Summary of Sampling, Line
Codes and PCM
Prepared for ELE 745
Xavier Fernando
Ryerson Communications Lab
Signal Sampling
• Sampling is converting a continuous time
signal into a discrete time signal
• Categories:
– Impulse (ideal) sampling
– Natural Sampling
– Sample and Hold operation
Impulse Sampling
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
The Aliasing Effect
fs > 2fm
fs < 2fm
Aliasing happens
Aliasing
Under sampling will result in
aliasing that will create spectral overlap
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
Natural Sampling
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
Different Sampling Models
Quantization
Discrete
Time &
Discrete
Ampl
Signal
Mapping
Discrete
Time
Cont.
Ampl.
Signal
Quantization
Analog
Signal
Sampling
• Quantization is done to make the signal
amplitude discrete
Binary
Sequence
Linear Quantization
L levels
(L-1)q = 2Vp = Vpp
For large L
Lq ≈ Vpp
PCM Mapping
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
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
Uniform
Non-Uniform
Non-uniform Quantization
Compress the signal first
Then perform linear quantization
 Result in nonlinear quantization
µ-law and A-law
Widely used compression algorithms
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
Line Coded Waveforms - I
NRZ – Non Return to Zero
-Level
NRZ – Non Return to Zero
-Mark (0no change,
1 change)
NRZ – Non Return to Zero
-Space (1no change,
0 change)
Bipolar Return to Zero
AMI – Alternate Mark Inversion
(zero  zero,
1 alternating pulse)
1
0 1 1 0 0 0 1 1 0 1
Bi-Phase level
(1 +v-v, 0 -v+v)
Bi Phase Mark
Bi-Phase Space
Delay Modulation
Dicode NRZ
Dicode RZ
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)
Some Power Spectral Densities
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
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
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
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.
• 00000000000V10V1
• Example:
(-) 0 0 0 - + 0 + - OR
•
(+) 0 0 0 + - 0 - +
• B8ZS is used in the North American telephone
systems at the T1 rate
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
11000011000000
+ - 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:
+-+00+-+-00-00
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
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)
M-ary Signaling
8-level signaling
2-level signaling
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)
Inter Symbol Interference (ISI)
• Unwanted interference from adjacent (usually
previous) symbols
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 .... )).
Min. BW Pulse satisfying the first criteria
Zero ISI Pulse
Vestigial Spectrum
Raised Cosine Pulse
r=0 (fx=0)
r=0.5 (fx=Rb/4)
r=1 (fx=Rb/2)
Raised Cosine Filter Transfer Function
in the f domain
Raised Cosine Filter Impulse Response
(time domain)
Note pulse
rapidly decays
for r = 1
Equalization
• The residual ISI can be
removed by equalization
• Estimate the amount of
ISI at each sampling
instance and subtract it
Eye Diagram
• Ideal (perfect)
signal
• Real (average)
signal
• Bad signal
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
Time Division Multiplexing (TDM)
• TDM is widely used in digital communication
systems to maximum use the channel capacity
Digit Interleaving
TDM – Word Interleaving
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
North American PCM Telephony
• Twenty four T1 carriers (64kb/s) are
multiplexed to generate one DS1 carrier
(1.544 Mb/s)
Each channel has 8 bits – 24 Channels
• Each frame has 24 X 8 = 192 information bits
• Frame time = 1/8000 = 125 μs.
T1 System Signalling Format
193 framing bits plus more signalling bits final bit rate = 1.544 Mb/s
North American Digital Hierarchy
Delta Modulation
Why transmit every sample?
You know the next amplitude will differ by only ‘delta’
Delta Modulation
Why transmit every sample?
You know the next amplitude will differ by only ‘delta’
Only transmit the error
LPC Coding
Transmit only few gain coefficients!
• In modern
communicatio
n system, the
voice is
artificially
generated at
the receiver
mimicking the
original voice
using the
appropriate
coefficients
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?
Solution
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
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.