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EI331 Signals and Systems Lecture 17 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 17 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University April 23, 2019 Contents 1. Sampling Theorem 2. Zero-order Hold and Linear Interpolation 3. Aliasing 1/30 Sampling CT


  1. EI331 Signals and Systems Lecture 17 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University April 23, 2019

  2. Contents 1. Sampling Theorem 2. Zero-order Hold and Linear Interpolation 3. Aliasing 1/30

  3. Sampling CT Signals Sampling converts CT signals to DT signals CT signal DT signal x ( t ) x [ n ] = x ( nT ) n 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 T 9 T 10 T 0 1 2 3 4 5 6 7 8 9 10 t T = sampling period, uniform sampling most common Allows use of digital electronics to process, record, transmit, store, and retrieve CT signals • MP3, digital camera, printer 2/30

  4. Sampling CT Signals Sampling loses information, different signals may have same samples • x 1 ( t ) = cos( π 3 t ) , x 2 ( t ) = cos( 7 π 3 t ) , different • x 1 [ n ] = cos( π 3 n ) = x 2 [ n ] = cos( 7 π 3 n ) , identical x t n Under what conditions can we recover signal from samples? 3/30

  5. Impulse-train Sampling Time domain x ( t ) CT signal x ( t ) t 0 impulse train p ( t ) 1 ∞ � p ( t ) = δ ( t − nT ) t − 5 T − 4 T − 3 T − 2 T − T 0 T 2 T 3 T 4 T 5 T n = −∞ x p ( t ) x p ( t ) = x ( t ) p ( t ) ∞ � = x ( nT ) δ ( t − nT ) t − 5 T − 4 T − 3 T − 2 T − T 0 T 2 T 3 T 4 T 5 T n = −∞ 4/30

  6. Impulse-train Sampling X ( j ω ) Frequency domain A CT signal X ( j ω ) ω − ω M ω M P ( j ω ) impulse train ω s = 2 π ∞ T � P ( j ω ) = ω s δ ( ω − k ω s ) k = −∞ ω − ω s 0 ω s 2 ω s X p ( j ω ) = 1 2 π ( X ∗ P )( ω ) X p ( j ω ) ω s − ω M A ∞ = 1 T � X ( j ( ω − k ω s )) T ω − ω s 0 ω s 2 ω s k = −∞ − ω M ω M 5/30

  7. Impulse-train Sampling X ( j ω ) Frequency domain A band-limited CT signal X ( j ω ) = 0 for | ω | > ω M ω − ω M ω M Sampling frequency ω s = 2 π T X p ( j ω ) ω s − ω M A Case 1: ω s > 2 ω M T no overlap between replicas ω − ω s ω s can recover X by lowpass filtering − ω M ω M X p ( j ω ) A Case 2: ω s < 2 ω M T replicas overlap ω − ω s − 2 ω s ω s 2 ω s cannot recover X − ω M ω M ω s − ω M 6/30

  8. Sampling Theorem Band-limited CT signal x ( t ) whose spectrum X ( j ω ) = 0 for | ω | > ω M is uniquely determined by its samples x ( nT ) , n ∈ Z if ω s � 2 π T > 2 ω M , 2 ω M called Nyquist rate Given { x ( nT ) : n ∈ Z } , x ( t ) can be reconstructed as follows ∞ � 1. construct x p ( t ) = x ( nT ) δ ( t − nT ) n = −∞ 2. send x p through lowpass filter with gain T and cutoff frequency ω c ∈ ( ω M , ω s − ω M ) , i.e. H ( j ω ) = T [ u ( ω + ω c ) − u ( ω − ω c )] 3. filter output x r ( t ) with X r ( j ω ) = X p ( j ω ) H ( j ω ) is same as x ( t ) 7/30

  9. Reconstruction in Frequency Domain X ( j ω ) x ( t ) × H ( j ω ) x r ( t ) A x p ( t ) ω − ω M ω M ∞ X p ( j ω ) � p ( t ) = δ ( t − nT ) ω s − ω M A n = −∞ T ω − ω s ω s − ω M ω M • Nyquist frequency ω M H ( j ω ) • Nyquist rate 2 ω M T Lowpass filter ω − ω c ω c • gain T X r ( j ω ) • cutoff frequency A ω M < ω c < ω s − ω M ω − ω M ω M 8/30

