6.003: Signals and Systems

Transcription

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

2 Mid-term Examination #3 Wednesday, November 16, 7:30-9:30pm, 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 ( inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1-hour exam; two hours to complete. Review session Monday at 3pm and at open office hours. Prior term midterm exams have been posted on the website. Conflict? Contact before Friday, Nov. 11, 5pm. 2

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

4 Signal Processing: Acoustical Mechano-acoustic components to optimize frequency response of loudspeakers: e.g., bass-reflex system. driver reflex port 4

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

6 6 Signal Processing: Electronic Low-cost electronics new ways to overcome frequency limitations. 110 Magnitude (db) Frequency (Hz) Small speakers (4 inch): eight facing wall, one facing listener. Electronic equalizer compensated for limited frequency response.

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

8 Signal Processing Modern audio systems process sounds digitally. AV SS(REF) V REFM V REFP V RFILT AV DD AV SS DV DD DV SS AINRP AINRM RINA RINB Voltage Reference Analog Supplies Digital Supplies AINRP Analog Control Register Texas Instruments TAS3004 AINLP AINLM LINA LINB AINRM 24 Bit Stereo ADC AINLP Output Format Control Logic SDOUT0 2 channels 24 bit ADC, 24 bit DAC 48 khz sampling rate ALLPASS INPA GPI5 GPI4 GPI3 GPI2 GPI1 Controller AINLM 24 Bit Stereo DAC VCOM AOUTL AOUTR 100 MIPS $9.63 ($5.20 in bulk) GPI0 CS1 SDA SCL I 2 C Control L+R 32 Bit Audio Signal Processor L+R SDOUT2 SDOUT1 PWR_DN RESET TEST Control L R SDATA Control 32 Bit Audio Signal Processor OSC/CLK Select PLL SDIN1 SDIN2 LRCLK/O SCLK/O IFM/S CLKSEL XTALI/ MCLK XTALO MCLKO CAP_PLL Figure 1 1. TAS3004 Block Diagram 8 Courtesy of Texas Instruments. Used with permission.

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

10 10 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) z n h[n] H(z) z n e st h(t) H(s) e st

11 11 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] + 3y[n 1] + 4y[n] = 2x[n 2] + 7x[n 1] + 8x[n] 2z 2 + 7z z + 8z 2 N(z) H(z) = = z 2 + 3z z + 4z2 D(z)

12 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 (z0 p 0 )(z 0 p 1 )(z 0 p 2 ) z 0 z 0 q 0 z 0 z-plane 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

13 DT Vector Diagrams Value of H(z) at z = z 0 can be determined by combining the contributions 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 (z0 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 (z0 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

14 14 DT Frequency Response Response to eternal sinusoids. Let x[n] = cos Ω 0 n (for all time): 1 jω 0 n + e jω 0n 1 n n x[n] = e = z 0 + z jω 0 jω 0 where z0 = 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 jω 0 n = H(e jω 0 ) e jω 0n + H(e jω 0 ) e 2

15 15 Conjugate Symmetry For physical systems, the complex conjugate of H(e jω ) is H(e jω ). The system function is the Z transform of the unit-sample response: 0 n H(z) = h[n]z n= where h[n] is a real-valued function of n for physical systems. jω 0 jωn H(e ) = h[n]e n= 0 jω jωn H(e ) = h[n]e H(e jω ) n=

16 16 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] = e jω 0n + e jω 0n. 2 Then 1 ( ) y[n] = H(e jω 0 )e jω 0n + H(e jω 0 )e jω 0n 2 = Re H(e jω 0 )e jω 0n = Re H(e jω 0 ) e j H(e jω 0 ) e jω 0n jω 0 n+j H(e jω 0 ) = H(e jω 0 ) Re e ( ) y[n] = H(e jω 0 ) cos Ω 0 n + H(e jω 0 )

17 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. cos(ωn) H(z) ( ) H(e jω ) cos Ωn + H(e jω ) H(e jω ) = H(z) z=e jω 17

18 18 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

19 19 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

20 20 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

21 21 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

22 22 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

23 23 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

24 24 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

25 25 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

26 26 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

27 27 Finding Frequency Response with Vector Diagrams H(z) = z q H(e jω ) 1 z p 1 z-plane 1 π 0 π H(e jω ) π/2 π π π/2

