6.003: Signals and Systems Lecture 14 March 30, 201016.003: Signals and SystemsFourier RepresentationsMarch 30, 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.Fourier RepresentationsFourier series represent signals in terms of sinusoids.→ leads to a new representation for systems as filters.Fourier SeriesRepresenting signals by their harmonic components.ωω02ω03ω04ω05ω06ω00 1 2 345 6← harmonic #DC →fundamental →second harmonic →third harmonic →fourth harmonic →fifth harmonic →sixth harmonic →Musical InstrumentsHarmonic content is natural way to describe some kinds of signals.Ex: musical instruments (http://theremin.music.uiowa.edu/MIS)pianotcellotbassoontoboethorntaltosaxtviolintbassoont1252secondsMusical InstrumentsHarmonic content is natural way to describe some kinds of signals.Ex: musical instruments (http://theremin.music.uiowa.edu/MIS)pianokcellokbassoonkoboekhornkaltosaxkviolink6.003: Signals and Systems Lecture 14 March 30, 20102Musical InstrumentsHarmonic content is natural way to describe some kinds of signals.Ex: musical instruments (http://theremin.music.uiowa.edu/MIS)pianotpianokviolintviolinkbassoontbassoonkHarmonicsHarmonic structure determines consonance and dissonance.octave (D+D’) fifth (D+A) D+E[time(periods of "D")harmonics0 1 2 3 4 5 6 7 8 9 1011120 1 2 3 4 5 6 7 8 9 1011120 1 2 3 4 5 6 7 8 9 101112–101DD'DADE0 1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7Harmonic RepresentationsWhat signals can be represented by sums of harmonic components?ωω02ω03ω04ω05ω06ω0T =2πω0tT =2πω0tOnly periodic signals: all harmonics of ω0are periodic in T = 2π/ω0.Harmonic RepresentationsIs it possible to represent ALL periodic signals with harmonics?What about discontinuous signals?2πω0t2πω0tFourier claimed YES — even though all harmonics are continuous!Lagrange ridiculed the idea that a discontinuous signal could bewritten as a sum of continuous signals.We will assume the answer is YES and see if the answer makes sense.Separating harmonic componentsUnderlying properties.1. Multiplying two harmonics produces a new harmonic with thesame fundamental frequency:ejkω0t× ejlω0t= ej(k+l)ω0t.2. The integral of a harmonic over any time interval with lengthequal to a period T is zero unless the harmonic is at DC:Zt0+Tt0ejkω0tdt ≡ZTejkω0tdt =0, k 6= 0T, k = 0= T δ[k]Separating harmonic componentsAssume that x(t) is periodic in T and is composed of a weighted sumof harmonics of ω0= 2π/T .x(t) = x(t + T ) =∞Xk=−∞akejω0ktThenZTx(t)e−jlω0tdt =ZT∞Xk=−∞akejω0kte−jω0ltdt=∞Xk=−∞akZTejω0(k−l)tdt=∞Xk=−∞akT δ[k − l] = T alThereforeak=1TZTx(t)e−jω0ktdt =1TZTx(t)e−j2πTktdt6.003: Signals and Systems Lecture 14 March 30, 20103Fourier SeriesDetermining harmonic components of a periodic signal.ak=1TZTx(t)e−j2πTktdt (“analysis” equation)x(t)= x(t + T ) =∞Xk=−∞akej2πTkt(“synthesis” equation)Check YourselfLet akrepresent the Fourier series coefficients of the followingsquare wave.t12−120 1How many of the following statements are true?1. ak= 0 if k is even2. akis real-valued3. |ak| decreases with k24. there are an infinite number of non-zero ak5. all of the aboveFourier Series PropertiesIf a signal is differentiated in time, its Fourier coefficients are multi-plied by j2πTk.Proof: Letx(t) = x(t + T ) =∞Xk=−∞akej2πTktthen˙x(t) = ˙x(t + T ) =∞Xk=−∞j2πTk akej2πTktCheck YourselfLet bkrepresent the Fourier series coefficients of the followingtriangle wave.t18−1801How many of the following statements are true?1. bk= 0 if k is even2. bkis real-valued3. |bk| decreases with k24. there are an infinite number of non-zero bk5. all of the aboveFourier SeriesOne can visualize convergence of the Fourier Series by incrementallyadding terms.Example: triangle waveformt5Xk = −5k odd−12k2π2ej2πkt18−1801Fourier SeriesOne can visualize convergence of the Fourier Series by incrementallyadding terms.Example: triangle waveformt39Xk = −39k odd−12k2π2ej2πkt18−1801Fourier series representations of functions with discontinuous slopesconverge toward functions with discontinuous slopes.6.003: Signals and Systems Lecture 14 March 30, 20104Fourier SeriesOne can visualize convergence of the Fourier Series by incrementallyadding terms.Example: square wavet5Xk = −5k odd1jkπej2πkt12−120 1Fourier SeriesOne can visualize convergence of the Fourier Series by incrementallyadding terms.Example: square wavet39Xk = −39k odd1jkπej2πkt12−120 1Fourier SeriesPartial sums of Fourier series of discontinuous functions “ring” neardiscontinuities: Gibb’s phenomenon.t12−120 19%This ringing results because the magnitude of the Fourier coefficientsis only decreasing as1k(while they decreased as1k2for the triangle).You can decrease (and even eliminate the ringing) by decreasing themagnitudes of the Fourier coefficients at higher frequencies.Fourier Series: SummaryFourier 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 waveHowever, convergence as # harmonics increases can be complicated.FilteringThe output of an LTI system is a “filtered” version of the input.Input: Fourier series → sum of complex exponentials.x(t) = x(t + T ) =∞Xk=−∞akej2πTktComplex exponentials: eigenfunctions of LTI systems.ej2πTkt→ H(j2πTk)ej2πTktOutput: same eigenfunctions, amplitudes/phases set by system.x(t) =∞Xk=−∞akej2πTkt→ y(t) =∞Xk=−∞akH(j2πTk)ej2πTktFilteringNotion 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+−vi+vo−RC6.003: Signals and Systems Lecture 14 March 30, 20105Lowpass 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 +
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