ECE 4371, Fall, 2014
Introduction to Telecommunication
Engineering/Telecommunication Laboratory
Zhu Han
Department of Electrical and Computer Engineering
Class 13
Oct. 8th, 2014
Outline

Partial Response

Carrier systems
– ASK, OOK, MASK
– FSK, MFSK
– BPSK, DBPSK, MPSK
– MQAM, MQPR
– OQPSK,
– Continuous phase modulation (CPM): MSK, GMSK
Partial Response Signals

Previous classes: Sy(w)=|P(w)|^2 Sx(w)
– Control signal generation methods to reduce Sx(w)
– Raise Cosine function for better |P(w)|^2

This class: improve the bandwidth efficiency
– Widen the pulse, the smaller the bandwidth.
– But there is ISI. For binary case with two symbols, there is only
few possible interference patterns.
– By adding ISI in a controlled manner, it is possible to achieve a
signaling rate equal to the Nyquist rate (2W symbols/sec) in a
channel of bandwidth W Hertz.
Example

Duobinary Pulse
– p(nTb)=1, n=0,1
– p(nTb)=1, otherwise

Interpretation of received signal
– 2: 11
– -2: 00
– 0: 01 or 10 depends on the previous transmission
Duobinary signaling

Duobinary signaling (class I partial response)
Duobinary signal and Nyguist Criteria

Nyguist second criteria: but twice the bandwidth
Differential Coding

The response of a pulse is spread over more than one signaling
interval.

The response is partial in any signaling interval.

Detection :
– Major drawback : error propagation.

To avoid error propagation, need deferential coding (precoding).
Modified duobinary signaling

Modified duobinary signaling
– In duobinary signaling, H(f) is nonzero at the origin.
– We can correct this deficiency by using the class IV partial
response.
Modified duobinary signaling

Spectrum
Modified duobinary signaling

Time Sequency: interpretation of receiving 2, 0, and -2?
Pulse Generation

Generalized form of
correlative-level
coding
(partial response signaling)
Tradeoffs

Binary data transmission over a physical baseband channel can
be accomplished at a rate close to the Nyquist rate, using
realizable filters with gradual cutoff characteristics.

Different spectral shapes can be produced, appropriate for the
application at hand.

However, these desirable characteristics are achieved at a price :
– A large SNR is required to yield the same average probability of
symbol error in the presence of noise.
Exercise

What is the differential Duobinary output

What is the modified Duobinary output and decoded signal?
Interpretation of receiving 2, 0, and -2?

What is the modified differential Duobinary output
Other types of partial response signals
paper
Type
r0
r1
r2
r3
r4
ideal
1
I
1
1
II
1
2
1
5
III
2
1
-1
6
IV
1
0
-1
3
V
-1
0
2
p(t)
P(W)
Levels
2
3
0
-1
5
ASK, OOK, MASK

The amplitude (or height) of the sine wave varies to transmit the
ones and zeros

One amplitude encodes a 0 while another amplitude encodes a 1
(a form of amplitude modulation)
Binary amplitude shift keying, Bandwidth

d ≥ 0  related to the condition of the line
B = (1+d) x S = (1+d) x N x 1/r
Implementation of binary ASK
OOK and MASK

OOK (On-OFF Key)
– 0 silence.
– Sensor networks: battery life, simple implementation

MASK: multiple amplitude levels
Pro, Con and Applications

Pro
– Simple implementation

Con
– Major disadvantage is that telephone lines are very susceptible to
variations in transmission quality that can affect amplitude
– Susceptible to sudden gain changes
– Inefficient modulation technique for data

Applications
– On voice-grade lines, used up to 1200 bps
– Used to transmit digital data over optical fiber
– Morse code
– Laser transmitters
Example

We have an available bandwidth of 100 kHz which spans from
200 to 300 kHz. What are the carrier frequency and the bit
rate if we modulated our data by using ASK with d = 1?

