6.003: Signals and Systems
Fourier Representations
October 27, 2011
1
Fourier Representations
Fourier series represent signals in terms of sinusoids.
→ leads to a new representation for systems as filters.
2
Fourier Series
Representing signals by their harmonic components.
1
2
3
4
5
fundamental →
second harmonic →
third harmonic →
fourth harmonic →
fifth harmonic →
6
sixth harmonic →
0
DC →
ω0 2ω0 3ω0 4ω0 5ω0 6ω0
3
ω
← harmonic #
Musical Instruments
Harmonic content is natural way to describe some kinds of signals.
Ex: musical instruments (http://theremin.music.uiowa.edu/MIS.html)
piano
cello
t
t
oboe
bassoon
horn
t
t
altosax
t
violin
bassoon
t
4
t
t
1
seconds
252
Musical Instruments
Harmonic content is natural way to describe some kinds of signals.
Ex: musical instruments (http://theremin.music.uiowa.edu/MIS.html)
piano
cello
k
oboe
bassoon
k
horn
k
altosax
k
violin
k
5
k
k
Musical Instruments
Harmonic content is natural way to describe some kinds of signals.
Ex: musical instruments (http://theremin.music.uiowa.edu/MIS.html )
piano
piano
t
k
violin
violin
t
k
bassoon
bassoon
t
k
6
Harmonics
Harmonic structure determines consonance and dissonance.
octave (D+D’)
fifth (D+A)
D+E
0 1 2 3 4 5 6 7 8 9 101112
0 1 2 3 4 5 6 7 8 9 101112
0 1 2 3 4 5 6 7 8 9 101112
1
0
–1
ti me( per iods of "D")
D'
A
E
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
0123456789
0123456789
0123456789
D
D
D
0
1
2
3
4
5
6
7
har monics
7
Harmonic Representations
What signals can be represented by sums of harmonic components?
ω0
ω
2ω0 3ω0 4ω0 5ω0 6ω0
t
2π
T= ω
0
t
T=
2π
ω0
Only periodic signals: all harmonics
of ω0 are periodic in T = 2π/ω0 .
8
Harmonic Representations
Is it possible to represent ALL periodic signals with harmonics?
What about discontinuous signals?
t
2π
ω0
t
2π
ω0
Fourier claimed YES — even though all harmonics are continuous!
Lagrange ridiculed the idea that a discontinuous signal could be
written as a sum of continuous signals.
We will assume the answer is YES and see if the answer makes sense.
9
Separating harmonic components
Underlying properties.
1. Multiplying two harmonics produces a new harmonic with the
same fundamental frequency:
e jkω0 t × e jlω0 t = e j(k+l)ω0 t .
2. The integral of a harmonic over any time interval with length
equal to a period T is zero unless the harmonic is at DC:
t0 +T
t0
e jkω0 t dt ≡
e jkω0 t dt =
T
0, k = 0
T, k = 0
= T δ[k]
10
Separating harmonic components
Assume that x(t) is periodic in T and is composed of a weighted sum
of harmonics of ω0 = 2π/T .
∞
f
x(t) = x(t + T ) =
ak e jω0 kt
k=−∞
Then
Z
x(t)e
−jlω0 t
Z
dt =
T
=
=
∞
f
ak e jω0 kt e−jω0 lt dt
T k=−∞
Z
∞
f
ak
k=−∞
∞
f
e jω0 (k−l)t dt
T
ak T δ[k − l] = T al
k=−∞
Therefore Z
1
ak =
x(t)e−jω0 kt dt
T T
=
Z
2π
1
x(t)e−j T kt dt
T T
11
Fourier Series
Determining harmonic components of a periodic signal.
Z
2π
1
ak =
x(t)e−j T kt dt
T T
x(t)= x(t + T ) =
∞
f
ak e j
(“analysis” equation)
2π kt
T
(“synthesis” equation)
k=−∞
12
Check Yourself
Let ak represent the Fourier series coefficients of the following
square wave.
1
2
0
1
t
− 12
How many of the following statements are true?
1.
2.
3.
4.
5.
ak = 0 if k is even
ak is real-valued
|ak | decreases with k 2
there are an infinite number of non-zero ak
all of the above
13
Check Yourself
Let ak represent the Fourier series coefficients of the following square
wave.
1
2
0
1
− 12
Z 1
Z
1 0 −j2πkt
1 2 −j2πkt
ak =
x(t)e
dt = −
e
dt +
e
dt
2 −1
2 0
T
2
�
�
1
=
2 − e jπk − e−jπk
j4πk
⎧
⎨ 1 ;
if k is odd
= jπk
⎩
0 ;
otherwise
Z
−j 2π
T kt
14
t
Check Yourself
Let ak represent the Fourier series coefficients of the following square
wave.
1
;
if k is odd
ak = jπk
0 ;
otherwise
How many of the following statements
√ are true?
1. ak = 0 if k is even
2. ak is real-valued
X
3. |ak | decreases with k 2
X
4. there are an infinite number of non-zero ak
5. all of the above
X
15
√
Check Yourself
Let ak represent the Fourier series coefficients of the following
square wave.
1
2
0
t
1
− 12
How many of the following statements are true?
1.
2.
3.
4.
5.
2
√
ak = 0 if k is even
ak is real-valued
X
|ak | decreases with k 2
X
there are an infinite number of non-zero ak
all of the above
X
16
√
Fourier Series Properties
If a signal is differentiated in time, its Fourier coefficients are multi­
plied by j 2π
T k.
Proof: Let
x(t) = x(t + T ) =
∞
f
ak e j
2π kt
T
k=−∞
then
ẋ(t) = ẋ(t + T ) =
∞
f
k=−∞
j
2π
2π
k ak e j T kt
T
17
Check Yourself
Let bk represent the Fourier series coefficients of the following
triangle wave.
1
8
0
1
t
− 18
How many of the following statements are true?
1.
2.
3.
4.
5.
bk = 0 if k is even
bk is real-valued
|bk | decreases with k 2
there are an infinite number of non-zero bk
all of the above
18
Check Yourself
The triangle waveform is the integral of the square wave.
1
2
0
1
t
− 12
1
8
0
1
t
− 18
Therefore the Fourier coefficients of the triangle waveform are
times those of the square wave.
1
1
−1
bk =
×
= 2 2 ; k odd
jkπ j2πk
2k π
19
1
j2πk
Check Yourself
Let bk represent the Fourier series coefficients of the following tri­
angle wave.
−1
bk = 2 2 ; k odd
2k π
How many of the following statements
√ are true?
1. bk = 0 if k is even √
2. bk is real-valued
√
3. |bk | decreases with k 2
4. there are an infinite√number of non-zero bk
5. all of the above
20
√
Check Yourself
Let bk represent the Fourier series coefficients of the following
triangle wave.
1
8
0
t
1
− 18
How many of the following statements are true?
1.
2.
3.
4.
5.
5
√
bk = 0 if k is even √
bk is real-valued
√
|bk | decreases with k 2
there are an infinite√number of non-zero bk
all of the above
21
√
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
0
X
k = −0
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
− 18
22
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
1
X
k = −1
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
− 18
23
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
3
X
k = −3
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
− 18
24
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
5
X
k = −5
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
− 18
25
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
7
X
k = −7
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
− 18
26
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
9
X
k = −9
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
− 18
27
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
19
X
k = −19
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
− 18
28
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
29
X
k = −29
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
− 18
29
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: triangle waveform
39
X
k = −39
k odd
1
8
0
−1 j2πkt
e
2k 2 π 2
1
t
− 18
Fourier series representations of functions with discontinuous slopes
converge toward functions with discontinuous slopes.
30
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
0
X
k = −0
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
31
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
1
X
k = −1
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
32
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
3
X
k = −3
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
33
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
5
X
k = −5
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
34
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
7
X
k = −7
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
35
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
9
X
k = −9
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
36
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
19
X
k = −19
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
37
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
29
X
k = −29
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
38
t
Fourier Series
One can visualize convergence of the Fourier Series by incrementally
adding terms.
