ei331 signals and systems
play

EI331 Signals and Systems Lecture 20 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 20 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University May 7, 2019 Contents 1. Magnitude-phase Representation of Fourier Transform 2. Uncertainty Principle 3. Relations Among


  1. EI331 Signals and Systems Lecture 20 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University May 7, 2019

  2. Contents 1. Magnitude-phase Representation of Fourier Transform 2. Uncertainty Principle 3. Relations Among Fourier Representations 1/22

  3. Magnitude-phase Representation of Fourier Transform X ( e j ω ) = | X ( e j ω ) | e j arg X ( e j ω ) X ( j ω ) = | X ( j ω ) | e j arg X ( j ω ) , Recall Fourier transform is decomposition of signal into superposition of complex exponentials (“waves”) • | X | gives magnitudes of components • arg X gives phases of components Phase arg X contains substantial information about signal • determines whether components add constructively or destructively • small change can lead to very differential-looking signals for same magnitude spectrum 2/22

  4. Importance of Phase Information x ( t ) = 1 + 1 2 cos( 2 π t + φ 1 ) + cos( 4 π t + φ 2 ) + 2 3 cos( 6 π t + φ 3 ) x 1 ( t ) φ 1 = 0 , φ 2 = 0 , φ 3 = 0 t x 2 ( t ) φ 1 = 4 , φ 2 = 8 , φ 3 = 12 t x 3 ( t ) φ 1 = 6 , φ 2 = − 2 . 7 , φ 3 = 0 . 93 t x 4 ( t ) φ 1 = 1 . 2 , φ 2 = 4 . 1 , φ 3 = − 7 . 02 t 3/22

  5. 6000000 5000000 4000000 3000000 2000000 1000000 0 0 200 400 600 800 4 3 2 1 0 1 2 3 4 0 200 400 600 800 frequency (Hz) Magnitude vs. Phase Waveform x for Chinese word “ ” Magnitude and phase spectra | X | , arg X (DFT) 6000000 5000000 4000000 3000000 2000000 1000000 0 0 5000 10000 15000 20000 4 3 2 1 0 1 2 3 4 0 5000 10000 15000 20000 frequency (Hz) 4/22

  6. Magnitude vs. Phase Waveform x for Chinese word “ ” Magnitude and phase spectra | X | , arg X (DFT) 6000000 6000000 5000000 5000000 4000000 4000000 3000000 3000000 2000000 2000000 1000000 1000000 0 0 0 2000 4000 6000 8000 10000 0 200 400 600 800 4 4 3 3 2 2 1 1 0 0 1 1 2 2 3 3 4 4 0 2000 4000 6000 8000 10000 0 200 400 600 800 frequency (Hz) frequency (Hz) 4/22

  7. Magnitude vs. Phase Waveform x for Chinese word “ ” Waveform reconstructed by magnitude spectra only F − 1 {| X |} Waveform reconstructed by phase spectra only F − 1 { e j arg X } 5/22

  8. Magnitude vs. Phase Top row X , | X | , arg X Bottom row F − 1 {| X |} F − 1 { e j arg X } 6/22

  9. Magnitude-phase Representation of Frequency Response For LTI systems Y ( e j ω ) = H ( e j ω ) X ( e j ω ) Y ( j ω ) = H ( j ω ) X ( j ω ) , Thus | Y | = | H | · | X | , | H | called gain of system and arg Y = arg H + arg X , arg H called phase shift of system Effects of LTI system may or may not be desirable • want specific effects for filtering • if undesirable, effects called distortion Example. Distortionless transmission • ideally, H ( j ω ) = 1 , but noncausal • H ( j ω ) = Ke − j ω t 0 , preserves shape, only scaling + delay 7/22

  10. Linear Phase For CT LTI system with unit gain and linear phase H ( j ω ) = e − j ω t 0 = ⇒ y ( t ) = x ( t − t 0 ) output is delayed version of input For DT LTI system with unit gain and linear phase H ( e j ω ) = e − j ω n 0 , output � π ∞ y [ n ] = 1 � X ( e j ω ) e − j ω n 0 e j ω n d ω = x [ m ] sinc( n − n 0 − m ) 2 π − π m = −∞ • for integer n 0 , y [ n ] = x [ n − n 0 ] is delayed version of input • for non-integer n 0 , y [ n ] = y c ( n − n 0 ) is sample of delayed ∞ version of envelope y c ( t ) = � x [ m ] sinc( t − m ) of x m = −∞ 8/22

  11. Linear Phase H ( j ω ) = e − j ω/ 2 For input 3 � x ( t ) = cos( 2 k π t ) = 1 + cos( 2 π t ) + cos( 4 π t ) + cos( 6 π t ) k = 0 output is 3 cos( 2 k π t − k π ) = x ( t − 1 � y ( t ) = 2 ) k = 0 x ( t ) t y ( t ) t 9/22

  12. Linear Phase H ( e j ω ) = e − j ω/ 2 Half-sample delay For input x [ n ] = cos( π 3 n ) output is ∞ cos( π 3 m ) sinc( n − n 0 − m ) = cos( π 3 [ n − 1 � y [ n ] = 2 )] m = −∞ x [ n ] n y [ n ] n 10/22

  13. Nonlinear Phase H ( j ω ) = e − j arctan ω For input 3 � x ( t ) = cos( 2 k π t ) = 1 + cos( 2 π t ) + cos( 4 π t ) + cos( 6 π t ) k = 0 output is 3 � y ( t ) = cos[ 2 k π t − arctan( 2 k π )] k = 0 x ( t ) t y ( t ) t 11/22

  14. Group Delay For narrowband input x centered at ω 0 , i.e. X ( j ω ) = 0 , for | ω − ω 0 | > ∆ ω, where ∆ ω ≪ 1 Use linear approximation for phase arg H ( j ω ) ≈ arg H ( j ω 0 ) − τ ( ω 0 )( ω − ω 0 ) = φ 0 − τ ( ω 0 ) ω where group delay at ω is τ ( ω ) = − d d ω arg H ( j ω ) If | H ( j ω ) | ≈ | H ( j ω 0 ) | for | ω − ω 0 | ≤ ∆ ω , Y ( j ω ) ≈ | H ( j ω 0 ) | X ( j ω ) e j φ 0 − j τ ( ω 0 ) ω = ⇒ y ( t ) ≈ | H ( j ω 0 ) | e j φ 0 x ( t − τ ( ω 0 )) 12/22

  15. Group Delay ⇒ τ ( ω ) = d 1 H ( j ω ) = e − j arctan ω = d ω arctan( ω ) = 1 + ω 2 For input (sum of two narrowband signals z , ¯ z centered at ± π ) 11 11 cos( k π 10 t ) = Re z ( t ) = 1 2 z ( t )+ 1 e j k π � � 10 t x ( t ) = 2 ¯ z ( t ) , where z ( t ) = k = 9 k = 9 output y ( t ) ≈ 1 2 e j φ 0 z ( t − τ 0 ) + 1 2 e − j φ 0 ¯ e j φ 0 z ( t − τ 0 ) � � z ( t − τ 0 ) = Re π 1 where φ 0 = − arctan π + 1 + π 2 , τ 0 = 1 + π 2 . x ( t ) t y ( t ) t 13/22

  16. Contents 1. Magnitude-phase Representation of Fourier Transform 2. Uncertainty Principle 3. Relations Among Fourier Representations 14/22

  17. Uncertainty Principle Assume CT signal x ∈ L 2 ( R ) , so X = F { x } ∈ L 2 ( R ) Define normalized power density in time and frequency | x ( t ) | 2 | X ( j ω ) | 2 p ( t ) = P ( ω ) = R | x ( τ ) | 2 d τ, � � R | X ( j θ ) | 2 d θ NB. p and P can be interpreted as probability densities, as done in quantum mechanics x and X are centered at t 0 and ω 0 resp. in the sense � � t 0 = tp ( t ) dt , ω 0 = ω P ( ω ) d ω R R “Standard deviation” measures energy spread around center � 1 � 1 �� �� 2 2 ( t − t 0 ) 2 p ( t ) dt ( ω − ω 0 ) 2 P ( ω ) d ω ∆ t = ∆ ω = , R R 15/22

