4.1 Frequency Analysis of Continuous-Time Signals
Reference
Continuous-Time Frequency Analysis
Reference:
Section 4.1 of
Professor Deepa Kundur
John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing:
Principles, Algorithms, and Applications, 4th edition, 2007.
University of Toronto
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Professor Deepa Kundur (University of Toronto) Continuous-Time Frequency Analysis
4.1 Frequency Analysis of Continuous-Time Signals
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4.1 Frequency Analysis of Continuous-Time Signals
Frequency Analysis
Frequency Synthesis
Scientific Designation
C1
C2
C3
C4 (middle C)
C5
C6
C7
C8
prism
spectrum
white
light
C1
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C2
C3
Frequency (Hz)
32.703
65.406
130.813
261.626
523.251
1046.502
2093.005
4186.009
C4
C5
Professor Deepa Kundur (University of Toronto) Continuous-Time Frequency Analysis
k for F0 = 8.176
4
8
16
32
64
128
256
512
C6
C7
C8
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
Complex Sinusoids
e
jΩt
= cos(Ωt) + j sin(Ωt)
Complex Sinusoids: as Eigenfunctions
≡
complex sinusoid
cos(2πft)
t
0
sin(2πft)
t
0
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4.1 Frequency Analysis of Continuous-Time Signals
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4.1 Frequency Analysis of Continuous-Time Signals
Complex Sinusoids: as Eigenfunctions
y (t) = h(t) ∗ e jΩt = H(Ω)e jΩt
The Continuous-Time Fourier Series
Av = λv
(CTFS)
Therefore, the set of functions {e jΩt } for Ω ∈ R represent
eigenfunctions of LTI systems.
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
Continuous-Time Fourier Series (CTFS)
For continuous-time periodic signals with period Tp =
I
CTFS: Intuition
1
F0
x(t) =
ck e j2πkF0 t
k=−∞
Synthesis equation:
∞
X
x(t) =
I
ck e
{e j2πkF0 t } forms an orthogonal set for k = 0, ±1, ±2, ±3, . . .
j2πkF0 t
k=−∞
< e j2πkF0 t , e j2πmF0 t > =
I
∞
X
Analysis equation:
Z
e j2πkF0 t (e j2πmF0 t )∗ dt
Tp
Z
1
ck =
Tp
=
Z
x(t)e
−j2πkF0 t
e
j2πkF0 t −j2πmF0 t
e
dt
=
e j2π(k−m)F0 t dt
dt =
Tp
Tp
Tp
Z
0

p
 t|T
0
k =m
Tp
e j2π(k−m)F0 t  j2π(k−m)F
0
k 6= m
=
Tp
0
k =m
k 6= m
0
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4.1 Frequency Analysis of Continuous-Time Signals
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4.1 Frequency Analysis of Continuous-Time Signals
CTFS: Intuition
CTFS: Dirichlet Conditions
x(t) =
∞
X
ck e j2πkF0 t
1
ck =
Tp
k=−∞
I
Professor Deepa Kundur (University of Toronto) Continuous-Time Frequency Analysis
Thus, x(t) is being broken down into a series of orthogonal basis functions
that are sinusoidal in nature.
Z
x̃(t) =
x(t)e −j2πkF0 t dt
Tp
∞
X
ck e j2πkF0 t
k=−∞
I
ck are the coefficients needed to represent x(t) in the basis set {e j2πkF0 t }.
I
There is a decoupling that takes place such that modifying the frequency
components of x(t) related to 2πkF0 will not affect those related to 2πmF0
for m 6= k.
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Q: For what conditions is x̃(t) equal to x(t)?
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
CTFS: Dirichlet Conditions
I
CTFS: Dirichlet Conditions
A: Sufficient conditions are given by Dirichlet conditions:
1. x(t) has a finite number of discontinuities in any period.
2. x(t) contains a finite number of maxima and minima during any
period.
3. x(t) is absolutely integrable in any period:
Z
|x(t)|dt < ∞
Tp
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4.1 Frequency Analysis of Continuous-Time Signals
I
Note: the Dirichlet conditions guarantee equality except at
values of t for which x(t) is discontinuous.
I
P
At discontinuities, ∀k ck e j2πkF0 t convergences to the midpoint
of the discontinuity.
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4.1 Frequency Analysis of Continuous-Time Signals
CTFS: Example
CTFS: Example
Find the CTFS of the following periodic square wave:
ck
x(t)
=
=
A
=
t
=
=
ck
1
Tp
Z
x(t)e −j2πkF0 t dt =
Tp
1
Tp
Z
Tp /2
x(t)e −j2πkF0 t dt
−Tp /2
τ /2
A e −j2πkF0 t A·e
dt =
Tp −j2πkF0 −τ /2
−τ /2
−j2πkF0 τ /2
A
e
− e +j2πkF0 τ /2
πkTp · F0
−2j
j2πkF0 τ /2
−j2πkF0 τ /2
A
e
−e
πk · 1
2j
A sin(πkF0 τ )
πk
1
Tp
Z
τ /2
−j2πkF0 t
c0
sinc
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3
4
5
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k
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
x(t)
CTFS: Example
CTFS: Example
A
For τ =
Tp
3
=
x̃(t) =
1
:
3F0
A sin(πk/3)
ck =
πk
∞
X
∞
X
A sin(πk/3) j2πkF0 t
e
πk
ck e j2πkF0 t =
k=−∞
t
k=−∞
x(t)
A
ck
A/2
t
c0
sinc
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3
-2 -1
0
1
4
5
2
k
c k t = τ /2),
Note: At square wave discontinuities (e.g.,
sinc
x(τ /2) =
∞
X
A sin(πk/3)
A
c0
e j2πkF0 (τ /2) =
πk
2
k=−∞
-5 -4 -3
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3
4
5
-2 -1 0 Frequency
1 2
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Analysis
4.1 Frequency Analysis of Continuous-Time Signals
k
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4.1 Frequency Analysis of Continuous-Time Signals
Continuous-Time Fourier Transform (CTFT)
For continuous-time aperiodic signals
I
The Continuous-Time Fourier Transform
Synthesis equation:
1
x(t) =
2π
(CTFT)
I
Z
∞
X (Ω)e jΩt dΩ
−∞
Analysis equation:
Z
∞
X (Ω) =
x(t)e −jΩt dt
−∞
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
Continuous-Time Fourier Transform (CTFT)
CTFT: Dirichlet Conditions
Cyclic frequency can also be used.
I
Synthesis equation:
I
1. x(t) has a finite number of finite discontinuities.
2. x(t) has a finite number of maxima and minima.
3. x(t) is absolutely integrable:
Z ∞
|x(t)|dt < ∞
∞
Z
x(t) =
X (F )e
j2πFt
dF
−∞
I
Analysis equation:
Z
Allowing Tp → ∞ in CTFS Dirichlet conditions:
∞
−∞
x(t)e −j2πFt dt
X (F ) =
−∞
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4.1 Frequency Analysis of Continuous-Time Signals
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4.1 Frequency Analysis of Continuous-Time Signals
CTFT: Example
CTFT: Example
Find the CTFS of the following periodic square wave:
x(t)
Z
X (Ω)
∞
=
x(t)e
−∞
A
t
−jΩt
Z
τ /2
dt =
Ae −jΩt dt
−τ /2
τ /2
e −jΩt sin(Ωτ /2)
= A
= 2A
−jΩ −τ /2
Ω
ck
c0
sinc
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4
5
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k
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
x(t)
CTFT: Example
CTFT: Intuition
A
Z
sin(Ωτ /2)
X (Ω) = 2A
Ω
∞
x(t)e −jΩt dt
X (Ω) =
t
−∞
I
Suppose a(t) and b(t) are continuous-time aperiodic signals. We define:
Z ∞
< a(t), b(t) > =
a(t)b ∗ (t)dt
−∞
X(0)
sinc
I
Therefore, we can interpret X (Ω) as follows:
0
X (Ω) = < x(t), e jΩt >
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4.1 Frequency Analysis of Continuous-Time Signals
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4.1 Frequency Analysis of Continuous-Time Signals
CTFT: Intuition
1
x(t) =
2π
Professor Deepa Kundur (University of Toronto) Continuous-Time Frequency Analysis
CTFT: Duality
Z
∞
Z ∞
1
x(t) =
X (Ω)e jΩt dΩ
2π
Z ∞ −∞
X (Ω) =
x(t)e −jΩt dt
X (Ω)e jΩt dΩ
−∞
I
We may consider x(t) as a linear combination of e jΩt for Ω ∈ R.
