6.003: Signals and Systems Discrete-Time Frequency Representations - PowerPoint PPT Presentation

6.003: Signals and Systems Discrete-Time Frequency Representations November 8, 2011 1

Mid-term Examination #3 Wednesday, November 16, 7:30­9:30pm, No recitations on the day of the exam. Coverage: Lectures 1–18 Recitations 1–16 Homeworks 1–10 Homework 10 will not be collected or graded. Solutions will be posted. Closed book: 3 pages of notes ( 8 1 2 × 11 inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1­hour exam; two hours to complete. Review session Monday at 3pm and at open office hours. Prior term midterm exams have been posted on the 6.003 website. Conflict? Contact before Friday, Nov. 11, 5pm. 2

Signal Processing: From CT to DT Signal­processing problems first conceived & addressed in CT: • audio − radio (noise/static reduction, automatic gain control, etc.) − telephone (equalizers, echo­suppression, etc.) − hi­fi (bass, treble, loudness, etc.) • imaging − television (brightness, tint, etc.) − photography (image enhancement, gamma) − x­rays (noise reduction, contrast enhancement) − radar and sonar (noise reduction, object detection) Such problems are increasingly solved with DT signal processing: • MP3 • JPEG • MPEG 3

Signal Processing: Acoustical Mechano­acoustic components to optimize frequency response of loudspeakers: e.g., “bass­reflex” system. driver reflex port 4

Signal Processing: Acoustico-Mechanical Passive radiator for improved low­frequency preformance. driver passive radiator 5

Signal Processing: Electronic Low­cost electronics → new ways to overcome frequency limitations. ✶✶✵ ▼❛❣♥✐t✉❞❡ ✭❞❇✮ ✶✵✵ ✾✵ ✽✵ ✼✵ ✶ ✷ ✺ ✸ ✹ ✶✵ ✶✵ ✶✵ ✶✵ ✶✵ ❋r❡q✉❡♥❝② ✭❍③✮ Small speakers (4 inch): eight facing wall, one facing listener. Electronic “equalizer” compensated for limited frequency response. 6

Signal Processing Modern audio systems process sounds digitally. x [ n ] y [ n ] x ( t ) A/D D/A y ( t ) DT filter 7

Signal Processing Modern audio systems process sounds digitally. ❆❱❙❙✭❘❊❋✮ ❱❘❊❋▼ ❱❘❋■▲❚ ❱❘❊❋P ❆❱❉❉ ❆❱❙❙ ❉❱❉❉ ❉❱❙❙ ❆■◆❘P ❆■◆❘▼ ❱♦❧t❛❣❡ ❆♥❛❧♦❣ ❉✐❣✐t❛❧ ❘■◆❆ ❘❡❢❡r❡♥❝❡ ❙✉♣♣❧✐❡s ❙✉♣♣❧✐❡s ❘■◆❇ ❆♥❛❧♦❣ ❆■◆❘P ❈♦♥tr♦❧ ❘❡❣✐st❡r ❆■◆❘▼ ❖✉t♣✉t ❋♦r♠❛t ✷✹➢❇✐t ❙❉❖❯❚✵ ❈♦♥tr♦❧ ❆■◆▲P ❙t❡r❡♦ Texas Instruments TAS3004 ▲♦❣✐❝ ❆■◆▲▼ ❆❉❈ ▲■◆❆ ❆■◆▲P ▲■◆❇ • 2 channels ❆■◆▲▼ ❆▲▲P❆❙❙ ❱❈❖▼ ■◆P❆ 24 bit ADC, 24 bit DAC • ❆❖❯❚▲ ●P■✺ ❈♦♥tr♦❧❧❡r ❆❖❯❚❘ ●P■✹ ●P■✸ ✷✹➢❇✐t 48 kHz sampling rate • ●P■✷ ❙t❡r❡♦ ❉❆❈ ●P■✶ ●P■✵ ▲✰❘ • 100 MIPS ▲✰❘ ❙❉❖❯❚✷ ❈❙✶ ❈♦♥tr♦❧ ✸✷➢❇✐t ❆✉❞✐♦ ❙✐❣♥❛❧ ■✷❈ ❙❉❆ Pr♦❝❡ss♦r $9.63 ($5.20 in bulk) • ❙❈▲ ❙❉❖❯❚✶ ✸✷➢❇✐t ❆✉❞✐♦ ❙✐❣♥❛❧ P❲❘❴❉◆ ❈♦♥tr♦❧ ▲ ❘ Pr♦❝❡ss♦r ❘❊❙❊❚ ❚❊❙❚ ❙❉❆❚❆ ❖❙❈✴❈▲❑ P▲▲ ❈♦♥tr♦❧ ❙❡❧❡❝t ❙❉■◆✶ ❙❉■◆✷ ▲❘❈▲❑✴❖ ❙❈▲❑✴❖ ■❋▼✴❙ ❈▲❑❙❊▲ ❳❚❆▲■✴ ▼❈▲❑ ▼❈▲❑❖ ❈❆P❴P▲▲ ❳❚❆▲❖ ❋✐❣✉r❡ ✶➢ ✶✳ ❚❆❙✸✵✵✹ ❇❧♦❝❦ ❉✐❛❣r❛♠ 8 Courtesy of Texas Instruments. Used with permission.