  18. Uncertainty Principle Theorem. If x ( t ) ∈ L 2 ( R ) with Fourier transform X ( j ω ) , then ∆ t ∆ ω ≥ 1 2 with equality iff x is Gaussian In fact, the following slightly more general relation holds D a ( x ) D b ( X ) ≥ 1 2 � x � 2 · � X � 2 where for g ∈ L 2 ( R ) and a ∈ R , � 1 �� 2 ( ξ − a ) 2 | g ( ξ ) | 2 d ξ D a ( g ) = R NB. Roughly speaking, signals cannot be localized in both time and frequency; short pulse has large bandwidth, narrowband signal has long duration 16/22

  19. Proof of Uncertainty Principle First assume a = b = 0 . � 1 � 1 �� �� 2 2 ω 2 | X ( j ω ) | 2 dt | j ω X ( j ω ) | 2 d ω D 0 ( X ) = = R R F Since x ′ ( t ) ← − − → j ω X ( j ω ) , Parseval’s identity yields � 1 � � 2 | x ′ ( t ) | 2 dt D 0 ( X ) = 2 π R By Cauchy-Schwarz inequality � 1 � 1 √ �� 2 �� 2 | tx ( t ) | 2 dt | x ′ ( t ) | 2 dt D 0 ( x ) D 0 ( X ) = 2 π R R √ √ � � � � � � � � � � tx ∗ ( t ) x ′ ( t ) dt tx ∗ ( t ) x ′ ( t ) dt ≥ 2 π � ≥ 2 π � Re � � � � � � R R 17/22

  20. Proof of Uncertainty Principle Note � � � � td | x ( t ) | 2 tx ∗ ( t ) x ′ ( t ) dt = tx ∗ ( t ) x ′ ( t ) dt + tx ( t ) x ′ ( t ) dt = 2 Re R R R R Integration by parts yields � � tx ∗ ( t ) x ′ ( t ) dt = t | x ( t ) | 2 � ∞ | x ( t ) | 2 dt 2 Re −∞ − � � R R Since inequality is trivial if D 0 ( x ) = ∞ , can assume D 0 ( x ) < ∞ , so t | x ( t ) | 2 → 0 as t → ∞ . Thus √ 2 π 2 = 1 � x � 2 D 0 ( x ) D 0 ( X ) ≥ 2 � x � 2 · � X � 2 2 For a , b � = 0 , note D a ( x ) = D 0 ( y ) and D b ( X ) = D 0 ( Y ) for F → Y ( j ω ) = X ( j ( ω + b )) e j ( ω + b ) a y ( t ) = x ( t + a ) e − jbt ← − − 18/22

  21. Contents 1. Magnitude-phase Representation of Fourier Transform 2. Uncertainty Principle 3. Relations Among Fourier Representations 19/22

  22. Four Fourier Representations CT Fourier series CT Fourier transform x [ k ] = 1 � � x ( t ) e − j 2 π T kt dt x ( t ) e − j ω t dt ˆ X ( j ω ) = T T R x ( t ) = 1 � � x [ k ] e j 2 π T kt x ( t ) = x ( t + T ) = ˆ X ( j ω ) e j ω t d ω 2 π k ∈ Z R DT Fourier series DT Fourier transform x [ k ] = 1 � X ( e j ω ) = x [ n ] e − j ω n � x [ n ] e − j 2 π N kn ˆ N n ∈ Z n ∈ [ N ] � x [ n ] = 1 x [ k ] e j 2 π X ( e j ω ) e j ω n d ω � N kn x [ n ] = x [ n + N ] = ˆ 2 π 2 π k ∈ [ N ] DFT is one period of DTFS 20/22

  23. Relations among Four Fourier Representations time frequency CTFS continuous periodic discrete aperiodic CTFT continuous aperiodic continuous aperiodic DTFS discrete periodic discrete periodic DTFT discrete aperiodic continuous periodic Observations • periodic in one domain ⇐ ⇒ discrete in other domain • discretization by sampling in one domain ⇐ ⇒ periodic extension in other domain • continualization by interpolation in one domain ⇐ ⇒ extraction of one period in other domain 21/22

  24. Relations among Four Fourier Representations T → ∞ (extract one period) CTFS CTFT periodic extension (sampling in frequency) sample interpolate sample interpolate N → ∞ (extract one period) DTFS DTFT periodic extension (sampling in frequency) NB. Conditions apply in some cases. 22/22

Recommend


More recommend