6.003: Signals and Systems Fourier Series November 1, 2011 1
Last Time: Describing Signals by Frequency Content Harmonic content is natural way to describe some kinds of signals. .html Ex: musical instruments (http://theremin.music.uiowa.edu/MIS ) piano piano t k violin violin t k bassoon bassoon t k 2
Last Time: Fourier Series Determining harmonic components of a periodic signal. 1 � 2 π − j kt a k = x ( t ) e dt (“analysis” equation) T T T ∞ 2 π j kt 0 x ( t )= x ( t + T ) = a k e (“synthesis” equation) T k = −∞ We can think of Fourier series as an orthogonal decomposition . 3
Orthogonal Decompositions Vector representation of 3-space: let r ¯ represent a vector with components { x , y , and z } in the { x ˆ , y ˆ , and z ˆ } directions, respectively. x = ¯ r · x ˆ y = ¯ r · y ˆ (“analysis” equations) z = ¯ r · z ˆ r ¯ = xx ˆ + yy ˆ + zz ˆ (“synthesis” equation) Fourier series: let x ( t ) represent a signal with harmonic components 2 π 2 π j t j kt j 0 t { a 0 , a 1 , . . . , a k } for harmonics { e , e , . . . , e } respectively. T T 1 � � 2 π − j kt a k = x ( t ) e dt (“analysis” equation) T T T ∞ j π 2 0 T kt x ( t )= x ( t + T ) = a k e (“synthesis” equation) k = −∞ 4
Orthogonal Decompositions Vector representation of 3-space: let r ¯ represent a vector with components { x , y , and z } in the { x ˆ , y ˆ , and z ˆ } directions, respectively. x = ¯ r · x ˆ y = ¯ r · y ˆ (“analysis” equations) z = ¯ r · z ˆ r ¯ = x ˆ x + y ˆ y + z ˆ z (“synthesis” equation) Fourier series: let x ( t ) represent a signal with harmonic components 2 π 2 π j t j kt j 0 t { a 0 , a 1 , . . . , a k } for harmonics { e , e , . . . , e } respectively. T T 1 � � 2 π − j kt a k = x ( t ) e dt (“analysis” equation) T T T ∞ j π 2 0 T kt x ( t )= x ( t + T ) = a k e (“synthesis” equation) k = −∞ 5
Orthogonal Decompositions Vector representation of 3-space: let r ¯ represent a vector with components { x , y , and z } in the { x ˆ , y ˆ , and z ˆ } directions, respectively. x = ¯ r · ˆ x y = ¯ r · ˆ y (“analysis” equations) z = ¯ r · ˆ z r ¯ = xx ˆ + yy ˆ + zz ˆ (“synthesis” equation) Fourier series: let x ( t ) represent a signal with harmonic components 2 π 2 π j t j kt j 0 t { a 0 , a 1 , . . . , a k } for harmonics { e , e , . . . , e } respectively. T T a k = 1 � � − j 2 π T kt dt x ( t ) e (“analysis” equation) T T ∞ j π 2 0 T kt x ( t )= x ( t + T ) = a k e (“synthesis” equation) k = −∞ 6
Orthogonal Decompositions Integrating over a period sifts out the k th component of the series. Sifting as a dot product: x = r ¯ · x ˆ ≡ | r ¯ || x ˆ | cos θ Sifting as an inner product: � 1 � 2 π 2 π j kt − j kt a k = e · x ( t ) ≡ x ( t ) e dt T T T T where � 1 � ∗ ( t ) b ( t ) dt . a ( t ) · b ( t ) = a T T The complex conjugate ( ∗ ) makes the inner product of the k th and m th components equal to 1 iff k = m : � 1 � 1 � � � 1 if k = m 2 π � ∗ � 2 π 2 π 2 π � � j kt j mt − j kt j mt e e dt = e e dt = T T T T T T 0 otherwise T T 7
Check Yourself How many of the following pairs of functions are orthogonal ( ⊥ ) in T = 3 ? 1. cos 2 πt ⊥ sin 2 πt ? 2. cos 2 πt ⊥ cos 4 πt ? 3. cos 2 πt ⊥ sin πt ? 4. cos 2 πt ⊥ e j 2 πt ? 8
Check Yourself How many of the following are orthogonal ( ⊥ ) in T = 3 ? cos 2 πt ⊥ sin 2 πt ? cos 2 πt t 1 2 3 sin 2 πt t 1 2 3 cos 2 πt sin 2 πt = 1 2 sin 4 πt t 1 2 3 3 � � dt = 0 therefore YES 0 9
Check Yourself How many of the following are orthogonal ( ⊥ ) in T = 3 ? cos 2 πt ⊥ cos 4 πt ? cos 2 πt t 1 2 3 cos 4 πt t 1 2 3 cos 2 πt cos 4 πt = 1 2 cos 6 πt + 1 2 cos 2 πt t 1 2 3 3 � � dt = 0 therefore YES 0 10
Check Yourself How many of the following are orthogonal ( ⊥ ) in T = 3 ? cos 2 πt ⊥ sin πt ? cos 2 πt t 1 2 3 sin πt t 1 2 3 cos 2 πt sin πt = 1 2 sin 3 πt − 1 2 sin πt t 1 2 3 3 � � dt = 0 therefore NO 0 11
