectrum of Aperiodic Signals
Example
Determine and sketch energy density spectrum Sxx (ω) of signal
x(n) = anu(n), −1 < a < 1
Applying Fourier transform
X(ω) =
∞
n=−∞
x(n)e−jωn
=
∞
n=0
an
e−jωn
=
∞
n=0
(ae−jω
)n
Since |ae−jω| = |a| < 1, using geometric series
X(ω) =
1
1 − ae−jω
Sxx (ω) = |X(ω)|2
= X(ω)X∗
(ω) =
1
(1 − ae−jω)(1 − aejω)
=
1
1 − 2a cos ω + a2
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Energy Density Spectrum of Aperiodic Signals
Figure 14: (a) Sequence x(n) = (1
2 )n
u(n) and x(n) = (−1
2 )n
u(n); (b) their
energy spectra. For a = −0.5 the signal has more rapid variations and as a
result its spectrum has stronger high frequencies.
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Energy Density Spectrum of Aperiodic Signals
Example
Determine Fourier transform and energy density spectrum of sequence
x(n) =
A, 0 ≤ n ≤ L − 1
0, otherwise
Figure 15: Discrete-time rectangular pulse.
Fourier transform of this signal is
X(ω) =
∞
n=−∞
x(n)e−jωn
=
L−1
n=0
Ae−jωn
= A
1 − e−jωL
1 − e−jω
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Energy Density Spectrum of Aperiodic Signals
Example (continued)
X(ω) = Ae−j(ω/2)(L−1) sin(ωL/2)
sin(ω/2)
|X(ω)| =
|A|L, ω = 0
|A| sin(ωL/2)
sin(ω/2) , otherwise
X(ω) = A −
ω
2
(L − 1) +
sin(ωL/2)
sin(ω/2)
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Energy Density Spectrum of Aperiodic Signals
Figure 16: Magnitude and phase of Fourier transform of the discrete-time
rectangular pulse in Fig. 15, for the case A = 1 and L = 5.
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References
John G. Proakis, Dimitris G. Manolakis, Digital Signal
Processing: Principles, Algorithms, and Applications, Prentice
Hall, 2006.
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