Circuits and Systems I CSI lecture 3 2013

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 ßà60 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

Prof Dr Volkan Cevher

LIONS/ Laboratory for Information and Inference Systems lions @epfl

License Info for SPFirst Slides

•  This work released under a

Creative Commons License with the following

•  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

Lecture Objectives

•  Sinusoids with DIFFERENT frequencies

–  SYNTHESIZE by Adding Sinusoids

=

N

k

k k

k f t A

A t

Another FREQ Diagram

•  Plot Complex Amplitude vs Freq

–100 –250

f (in Hz)

3 /

7 e j π 7 e − j π / 3

2 /

=

N

k

k k

k f t A

A t

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

§  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)

) sin(

) cos( t j t

) sin(

)

e − j ω t = − ω + − ω

) sin(

) cos( t j t

e − j ω t = ω − ω

) cos(

e

e j ω t + − j ω t = ω

) (

) cos( ω t = 2 1 e j ω t + e − j ω t

INVERSE Euler’s Formula

•  Solve for cosine (or sine)

) (

) cos( ω t = 1 2 e j ω t + e − j ω t

) (

) sin( ω t = 2 1 j e j ω t − e − j ω t

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

t j

A t

j

t

A cos( 7 ) = 2 7 + 2 − 7

SPECTRUM of SINE

•  Sine = sum of 2 complex exponentials:

–  Positive freq has phase = -0.5π

t j j

t j j

t

j j

A t

j j

A

e Ae

e Ae

e e

t

A

7 5

.

0 2

1 7

5

0 2

1

7 2

7 2

) 7

π

5 0

j = j = e

−

Negative Frequency

•  Is negative frequency real?

•  Doppler Radar provides an example

–  Police radar measures speed by using the Doppler shift principle

–  Let’s assume 400Hz ßà60 mph

–   +400Hz means towards the radar

–   -400Hz means away (opposite direction )

–  Think of a train whistle

t j

j e Ae e Ae

t

A sin( 7 ) = 21 − 0.5π 7 + 21 0.5π − 7

π

5

0 2

SPECTRUM 2 SINUSOID

•  Add the spectrum components:

What is the formula for the signal x(t)?

–100 –250

f (in Hz)

3 /

7 e j π 7 e − j π / 3

2 /

10

Gather (A,ω,φ) information

DC is another name for zero-freq component

DC component always has φ=0 or π (for real x(t) )

Note the conjugate phase

Add Spectrum Components-1

•  Amplitude & Phase

j t

j j

t j

j t

j j

e e

e e

e e

e e

t

x

) 250 (

2 2

/ )

250 (

2 2

/

) 100 (

2 3

/ )

100 (

2 3

/

4 4

7 7

10 )

(

π π

π π

π π

π π

=

Add Spectrum Components-2

t j

j t

j j

t j

j t

j j

e e

e e

e e

e e

t

x

) 250 (

2 2

/ )

250 (

2 2

/

) 100 (

2 3

/ )

100 (

2 3

/

4 4

7 7

10 )

(

π π

π π

π π

π π

+

=

–100 –250

f (in Hz)

3 /

7 e j π 7 e − j π / 3

2 /

10

Use Euler’s Formula to get REAL sinusoids:

Simplify Components

t j

j t

j j

t j

j t

j j

e e

e e

e e

e e

t

x

) 250 (

2 2

/ )

250 (

2 2

/

) 100 (

2 3

/ )

100 (

2 3

/

4 4

7 7

10 )

(

π π

π π

π π

π π

+

=

t j j

t j

j e Ae e Ae

N k

k k

A A

250 (

2 cos(

8

) 3 / )

100 (

2 cos(

14 10

)

(

π π

π

π

+ +

− +

=

t

t t

x

Summary: General Form

∗+

=

ℜ e { z } 21 z 21 z k

j k

k

f

e A

=

= Frequency

=

N

k

t f

j

ke kX

e X

t

x

1

20

j k

t f

j

ke k X e kX

X t

x

1

22

1

22

10

=

N

k

k k

k f t A

A t

Example: Synthetic Vowel

•  Sum of 5 Frequency Components

SPECTRUM of VOWEL

–  Conjugates in negative frequency

SPECTRUM of VOWEL (Polar

Format)

φ k

0.5Ak

Vowel Waveform (sum of all 5 components)

Problem Solving Skills

•   MATLAB

–  Numerical –  Computation –  Plotting list of numbers

Lecture Objectives

•  Signals with HARMONIC Frequencies

FREQUENCY can change vs TIME

=

N k

k

A A

–100 –250

f (in Hz)

3 /

7 e j π 7 e − j π / 3

2 /

10

) 2 / )

250 (

2 cos(

8

) 3 / )

100 (

2 cos(

14 10

)

(

π π

π

π

+ +

− +

=

t

t t

x

k

j k

Spectrum for Periodic Signals?

