TELE3113 Analogue and Digital Communications – PCM Wei Zhang firstname.lastname@example.org School of Electrical Engineering and Telecommunications The University of New South WalesTELE3113 - PCM 16 Sept. 2009 p. -1
Analog-to-Digital Conversion (PCM) 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 16 Sept. 2009 p. -2
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 16 Sept. 2009 p. -3
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. -4TELE3113 - PCM 16 Sept. 2009
PCM - Encoding Each quantized level is translated into a binary code (digital). For L (=2n) quantized levels, number of bits per code is n (=log2L). The binary code is then converted into a sequential string of pulses for transmission Pulse code modulation (PCM) Code Code Quantized number level 4 x(t) 111 7 3.5 3 110 6 2.5 2 101 5 1.5 1 100 4 0.5 0 011 3 -0.5 time -1 010 2 -1.5 -2 001 1 -2.5 -3 000 0 -3.5 -4 Sampler Sampled value -0.25 3.3 1.2 -2.8 -3.8 -2.1 Quantizer Quantized value -0.5 3.5 1.5 -2.5 -3.5 -2.5 Code number 3 7 5 1 0 1 Encoder Binary code 011 111 101 001 000 001 Parallel to-serial converterTELE3113 - PCM 16 Sept. 2009 p. -5 011 111 101 001 000 001
PCM Signal - Demodulation At the receiver, the received pulses are distorted, need regeneration to restore the ideal pulse (rectangular). Regenerated Decision signal threshold Distorted and time time Re-shaped, time noise corrupted Sampling time Re-amplified, (Re-timed) Regeneration The regenerated pulse stream will be separated into codes using serial-to- parallel converter. 000 011 000011001011111001 001 3-bit 011 codes 111 001 The recovered codes are translated back into respective quantized levels. The quantized levels are interpolated using low-pass filter to form the recovered signal.TELE3113 - PCM 16 Sept. 2009 p. -6
PCM Signal - Sampling Rate For L (=2n) quantized levels, each sample point of the message signal is quantized into n bits. Then, n (=log2L) binary pulses must be transmitted for each sample point of the message signal. If the message bandwidth is B and the sampling rate is fs (≥2B), then nfs binary pulses must be transmitted per second as each sample point will be translated into a n-bit code. minimum sampling rate for the PCM signal is R=nfs 1 Ts (at least one sampling point for each bit or pulse, bit period= = ) nf s n minimum bandwidth for PCM signal = nfs /2 ≥ nB Minimum required bandwidth for PCM is proportional to the message signal bandwidth, B, and the number of bits per code (symbol), n.TELE3113 - PCM 16 Sept. 2009 p. -7
PCM Signal – SNR Analysis 2 Vmax With L =2n, average SNRx (dB) = 4.77 + 20 log L − 10 log x 2 (t ) Vmax 2 = 4.77 + 20 log 2 − 10 log n x (t ) 2 V2 average SNRx (dB) = 4.77 + 6.02n − 10 log 2max x (t ) 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) = 1.76 + 20 log L = (1.76 + 6.02n ) dB If x(t) is uniformly distributed in the range [-Vmax,+Vmax], then pdf f(x)=1/(2Vmax), Thus, average SNRx (dB) = 20 log L = 6.02n dB The SNR can be improved by 6dB when one more bit is used in the code, but the bandwidth required for the PCM signal will be getting larger (as minimum bandwidth for PCM signal is nB where B is the message signal bandwidth).TELE3113 - PCM 16 Sept. 2009 p. -8
PCM - Applications Telephone System: Voice bandwidth ~ 4kHz minimum sampling rate: 8kHz 8000 samples per second 8-bit PCM is used 8 bits per sample 8x8000 =64kbits per second For modem application, only 7 bits are for data, the other is an overhead bit 56kbit/s Compact Disk (CD) Audio: Each of the two stereo signals is sampled at 44.1kHz 16-bit PCM is used 16x44.1k = 0.7056Mbits per second per stereo channel 100% overhead (error correction code) 1.411Mbit/s per stereo channel One CD can record 1 hour music, total number of bits (for 2 stereo channels) = 1.411M x 2 x 3600 = 10.16GbitsTELE3113 - PCM 16 Sept. 2009 p. -9
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