Check Yourself How many of the following are orthogonal ( ⊥ ) in T = 3 ? 2 πt ? cos 2 πt ⊥ e 2 πt = cos 2 πt + j sin 2 πt e cos 2 πt ⊥ sin 2 πt but not cos 2 πt Therefore NO 12
Check Yourself How many of the following pairs of functions are orthogonal ( ⊥ ) in T = 3 ? 2 √ 1. cos 2 πt ⊥ sin 2 πt ? √ 2. cos 2 πt ⊥ cos 4 πt ? 3. cos 2 πt ⊥ sin πt ? X 4. cos 2 πt ⊥ e j 2 πt ? X 13
Speech Vowel sounds are quasi-periodic. bat bait bet beet t t t t bought bit bite boat t t t t but boot t t 14
Speech Harmonic content is natural way to describe vowel sounds. bat bait bet beet k k k k bought bit bite boat k k k k but boot k k 15
Speech Harmonic content is natural way to describe vowel sounds. bat bat t k beet beet t k boot boot t k 16
Speech Production Speech is generated by the passage of air from the lungs, through the vocal cords, mouth, and nasal cavity. Nasal cavity Hard palate Soft palate Lips (velum) Tongue Pharynx Epiglottis Larynx Esophogus Vocal cords (glottis) Trachea Stomach Lungs Adapted from T.F. Weiss 17
Speech Production Controlled by complicated muscles, vocal cords are set in vibration by the passage of air from the lungs. Looking down the throat: Vocal cords open Glottis Vocal cords closed Vocal cords Gray's Anatomy Adapted from T.F. Weiss 18
Speech Production Vibrations of the vocal cords are “filtered” by the mouth and nasal cavities to generate speech. 19
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 R + + v i v o C − − 20
Lowpass Filter Calculate the frequency response of an RC circuit. KVL: v i ( t ) = Ri ( t ) + v o ( t ) R C: i ( t ) = Cv ˙ o ( t ) + Solving: v i ( t ) = RC v ˙ o ( t ) + v o ( t ) + v i v o C − V i ( s ) = (1 + sRC ) V o ( s ) − V o ( s ) 1 H ( s ) = = V i ( s ) 1 + sRC 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 21
Lowpass Filtering Let the input be a square wave. 1 2 t 0 T − 1 2 1 2 π 0 jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | X ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ X ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 22
Lowpass Filtering Low frequency square wave: ω 0 << 1 /RC . 1 2 t 0 T − 1 2 1 2 π 0 jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 23
Lowpass Filtering Higher frequency square wave: ω 0 < 1 /RC . 1 2 t 0 T − 1 2 1 2 π 0 jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 24
Lowpass Filtering Still higher frequency square wave: ω 0 = 1 /RC . 1 2 t 0 T − 1 2 1 2 π 0 jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 25
Lowpass Filtering High frequency square wave: ω 0 > 1 /RC . 1 2 t 0 T − 1 2 1 2 π 0 jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 26
Source-Filter Model of Speech Production Vibrations of the vocal cords are “filtered” by the mouth and nasal cavities to generate speech. throat and buzz from speech vocal cords nasal cavities 27
Speech Production X-ray movie showing speech in production. Courtesy of Kenneth N. Stevens. Used with permission. 28
Demonstration Artificial speech. throat and buzz from speech vocal cords nasal cavities 29
Formants Resonant frequencies of the vocal tract. F1 F2 amplitude F3 frequency 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
Speech Production Same glottis signal + different formants → different vowels. glottis signal vocal tract filter vowel sound a k b k a k b k We detect changes in the filter function to recognize vowels. 31
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
Speech Production We detect changes in the filter function to recognize vowels. low intermediate high 33
Speech Production We detect changes in the filter function to recognize vowels. low intermediate high 34
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