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6.003: Signals and Systems
Fourier Series
November 1, 2011 1
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Last Time: Describing Signals by Frequency Content
Harmonic content is natural way to describe some kinds of signals.
Ex: musical instruments (http://theremin.music.uiowa.edu/MIS )
piano
t
piano
k
violin
t
violin
k
bassoon
t
bassoon
k2
.html
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Last Time: Fourier Series
Determining harmonic components of a periodic signal.
1 2π−j ktx(t)e (“analysis” equation) ak = dtTT T
∞0 2πj kt x(t)= x(t + T ) = (“synthesis” equation) Take k=−∞
We can think of Fourier series as an orthogonal decomposition.
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Orthogonal Decompositions
Vector representation of 3-space: let r represent a vector with
components {x, y, and z} in the {x, y, and z} directions, respectively.
x = r · xy = r · y (“analysis” equations)
z = r · z
r = xx + yy + zz (“synthesis” equation)
Fourier series: let x(t) represent a signal with harmonic components 2
j0t j 2tTπ πj kt. . ., ak} for harmonics {e } respectively. {a0, a1, T, e , . . ., e
1 2π−j ktx(t)e (“analysis” equation) ak = dtTT T
∞0 2x(t)= x(t + T ) = ake
k=−∞
j π T kt (“synthesis” equation)
4
∫
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Orthogonal Decompositions
Vector representation of 3-space: let r represent a vector with
components {x, y, and z} in the {x, y, and z} directions, respectively.
x = r · xy = r · y (“analysis” equations)
z = r · z
r = xˆ z (“synthesis” equation) x + yy + zˆ
Fourier series: let x(t) represent a signal with harmonic components 2
j0t j 2tTπ πj kt. . ., ak} for harmonics {e } respectively. {a0, a1, T, e , . . ., e
1 2π−j ktx(t)e (“analysis” equation) ak = dtTT T
x(t)= x(t + T ) = ∞0
k=−∞
ake 2j π T kt (“synthesis” equation)
5
∫
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Orthogonal Decompositions
Vector representation of 3-space: let r represent a vector with
components {x, y, and z} in the {x, y, and z} directions, respectively.
(“analysis” equations)
r = xx + yy + zz (“synthesis” equation)
x = r · x
y = r · y
z = r · z
Fourier series: let x(t) represent a signal with harmonic components 2
j0t j 2tTπ πj kt. . ., ak} for harmonics {e } respectively. {a0, a1, T, e , . . ., e
ak = 1 T T
x(t)e −j 2π T ktdt (“analysis” equation)
∞0 2x(t)= x(t + T ) = ake
k=−∞
j π T kt (“synthesis” equation)
6
∫
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Orthogonal Decompositions
Integrating over a period sifts out the kth component of the series.
Sifting as a dot product:
x = r · x ≡ |r||x| cos θ
Sifting as an inner product: 1 · x(t) ≡ x(t)e
2 2π π−jj kt kt=ak e dtT TT T
where 1
a(t) · b(t) = a ∗(t)b(t)dt . T T
The complex conjugate (∗) makes the inner product of the kth and
mth components equal to 1 iff k = m: 1 1 ∗ 12 2 2 2 if k = mπ π π π−jj kt je kt je mt mtdt = dt =T T T Te e 0T TT otherwiseT
7
∫∫
∫∫
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Check Yourself
How many of the following pairs of functions are
orthogonal (⊥) in T = 3?
1. cos 2πt ⊥ sin 2πt ?
2. cos 2πt ⊥ cos 4πt ?
3. cos 2πt ⊥ sin πt ?
4. cos 2πt ⊥ e j2πt ?
8
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Check Yourself
How many of the following are orthogonal (⊥) in T = 3?
cos 2πt ⊥ sin 2πt ?
1 2 3t
cos 2πt
1 2 3t
sin 2πt
1 2 3t
cos 2πt sin 2πt = 12 sin 4πt
3 dt = 0 therefore YES
0 9
∫
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Check Yourself
How many of the following are orthogonal (⊥) in T = 3?
cos 2πt ⊥ cos 4πt ?
1 2 3t
cos 2πt
1 2 3t
cos 4πt
1 2 3t
cos 2πt cos 4πt = 12 cos 6πt+ 1
2 cos 2πt
3 dt = 0 therefore YES
0 10
∫
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Check Yourself
How many of the following are orthogonal (⊥) in T = 3?
cos 2πt ⊥ sin πt ?
1 2 3t
cos 2πt
1 2 3t
sin πt
1 2 3t
cos 2πt sin πt = 12 sin 3πt− 1
2 sin πt
3 dt = 0 therefore NO
0 11
∫
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Check Yourself
How many of the following are orthogonal (⊥) in T = 3?
2πt ?cos 2πt ⊥ e
e 2πt = cos 2πt + j sin 2πt
cos 2πt ⊥ sin 2πt but not cos 2πt
Therefore NO
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Check Yourself
How many of the following pairs of functions
are orthogonal (⊥) in T = 3? 2
1. cos 2πt ⊥ sin 2πt ? √
2. cos 2πt ⊥ cos 4πt ? √
3. cos 2πt ⊥ sin πt ? X
4. cos 2πt ⊥ e j2πt ? X
13
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Speech
Vowel sounds are quasi-periodic.
bat
t
bait
t
bet
t
beet
t
bit
t
bite
t
bought
t
boat
t
but
t
boot
t
14
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Speech
Harmonic content is natural way to describe vowel sounds.
bat
k
bait
k
bet
k
beet
k
bit
k
bite
k
bought
k
boat
k
but
k
boot
k
15
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Speech
Harmonic content is natural way to describe vowel sounds.
bat
t
bat
k
beet
t
beet
k
boot
t
boot
k
16
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Speech Production
Speech is generated by the passage of air from the lungs, through
the vocal cords, mouth, and nasal cavity.
