Circuits and Systems I LECTURE 3 The Spectrum, Periodic Signals, and the TimeVarying Spectrum Prof. Dr. Volkan Cevher lionsepfl LIONSLaboratory for Information and Inference SystemsLicense Info for SPFirst Slides • This work released under a Creative Commons License with the following terms: • Attribution § The licensor permits others to copy, distribute, display, and perform the work. In return, licensees must give the original authors credit. • NonCommercial § The licensor permits others to copy, distribute, display, and perform the work. In return, licensees may not use the work for commercial purposes—unless they get the licensors permission. • Share Alike § The licensor permits others to distribute derivative works only under a license identical to the one that governs the licensors work. • Full Text of the License • This (hidden) page should be kept with the presentationOutline Today • Today Section 31 – 33 Section 37 Section 38 Next week Section 34 Section 35 Section 36 Lab 2 CSI Progress Level: READLecture Objectives • Sinusoids with DIFFERENT frequencies – SYNTHESIZE by Adding Sinusoids • SPECTRUM Representation – Graphical Form shows DIFFERENT Freqs ∑= = + + N k x t A Ak fkt k 1 ( ) 0 cos(2π ϕ ) CSI Progress Level:FREQUENCY DIAGRAM Frequency is the vertical axis Time is the horizontal axis A440Another FREQ. Diagram • Plot Complex Amplitude vs. Freq –250 –100 0 100 250 f (in Hz) 3 7e jπ 7e− jπ 3 2 4e− jπ 4e jπ 2 10 ∑= = + + N k x t A Ak fkt k 1 ( ) 0 cos(2π ϕ )Motivation • Synthesize Complicated Signals – Musical Notes § Piano uses 3 strings for many notes § Chords: play several notes simultaneously – Human Speech § Vowels have dominant frequencies § Application: computer generated speech – Can all signals be generated this way? § Sum of sinusoids?Fur Elise WAVEFORM Beat NotesSpeech Signal: BAT • Nearly Periodic in Vowel Region – Period is (Approximately) T = 0.0065 secEuler’s Formula Reversed • Solve for cosine (or sine) e jωt = cos(ωt) + j sin(ωt) e− jωt = cos(−ωt) + jsin(−ωt) e− jωt = cos(ωt) − j sin(ωt) e jωt + e− jωt = 2cos(ωt) cos(ωt) = 12 (e jωt + e− jωt)INVERSE Euler’s Formula • Solve for cosine (or sine) cos(ωt) = 12 (e jωt + e− jωt) sin(ωt) = 21j (e jωt − e− jωt)SPECTRUM Interpretation • Cosine = sum of 2 complex exponentials: One has a positive frequency The other has negative freq. Amplitude of each is half as big A j t A j t A t e e 7 2 7 cos(7 ) = 2 + −SPECTRUM of SINE • Sine = sum of 2 complex exponentials: – Positive freq. has phase = 0.5π – Negative freq. has phase = +0.5π j j t j j t j t Aj j t Aj Ae e Ae e A t e e 0.5 7 12 0.5 7 12 7 2 7 sin(7 ) 2 − − − = + = − π π 1 j0.5π −j = j = eNegative Frequency • Is negative frequency real? • Doppler Radar provides an example – Police radar measures speed by using the Doppler shift principle – Let’s assume 400Hz ßà60 mph – +400Hz means towards the radar – 400Hz means away (opposite direction) – Think of a train whistleGraphical Spectrum EXAMPLE of SINE AMPLITUDE, PHASE FREQUENCY are shown 7 0 7 ω j j t j j t A t Ae e Ae 0.5 e 7 12 0.5 7 12 sin(7 ) = − π + π − 0.5π 12 ( A)e j (12 A)e− j0.5πSPECTRUM2SINUSOID • Add the spectrum components: What is the formula for the signal x(t)? –250 –100 0 100 250 f (in Hz) 3 7e jπ 7e− jπ 3 2 4e− jπ 4e jπ 2 10Gather (A,ω,φ) information • Frequencies: – 250 Hz – 100 Hz – 0 Hz – 100 Hz – 250 Hz • Amplitude Phase – 4 π2 – 7 +π3 – 10 0 – 7 π3 – 4 +π2 DC is another name for zerofreq component DC