Presentation is loading. Please wait.

Presentation is loading. Please wait.

Summary of Sampling, Line Codes and PCM

Similar presentations

Presentation on theme: "Summary of Sampling, Line Codes and PCM"— Presentation transcript:

1 Summary of Sampling, Line Codes and PCM
Prepared for ELE 745 Xavier Fernando Ryerson Communications Lab

2 Signal Sampling Sampling is converting a continuous time signal into a discrete time signal Categories: Impulse (ideal) sampling Natural Sampling Sample and Hold operation

3 Impulse Sampling

4 Impulse Sampling Impulse train spaced at Ts multiplies the signal x(t) in time domain, creating discrete time, continuous amplitude signal xs(t) Impulse train spaced at fs convolutes the signal X(f) in frequency domain, creating Repeating spectrum Xs(f) spaced at fs

5 The Aliasing Effect fs > 2fm fs < 2fm Aliasing happens

6 Aliasing Under sampling will result in
aliasing that will create spectral overlap

7 Ideal Sampling and Aliasing
Sampled signal is discrete in time domain with spacing Ts Spectrum will repeat for every fs Hz Aliasing (spectral overlapping) if fs is too small (fs < 2fm) Nyquist sampling rate fs = 2fm Generally oversampling is done  fs > 2fm

8 Natural Sampling

9 Natural Sampling Sampling pulse train has a finite width τ
Sampled spectrum will repeat itself with a ‘Sinc’ envelope More realistic modeling Distortion after recovery depends on τ/Ts

10 Different Sampling Models

11 Time & Discrete Ampl Signal
Quantization Quantization is done to make the signal amplitude discrete Quantization Sampling Mapping Discrete Time Cont. Ampl. Signal Discrete Time & Discrete Ampl Signal Analog Signal Binary Sequence

12 Linear Quantization L levels (L-1)q = 2Vp = Vpp For large L Lq ≈ Vpp

13 PCM Mapping

14 Linear Quantization Summary
Mean Squared Error (MSE) = q2/12 Mean signal power = E[m2(t)] Mean SNR = 12 E[m2(t)]/q2 For binary PCM, L = 2n  n bits/sample Let signal bandwidth = B Hz If Nyquist sampling  2B samples/sec If 20% oversampling  1.2(2B) samples/sec Bit rate = 2nB bits/sec Required channel bandwidth = nB Hz

15 Non-Uniform Quantization
In speech signals, very low speech volumes predominates Only 15% of the time, the voltage exceeds the RMS value These low level signals are under represented with uniform quantization Same noise power (q2/12) but low signal power The answer is non uniform quantization

16 Uniform Non-Uniform

17 Non-uniform Quantization
Compress the signal first Then perform linear quantization  Result in nonlinear quantization

18 µ-law and A-law Widely used compression algorithms

19 Line Coding Digital output of the PCM coder is converted to an appropriate waveform for transmission over channel  line coding or transmission coding Different line codes have different attributes Best line code has to be selected for a given application and channel condition

20 Line Coded Waveforms - I
NRZ – Non Return to Zero -Level -Mark (0no change, 1 change) -Space (1no change, 0 change) Bipolar Return to Zero AMI – Alternate Mark Inversion (zero  zero, 1 alternating pulse)

21 Bi-Phase level (1 +v-v, 0 -v+v) Bi Phase Mark Bi-Phase Space Delay Modulation Dicode NRZ Dicode RZ

22 Line Coding Requirements
Favorable power spectral density (PSD) Low bandwidth (multilevel codes better) No/little DC power Error detection and/or correction capability Self clocking (Ex. Manchester) Transparency in generating the codes (dependency on the previous bit?) Differential encoding (polarity reversion) Noise immunity (BER for a given SNR)

23 Some Power Spectral Densities

24 Polar Signalling {p(t) or –p(t)}
Polar signalling is not bandwidth efficient (best case BW = Rb . Theoretical min is Rb/2) Non-zero DC No error detection (each bit is independent) Efficient in power requirement Transparent Clock can be recovered by rectifying the received signal

25 On-Off Signalling On-off is a sum of polar signal and periodic clock signal (Fig. 7.2)  spectrum has discrete freq. Components Polar amplitude is A/2  PSD is scaled by ¼ No error detection Excessive zeros cause error in timing extraction Excessive BW Excessive DC

26 AMI (bipolar) Signalling
DC null Single error detection (violation) capability Clock extraction possible Twice as much power as polar signalling Not transparent Excessive zeros cause timing extraction error  HDB or B8ZS schemes used to overcome this issue

27 Bipolar with 8 Zeros Substitution
B8ZS uses violations of the Alternate Mark Inversion (AMI) rule to replace a pattern of eight zeros in a row.  V 1 0 V 1 Example: (-) OR (+) B8ZS is used in the North American telephone systems at the T1 rate

