sampling
DESCRIPTION
code sampling, PCMTRANSCRIPT
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Summary of Sampling, Line Codes and PCM
Prepared for ELE 745Xavier Fernando
Ryerson Communications Lab
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Signal Sampling
• Sampling is converting a continuous time signal into a discrete time signal
• Categories:– Impulse (ideal) sampling– Natural Sampling– Sample and Hold operation
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Impulse Sampling
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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
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The Aliasing Effect
fs > 2fm
fs < 2fm
Aliasing happens
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Aliasing
Under sampling will result in aliasing that will create spectral overlap
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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
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Natural Sampling
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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
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Different Sampling Models
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Quantization• Quantization is done to make the signal
amplitude discrete
Analog Signal
Discrete TimeCont. Ampl. Signal
Discrete Time & Discrete
Ampl Signal
Binary Sequence
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Linear Quantization
L levels (L-1)q = 2Vp = Vpp
For large LLq ≈ Vpp
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PCM Mapping
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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
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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
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Uniform Non-Uniform
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Non-uniform Quantization
Compress the signal first Then perform linear quantization Result in nonlinear quantization
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µ-law and A-lawWidely used compression algorithms
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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
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Line Coded Waveforms - I
NRZ – Non Return to Zero-Level
NRZ – Non Return to Zero-Mark (0no change, 1 change)
NRZ – Non Return to Zero-Space (1no change, 0 change)
Bipolar Return to Zero
AMI – Alternate Mark Inversion (zero zero, 1 alternating pulse)
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Bi-Phase level (1 +v-v, 0 -v+v)
Bi Phase Mark
Bi-Phase Space
Delay Modulation
Dicode NRZ
Dicode RZ
1 0 1 1 0 0 0 1 1 0 1
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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)
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Some Power Spectral Densities
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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
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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
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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
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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.
• 0 0 0 0 0 0 0 0 0 0 0 V 1 0 V 1• Example: (-) 0 0 0 - + 0 + - OR• (+) 0 0 0 + - 0 - +• B8ZS is used in the North American telephone
systems at the T1 rate
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High Density Bipolar 3 code
• HDB3 encodes any pattern of more than four bits as B00V (or 100V; 1B (Bit))
• Ex: The pattern of bits1 1 0 0 0 0 1 1 0 0 0 0 0 0+ - 0 0 0 0 + - 0 0 0 0 0 0 (AMI) • Encoded in HDB3 is:+ - B 0 0 V - + B 0 0 V 0 0, which is:+ - + 0 0 + - + - 0 0 - 0 0
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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
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M-Ary Coding
• Advantages:– Required transmission rate is low (bit rate/M)– Low bandwidth
• Disadvantages:– Low signal to noise ratio (due to multiple
amplitude pulses)
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M-ary Signaling
8-level signaling
2-level signaling
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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)
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Inter Symbol Interference (ISI)
• Unwanted interference from adjacent (usually previous) symbols
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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. 2. 3 .... )).
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Min. BW Pulse satisfying the first criteria
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Zero ISI Pulse
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Vestigial Spectrum
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Raised Cosine Pulse
r=0 (fx=0)
r=0.5 (fx=Rb/4)
r=1 (fx=Rb/2)
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Raised Cosine Filter Transfer Functionin the f domain
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Raised Cosine Filter Impulse Response (time domain)
Note pulse rapidly decays for r = 1
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Equalization• The residual ISI can be
removed by equalization• Estimate the amount of
ISI at each sampling instance and subtract it
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Eye Diagram
• Ideal (perfect) signal
• Real (average) signal
• Bad signal
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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
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Time Division Multiplexing (TDM)• TDM is widely used in digital communication
systems to maximum use the channel capacity
Digit Interleaving
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TDM – Word Interleaving
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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
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North American PCM Telephony• Twenty four T1 carriers (64kb/s) are multiplexed
to generate one DS1 carrier (1.544 Mb/s)
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Each channel has 8 bits – 24 Channels
• Each frame has 24 X 8 = 192 information bits• Frame time = 1/8000 = 125 μs.
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T1 System Signalling Format
193 framing bits plus more signalling bits final bit rate = 1.544 Mb/s
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North American Digital Hierarchy
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Delta Modulation
Why transmit every sample? You know the next amplitude will differ by only ‘delta’
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Delta Modulation
Why transmit every sample? You know the next amplitude will differ by only ‘delta’Only transmit the error
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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!
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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 1.375 MHz?
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Solution
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Example - 2Sklar 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
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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.
(a)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.
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