component always has φ=0 or π (for real x(t) ) Note the conjugate phaseAdd Spectrum Components1 • Amplitude Phase – 4 π2 – 7 +π3 – 10 0 – 7 π3 – 4 +π2 • Frequencies: – 250 Hz – 100 Hz – 0 Hz – 100 Hz – 250 Hz j j t j j t j j t j j t e e e e e e e e x t 2 2 (250) 2 2 (250) 3 2 (100) 3 2 (100) 4 4 7 7 ( ) 10 π π π π π π π π − − − − + + = +Add Spectrum Components2 j j t j j t j j t j j t e e e e e e e e x t 2 2 (250) 2 2 (250) 3 2 (100) 3 2 (100) 4 4 7 7 ( ) 10 π π π π π π π π − − − − + + = + –250 –100 0 100 250 f (in Hz) 3 7e jπ 7e− jπ 3 2 4e− jπ 4e jπ 2 10Use Euler’s Formula to get REAL sinusoids: Simplify Components j j t j j t j j t j j t e e e e e e e e x t 2 2 (250) 2 2 (250) 3 2 (100) 3 2 (100) 4 4 7 7 ( ) 10 π π π π π π π π − − − − + + = + j j t j j t Acos(ωt +ϕ) = 12 Ae+ ϕe ω + 12 Ae− ϕe− ωFinal Answer So, we get the general form: ∑= = + + N k x t A Ak fkt k 1 ( ) 0 cos(2π ϕ ) 8cos(2 (250) 2) ( ) 10 14 cos(2 (100) 3) π π π π + + = + − tt x tSummary: General Form ∗ ℜe z = z + z 12 12 { } k j k k f X A e k = = Frequency ∑ { } ϕ = = + ℜ N k j f t k x t X e X e k 1 2 ( ) 0 π ∑{ } = ∗ − = + + N k j f t k j f t k x t X X e k X e k 1 2 12 2 12 ( ) 0 π π ∑= = + + N k x t A Ak fkt k 1 ( ) 0 cos(2π ϕ )Example: Synthetic Vowel • Sum of 5 Frequency ComponentsSPECTRUM of VOWEL – Note: Spectrum has 0.5Xk (except XDC) – Conjugates in negative frequencySPECTRUM of VOWEL (Polar Format) φk 0.5A kVowel Waveform (sum of all 5 components)Problem Solving Skills • Math Formula – Sum of Cosines – Amp, Freq, Phase • Recorded Signals – Speech – Music – No simple formula • Plot Sketches – S(t) versus t – Spectrum • MATLAB – Numerical – Computation – Plotting list of numbersLecture Objectives • Signals with HARMONIC Frequencies – Add Sinusoids with fk = kf0 FREQUENCY can change vs. TIME Chirps: Introduce Spectrogram Visualization (specgram.m) (plotspec.m) x(t)= cos(αt2) ∑= = + + N k x t A Ak kf t k 1 ( ) 0 cos(2π 0 ϕ )Spectrum Diagram • Recall Complex Amplitude vs. Freq 12 Xk = ak –250 –100 0 100 250 f (in Hz) 3 7e jπ 7e− jπ 3 2 4e− jπ 4e jπ 2 10 8cos(2 (250) 2) ( ) 10 14 cos(2 (100) 3) π π π π + + = + − tt x t j k Xk = Ake ϕ ∗k X 12Spectrum for Periodic Signals? • Nearly Periodic in the Vowel Region – Period is (Approximately) T = 0.0065 secPeriodic Signals • Repeat every T secs – Definition – Example: – Speech can be “quasiperiodic” x(t) = x(t + T) x(t) = cos2(3t) T = ? π3 T = 3 2π T =Period of Complex Exponentials Definition: Period is T k = integer j t T j t e e ω ω = ( + ) ( ) ( ) ? ( ) x t T x t x t e j t + = = ω e j2πk = 1 ⇒ e jωT =1 ⇒ωT = 2πk k k T T k 0 2 2 ω π π ω = ⎞⎟⎠ ⎛⎜⎝ = =∑{ } ∑ = ∗ − = = + + = = + + N k j kf t k j kf t k j k k N k k k x t X X e X e X A e x t A A kf t k 1 2 12 2 12 0 1 0 0 ( ) 0 0 ( ) cos(2 ) π π ϕ π ϕ Harmonic Signal Spectrum Periodic signal can only have : fk = k f0 T f0 = 1Define Fundamental Frequency 0 0 1T f = fundamental Period (shortest) fundamental Frequency (largest) ( 2 ) ( ) cos(2 ) 0 0 0 0 0 1 0 0 = = = = = +∑ + = f T f k f f x t A A kf t k N k k k ω π π ϕWhat is the fundamental frequency? Harmonic Signal (3 Freqs) 3rd 5th 10 HzPOP QUIZ: Fundamental Freq. • Here’s another spectrum: What is the fundamental frequency? 100 Hz ? 