28 High Density Bipolar 3 code
HDB3 encodes any pattern of more than four bits as B00V (or 100V; 1B (Bit)) Ex: The pattern of bits (AMI) Encoded in HDB3 is: + - B 0 0 V - + B 0 0 V 0 0, which is:

29 M-Ary Coding (Signaling)
In binary coding: Data bit ‘1’ has waveform 1 Data bit ‘0’ has waveform 2 Data rate = bit rate = symbol rate In M-ary coding, take M bits at a time (M = 2k) and create a waveform (or symbol). ‘00’  waveform (symbol) 1 ‘01’  waveform (symbol) 2 ‘10’  waveform (symbol) 3 ‘11’  waveform (symbol) 2 Symbol rate = bit rate/k

30 M-Ary Coding Advantages: Disadvantages:
Required transmission rate is low (bit rate/M) Low bandwidth Disadvantages: Low signal to noise ratio (due to multiple amplitude pulses)

31 M-ary Signaling 8-level signaling 2-level signaling

32 M-ary (Multilevel) Signaling
M-ary signals reduce required bandwidth Instead of transmitting one pulse for each bit (binary PCM), we transmit one multilevel pulse a group of k-bits (M=2k) Bit rate = Rb bits/s  min BW = Rb/2 Symbol rate = R/k sym/s  min BW = Rb/2k Needed bandwidth goes down by k Trade-off is relatively high bit error rate (BER)

33 Inter Symbol Interference (ISI)
Unwanted interference from adjacent (usually previous) symbols

34 Nyquist's First Criterion for Zero lSI
In the first method Nyquist achieves zero lSI by choosing a pulse shape that has a nonzero amplitude at its center (t=0) and zero amplitudes at (t=±nT" (n = I )).

35 Min. BW Pulse satisfying the first criteria

36 Zero ISI Pulse

37 Vestigial Spectrum

38 Raised Cosine Pulse r=0 (fx=0) r=0.5 (fx=Rb/4) r=1 (fx=Rb/2)

39 Raised Cosine Filter Transfer Function in the f domain

40 Raised Cosine Filter Impulse Response (time domain)
Note pulse rapidly decays for r = 1

41 Equalization The residual ISI can be removed by equalization
Estimate the amount of ISI at each sampling instance and subtract it

42 Eye Diagram Ideal (perfect) signal Real (average) signal Bad signal

43 Eye Diagram Run the oscilloscope in the storage mode for overlapping pulses X-scale = pulse width Y-Scale = Amplitude Close Eye  bad ISI Open Eye  good ISI

44 Time Division Multiplexing (TDM)
TDM is widely used in digital communication systems to maximum use the channel capacity Digit Interleaving

45 TDM – Word Interleaving

46 TDM When each channel has Rb bits/sec bit rate and N such channels are multiplexed, total bit rate = NRb (assuming no added bits) Before Multiplexing the bit period = Tb After Multiplexing the bit period = Tb/N Timing and bit rate would change if you have any added bits

47 North American PCM Telephony
Twenty four T1 carriers (64kb/s) are multiplexed to generate one DS1 carrier (1.544 Mb/s)

48 Each channel has 8 bits – 24 Channels
Each frame has 24 X 8 = 192 information bits Frame time = 1/8000 = 125 μs.

49 T1 System Signalling Format
193 framing bits plus more signalling bits final bit rate = Mb/s

50 North American Digital Hierarchy

51 Delta Modulation Why transmit every sample?
You know the next amplitude will differ by only ‘delta’

52 Delta Modulation Why transmit every sample?
You know the next amplitude will differ by only ‘delta’ Only transmit the error

53 LPC Coding In modern communication system, the voice is artificially generated at the receiver mimicking the original voice using the appropriate coefficients Transmit only few gain coefficients!

54 Example -1 Sklar 3.8: (a) What is the theoretical minimum system bandwidth needed for a 10 Mb/s signal using 16-level PAM without ISI? (b) How large can the filter roll-off factor (r) be if the applicable system bandwidth is MHz?

55 Solution

56 Example - 2 Sklar 3.10: Binary data at 9600 bits/s are transmitted using 8-ary PAM modulation with a system using a raised cosine roll-off filter characteristics. The system has a frequency response out to 2.4 kHz. (a) What is the symbol rate (b) What is the roll o® factor r

57 Example 3 Sklar 3.11: A voice signal in the range 300 to 3300 Hz is sampled at 8000 samples/s. We may transmit these samples directly as PAM pulses or we may first convert each sample to a PCM format and use binary (PCM) waveform for transmission. What is the minimum system bandwidth required for the detection of PAM with no ISI and with a filter roll-off factor of 1. (b) Using the same roll-off, what is the minimum bandwidth required for the detection of binary PCM waveform if the samples are quantized to 8-levels (c) Repeat part (b) using 128 quantization levels.


Download ppt "Summary of Sampling, Line Codes and PCM"

Similar presentations

Ads by Google