6 003 signals and systems
play

6.003: Signals and Systems Applications of Fourier Transforms - PowerPoint PPT Presentation

6.003: Signals and Systems Applications of Fourier Transforms November 17, 2011 1 Filtering Notion of a filter. LTI systems cannot create new frequencies. can only scale magnitudes and shift phases of existing components. Example: LowPass


  1. 6.003: Signals and Systems Applications of Fourier Transforms November 17, 2011 1

  2. Filtering Notion of a filter. LTI systems • cannot create new frequencies. • can only scale magnitudes and shift phases of existing components. Example: Low­Pass Filtering with an RC circuit R + + v i v o C − − 2

  3. Lowpass Filter Calculate the frequency response of an RC circuit. KVL: v i ( t ) = Ri ( t ) + v o ( t ) R C: i ( t ) = Cv ˙ o ( t ) + Solving: v i ( t ) = RC v ˙ o ( t ) + v o ( t ) + v i v o C − V i ( s ) = (1 + sRC ) V o ( s ) − V o ( s ) 1 H ( s ) = = V i ( s ) 1 + sRC 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 3

  4. Lowpass Filtering Let the input be a square wave. 1 2 t 0 T − 1 2 1 2 π jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | X ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ X ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 4

  5. Lowpass Filtering Low frequency square wave: ω 0 << 1 /RC . 1 2 t 0 T − 1 2 1 2 π jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 5

  6. Lowpass Filtering Higher frequency square wave: ω 0 < 1 /RC . 1 2 t 0 T − 1 2 1 2 π jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 6

  7. Lowpass Filtering Still higher frequency square wave: ω 0 = 1 /RC . 1 2 t 0 T − 1 2 1 2 π jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 7

  8. Lowpass Filtering High frequency square wave: ω 0 > 1 /RC . 1 2 t 0 T − 1 2 1 2 π jω kt x ( t ) = e ; ω = 0 0 jπk T k odd 1 | H ( jω ) | 0 . 1 0 . 01 ω 1 /RC 0 . 01 0 . 1 1 10 100 ∠ H ( jω ) | 0 − π ω 2 1 /RC 0 . 01 0 . 1 1 10 100 8

  9. Source-Filter Model of Speech Production Vibrations of the vocal cords are “filtered” by the mouth and nasal cavities to generate speech. throat and buzz from speech vocal cords nasal cavities 9

  10. Filtering LTI systems “filter” signals based on their frequency content. Fourier transforms represent signals as sums of complex exponen- tials. 1 ∞ jωt dω x ( t ) = X ( jω ) e 2 π −∞ Complex exponentials are eigenfunctions of LTI systems. jωt → H ( jω ) e jωt e LTI systems “filter” signals by adjusting the amplitudes and phases of each frequency component. 1 ∞ 1 ∞ jωt dω jωt dω x ( t ) = X ( jω ) e → y ( t ) = H ( jω ) X ( jω ) e 2 π 2 π −∞ −∞ 10

  11. Filtering Systems can be designed to selectively pass certain frequency bands. Examples: low­pass filter (LPF) and high­pass filter (HPF). LPF HPF ω 0 LPF t t HPF t 11

  12. Filtering Example: Electrocardiogram An electrocardiogram is a record of electrical potentials that are generated by the heart and measured on the surface of the chest. x ( t ) [mV] 2 1 0 t [s] − 1 0 10 20 30 40 50 60 ECG and analysis by T. F. Weiss 12

  13. Filtering Example: Electrocardiogram In addition to electrical responses of heart, electrodes on the skin also pick up other electrical signals that we regard as “noise.” We wish to design a filter to eliminate the noise. x ( t ) y ( t ) filter 13

  14. Filtering Example: Electrocardiogram We can identify “noise” using the Fourier transform. x ( t ) [mV] 2 1 0 t [s] − 1 0 10 20 30 40 50 60 1000 60 Hz 100 | X ( jω ) | [ µ V] 10 1 0 . 1 0 . 01 low­freq. noise cardiac 0 . 001 signal high­freq. noise 0 . 0001 0 . 01 0 . 1 1 10 100 f = ω 2 π [Hz] 14

  15. Filtering Example: Electrocardiogram Filter design: low­pass flter + high­pass filter + notch. 1 | H ( jω ) | 0 . 1 0 . 01 0 . 001 0 . 01 0 . 1 1 10 100 f = ω 2 π [Hz] 15

  16. Electrocardiogram: Check Yourself Which poles and zeros are associated with • the high­pass filter? • the low­pass filter? • the notch filter? s ­plane 2 2 2 ( ) ( ) ( ) 16

  17. Electrocardiogram: Check Yourself Which poles and zeros are associated with • the high­pass filter? • the low­pass filter? • the notch filter? s ­plane notch low­pass high­pass 2 2 2 ( ) ( ) ( ) notch 17

  18. Filtering Example: Electrocardiogram Filtering is a simple way to reduce unwanted noise. Unfiltered ECG 2 x ( t ) [ mV ] 1 0 t [ s ] 0 10 20 30 40 50 60 Filtered ECG y ( t ) [ mV ] 1 t [ s ] 0 0 10 20 30 40 50 60 18

  19. Fourier Transforms in Physics: Diffraction A diffraction grating breaks a laser beam input into multiple beams. Demonstration. 19

  20. Fourier Transforms in Physics: Diffraction Multiple beams result from periodic structure of grating (period D ). grating λ sin θ = λ D D θ Viewed at a distance from angle θ , scatterers are separated by D sin θ . nλ Constructive interference if D sin θ = nλ , i.e., if sin θ = D → periodic array of dots in the far field 20

  21. Fourier Transforms in Physics: Diffraction CD demonstration. 21

  22. Check Yourself CD demonstration. 3 feet 1 feet laser pointer λ = 500 nm CD screen What is the spacing of the tracks on the CD? 1. 160 nm 2. 1600 nm 3. 16 µ m 4. 160 µ m 22

  23. Check Yourself What is the spacing of the tracks on the CD? 500 nm grating tan θ θ sin θ D = manufacturing spec. sin θ 1 CD 0 . 32 0 . 31 1613 nm 1600 nm 3 23

  24. Check Yourself Demonstration. 3 feet 1 feet laser pointer λ = 500 nm CD screen What is the spacing of the tracks on the CD? 2. 1. 160 nm 2. 1600 nm 3. 16 µ m 4. 160 µ m 24

  25. Fourier Transforms in Physics: Diffraction DVD demonstration. 25

  26. Check Yourself DVD demonstration. 1 feet 1 feet laser pointer λ = 500 nm DVD screen What is track spacing on DVD divided by that for CD? 1 1 1. 4 × 2. 2 × 3. 2 × 4. 4 × 26

  27. Check Yourself What is spacing of tracks on DVD divided by that for CD? 500 nm grating tan θ θ sin θ D = manufacturing spec. sin θ 1 CD 0 . 32 0 . 31 1613 nm 1600 nm 3 DVD 1 0 . 78 0 . 71 704 nm 740 nm 27

  28. Check Yourself DVD demonstration. 1 feet 1 feet laser pointer λ = 500 nm DVD screen What is track spacing on DVD divided by that for CD? 3 1 1 1. 4 × 2. 2 × 3. 2 × 4. 4 × 28

  29. Fourier Transforms in Physics: Diffraction Macroscopic information in the far field provides microscopic (invis- ible) information about the grating. λ sin θ = λ D D θ 29

  30. Fourier Transforms in Physics: Crystallography What if the target is more complicated than a grating? target image? 30

  31. Fourier Transforms in Physics: Crystallography Part of image at angle θ has contributions for all parts of the target. θ target image? 31

  32. Fourier Transforms in Physics: Crystallography The phase of light scattered from different parts of the target un- dergo different amounts of phase delay. x sin θ θ x Phase at a point x is delayed (i.e., negative) relative to that at 0 : x sin θ φ = − 2 π λ 32

  33. Fourier Transforms in Physics: Crystallography Total light F ( θ ) at angle θ is integral of light scattered from each part of target f ( x ) , appropriately shifted in phase. − j 2 π x sin θ F ( θ ) = f ( x ) e λ dx Assume small angles so sin θ ≈ θ . Let ω = 2 π θ , then the pattern of light at the detector is λ − jωx dx F ( ω ) = f ( x ) e which is the Fourier transform of f ( x ) ! 33

  34. Fourier Transforms in Physics: Diffraction Fourier transform relation between structure of object and far­field intensity pattern. grating ≈ impulse train with pitch D · · · · · · t 0 D far­field intensity ≈ impulse train with reciprocal pitch ∝ λ D · · · · · · ω 0 2 π D 34

  35. Impulse Train The Fourier transform of an impulse train is an impulse train. ∞ x ( t ) = δ ( t − kT ) k = −∞ 1 · · · · · · t 0 T a k = 1 ∀ k T 1 T · · · · · · k ∞ 2 π T δ ( ω − k 2 π X ( jω ) = T ) k = −∞ 2 π T · · · · · · ω 0 2 π T 35

  36. Two Dimensions Demonstration: 2D grating. 36

  37. An Historic Fourier Transform Taken by Rosalind Franklin, this image sparked Watson and Crick’s insight into the double helix. Reprinted by permission from Macmillan Publishers Ltd: Nature. Source: Franklin, R., and R. G. Gosling. "Molecular Configuration in Sodium Thymonucleate." Nature 171 (1953): 740-741. (c) 1953. 37

  38. An Historic Fourier Transform This is an x­ray crystallographic image of DNA, and it shows the Fourier transform of the structure of DNA. Reprinted by permission from Macmillan Publishers Ltd: Nature. Source: Franklin, R., and R. G. Gosling. "Molecular Configuration in Sodium Thymonucleate." Nature 171 (1953): 740-741. (c) 1953. 38

  39. An Historic Fourier Transform High­frequency bands indicate repeating structure of base pairs. b 1 /b Reprinted by permission from Macmillan Publishers Ltd: Nature. Source: Franklin, R., and R. G. Gosling. "Molecular Configuration in Sodium Thymonucleate." Nature 171 (1953): 740-741. (c) 1953. 39

Recommend


More recommend