### Lecture 5: Analog Transmission

```Analog Transmission
5.2
DIGITAL-TO-ANALOG CONVERSION
Digital-to-analog conversion is the process of
changing one of the characteristics of an analog
signal based on the information in digital data.
Topics discussed in this lecture
Aspects of Digital-to-Analog Conversion
Amplitude Shift Keying
Frequency Shift Keying
Phase Shift Keying
5.3
Digital-to-analog conversion
Digital to analog conversion is the process of changing one of the
characteristics of an analog signal based on the information in digital data
By changing one
of the sine wave
characteristic, we
can represent
digital data
Digital-to-analog conversion
• Bit rate is the number of bits per second. Baud rate is the number
of signal elements per second.
• In the analog transmission of digital data, the baud rate is less
than or equal to the bit rate.
Example
An analog signal carries 4 bits per signal element. If 1000 signal
elements are sent per second, find the bit rate.
Solution
In this case, r = 4, S = 1000, and N is unknown. We can
find the value of N from
5.5
Digital-to-analog conversion
Example
An analog signal has a bit rate of 8000 bps and a baud rate of 1000
baud. How many data elements are carried by each signal element?
How many signal elements do we need?
Solution
In this example, S = 1000, N = 8000, and r and L are unknown. We
find first the value of r and then the value of L.
5.6
Binary amplitude shift keying
5.7
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).
5.8
Example
In data communications, we normally use full-duplex links
with communication in both directions. We need to divide
the bandwidth into two with two carrier frequencies. The
available bandwidth for each direction is 50 kHz, which
leaves us with a data rate of 25 kbps in each direction.
5.9
Binary frequency shift keying
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
The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this
means
5.10
Bandwidth of MFSK
Implementation
B=LxS
Example
5.11
We need to send data 3 bits at a time at a bit rate of 3 Mbps. The
carrier frequency is 10 MHz. Calculate the number of levels
(different frequencies), the baud rate, and the bandwidth.
Solution
We can have L = 23 = 8. The baud rate is S = 3 MHz/3 = 1000
Mbaud. This means that the carrier frequencies must be 1 MHz apart
(2Δf = 1 MHz). The bandwidth is B = 8 × 1000 = 8000.
Binary phase shift keying
• PSK uses a finite number of phases, each assigned a
unique pattern of binary digits.
• Each pattern of bits forms the symbol that is represented
by the particular phase.
There are two fundamental ways of utilizing the phase of a
signal in this way:
• By viewing the phase itself as conveying the information, in which
case the demodulator must have a reference signal to compare the
received signal's phase against; (Coherent PSK)
• By viewing the change in the phase as conveying information —
differential schemes, some of which do not need a reference carrier.
(Differential PSK)
A convenient way to represent PSK schemes is on a constellation
diagram
Concept of a constellation diagram
• Represents the possible symbols that may be
selected by a given modulation scheme as points
in the complex plane.
• By representing a transmitted symbol as a
complex number and modulating a cosine and
sine carrier signal with the real and imaginary
parts (respectively), the symbol can be sent with
two carriers on the same frequency.
• The real and imaginary axes are often called the
in phase, or I-axis and the quadrature, or Q-axis.
Plotting several symbols in a
scatter diagram produces the
constellation diagram. The points
on a constellation diagram are
called constellation points. They
are a set of modulation symbols
which comprise the modulation
alphabet
5.14
Constellation diagrams for some QAMs
The constellation diagrams for an ASK (OOK), BPSK, and QPSK signals.
Application
• The wireless LAN standard, IEEE 802.11b-1999, uses a
variety of different PSKs.
• 1 Mbit/s  differential BPSK
• 2 Mbit/s  DQPSK
• 5.5 Mbit/s and the full-rate of 11 Mbit/s  QPSK
• Bluetooth 2
• DQPSK at its lower rate (2 Mbit/s)
• 8-DPSK at its higher rate (3 Mbit/s)
• IEEE 802.15.4 (the wireless standard used by ZigBee)
also relies on PSK
Binary phase shift keying
• Uses two phases which are separated by 180°
• It does not matter exactly where the constellation
points are positioned
• In the presence of an arbitrary phase-shift
introduced by the communications channel, the
demodulator is unable to tell which constellation
point is which. As a result, the data is often
differentially encoded prior to modulation.