Solution
– The middle of the bandwidth is located at 250 kHz. This
means that our carrier frequency can be at fc = 250 kHz.
We can use the formula for bandwidth to find the bit rate
(with d = 1 and r = 1).
Frequency Shift Keying

One frequency encodes a 0 while another frequency encodes a 1
(a form of frequency modulation)

Represent each logical value with another frequency (like FM)

 A cos2f1t 
s t   

 A cos2f 2t 
binary1
binary 0
FSK Bandwidth



Limiting factor: Physical capabilities of the carrier
Not susceptible to noise as much as ASK
Applications
– On voice-grade lines, used up to 1200bps
– Used for high-frequency (3 to 30 MHz) radio transmission
– used at higher frequencies on LANs that use coaxial cable
Example

We have an available bandwidth of 100 kHz which spans from
200 to 300 kHz. What should be the carrier frequency and the
bit rate if we modulated our data by using FSK with d = 1?

Solution
– This problem is similar to Example 5.3, but we are modulating
by using FSK. The midpoint of the band is at 250 kHz. We
choose 2Δf to be 50 kHz; this means
Multiple Frequency-Shift Keying (MFSK)

More than two frequencies are used

More bandwidth efficient but more susceptible to error
si t   A cos2f i t
1 i  M
f i = f c + (2i – 1 – M)f d
 f c = the carrier frequency
 f d = the difference frequency
L
 M = number of different signal elements = 2
 L = number of bits per signal element

Phase Shift Keying

One phase change encodes a 0 while another phase change
encodes a 1 (a form of phase modulation)

binary1
 A cos2f ct 
s t   
binary 0

 A cos2f ct   
DBPSK, QPSK

Differential BPSK
– 0 = same phase as last signal element
– 1 = 180º shift from last signal element

Four Level: QPSK



s t   





A cos 2f c t 

4

3 

A cos 2f c t 

4 

3 

A cos 2f c t 

4 

 

A cos 2f c t 

4

11
01
00
10
QPSK Example
Bandwidth

Min. BW requirement: same as ASK!

Self clocking (most cases)
Concept of a constellation diagram
MPSK

Using multiple phase angles with each angle having more than
one amplitude, multiple signals elements can be achieved
R
R
D 
L log2 M
– D = modulation rate, baud
– R = data rate, bps
– M = number of different signal elements = 2L
– L = number of bits per signal element
QAM – Quadrature Amplitude Modulation

Modulation technique used in the cable/video networking world

Instead of a single signal change representing only 1 bps –
multiple bits can be represented buy a single signal change

Combination of phase shifting and amplitude shifting (8 phases, 2
amplitudes)
QAM

QAM
– As an example of QAM, 12
different phases are combined
with two different amplitudes
– Since only 4 phase angles have 2
different amplitudes, there are a
total of 16 combinations
– With 16 signal combinations, each
baud equals 4 bits of information
(2 ^ 4 = 16)
– Combine ASK and PSK such that
each signal corresponds to
multiple bits
– More phases than amplitudes
– Minimum bandwidth requirement
same as ASK or PSK
QAM and QPR

QAM is a combination of ASK and PSK
– Two different signals sent simultaneously on the same carrier frequency
st   d1 t cos2f ct  d2 t sin 2f ct
– M=4, 16, 32, 64, 128, 256

Quadrature Partial Response (QPR)
– 3 levels (+1, 0, -1), so 9QPR, 49QPR
Offset quadrature phase-shift keying (OQPSK)

QPSK can have 180 degree jump, amplitude fluctuation

By offsetting the timing of the odd and even bits by one bit-period, or half a
symbol-period, the in-phase and quadrature components will never change at
the same time.
Continuous phase modulation (CPM)

CPM the carrier phase is modulated in a continuous manner

constant-envelope waveform

yields excellent power efficiency

high implementation complexity required for an optimal
receiver

minimum shift keying (MSK)
– Similarly to OQPSK, MSK is encoded with bits alternating between
quarternary components, with the Q component delayed by half a bit
period. However, instead of square pulses as OQPSK uses, MSK encodes
each bit as a half sinusoid. This results in a constant-modulus signal,
which reduces problems caused by non-linear distortion.
ECE 4371 Fall 2008
Gaussian minimum shift keying

GMSK is similar to MSK except it incorporates a premodulation Gaussian
LPF

Achieves smooth phase transitions between signal states which can
significantly reduce bandwidth requirements

There are no well-defined phase transitions to detect for bit synchronization
at the receiving end.

With smoother phase transitions, there is an increased chance in intersymbol
interference which increases the complexity of the receiver.

Used extensively in 2nd generation digital cellular and cordless telephone
apps. such as GSM