6.003: Signals and SystemsFourier SeriesApril 1, 2010Mid-term Examination #2Wednesday, April 7, 7:30-9:30pm, 34-101.No recitations on the day of the exam.Coverage: Lectures 1–15Recitations 1–15Homeworks 1–8Homework 8 will not collected or graded. Solutions will be posted.Closed book: 2 pages of notes (812× 11 inches; front and back).Designed as 1-hour exam; two hours to complete.Review sessions during open office hours.Conflict? Contact [email protected] before Friday, April 2, 5pm.Last Time: Describing Signals by Frequency ContentHarmonic content is natural way to describe some kinds of signals.Ex: musical instruments (http://theremin.music.uiowa.edu/MIS)pianotpianokviolintviolinkbassoontbassoonkLast Time: Fourier SeriesDetermining harmonic components of a periodic signal.ak=1TZTx(t)e−j2πTktdt (“analysis” equation)x(t)= x(t + T ) =∞Xk=−∞akej2πTkt(“synthesis” equation)Last Time: Fourier SeriesDetermining harmonic components of a periodic signal.ak=1TZTx(t)e−j2πTktdt (“analysis” equation)x(t)= x(t + T ) =∞Xk=−∞akej2πTkt(“synthesis” equation)We can think of Fourier series as an orthogonal decomposition.Orthogonal DecompositionsVector representation of 3-space: let ¯r represent a vector withcomponents {x, y, and z} in the {ˆx, ˆy, and ˆz} directions, respectively.x =¯r · ˆxy = ¯r · ˆyz = ¯r · ˆz(“analysis” equations)¯r = xˆx + yˆy + zˆz (“synthesis” equation)ak=1TZTx(t)e−j2πTktdt (“analysis” equation)x(t)= x(t + T ) =∞Xk=−∞akej2πTkt(“synthesis” equation)Orthogonal DecompositionsVector representation of 3-space: let ¯r represent a vector withcomponents {x, y, and z} in the {ˆx, ˆy, and ˆz} directions, respectively.x =¯r · ˆxy = ¯r · ˆyz = ¯r · ˆz(“analysis” equations)¯r = xˆx + yˆy + zˆz (“synthesis” equation)Fourier series: let x(t) represent a signal with harmonic components{a0, a1, . . ., ak} for harmonics {ej0t, ej2πTt, . . ., ej2πTkt} respectively.ak=1TZTx(t)e−j2πTktdt (“analysis” equation)x(t)= x(t + T ) =∞Xk=−∞akej2πTkt(“synthesis” equation)Orthogonal DecompositionsVector representation of 3-space: let ¯r represent a vector withcomponents {x, y, and z} in the {ˆx, ˆy, and ˆz} directions, respectively.x =¯r · ˆxy = ¯r · ˆyz = ¯r · ˆz(“analysis” equations)¯r = xˆx + yˆy + zˆz (“synthesis” equation)Fourier series: let x(t) represent a signal with harmonic components{a0, a1, . . ., ak} for harmonics {ej0t, ej2πTt, . . ., ej2πTkt} respectively.ak=1TZTx(t)e−j2πTktdt (“analysis” equation)x(t)= x(t + T ) =∞Xk=−∞akej2πTkt(“synthesis” equation)Orthogonal DecompositionsVector representation of 3-space: let ¯r represent a vector withcomponents {x, y, and z} in the {ˆx, ˆy, and ˆz} directions, respectively.x =¯r · ˆxy = ¯r · ˆyz = ¯r · ˆz(“analysis” equations)¯r = xˆx + yˆy + zˆz (“synthesis” equation)Fourier series: let x(t) represent a signal with harmonic components{a0, a1, . . ., ak} for harmonics {ej0t, ej2πTt, . . ., ej2πTkt} respectively.ak=1TZTx(t)e−j2πTktdt (“analysis” equation)x(t)= x(t + T ) =∞Xk=−∞akej2πTkt(“synthesis” equation)Orthogonal DecompositionsIntegrating over a period sifts out the kthcomponent of the