DT Fourier Series and Frequency Response Today: frequency representations for DT signals and systems. 9

Review: Complex Geometric Sequences Complex geometric sequences are eigenfunctions of DT LTI systems. n Find response of DT LTI system ( h [ n ] ) to input x [ n ] = z . ∞ ∞ 0 0 n − k n − k n y [ n ] = ( h ∗ x )[ n ] = h [ k ] z = z h [ k ] z = H ( z ) z . k = −∞ k = −∞ Complex geometrics (DT): analogous to complex exponentials (CT) H ( z ) z n z n h [ n ] H ( s ) e st e st h ( t ) 10

Review: Rational System Functions A system described by a linear difference equation with constant coefficients → system function that is a ratio of polynomials in z . Example: y [ n − 2] + 3 y [ n − 1] + 4 y [ n ] = 2 x [ n − 2] + 7 x [ n − 1] + 8 x [ n ] 2 z − 2 + 7 z − 1 + 8 2 + 7 z + 8 z 2 N ( z ) H ( z ) = = 2 ≡ z − 2 + 3 z − 1 + 4 1 + 3 z + 4 z D ( z ) 11

DT Vector Diagrams Factor the numerator and denominator of the system function to make poles and zeros explicit. ( z 0 − q 0 )( z 0 − q 1 )( z 0 − q 2 ) · · · H ( z 0 ) = K ( z 0 − p 0 )( z 0 − p 1 )( z 0 − p 2 ) · · · z 0 z ­plane z 0 z 0 − q 0 q 0 q 0 Each factor in the numerator/denominator corresponds to a vector from a zero/pole (here q 0 ) to z 0 , the point of interest in the z ­plane. Vector diagrams for DT are similar to those for CT. 12

DT Vector Diagrams Value of H ( z ) at z = z 0 can be determined by combining the contri- butions of the vectors associated with each of the poles and zeros. ( z 0 − q 0 )( z 0 − q 1 )( z 0 − q 2 ) · · · H ( z 0 ) = K ( z 0 − p 0 )( z 0 − p 1 )( z 0 − p 2 ) · · · The magnitude is determined by the product of the magnitudes. | ( z 0 − q 0 ) || ( z 0 − q 1 ) || ( z 0 − q 2 ) | · · · | H ( z 0 ) | = | K | | ( z 0 − p 0 ) || ( z 0 − p 1 ) || ( z 0 − p 2 ) | · · · The angle is determined by the sum of the angles. ∠ H ( z 0 ) = ∠ K + ∠ ( z 0 − q 0 )+ ∠ ( z 0 − q 1 )+ · · ·− ∠ ( z 0 − p 0 ) − ∠ ( z 0 − p 1 ) −· · · 13

DT Frequency Response Response to eternal sinusoids. Let x [ n ] = cos Ω 0 n (for all time): 1 1 � � j Ω 0 n + e − j Ω 0 n n n x [ n ] = e = z 0 + z 1 2 2 j Ω 0 − j Ω 0 where z 0 = e and z 1 = e . The response to a sum is the sum of the responses: 1 � n n y [ n ] = H ( z 0 ) z 0 + H ( z 1 ) z 1 2 1 � H ( e j Ω 0 ) e j Ω 0 n + H ( e − j Ω 0 ) e − j Ω 0 n = 2 14

Conjugate Symmetry For physical systems, the complex conjugate of H ( e j Ω ) is H ( e − j Ω ) . The system function is the Z transform of the unit­sample response: ∞ 0 − n H ( z ) = h [ n ] z n = −∞ where h [ n ] is a real­valued function of n for physical systems. ∞ j Ω ) = 0 − j Ω n H ( e h [ n ] e = n −∞ ∞ ∗ � − j Ω ) = 0 j Ω n ≡ H ( e j Ω ) H ( e h [ n ] e n = −∞ 15

DT Frequency Response Response to eternal sinusoids. Let x [ n ] = cos Ω 0 n (for all time), which can be written as 1 � � x [ n ] = 2 e j Ω 0 n + e − j Ω 0 n . Then 1 � � y [ n ] = 2 H ( e j Ω 0 ) e j Ω 0 n + H ( e − j Ω 0 ) e − j Ω 0 n H ( e j Ω 0 ) e j Ω 0 n = Re � | H ( e j Ω 0 ) | e j ∠ H ( e j Ω0 ) e j Ω 0 n = Re � j Ω 0 n + j ∠ H ( e j Ω0 ) = | H ( e j Ω 0 ) | Re e � � � � y [ n ] = � H ( e j Ω 0 ) � cos Ω 0 n + ∠ H ( e j Ω 0 ) � � 16

DT Frequency Response The magnitude and phase of the response of a system to an eternal cosine signal is the magnitude and phase of the system function evaluated on the unit circle. � | H ( e j Ω ) | cos Ω n + ∠ H ( e j Ω ) cos(Ω n ) H ( z ) H ( e j Ω ) = H ( z ) | z = e j Ω 17

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 18

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 19

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 20

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 21

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 22

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 23

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 24

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 25

Finding Frequency Response with Vector Diagrams � � � H ( e j Ω ) � � H ( z ) = z − q 1 � z − p 1 1 z ­plane π − π 0 ∠ H ( e j Ω ) π/ 2 π − π − π/ 2 26