Example: square wave
39
X
k = −39
k odd
1
2
0
1 j2πkt
e
jkπ
1
− 12
39
t
Fourier Series
Partial sums of Fourier series of discontinuous functions “ring” near
discontinuities: Gibb’s phenomenon.
9%
1
2
0
t
1
− 12
This ringing results because the magnitude of the Fourier coefficients
is only decreasing as k1 (while they decreased as 12 for the triangle).
k
You can decrease (and even eliminate the ringing) by decreasing the
magnitudes of the Fourier coefficients at higher frequencies.
40
Fourier Series: Summary
Fourier series represent periodic signals as sums of sinusoids.
• valid for an extremely large class of periodic signals
• valid even for discontinuous signals such as square wave
However, convergence as # harmonics increases can be complicated.
41
Filtering
The output of an LTI system is a “filtered” version of the input.
Input: Fourier series → sum of complex exponentials.
∞
f
2π
x(t) = x(t + T ) =
ak e j T kt
k=−∞
Complex exponentials: eigenfunctions of LTI systems.
2π
2π
2π
e j T kt → H(j k)e j T kt
T
Output: same eigenfunctions, amplitudes/phases set by system.
∞
∞
f
f
2π
2π
j 2Tπ kt
x(t) =
ak e
→ y(t) =
ak H(j k)e j T kt
T
k=−∞
k=−∞
42
Filtering
Notion of a filter.
LTI systems
• cannot create new frequencies.
• can scale magnitudes and shift phases of existing components.
Example: Low-Pass Filtering with an RC circuit
R
+
vi
+
−
C
vo
−
43
Lowpass Filter
Calculate the frequency response of an RC circuit.
R
+
vi (t)
= Ri(t) + vo (t)
C:
i(t)
= Cv̇o (t)
Solving: vi (t)
vo
C
= RC v̇o (t) + vo (t)
Vi (s) = (1 + sRC)Vo (s)
1
Vo (s)
H(s) =
=
1 + sRC
Vi (s)
−
1
|H(jω)|
+
−
0.1
0.01
∠H(jω)|
vi
KVL:
0.01
0.1
1
10
ω
100 1/RC
10
ω
100 1/RC
0
− π2
0.01
0.1
1
44
Lowpass Filtering
Let the input be a square wave.
1
2
0
− 12
1 jω0 kt
e
;
jπk
k odd
1
|X(jω)|
f
ω0 =
2π
T
0.1
0.01
∠X(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω
100 1/RC
10
ω
1/RC
100
0
− π2
0.01
0.1
1
45
Lowpass Filtering
Low frequency square wave: ω0 << 1/RC.
1
2
0
− 12
1 jω0 kt
e
;
jπk
k odd
1
|H(jω)|
f
ω0 =
2π
T
0.1
0.01
∠H(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω
100 1/RC
10
ω
1/RC
100
0
− π2
0.01
0.1
1
46
Lowpass Filtering
Higher frequency square wave: ω0 < 1/RC.
1
2
0
− 12
1 jω0 kt
e
;
jπk
k odd
1
|H(jω)|
f
ω0 =
2π
T
0.1
0.01
∠H(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω
100 1/RC
10
ω
1/RC
100
0
− π2
0.01
0.1
1
47
Lowpass Filtering
Still higher frequency square wave: ω0 = 1/RC.
1
2
0
− 12
1 jω0 kt
e
;
jπk
k odd
1
|H(jω)|
f
ω0 =
2π
T
0.1
0.01
∠H(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω
100 1/RC
10
ω
1/RC
100
0
− π2
0.01
0.1
1
48
Lowpass Filtering
High frequency square wave: ω0 > 1/RC.
1
2
0
− 12
1 jω0 kt
e
;
jπk
k odd
1
|H(jω)|
f
ω0 =
2π
T
0.1
0.01
∠H(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω
100 1/RC
10
ω
1/RC
100
0
− π2
0.01
0.1
1
49
Fourier Series: Summary
Fourier series represent signals by their frequency content.
Representing a signal by its frequency content is useful for many
signals, e.g., music.
Fourier series motivate a new representation of a system as a filter.
50
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6.003 Signals and Systems
Fall 2011
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