Adapted from T.F. Weiss
Lips
Nasalcavity
Hard palate
TongueSoft palate
(velum)
Pharynx
Vocal cords(glottis)
Esophogus
Epiglottis
LungsStomach
Larynx
Trachea
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Speech Production
Controlled by complicated muscles, vocal cords are set in vibration
by the passage of air from the lungs.
Looking down the throat:
Gray's Anatomy Adapted from T.F. Weiss
Glottis
Vocalcords
Vocal cords open
Vocal cords closed
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Speech Production
Vibrations of the vocal cords are “filtered” by the mouth and nasal
cavities to generate speech.
19
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Filtering
Notion of a filter.
LTI systems
• cannot create new frequencies.
• can only scale magnitudes & shift phases of existing components.
Example: Low-Pass Filtering with an RC circuit
+−
vi
+
vo
−
R
C
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0.01
0.1
1
0.01 0.1 1 10 100ω
1/RC
|H(jω
)|
0
−π20.01 0.1 1 10 100
ω
1/RC
∠H
(jω
)|
Lowpass Filter
Calculate the frequency response of an RC circuit.
+−
vi
+
vo
−
R
C
KVL: vi(t) = Ri(t) + vo(t)
C: i(t) = Cvo(t)
Solving: vi(t) = RC vo(t) + vo(t)
Vi(s) = (1 + sRC)Vo(s) Vo(s) 1
H(s) = = Vi(s) 1 + sRC
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Lowpass Filtering
x(t) =
Let the input be a square wave.
t
12
−12
0 T
T π2=0ω;kt 0jωe
jπk 1
odd k
0
0.01
0.1
1
0.01 0.1 1 10 100ω
1/RC
|X(jω
)|
0
−π20.01 0.1 1 10 100
ω
1/RC
∠X
(jω
)|
22
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Lowpass Filtering
x(t) =
Low frequency square wave: ω0 << 1/RC.
t
12
−12
0 T
T π2=0ω;kt 0jωe
jπk 1
odd k
0
0.01
0.1
1
0.01 0.1 1 10 100ω
1/RC
|H(jω
)|
0
−π20.01 0.1 1 10 100
ω
1/RC
∠H
(jω
)|
23
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Lowpass Filtering
x(t) =
Higher frequency square wave: ω0 < 1/RC.
t
12
−12
0 T
T π2=0ω;kt 0jωe
jπk 1
odd k
0
0.01
0.1
1
0.01 0.1 1 10 100ω
1/RC
|H(jω
)|
0
−π20.01 0.1 1 10 100
ω
1/RC
∠H
(jω
)|
24
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Lowpass Filtering
x(t) =
Still higher frequency square wave: ω0 = 1/RC.
t
12
−12
0 T
T π2=0ω;kt 0jωe
jπk 1
odd k
0
0.01
0.1
1
0.01 0.1 1 10 100ω
1/RC
|H(jω
)|
0
−π20.01 0.1 1 10 100
ω
1/RC
∠H
(jω
)|
25
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Lowpass Filtering
x(t) =
High frequency square wave: ω0 > 1/RC.
t
12
−12
0 T
T π2=0ω;kt 0jωe
jπk 1
odd k
0
0.01
0.1
1
0.01 0.1 1 10 100ω
1/RC
|H(jω
)|
0
−π20.01 0.1 1 10 100
ω
1/RC
∠H
(jω
)|
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Source-Filter Model of Speech Production
Vibrations of the vocal cords are “filtered” by the mouth and nasal
cavities to generate speech.
buzz fromvocal cords
speechthroat and
nasal cavities27
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Speech Production
X-ray movie showing speech in production.
28
Courtesy of Kenneth N. Stevens. Used with permission.
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Demonstration
Artificial speech.
buzz fromvocal cords
speechthroat and
nasal cavities29
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Formants
Resonant frequencies of the vocal tract.
frequency
am
plitu
de
F1F2
F3
Formant heed head had hod haw’d who’d
Men F1 270 530 660 730 570 300 F2 2290 1840 1720 1090 840 870 F3 3010 2480 2410 2440 2410 2240
Women F1 310 610 860 850 590 370 F2 2790 2330 2050 1220 920 950 F3 3310 2990 2850 2810 2710 2670
Children F1 370 690 1010 1030 680 430 F2 3200 2610 2320 1370 1060 1170 F3 3730 3570 3320 3170 3180 3260
http://www.sfu.ca/sonic-studio/handbook/Formant.html 30
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Speech Production
Same glottis signal + different formants → different vowels.
glottis signal vocal tract filter vowel sound
ak
ak
bk
bk
We detect changes in the filter function to recognize vowels. 31
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Singing
We detect changes in the filter function to recognize vowels
... at least sometimes.
Demonstration.
“la” scale.
“lore” scale.
“loo” scale.
“ler” scale.
“lee” scale.
Low Frequency: “la” “lore” “loo” “ler” “lee”.
High Frequency: “la” “lore” “loo” “ler” “lee”.
http://www.phys.unsw.edu.au/jw/soprane.html 32
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Speech Production
We detect changes in the filter function to recognize vowels.
low
intermediate
high
33
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Speech Production
We detect changes in the filter function to recognize vowels.
low
intermediate
high
34
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Continuous-Time 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.
Representing a system as a filter is useful for many systems, e.g.,
speech synthesis.
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6.003 Signals and SystemsFall 2011
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