28 Comparision of CT and DT Frequency Responses CT frequency response: H(s) on the imaginary axis, i.e., s = jω. DT frequency response: H(z) on the unit circle, i.e., z = e jω. ω s-plane z-plane σ H(jω) H(e jω ) π 0 π

29 DT Periodicity DT frequency responses are periodic functions of Ω, with period 2π. If Ω 2 = Ω 1 + 2πk where k is an integer then H(e jω 2 ) = H(e j(ω 1+2πk) ) = H(e jω 1 e j2πk ) = H(e jω 1 ) jω The periodicity of H(e jω ) results because H(e jω ) is a function of e, which is itself periodic in Ω. Thus DT complex exponentials have many aliases. e jω 2 = e j(ω 1 +2πk) = e jω 1 e j2πk = e jω 1 Because of this aliasing, there is a highest DT frequency: Ω = π. 29

30 Comparision of CT and DT Frequency Responses CT frequency response: H(s) on the imaginary axis, i.e., s = jω. DT frequency response: H(z) on the unit circle, i.e., z = e jω. ω s-plane z-plane σ H(jω) H(e jω ) π 0 π

31 31 Check Yourself Consider 3 CT signals: x 1 (t) = cos(3000t) ; x 2 (t) = cos(4000t) ; x 3 (t) = cos(5000t) Each of these is sampled so that x 1 [n] = x 1 (nt ) ; x 2 [n] = x 2 (nt ) ; x 3 [n] = x 3 (nt ) where T = Which list goes from lowest to highest DT frequency? 0. x 1 [n] x 2 [n] x 3 [n] 1. x 1 [n] x 3 [n] x 2 [n] 2. x 2 [n] x 1 [n] x 3 [n] 3. x 2 [n] x 3 [n] x 1 [n] 4. x 3 [n] x 1 [n] x 2 [n] 5. x 3 [n] x 2 [n] x 1 [n]

32 Check Yourself The discrete signals are x 1 [n] = cos[3n] x 2 [n] = cos[4n] x 3 [n] = cos[5n] and the corresponding discrete frequencies are Ω = 3, 4 and 5, represented below with marking e jω and o marking e jω )

33 33 Check Yourself Ω = 0.25 x[n] = cos(0.25n) n

34 34 Check Yourself Ω = 0.5 x[n] = cos(0.5n) n

35 35 Check Yourself Ω = 1 x[n] = cos(n) n

36 36 Check Yourself Ω = 2 x[n] = cos(2n) n

37 37 Check Yourself Ω = 3 x[n] = cos(3n) n

38 38 Check Yourself Ω = 4 x[n] = cos(4n) = cos(2π 4n) cos(2.283n) n

39 39 Check Yourself Ω = 5 x[n] = cos(5n) = cos(2π 5n) cos(1.283n) n

40 40 Check Yourself Ω = 6 x[n] = cos(6n) = cos(2π 6n) cos(0.283n) n

41 Check Yourself The discrete signals are x 1 [n] = cos[3n] x 2 [n] = cos[4n] x 3 [n] = cos[5n] and the corresponding discrete frequencies are Ω = 3, 4 and 5, represented below with marking e jω and o marking e jω )

42 42 Check Yourself Consider 3 CT signals: x 1 (t) = cos(3000t) ; x 2 (t) = cos(4000t) ; x 3 (t) = cos(5000t) Each of these is sampled so that x 1 [n] = x 1 (nt ) ; x 2 [n] = x 2 (nt ) ; x 3 [n] = x 3 (nt ) where T = Which list goes from lowest to highest DT frequency? 5 0. x 1 [n] x 2 [n] x 3 [n] 1. x 1 [n] x 3 [n] x 2 [n] 2. x 2 [n] x 1 [n] x 3 [n] 3. x 2 [n] x 3 [n] x 1 [n] 4. x 3 [n] x 1 [n] x 2 [n] 5. x 3 [n] x 2 [n] x 1 [n]

43 Check Yourself What kind of filtering corresponds to the following? z-plane 1. high pass 2. low pass 3. band pass 4. band stop (notch) 5. none of above 43

44 Check Yourself What kind of filtering corresponds to the following? 1 z-plane 1. high pass 2. low pass 3. band pass 4. band stop (notch) 5. none of above 44

45 45 DT Fourier Series DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components. 0 jkω 0 n x[n] = a k e The period N of all harmonic components is the same (as in CT).