The general form for BPSK follows the equation:
Tb = Bit duration
Eb = Energy per bit
This yields two phases, 0 and π.
for binary “0”
for binary “1”
Basis function
5.17
Binary phase shift keying
• QPSK uses four points on the constellation
diagram, equispaced around a circle. With four
phases, QPSK can encode two bits per symbol,
shown in the diagram with gray coding to minimize
the bit error rate.

The mathematical analysis shows that QPSK can be used either
to double the data rate compared with a BPSK system while
maintaining the same bandwidth of the signal, or to maintain the
data-rate of BPSK but halving the bandwidth needed.
Writing the symbols in the constellation diagram in terms of the sine
and cosine waves used to transmit them:
This yields the four phases π/4, 3π/4, 5π/4 and 7π/4 as needed.
Ts Symbol duration, Es  Energy-per-symbol = with n bits per symbol
5.19
is a combination of ASK and PSK.
• Two-dimensional signal space with unit basis functions
• The signal constellation consists of the signal-space 4 points
Transmitter
QPSK signal in the time domain
• The odd-numbered bits have been assigned to the in-
phase component and the even-numbered bits to the

Jumps in phase can be seen as the PSK changes the phase on
each component at the start of each bit-period.
Differential phase-shift keying (DPSK)
• In differentially-encoded QPSK (DQPSK), the phase-shifts are
0°, 90°, 180°, -90° corresponding to data '00', '01', '11', '10'.
This kind of encoding may be demodulated in the same way as
for non-differential PSK but the phase ambiguities can be
ignored.

In general, channel introduces an unknown phase-shift to the
PSK signal; in these cases the differential schemes can yield a
better error-rate than the ordinary schemes which rely on precise
phase information
• The amplitude of two waves, 90 degrees out-of-phase with each
other (in quadrature) are changed to represent the data signal.
Amplitude modulating two carriers in quadrature can be
equivalently viewed as both amplitude modulating and phase
modulating a single carrier.
• When transmitting two signals by modulating them with QAM, the
transmitted signal will be of the form:
• At the receiver, these two modulating signals can be demodulated
using a coherent demodulator.
• In the ideal case I(t) is demodulated by multiplying the transmitted
signal with a cosine signal:
Removed by Low-pass filter
Quantized QAM
• If data-rates beyond those offered by 8-PSK are required, it is
more usual to move to QAM since it achieves a greater
distance between adjacent points in the I-Q plane by
distributing the points more evenly.
• The complicating factor is that the points are no longer all
the same amplitude and so the demodulator must now
correctly detect both phase and amplitude, rather than
just phase.

Communication systems designed to achieve very high levels
of spectral efficiency usually employ very dense QAM
constellations
Quantized QAM: Ideal structure
Transmitter
Quantized QAM: Example
• We'll use a signal that is transmitting at 3600 bps, or 3 bits per
baud. This means that we can represent 8 binary combinations.
• We'll use 2 measures of amplitude, 1 and 2, just we did before.
We'll also use 4 possible phase shifts.
• Combining the two, we have 8 possible waves that we can send.
Bit value
Amplitude
Phase
shift
000
1
None
001
2
None
010
1
1/4
011
2
1/4
100
1
1/2
101
2
1/2
110
1
3/4
111
2
3/4
Let's encode a big bit stream:
001010100011101000011110
First, we break it up into 3-bit triads:
001-010-100-011-101-000-011-110
Example
5.27
Find the bandwidth for a signal transmitting at 12 Mbps for
QPSK. The value of d = 0.
Solution
For QPSK, 2 bits is carried by one signal element. This means that
r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud.
With a value of d = 0, we have B = (1+d)S=S = 6 MHz.
ANALOG AND DIGITAL
Analog-to-analog conversion is the representation of
Why we need to modulate an analog signal; it is
Modulation is needed if the medium is bandpass in
nature or if only a bandpass channel is available to us.
Amplitude Modulation
Frequency Modulation
Phase Modulation
5.29
Types of analog-to-analog modulation
5.30
Amplitude modulation
The total bandwidth required for AM can be determined
from the bandwidth of the audio signal: BAM = 2B.
AM band
allocation
5.31
Frequency modulation
The total bandwidth required for FM can be determined
from the bandwidth of the audio signal: BFM = 2(1 + β)B.
FM band
allocation
5.32
Phase modulation
The total bandwidth required for PM can be determined from
the bandwidth and maximum amplitude of the modulating
signal: BPM = 2(1 + β)B.
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