I
The larger |X (Ω)|, the more x(t) will look like a sinusoid with Ω.
−∞
F
x(t) ←→
rectangle
sinc
F
←→
F
←→
F
convolution ←→
F
multiplication ←→
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X (Ω)
sinc
rectangle
multiplication
convolution
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
CTFT: Duality
CTFT: Magnitude and Phase
Z ∞
1
x(t) =
X (Ω)e jΩt dΩ
2π −∞
Z ∞
X (Ω) =
x(t)e −jΩt dt
x(t)
−∞
Z ∞
1
X (Ω)e jΩt dΩ
2π −∞
Z ∞
1
|X (Ω)|e j∠X (Ω) e jΩt dΩ
=
2π −∞
Z ∞
=
|X (Ω)|e j(Ωt+∠X (Ω)) df
=
∞
F
Shape A ←→ Shape B
F
Shape B ←→ Shape A
I
|X (Ω)| dictates the relative presence of the sinusoid of frequency Ω in x(t).
I
∠X (Ω) dictates the relative alignment of the sinusoid of frequency Ω in
x(t).
F
Operation A ←→ Operation B
F
Operation B ←→ Operation A
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Professor Deepa Kundur (University of Toronto) Continuous-Time Frequency Analysis
4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
CTFT: Magnitude and Phase
CTFT: Audio Example
Q: Which is more important for a given signal?
I
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Does one component (magnitude or phase) contain more
information than another?
I
An audio signal is represented by a real function x(t).
I
The function x(−t) represents playing the audio signal backwards.
I
Since x(t) is real:
X (Ω) = X ∗ (−Ω)
|X (Ω)| =
I
When filtering, if we had to preserve on component (magnitude
or phase) more, which one would it be?
I
|X ∗ (−Ω)| = |X (−Ω)| since |c| = |c ∗ | for c ∈ C
Therefore,
|X (Ω)|
=
|X (−Ω)|
That is, the CTFT magnitude is even for x(t) real.
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
CTFT: Audio Example
I
Recall,
I
Therefore,
F
x(t) ←→ X (Ω)
Example: Still Image x(t1, t2)
F
x(−t) ←→ X (−Ω)
spectrum magnitude of x(−t)
|X (Ω)|
| {z }
=
z }| {
|X (−Ω)|
spectrum magnitude of x(t)
Therefore, the magnitude of the FT of an audio signal played forward
and backward is the same!
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4.1 Frequency Analysis of Continuous-Time Signals
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4.1 Frequency Analysis of Continuous-Time Signals
Example: |X (Ω1, Ω2)|
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Professor Deepa Kundur (University of Toronto) Continuous-Time Frequency Analysis
Example: ∠X (Ω1, Ω2)
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4.1 Frequency Analysis of Continuous-Time Signals
4.1 Frequency Analysis of Continuous-Time Signals
Reconstruction using
magnitude only
Top Left Photo: Ralph’s
magnitude is the same,
Phase = 0
Top Right Photo: Meg’s
magnitude is the same,
Phase = 0
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4.1 Frequency Analysis of Continuous-Time Signals
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4.1 Frequency Analysis of Continuous-Time Signals
Reconstruction swapping
magnitude and phase
of the images.
Top Left Photo: Ralph’s
phase + Meg’s magnitude
Top Right Photo: Meg’s
phase + Ralph’s magnitude
Reconstruction using
phase only
Top Left Photo: Ralph’s
magnitude normalized to
one, Phase is the same
Top Right Photo: Meg’s
magnitude normalized to
one, Phase is the same
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4.1 Frequency Analysis of Continuous-Time Signals
CTFT: Magnitude and Phase
Q: Which is more important for a given signal? A: Phase
I
Does one component (magnitude or phase) contain more
information than another? A: Yes, phase
I
When filtering, if we had to preserve on component (magnitude
or phase) more, which one would it be?
A: Phase
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