50 Hz ? –250 –100 0 100 250 f (in Hz) 3 7e jπ 7e− jπ 3 2 4e− jπ 4e jπ 2 10SPECIAL RELATIONSHIP to get a PERIODIC SIGNAL IRRATIONAL SPECTRUMHarmonic Signal (3 Freqs) T=0.1NONHarmonic Signal NOT PERIODICFrequency Analysis • Now, a much HARDER problem • Given a recording of a song, have the computer write the music § Can a machine extract frequencies? § Yes, if we COMPUTE the spectrum for x(t) § During short intervalsTimeVarying FREQUENCIES Diagram Frequency is the vertical axis Time is the horizontal axis A440A Simple Test Signal • Cmajor SCALE: stepped frequencies – Frequency is constant for each note IDEALRrated: ADULTS ONLY • SPECTROGRAM Tool – MATLAB function is specgram.m – SPFirst has plotspec.m spectgr.m • ANALYSIS program – Takes x(t) as input – Produces spectrum values Xk – Breaks x(t) into SHORT TIME SEGMENTS § Then uses the FFT (Fast Fourier Transform)Rrated: ADULTS ONLY • SPECTROGRAM Tool – MATLAB function is specgram.m – SPFirst has plotspec.m spectgr.m • ANALYSIS program – Takes x(t) as input – Produces spectrum values Xk – Breaks x(t) into SHORT TIME SEGMENTS § Then uses the FFT (Fast Fourier Transform) CSI Progress Level:Spectrogram Example • Two Constant Frequencies: Beats cos(2π (660)t)sin(2π (12)t)12 (e j2π (660)t + e− j2π (660)t )21j (e j2π (12)t − e− j2π (12)t ) AM Radio Signal • Same as BEAT Notes cos(2π (660)t)sin(2π (12)t) 12 cos(2π(672)t − π2 ) + 12 cos(2π(648)t + π2 ) 41j (e j2π (672)t − e− j2π (672)t − e j2π (648)t + e− j2π (648)t)Spectrum of AM (Beat) • 4 complex exponentials in AM: What is the fundamental frequency? 648 Hz ? 24 Hz ? 0 648 672 –672 –648 f (in Hz) 2 14 jπ e 2 14 jπ e 2 − 14 jπ e 2 − 14 jπ eStepped Frequencies • Cmajor SCALE: successive sinusoids – Frequency is constant for each note IDEALSpectrogram of CScale ARTIFACTS at Transitions Sinusoids ONLY From SPECGRAM ANALYSIS PROGRAMSpectrogram of LAB SONG ARTIFACTS at Transitions Sinusoids ONLY Analysis Frame = 40msTimeVarying Frequency • Frequency can change vs. time – Continuously, not stepped • FREQUENCY MODULATION (FM) • CHIRP SIGNALS – Linear Frequency Modulation (LFM) x(t) = cos(2π fct + v(t)) VOICEx(t) = Acos(αt2 + 2π f0t +ϕ) New Signal: Linear FM • Called Chirp Signals (LFM) – Quadratic phase • Freq will change LINEARLY vs. time – Example of Frequency Modulation (FM) – Define “instantaneous frequency” QUADRATICInstantaneous Frequency • Definition • For Sinusoid: Derivative ( ) ( ) of the “Angle” ( ) cos( ( )) t t x t A t dt d ωi ψ ψ ⇒ = = Makes sense 0 0 0 ( ) ( ) 2 ( ) 2 ( ) cos(2 ) t t f t f t x t A f t dt d ωi ψ π ψ π ϕ π ϕ ⇒ = = = + = +Instantaneous Frequency of the Chirp • Chirp Signals have Quadratic phase • Freq will change LINEARLY vs. time ψ α β ϕ α β ϕ ⇒ = + + = + + t t t x t A t t 2 2 ( ) ( ) cos( ) ⇒ωi(t) = ddtψ(t) = 2αt + βChirp SpectrogramChirp WaveformOTHER CHIRPS • ψ(t) can be anything: • ψ(t) could be speech or music: – FM radio broadcast x(t) = Acos(α cos(βt) +ϕ) ⇒ωi(t) = dt d ψ(t) = −αβsin(βt)SineWave Frequency Modulation (FM) Look at CDROM Demos in Ch 3• Next week Section 34 Section 35 Section 36 Lab 2
Circuits and Systems I LECTURE #3 The Spectrum, Periodic Signals, and the Time-Varying Spectrum lions@epfl Prof Dr Volkan Cevher LIONS/Laboratory for Information and Inference Systems License Info for SPFirst Slides • This work released under a Creative Commons License with the following terms: ã Attribution Đ The licensor permits others to copy, distribute, display, and perform the work In return, licensees must give the original authors credit ã Non-Commercial Đ The licensor permits others to copy, distribute, display, and perform the work In return, licensees may not use the work for commercial purposes—unless they get the licensor's