  10. Reconstruction in Time Domain ∞ p ( t ) = � δ ( t − nT ) H ( j ω ) n = −∞ T x p ( t ) x ( t ) × x r ( t ) − ω c ω c Impulse response of lowpass filter h ( t ) = T sin( ω c t ) π t x recovered by band-limited interpolation using sinc function ∞ x ( nT ) T sin( ω c ( t − nT )) � x ( t ) = x r ( t ) = ( x p ∗ h )( t ) = π ( t − nT ) n = −∞ 9/30

  11. Reconstruction in Time Domain x p ( t ) t − 2 T − T 0 T 2 T ω c = ω s 2 = π T x r ( t ) t − T T − 2 T 0 2 T 10/30

  12. Reconstruction in Time Domain Setting ω c = ω s 2 yields Whittaker-Shannon interpolation formula ∞ � t − nT � where sinc( t ) = sin( π t ) � x r ( t ) = x ( nT ) sinc , T π t n = −∞ Since � t − nT � → Te − jnT ω [ u ( ω + π T ) − u ( ω − π F sinc ← − − T )] T Parseval’s identity (or multiplication property) implies π � t − nT � � t − mT � dt = T 2 � � T e j ( m − n ) T ω d ω = T δ [ n − m ] sinc sinc 2 π T T R − π T Whittaker-Shannon formula is orthogonal expansion 11/30

  13. Reconstruction in Time Domain x p ( t ) t − 2 T − T 0 T 2 T ω c = ω s 4 = π 2 T > ω M x r ( t ) t − T T − 2 T 0 2 T 12/30

  14. Reconstruction in Time Domain x p ( t ) t − 2 T − T 0 T 2 T ω c = 3 ω s < ω s − ω M 5 x r ( t ) t − T T − 2 T 0 2 T 13/30

  15. Contents 1. Sampling Theorem 2. Zero-order Hold and Linear Interpolation 3. Aliasing 14/30

  16. Other Interpolation Methods x [ n ] x ( t ) Zero-order hold n 0 1 2 3 4 5 6 7 8 9 10 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 T 9 T 10 T t T = sampling period x ( t ) Linear interpolation 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 T 9 T 10 T t 15/30

  17. Zero-order Hold ∞ p ( t ) = � δ ( t − nT ) h 0 ( t ) n = −∞ 1 x p ( t ) x ( t ) × x 0 ( t ) t 0 T Reconstructed signal ∞ � x 0 ( t ) = ( x p ∗ h 0 )( t ) = x ( nT ) h 0 ( t − nT ) | H ( j ω ) | n = −∞ T ideal Zero-order hold filter zero-order hold H 0 ( j ω ) = e − j ω T / 2 2 sin( ω T / 2 ) − ω s − 2 ω s ω s 2 ω s − ω s ω s ω ω 2 2 16/30

  18. Zero-order Hold x p ( t ) t − 2 T − T 0 T 2 T x 0 ( t ) t − T T − 2 T 0 2 T 17/30

  19. Linear Interpolation (First-order Hold) ∞ p ( t ) = � δ ( t − nT ) h 1 ( t ) n = −∞ 1 x p ( t ) x ( t ) × x 0 ( t ) t − T T Reconstructed signal ∞ � x 1 ( t ) = ( x p ∗ h 1 )( t ) = x ( nT ) h 1 ( t − nT ) | H ( j ω ) | n = −∞ T ideal First-order hold filter first-order hold � 2 � sin( ω T / 2 ) H 1 ( j ω ) = 1 − ω s ω s − ω s ω s ω ω/ 2 T 2 2 18/30

  20. Linear Interpolation x p ( t ) t − T T − 2 T 0 2 T x 1 ( t ) t − 2 T − T 0 T 2 T 19/30

  21. Contents 1. Sampling Theorem 2. Zero-order Hold and Linear Interpolation 3. Aliasing 20/30

  22. Aliasing output frequency Aliasing wraps ω s 2 frequencies input frequency ω s 2 x ( t ) = cos( ω 0 t + φ ) X ( j ω ) ω 0 < 1 π e − j φ π e j φ 2 ω s ω − ω 0 ω 0 P ( j ω ) − ω s − 2 ω s 0 ω s 2 ω s X p ( j ω ) x r ( t ) = cos( ω 0 t + φ ) − ω s ω s − ω s − 2 ω s ω 0 ω s 2 ω s 2 2 21/30