series.Sifting as a dot product:x = ¯r · ˆx ≡ |¯r||ˆx|cos θSifting as an inner product:ak= ej2πTkt· x(t) ≡1TZTx(t)e−j2πTktdtwherea(t) · b(t) =1TZTa∗(t)b(t)dt .Orthogonal DecompositionsIntegrating over a period sifts out the kthcomponent of the series.Sifting as a dot product:x = ¯r · ˆx ≡ |¯r||ˆx|cos θSifting as an inner product:ak= ej2πTkt· x(t) ≡1TZTx(t)e−j2πTktdtwherea(t) · b(t) =1TZTa∗(t)b(t)dt .The complex conjugate (∗) makes the inner product of the kthandmthcomponents equal to 1 iff k = m:1TZTej2πTkt∗ej2πTmtdt =1TZTe−j2πTktej2πTmtdt =1 if k = m0 otherwiseCheck YourselfHow many of the following pairs of functions are orthog-onal (⊥) in T = 3?1. cos 2πt ⊥ sin 2πt ?2. cos 2πt ⊥ cos 4πt ?3. cos 2πt ⊥ sin πt ?4. cos 2πt ⊥ ej2πt?Check YourselfHow many of the following are orthogonal (⊥) in T = 3?cos 2πt ⊥ sin 2πt ?1 2 3tcos 2πt1 2 3tsin 2πt1 2 3tcos 2πt sin 2πtZ30dt = 0 therefore YESCheck YourselfHow many of the following are orthogonal (⊥) in T = 3?cos 2πt ⊥ cos 4πt ?1 2 3tcos 2πt1 2 3tcos 4πt1 2 3tcos 2πt cos 4πtZ30dt = 0 therefore YESCheck YourselfHow many of the following are orthogonal (⊥) in T = 3?cos 2πt ⊥ cos πt ?1 2 3tcos 2πt1 2 3tsin πt1 2 3tcos 2πt sin πtZ30dt 6= 0 therefore NOCheck YourselfHow many of the following are orthogonal (⊥) in T = 3?cos 2πt ⊥ e2πt?e2πt= cos 2πt + j sin 2πtcos 2πt ⊥ sin 2πt but not cos 2πtThereforeNOCheck YourselfHow many of the following pairs of functions are orthog-onal (⊥) in T = 3? 21. cos 2πt ⊥ sin 2πt ?√2. cos 2πt ⊥ cos 4πt ?√3. cos 2πt ⊥ sin πt ? X4. cos 2πt ⊥ ej2πt? XSpeechVowel sounds are quasi-periodic.battbaittbettbeettbittbitetboughttboattbuttboottSpeechHarmonic content is natural way to describe vowel sounds.batkbaitkbetkbeetkbitkbitekboughtkboatkbutkbootkSpeechHarmonic content is natural way to describe vowel sounds.battbatkbeettbeetkboottbootkSpeech ProductionSpeech is generated by the passage of air from the lungs, throughthe vocal cords, mouth, and nasal cavity.Adapted from T.F. WeissLipsNasalcavityHard palateTongueSoft palate(velum)PharynxVocal cords(glottis)EsophogusEpiglottisLungsStomachLarynxTracheaSpeech ProductionControlled by complicated muscles, the vocal cords are set into vi-brational motion by the passage of air from the lungs.Looking down the throat:Gray's Anatomy Adapted from T.F. WeissGlottisVocalcordsVocal cords openVocal cords closedSpeech ProductionVibrations of the vocal cords are “filtered” by the mouth and nasalcavities to generate speech.FilteringNotion of a filter.LTI systems• cannot create new frequencies.• can only scale magnitudes and shift phases of existing components.Example: Low-Pass Filtering with an RC circuit+−vi+vo−RCLowpass FilterCalculate the frequency response of an RC circuit.+−vi+vo−RCKVL: vi(t) = Ri(t) + vo(t)C: i(t) = C ˙vo(t)Solving: vi(t) = RC ˙vo(t) + vo(t)Vi(s) = (1 + sRC)Vo(s)H(s) =Vo(s)Vi(s)=11 + sRC0.010.110.01 0.1 1 10 100ω1/RC|H(jω)|0−π20.01 0.1 1 10 100ω1/RC∠H(jω)|Lowpass FilteringLet the input be a square wave.t12−120Tx(t) =Xk odd1jπkejω0kt; ω0=2πT0.010.110.01 0.1 1 10
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