46 46 DT Fourier Series There are (only) N distinct complex exponentials with period N. (There were an infinite number in CT!) jωn If y[n] = e is periodic in N then jωn jω(n+n) jωn jωn y[n] = e = y[n + N] = e = e e jωn and e must be 1, and e jω must be one of the N th roots of 1. Example: N = 8 z-plane

47 DT Fourier Series There are N distinct complex exponentials with period N. These can be combined via Fourier series to produce periodic time signals with N independent samples. Example: periodic in N=3 n 3 samples repeated in time 3 complex exponentials Example: periodic in N=4 n 4 samples repeated in time 4 complex exponentials 47

48 48 DT Fourier Series DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components. N 0 1 jkω 0 n 2π x[n] = x[n + N] = a k e ; Ω 0 = N k=0 N equations (one for each point in time n) in N unknowns (a k ). Example: N = 4 2π 2π 2π 2π x[0] j 0 0 j 1 0 j 2 0 j 3 0 e N e N e N e N a 0 2 π π N π 2 1 2π 3 1 x[1] j j j j = e N e N e e N a 1 x[2] j 2π 0 2 j 2π 1 2 j 2π 2 2 j 2π 3 2 e N e N e N e N a 2 2π 0 3 2π 1 3 2π 2 3 2π x[3] 3 3 e j N e j N e j N e j N a 3

49 49 DT Fourier Series DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components. N 0 1 jkω 0 n 2π x[n] = x[n + N] = a k e ; Ω 0 = N k=0 N equations (one for each point in time n) in N unknowns (a k ). Example: N = 4 x[0] a 0 x[1] 1 j 1 j = a 1 x[2] a 2 x[3] 1 j 1 j a 3

50 50 Orthogonality DT harmonics are orthogonal to each other (as were CT harmonics). N 0 1 N 0 1 jω 0 kn jω 0 ln jω 0 (k l)n e e = e n=0 n=0 N j 2π = (k l)n 1 e jω 0 (k l)n 1 e N jω 0 (k l) 2π 1 e j (k l) 1 e N ; k = l = = 0 ; k = l = Nδ[k l]

51 51 Sifting Use orthogonality property of harmonics to sift out FS coefficients. N 0 1 jkω 0 n Assume x[n] = a k e k=0 Multiply both sides by the complex conjugate of the l th harmonic, and sum over time. N 0 1 N 0 1 N 0 1 N 0 1 N 0 1 jlω 0 n jkω 0 n jlω 0 n jkω 0 n jlω 0 n x[n]e = a k e e = a k e e n=0 n=0 k=0 k=0 n=0 N 0 1 = a k Nδ[k l] = Na l k=0 a k = 1 N 0 1 jkω 0 n x[n]e N n=0

52 52 DT Fourier Series Since both x[n] and a k are periodic in N, the sums can be taken over any N successive indices. Notation. If f[n] is periodic in N, then 0 1 0N N+1 0 f[n] = f[n] = f[n] = = 0 f[n] n=0 n=1 n=2 n=<n> DT Fourier Series 1 0 jkω 0 n 2π a k = a k+n = x[n]e ; Ω 0 = ( analysis equation) N N n=<n> 0 jkω 0 n x[n]= x[n + N] = a k e ( synthesis equation) k=<n>

53 These matrices are inverses of each other. 53 DT Fourier Series DT Fourier series have simple matrix interpretations. 0 0 jk 2 π 0 jkω 0 n x[n] = x[n + 4] = a = 4 n = a k j kn k e a k e k=<4> k=<4> k=<4> x[0] a 0 x[1] 1 j 1 j = a 1 x[2] a 2 x[3] 1 j 1 j a π 1 0 jkω 0 n jk n x[n]e = = x[n]j kn ak = a k+4 = e N n=<4> n=<4> n=<4> a x[0] a j 1 j x[1] = a x[2] a 3 1 j 1 j x[3]

54 Discrete-Time Frequency Representations Similarities and differences between CT and DT. DT frequency response vector diagrams (similar to CT) frequency response on unit circle in z-plane (jω axis in CT) DT Fourier series represent signal as sum of harmonics (similar to CT) finite number of periodic harmonics (unlike CT) finite sum (unlike CT) The finite length of DT Fourier series make them especially useful for signal processing! (more on this next time) 54

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