  23. Aliasing output frequency Aliasing wraps ω s 2 frequencies input frequency ω s 2 x ( t ) = cos( ω 0 t + φ ) X ( j ω ) ω 0 < 1 π e − j φ π e j φ 2 ω s ω − ω 0 ω 0 P ( j ω ) − ω s − 2 ω s 0 ω s 2 ω s X p ( j ω ) x r ( t ) = cos( ω 0 t + φ ) − ω s ω s − ω s − 2 ω s ω 0 ω s 2 ω s 2 2 22/30

  24. Aliasing output frequency Aliasing wraps ω s 2 frequencies input frequency ω s 2 x ( t ) = cos( ω 0 t + φ ) X ( j ω ) ω 0 > 1 π e − j φ π e j φ 2 ω s ω − ω 0 ω 0 P ( j ω ) − ω s − 2 ω s 0 ω s 2 ω s X p ( j ω ) x r ( t ) = cos(( ω s − ω 0 ) t − φ ) − ω s ω s − ω s − 2 ω s ω 0 ω s 2 ω s ω s − ω 0 2 2 23/30

  25. Aliasing output frequency Aliasing wraps ω s 2 frequencies input frequency ω s 2 x ( t ) = cos( ω 0 t + φ ) X ( j ω ) ω 0 > 1 π e − j φ π e j φ 2 ω s ω − ω 0 ω 0 P ( j ω ) − ω s − 2 ω s 0 ω s 2 ω s X p ( j ω ) x r ( t ) = cos(( ω s − ω 0 ) t − φ ) − ω s ω s − ω s ω s − ω 0 − 2 ω s ω 0 ω s 2 ω s 2 2 24/30

  26. Aliasing • x ( t ) = cos( 5 π 3 t ) , x [ n ] = cos( 5 π 3 n ) = cos( π 3 n ) • x r ( t ) = cos( π 3 t ) x ( t ) t n x r ( t ) t n Example. In movies, wheels often appear to rotate more slowly than they actually do and even in wrong direction 25/30

  27. Aliasing output frequency Aliasing wraps ω s 2 frequencies input frequency ω s 2 x ( t ) = cos( ω 0 t + φ ) X ( j ω ) ω 0 = 1 π e − j φ π e j φ 2 ω s ω − ω 0 ω 0 P ( j ω ) − ω s − 2 ω s 0 ω s 2 ω s X p ( j ω ) x r ( t ) = (cos φ ) cos( ω 0 t ) − ω s ω s − ω s − 2 ω s ω s 2 ω s 2 2 26/30

  28. Aliasing Aliasing for more complex signals also wraps frequencies X ( j ω ) ω − ω 0 ω 0 P ( j ω ) − ω s − 2 ω s 0 ω s 2 ω s X p ( j ω ) − ω s − ω s ω s − 2 ω s ω 0 ω s 2 ω s 2 2 ω s − ω 0 27/30

  29. Aliasing Aliasing increases as sampling rate decreases X ( j ω ) ω − ω 0 ω 0 P ( j ω ) − ω s − 2 ω s 0 ω s 2 ω s X p ( j ω ) − ω s − ω s ω s − 2 ω s ω s 2 ω s 2 2 28/30

  30. Anti-aliasing Filter Filter out frequencies above ω s 2 before sampling X ( j ω ) X a ( j ω ) − ω s ω s − ω 0 ω 0 − ω 2 ω 2 2 2 2 2 P ( j ω ) − ω s − 2 ω s ω s 0 2 ω s X p ( j ω ) − ω s − ω s ω s − ω 0 ω s − 2 ω s ω 0 ω s 2 ω s 2 2 29/30

  31. Anti-aliasing Filter Filter out frequencies above ω s 2 before sampling X ( j ω ) X a ( j ω ) − ω s ω s − ω 0 ω 0 − ω 2 ω 2 2 2 2 2 P ( j ω ) − ω s − 2 ω s 0 ω s 2 ω s X p ( j ω ) − ω s − ω s ω s − ω 0 ω s − 2 ω s ω 0 ω s 2 ω s 2 2 30/30

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