permission • Share Alike § The licensor permits others to distribute derivative works only under a license identical to the one that governs the licensor's work • Full Text of the License • This (hidden) page should be kept with the presentation Outline - Today • Today Next week CSI Progress Level: Section 3-1 – 3-3 Section 3-7 Section 3-8 Section 3-4 Section 3-5 Section 3-6 Lab READ Lecture Objectives • Sinusoids with DIFFERENT frequencies – SYNTHESIZE by Adding Sinusoids N x(t ) = A0 + ∑ Ak cos( 2π f k t + ϕ k ) k =1 • SPECTRUM Representation – Graphical Form shows DIFFERENT Freqs CSI Progress Level: FREQUENCY DIAGRAM Frequency is the vertical axis A-440 Time is the horizontal axis Another FREQ Diagram • Plot Complex Amplitude vs Freq N x(t ) = A0 + ∑ Ak cos( 2π f k t + ϕ k ) 4e − jπ / –250 7e jπ / –100 k =1 10 7e − jπ / 100 4e jπ / 250 f (in Hz) Motivation • Synthesize Complicated Signals – Musical Notes § Piano uses strings for many notes § Chords: play several notes simultaneously – Human Speech § Vowels have dominant frequencies § Application: computer generated speech – Can all signals be generated this way? § Sum of sinusoids? Fur Elise WAVEFORM Beat Notes Speech Signal: BAT • Nearly Periodic in Vowel Region – Period is (Approximately) T = 0.0065 sec Euler’s Formula Reversed • Solve for cosine (or sine) e jω t = cos(ω t ) + j sin(ω t ) e − jω t = cos( −ω t ) + j sin(−ω t ) e − jω t = cos(ω t ) − j sin(ω t ) e jω t +e − jω t cos(ω t ) = = cos(ω t ) (e jω t +e − jω t ) Spectrogram Example • Two Constant Frequencies: Beats cos( 2π (660)t ) sin( 2π (12)t ) AM Radio Signal • Same as BEAT Notes cos( 2π (660)t ) sin( 2π (12)t ) (e 4j j 2π ( 660 ) t (e j 2π ( 672 )t + e − j 2π ( 660 ) t −e − j 2π ( 672 )t cos(2π (672)t ) (e 2j −e j 2π (12) t j 2π ( 648)t − e − j 2π (12)t ) − j 2π ( 648)t ) +e − π2 ) + 12 cos(2π (648)t + π2 ) Spectrum of AM (Beat) • complex exponentials in AM: e jπ / –672 e − jπ / –648 What is the fundamental frequency? 648 Hz ? 24 Hz ? e jπ / 648 e − jπ / 672 f (in Hz) Stepped Frequencies • C-major SCALE: successive sinusoids – Frequency is constant for each note IDEAL Spectrogram of C-Scale Sinusoids ONLY From SPECGRAM ANALYSIS PROGRAM ARTIFACTS at Transitions Spectrogram of LAB SONG Sinusoids ONLY Analysis Frame = 40ms ARTIFACTS at Transitions Time-Varying Frequency • Frequency can change vs time – Continuously, not stepped • FREQUENCY MODULATION (FM) x(t ) = cos(2π f c t + v(t )) VOICE • CHIRP SIGNALS – Linear Frequency Modulation (LFM) New Signal: Linear FM • Called Chirp Signals (LFM) – Quadratic phase QUADRATIC x(t ) = A cos(α t + 2π f t + ϕ ) • Freq will change LINEARLY vs time – Example of Frequency Modulation (FM) – Define “instantaneous frequency” Instantaneous Frequency • Definition x (t ) = A cos(ψ (t )) d ⇒ ωi (t ) = dt ψ (t ) Derivative of the “Angle” • For Sinusoid: x(t ) = A cos( 2π f 0t + ϕ ) ψ (t ) = 2π f 0t + ϕ ⇒ ωi ( t ) = d ψ (t ) dt = 2π f Makes sense Instantaneous Frequency of the Chirp • Chirp Signals have Quadratic phase • Freq will change LINEARLY vs time x(t ) = A cos(α t + β t + ϕ ) ⇒ ψ (t ) = α t + β t + ϕ ⇒ ωi (t ) = d ψ (t ) = 2α t + β dt Chirp Spectrogram Chirp Waveform OTHER CHIRPS • ψ(t) can be anything: x(t ) = A cos(α cos( β t ) + ϕ ) • ψ(t) could be speech or music: – FM radio broadcast ⇒ ωi (t ) = d ψ ( t ) dt = −αβ sin(β t ) Sine-Wave Frequency Modulation (FM) Look at CD-ROM Demos in Ch • Next week Section 3-4 Section 3-5 Section 3-6 Lab