6.003: Signals and Systems. Sampling and Quantization

Transcription

1 6.003: Signals and Systems Sampling and Quantization December 1, 2009

2 Last Time: Sampling and Reconstruction Uniform sampling (sampling interval T ): x[n] = x(nt ) t n Impulse reconstruction: x p (t) = x[n]δ(t nt ) n= t n Relation: x p (t) = x(t)p(t) where p(t) = δ(t nt ) n=

3 Fourier Representation Sampling followed by impulse reconstruction multiplication by an impulse train in time convolving by an impulse train in frequency X(jω) 1 W W ω ω s X p (jω) = 1 2π ω s P (jω) 2π T ω s ω (X(j ) P (j ))(ω) 1 T ωs = 2π T ω

4 Fourier Representation Bandlimited reconstruction: lowpass filter, remove high frequencies. X(jω) 1 W W ω ω s P (jω) 2π T ω s ω X p (jω) = 2π 1 (X(j ) P (j ))(ω) T 1 T ω s 2 ω s 2 ω

5 The Sampling Theorem If signal is bandlimited sample without loosing information. If x(t) is bandlimited so that X(jω) = 0 for ω > ω m then x(t) is uniquely determined by its samples x(nt ) if ω s = 2π T > 2ω m. The minimum sampling frequency, 2ω m, is called the Nyquist rate.

6 Aliasing Aliasing results if there are frequency components with ω > ωs 2. X(jω) 1 ω ω s P (jω) 2π T ω s ω X p (jω) = 2π 1 (X(j ) P (j ))(ω) ωs 2 ω s 2 1 T ω

7 Aliasing Aliasing results if there are frequency components with ω > ωs 2. X(jω) 1 ω ω s P (jω) 2π T ω s ω X p (jω) = 2π 1 (X(j ) P (j ))(ω) ωs 2 ω s 2 1 T ω

8 Aliasing Aliasing results if there are frequency components with ω > ωs 2. X(jω) 1 ω ω s P (jω) 2π T ω s ω X p (jω) = 2π 1 (X(j ) P (j ))(ω) ωs 2 ω s 2 1 T ω

9 Aliasing Aliasing results if there are frequency components with ω > ωs 2. X(jω) 1 ω ω s P (jω) 2π T X p (jω) = 2π 1 (X(j ) P (j ))(ω) ωs 2 ω s 2 ω s 1 T ω ω

10 Anti-Aliasing Filter To avoid aliasing, remove frequency components that alias before sampling. x(t) Anti-aliasing Filter 1 ωs 2 ω s 2 ω x p (t) Reconstruction Filter T ωs 2 ω s 2 ω x r (t) p(t)

11 Aliasing Aliasing increases as the sampling rate decreases. X(jω) 1 ω s P (jω) 2π T ω s ω ω X p (jω) = 2π 1 (X(j ) P (j ))(ω) ωs 2 ω s 2 1 T ω

12 Aliasing Aliasing increases as the sampling rate decreases. Anti-aliased X(jω) ω s P (jω) 2π T ω s ω ω X p (jω) = 2π 1 (X(j ) P (j ))(ω) ωs 2 ω s 2 1 T ω

13 Aliasing Aliasing increases as the sampling rate decreases. Anti-aliased X(jω) ω s P (jω) 2π T ω s ω ω X p (jω) = 2π 1 (X(j ) P (j ))(ω) ωs 2 ω s 2 1 T ω

14 Aliasing Aliasing increases as the sampling rate decreases. Anti-aliased X(jω) ω ω s P (jω) 2π T X p (jω) = 2π 1 (X(j ) P (j ))(ω) ωs 2 ω s 2 ω s 1 T ω ω

15 Anti-Aliasing Demonstration Sampling Music ω s = 2π T = 2πf s f s = 11 khz without anti-aliasing f s = 11 khz with anti-aliasing f s = 5.5 khz without anti-aliasing f s = 5.5 khz with anti-aliasing f s = 2.8 khz without anti-aliasing f s = 2.8 khz with anti-aliasing J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto Nathan Milstein, violin

16 Quantization Analog to digital conversion requires not only sampling in time but also quantization in amplitude. Output voltage bits 10 3 bits 4 bits Input voltage Input voltage Input voltage Time (second) Time (second) Time (second) Bit rate = (# bits/sample) (# samples/sec) e.g., CD quality audio: bit rate = (16 bits/sample) (44,100 samples/sec) 0.7 Mbits/sec

17 Check Yourself The original Milstein violin piece is taken from a CD. The sampling rate was 44.1 khz and each sample was quantized to 16 bits. If we re-quantize it to fewer bits, how few bits can we use without causing significant problems for lecture demos?

18 Quantization Demonstration Quantizing Music 16 bits/sample 8 bits/sample 4 bits/sample 3 bits/sample 2 bits/sample 1 bit/sample J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto Nathan Milstein, violin

19 Quantizing Images Converting an image from a continuous representation to a discrete representation involves the same sort of issues. This image has pixels, with brightness quantized to 8 bits.

20 Quantizing Images 8 bit image 7 bit image

21 Quantizing Images 8 bit image 6 bit image

22 Quantizing Images 8 bit image 5 bit image

23 Quantizing Images 8 bit image 4 bit image

24 Quantizing Images 8 bit image 3 bit image

25 Quantizing Images 8 bit image 2 bit image

26 Quantizing Images 8 bit image 1 bit image

27 Check Yourself What is the most objectionable artifact of coarse quantization? 8 bit image 3 bit image

28 Dithering One very annoying artifact is banding caused by clustering of pixels that quantize to the same level. Banding can be reduced by dithering. Dithering: adding a small amount (±1 quantum) of random noise to the image before quantizing. Since the noise is different for each pixel in the band, the noise causes some of the pixels to quantize to a higher value and some to a lower. But the average value of the brightness is preserved.

29 Quantizing Images with Dither 7 bit image 7 bits with dither

30 Quantizing Images with Dither 6 bit image 6 bits with dither

31 Quantizing Images with Dither 5 bit image 5 bits with dither

32 Quantizing Images with Dither 4 bit image 4 bits with dither

33 Quantizing Images with Dither 3 bit image 3 bits with dither

34 Quantizing Images with Dither 2 bit image 2 bits with dither

35 Quantizing Images with Dither 1 bit image 1 bit with dither

36 Check Yourself What is the most objectionable artifact of dithering? 3 bit image 3 bit dithered image

37 Robert s Technique One annoying feature of dithering is that it adds noise. The noise can be reduced using Robert s technique. Robert s technique: add a small amount (±1 quantum) of random noise before quantizing, then subtract that same amount of random noise.

38 Quantizing Images with Robert s Method 7 bits with dither 7 bits with Robert s method

39 Quantizing Images with Robert s Method 6 bits with dither 6 bits with Robert s method

40 Quantizing Images with Robert s Method 5 bits with dither 5 bits with Robert s method

41 Quantizing Images with Robert s Method 4 bits with dither 4 bits with Robert s method

42 Quantizing Images with Robert s Method 3 bits with dither 3 bits with Robert s method

43 Quantizing Images with Robert s Method 2 bits with dither 2 bits with Robert s method

44 Quantizing Images with Robert s Method 1 bits with dither 1 bit with Robert s method

45 Quantizing Images: 3 bits 8 bits 3 bits dither Robert s

46 Quantizing Images: 2 bits 8 bits 2 bits dither Robert s

47 Quantizing Images: 1 bit 8 bits 1 bit dither Robert s

48 Sampling and Quantization Combining sampling and quantization. Resampling (discrete-time sampling) and progressive refinement JPEG

49 Discrete-time Sampling DT sampling is much like CT sampling. x[n] x p [n] p[n] = k δ[n kn] x[n] 0 p[n] n 0 x p [n] n 0 n

50 Discrete-time Sampling As in CT, sampling introduces additional copies of X(e jω ). x[n] x p [n] p[n] = k δ[n kn] X(e jω ) 1 2π 0 2π P (e jω ) Ω 2π 4π 3 2π 3 2π 0 2π 4π 2π 3 X p (e jω 3 3 ) 1 3 Ω 2π 4π 3 2π 3 0 2π 3 4π 3 2π Ω

51 Discrete-time Sampling Sampling a finite sequence gives rise to a shorter sequence. x[n] 0 x p [n] n 0 x b [n] n 0 n X b (e jω ) = n x b [n]e jωn = n x p [3n]e jωn = k x p [k]e jωk/3 = X p (e jω/3 )

52 Discrete-time Sampling But the shorter sequence has a wider frequency representation. X(e jω ) 1 2π 0 2π Ω 2π X p (e jω ) π X b (e jω ) = X p (e jω/3 ) 1 3 Ω 2π 4π 3 2π 3 0 2π 4π 2π 3 3 Ω

53 Discrete-time Sampling

54 Discrete-time Sampling

55 Discrete-time Sampling

56 Discrete-time Sampling

57 Discrete-time Sampling

58 Discrete-time Sampling

59 Discrete-time Sampling

60 Discrete-time Sampling Insert zeros between samples to upsample the images. x[n] 0 x p [n] n 0 x b [n] n 0 n X b (e jω ) = n x b [n]e jωn = n x p [3n]e jωn = k x p [k]e jωk/3 = X p (e jω/3 )