•  Nearly Periodic in the Vowel Region

) ( t x t T

) 3 ( cos )

Period of Complex Exponentials

Definition: Period is T

k = integer

t j

T t

? ) ( )

(

)

(

t x T

t

x

e t

= +

e π

k T

⇒

k

k T

=

+ +

=

N k

t kf

j k

t kf

j k

j k k

N k

k k

e X

e X

X t

x

e A X

t kf A

A t

x

k

1

2 2

1

2 2

1 0

1

0 0

0 0

) (

) 2

cos(

) (

π π

ϕ

ϕ π

Harmonic Signal Spectrum

0

: have only

can signal

T

f 0 = 1

Define Fundamental Frequency

l fundamenta

(largest) Frequency

l fundamenta

) 2

(

) 2

0

1

0 0

k f

t kf A

A t

x

k

N k

k k

π ω

ϕ π

What is the fundamental frequency?

Harmonic Signal (3 Freqs)

3rd

5th

10 Hz

POP QUIZ: Fundamental Freq

•  Here’s another spectrum:

What is the fundamental frequency?

–100 –250

f (in Hz)

3 /

7 e j π 7 e − j π / 3

2 /

10

SPECIAL RELATIONSHIP

to get a PERIODIC SIGNAL

IRRATIONAL SPECTRUM

Harmonic Signal (3 Freqs)

T=0.1

NON-Harmonic Signal

NOT PERIODIC

§   Can a machine extract frequencies?

§   Yes, if we COMPUTE the spectrum for x(t)

§   During short intervals

A Simple Test Signal

•  C-major SCALE: stepped frequencies

–  Frequency is constant for each note

IDEAL

R-rated: ADULTS ONLY

•  SPECTROGRAM Tool

•  ANALYSIS program

–  Takes x(t) as input &

§  Then uses the FFT (Fast Fourier Transform)

R-rated: ADULTS ONLY

•  SPECTROGRAM Tool

•  ANALYSIS program

–  Takes x(t) as input &

§  Then uses the FFT (Fast Fourier Transform)

CSI

Progress

Level:

Spectrogram Example

•  Two Constant Frequencies: Beats

) ) 12 ( 2 sin(

) ) 660 (

2 cos( π t π t

( ) ( j t j t )

j

t j

t

e 2 (660) 2 (660) 21 2 (12) 2 (12)2

− +

AM Radio Signal

•  Same as BEAT Notes

) ) 12 (

2 sin(

) ) 660 (

2

) )

648 (

2 cos(

) )

672 (

2

22

1 π t − π + π t + π

j e 2 (672) e 2 (672) e 2 (648) e 2 (648)

Spectrum of AM (Beat)

•  4 complex exponentials in AM:

What is the fundamental frequency?

f (in Hz) –672 –648

2

/ 4

1 e jπ 41 e− jπ /2

2

/ 4

Stepped Frequencies

•  C-major SCALE: successive sinusoids

–  Frequency is constant for each note

IDEAL

Spectrogram of LAB SONG

ARTIFACTS at Transitions Sinusoids ONLY

Analysis Frame = 40ms

Time-Varying Frequency

•  Frequency can change vs time

–  Continuously, not stepped

•  FREQUENCY MODULATION (FM)

•  CHIRP SIGNALS

–  Linear Frequency Modulation (LFM)

)) (

2 cos(

)

VOICE

) 2

cos(

) ( t = A α t 2 + π f 0 t + ϕ x

New Signal: Linear FM

•  Called Chirp Signals (LFM)

–  Quadratic phase

•  Freq will change LINEARLY vs time

–  Example of Frequency Modulation (FM)

–  Define “instantaneous frequency”

QUADRATIC

)) (

cos(

)

(

t t

t A

t x

0

2 )

( )

(

2 )

(

) 2

cos(

) (

f t

t

t f t

t f A

t x

dt d

ω

ϕ π

ψ

ϕ π

Instantaneous Frequency of the

Chirp

•  Chirp Signals have Quadratic phase

•  Freq will change LINEARLY vs time

ϕ β

α ψ

ϕ β

α

+ +

=

⇒

+ +

=

t t

t

t t

A t

x

2

2

) (

) cos(

) (

β α

ψ

⇒ i ( t ) dt d ( t ) 2 t

Chirp Spectrogram

Chirp Waveform

) ( )

⇒

Sine-Wave Frequency Modulation (FM)

Look at CD-ROM Demos in Ch 3

•   Next week <> Section 3-4

Section 3-5 Section 3-6 Lab 2