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Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
Tele3113 wk7wed
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Tele3113 wk7wed

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  • 1. TELE3113 Analogue and Digital Communications – Quantization Wei Zhang w.zhang@unsw.edu.au School of Electrical Engineering and Telecommunications The University of New South WalesTELE3113 - PCM 2 Sept. 2009 p. -1
  • 2. Analog-to-Digital Conversion Goal: To transmit the analog signals by digital means better performance convert the analog signal into digital format (Pulse-Code Modulation) Analog sampler Quantizer Encoder Digital signal signal 1111111 1111110 1111101 1111100 1111101 1111001 1111001 1111011 1111101 1111110 1111101 1111001 1111011 1111010 c8 c7 c6 c5 c4 c3 c2 c1 x(t) 1111001 1111000 time time time Sampling: a continuous-time signal is sampled by measuring its amplitude at discrete time instants. Quantizing: represents the sampled values of the amplitude by a finite set of levels Encoding: designates each quantized level by a digital codeTELE3113 - PCM 2 Sept. 2009 p. -2
  • 3. Reconstruction of Sampled Signal 1111101 11110011111001 1111011 1111101 1111110 11111011111001 Received digital signal c8 c7 c6 c5 c4 c3 c2 c1 time decoding Recovered signal with discrete levels interpolation Recovered analog signalTELE3113 - PCM 2 Sept. 2009 p. -3
  • 4. Sampling Consider an analog signal x(t) which is bandlimited to B (Hz), that is: X ( f ) = 0 for | f |≥ B The sampling theorem states that x(t) can be sampled at intervals as large as 1/(2B) such that the it is possible to reconstruct x(t) from its samples, or the sampling rate fs=1/Ts can be as low as 2B. x(t) 1 f s ≥ 2B or Ts ≤ 2B Sampling rate fs=1/Ts time Sampling period Ts Minimum required sampling rate=2B (Nyquist rate) i.e. 2B samples per second Sampling rate should be equal or greater than twice the highest frequency in the baseband signal.TELE3113 - PCM 2 Sept. 2009 p. -4
  • 5. Analogue Pulse Modulation Pulse Amplitude Modulation Pulse Duration (Width) Modulation Pulse Position ModulationTELE3113 - PCM 2 Sept. 2009 p. -5
  • 6. Quantization After the sampling process, the sampled points will be transformed into a set of predefined levels (quantized level) Quantization Assume the signal amplitude of x(t) lies within [-Vmax ,+Vmax], we divide the total peak-to-peak range (2Vmax) into L levels in which the quantized levels mi (i=0,…,(L-1)) are defined as their respective mid-ways. Vmax xq(t) x(t) Output m7 ∆7 m7= 7∆/2 m6 ∆6 uniform m6= 5∆/2 m5 ∆5 m5= 3∆/2 m4 ∆4 m4= ∆/2 m3 ∆3 time −4∆ −3∆ −2∆ −∆ ∆ 2∆ 3∆ 4∆ Input m2 ∆2 −3∆/2 uniform m1 ∆1 −5∆/2 m0 ∆0 −7∆/2 -Vmax Sampling time Uniform quantizer 2Vmax (midrise type) For uniform quantization, ∆ i ( i =0 ,L,( L −1)) = ∆ = L p. -6TELE3113 - PCM 2 Sept. 2009
  • 7. Quantization Noise (1) The quantized signal, xq(t) is an approximation of the original message signal, x(t). −∆ ∆ Quantization error/noise: eq(t) ={x(t) - xq(t)} varies randomly within ≤ eq (t ) ≤ 2 2 x(t) xq(t) eq(t) ={x(t) - xq(t)}TELE3113 - PCM 2 Sept. 2009 p. -7
  • 8. Quantization Noise (2) Assume the quantization error varies uniformly within [-∆/2, ∆/2] with a pdf of f(eq)=1/∆, then ∆ 2 ∆ 2 2 q [ ] e (t ) = ∫ f (eq ) eq (t ) deq = 2 1 ∫ ∆ −∆ 2 [ 2 eq (t ) deq ] Q f (eq ) = 1 ∆ −∆ 2 3 ∆ 2 1 eq ∆2 Vmax 2 2Vmax = = = with ∆ = ∆ 3 −∆ 2 12 3L2 L To minimize eq(t), we can use smaller ∆ or more quantized levels L. 2 In general, the average power of a signal is x (t ) or x 2 (t ) x 2 (t ) 3L2 x 2 (t ) average SNRx = 2 = 2 eq (t ) Vmax  3L2 x 2 (t )   V2  average SNRx (dB) = 10 log   = 4.77 + 20 log L − 10 log 2max   Vmax 2   x (t )     TELE3113 - PCM 2 Sept. 2009 p. -8
  • 9. Quantization Noise (3) 2 Vmax If x(t) is a full-scale sinusoidal signal, i.e. x(t)=Vmaxcosωt , then x (t ) = x (t ) = 2 2 2 Thus, average SNRx (dB) = 4.77 + 20 log L − 10 log(2 ) = (1.76 + 20 log L ) dB If x(t) is uniformly distributed in the range [-Vmax,+Vmax], then pdf f(x)=1/(2Vmax), Vmax Vmax 1 1 x (t ) = ∫ f ( x)[x(t )] dx = ∫ [x(t )]2 dx 2 2 Q f (eq ) = −Vmax 2Vmax −Vmax 2Vmax Vmax 1 x3 2 Vmax = = 2Vmax 3 3 Thus, −Vmax average SNRx (dB) = 4.77 + 20 log L − 10 log(3) = 20 log L dBTELE3113 - PCM 2 Sept. 2009 p. -9
  • 10. Non-uniform Quantization (1) In some cases, uniform quantization is not efficient. In speech communication, it is found (statistically) that smaller amplitudes predominate in speech and that larger amplitudes are relatively rare. many quantized levels are rarely used (wasteful !) Non-uniform quantization is more efficient. xmax x(t) Quantized levels time -xmaxTELE3113 - PCM 2 Sept. 2009 p. -10
  • 11. Non-uniform Quantization (2) The non-uniform quantization can be achieved by first compressing the signal samples and then performing uniform quantization. Output 1 uniform ∆yi ∆ si 1 Input x s = Non-uniform x max Compressor There exists more quantized levels for small x and fewer levels for larger x.TELE3113 - PCM 2 Sept. 2009 p. -11
  • 12. Non-uniform Quantization (3) Input Uniform Communication Sampler Compressor Encoder signal Quantizer Channel Received Decoder Expander Interpolator signal Output Output Non-uniform uniform ∆ yi ∆si ∆si Input Input ∆ yi Non-uniform Uniform Compressor ExpanderTELE3113 - PCM 2 Sept. 2009 p. -12
  • 13. Non-uniform Quantization (4) Two common compression laws Α-law : 1 + ln  A x          xmax  sgn( x ) for 1 ≤ x ≤ 1 µ-law :   y ( x) =  1 + ln A A xmax x ln1 + µ     A xmax x x x 1 y ( x) =   sgn( x) for ≤1  sgn( x ) for 0 ≤ ≤ ln (1 + µ ) xmax 1 + ln A xmax  xmax A Digital telephone system in North Digital telephone system in Europe America and Japan (µ=255) (Α=87.6)TELE3113 - PCM 2 Sept. 2009 p. -13

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