61 Discrete-time Sampling Then filter out the additional copies in frequency. X(e jω ) 1 2π 0 2π Ω 2π X p (e jω ) π X b (e jω ) = X p (e jω/3 ) 1 3 Ω 2π 4π 3 2π 3 0 2π 4π 2π 3 3 Ω

62 Discrete-time Sampling: Progressive Refinement

63 Discrete-time Sampling: Progressive Refinement

64 Discrete-time Sampling: Progressive Refinement

65 Discrete-time Sampling: Progressive Refinement

66 Discrete-time Sampling: Progressive Refinement

67 Sampling and Quantization Combining sampling and quantization. Resampling (discrete-time sampling) and progressive refinement JPEG

68 JPEG Discrete-time Fourier representations are useful for coding digital images. Example: JPEG ( Joint Photographic Experts Group ) encodes images by a sequence of transformations: color encoding DCT (discrete cosine transform): a kind of Fourier series quantization to achieve perceptual compression (lossy) Huffman encoding: lossless information theoretic coding We will focus on the DCT and quantization of its components. the image is broken into 8 8 pixel blocks each block is represented by its 8 8 DCT coefficients each DCT coefficient is quantized, using higher resolutions for coefficients with greater perceptual importance

69 JPEG Discrete cosine transform (DCT) is similar to a Fourier series, but high-frequency artifacts are typically smaller. Example: imagine coding the following 8 8 block. For a two-dimensional transform, take the transforms of all of the rows, assemble those results into an image and then take the transforms of all of the columns of that image.

70 JPEG Periodically extend a row and represent it with a Fourier series. x[n] = x[n + 8] 0 8 n There are 8 distinct Fourier series coefficients. a k = 1 x[n]e jkω0n ; Ω 0 = 2π 8 8 n=<8>

71 JPEG DCT is based on a different periodic representation, shown below. y[n] = y[n + 16] 0 16 n

72 Check Yourself Which signal has greater high frequency content? x[n] = x[n + 8] 0 8 n y[n] = y[n + 16] 0 16 n

73 Check Yourself The first signal, x[n], has discontinuous amplitude. The second signal, y[n] is not discontinuous, but has discontinuous slope. x[n] = x[n + 8] 0 8 n y[n] = y[n + 16] 0 16 n The magnitude of its Fourier series coefficients decreases faster with k for the second than for the first.

74 JPEG Periodic extension of an 8 8 pixel block can lead to a discontinuous function even when the block was taken from a smooth image. original row 0 n 8 pixel block 0 n x[n] = x[n + 8] 0 8 n

75 JPEG Periodic extension of the type done for JPEG generates a continuous function from a smoothly varying image. original row 0 n 8 pixel block 0 n y[n] = y[n + 16] 0 16 n

76 JPEG Although periodic in N = 16, y[n] can be represented by just 8 distinct DCT coefficients. y[n] = y[n + 16] 0 16 n b k = 7 n=0 ( ( πk y[n] cos n + 1 )) N 2 This results because y[n] is symmetric about n = 1 2, and this symmetry introduces redundancy in the Fourier series representation. Notice also that the DCT of a real-valued signal is real-valued.

77 JPEG The magnitudes of the higher order DCT coefficients are smaller than those of the Fourier series. n Fourier series m n DCT m a k b k m m

78 JPEG Humans are less sensitive to small deviations in high frequency components of an image than they are to small deviations at low frequencies. Therefore, the DCT coefficients are quantized more coarsely at high frequencies. Divide coefficient b[m, n] by q[m, n] and round to nearest integer. q[m, n] m n

79 Check Yourself Which of the following tables of q[m, n] (top or bottom) will result in higher quality images? q[m, n] m n q[m, n] m n

80 JPEG Finally, encode the DCT coefficients for each block using runlength encoding followed by an information theoretic (lossless) Huffman scheme, in which frequently occuring patterns are represented by short codes. The quality of the image can be adjusted by changing the values of q[m, n]. Large values of q[m, n] result in large runs of zeros, which compress well.

81 JPEG: Results 1%: 1432 bytes 10%: 2457 bytes 20%: 3533 bytes 40%: 4795 bytes 80%: 8345 bytes 100%: 26k bytes

82 Summary Today we looked at a number of applications of sampling. Sampling and Quantizing Music Sampling and Quantizing Images Upsampling and Downsampling Images JPEG Image Coding