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

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend