Chapter 4
Handout #4
Dr. Clincy
Professor of CS
Dr. Clincy
Lecture 2
Slide 1
Four Data-to-Signal Cases
DATA
D
SIGNAL
D
A
DATA
A
SIGNAL
D
A
Dr. Clincy
Lecture 2
Already covered
Will cover today
Slide 2
ANALOG-TO-DIGITAL CONVERSION
We have seen in Chapter 3 that a digital signal is
superior to an analog signal. The tendency today
is to change an analog signal to digital data. In
this section we describe two techniques, pulse
code modulation and delta modulation.
Components of PCM encoder
PCM – Pulse Code Modulation
1st: analog signal is sampled
Analog signal is sampled
every Ts seconds
2nd: sampled signal is quantized
Sample rate is fs = 1/Ts
3rd: quantized values are encoded as
bit streams (or codes)
Three different sampling methods for PCM
High-speed
switch used –
able to retain
the shape of
the signal
Ideal but
complex
Sample-and-hold
method that creates
flat-top samples by
using a circuit
Sampling process also
called pulse amplitude
modulation (PAM)
Recovery of a sampled sine wave for different sampling rates
Nyquist theorem states that the sampling rate must be
at least 2 times the highest frequency of the signal
Catches the essence
of the signal
Doesn’t improve
the case
Doesn’t capture the
essence of the
signal
Quantization and encoding of a sampled signal
actual
amplitude
actual-amplitude/D
Quantization steps:
1. Determine Vmin and Vmax
2. Divide range into L zones,
each of height D
Norm.
Actual
Error
between
actual and
nornalized
Quant.
value for
code
Code that
represents the
voltage level
D = [Vmin - Vmax]/L
3. Assign quantized values of 0
to L-1 to midpoint of each
zone
4. Map the sample value to a
quantized value
Assume sample amplitudes between -20V and +20V
Let L = 8 (levels) – therefore, D = [20 - -20]/8 = 5
Quantization error can contribute to Shannon’s SNR: SNRdB = 6.02nb + 1.76
where nb is bits per sample
Bit rate = sampling rate x # of bits per sample = fs x nb
Components of a PCM decoder
Smooths out
the staircase
signal
PCM
decoder
recovers
the
original
signal
What is the minimum bandwidth of the filter the digitized signal will need ?
Bmin = c x nb x 2 x Banalog x 1/r (nb = # bits per sample)
If 1/r=1 and c=1/2, Bmin=nb x Banalog
If the data rate and number of signal levels are fixed, minimum bandwidth is
Bmin = N / [2 x log2 L]
The process of delta modulation
PCM is
more
complex
than Delta
modulation
PCM finds the amplitude of the signal; delta modulation simply finds
the change in the signal from the previous sample
Delta modulation doesn’t use codes – bits are sent one after another
Positive changes are encoded as 1; negative changes are encoded as 0
TRANSMISSION MODES
The transmission of binary data across a link can be accomplished in
either parallel or serial mode.
In parallel mode, multiple bits are sent with each clock tick.
In serial mode, 1 bit is sent with each clock tick.
While there is only one way to send parallel data, there are three
subclasses of serial transmission: asynch, syn and iso approaches
(asynchronous, synchronous, and isochronous.)
Parallel transmission
Serial transmission
Asynchronous transmission
In asynchronous transmission, we send 1 start bit (0) at the beginning and
1 or more stop bits (1s) at the end of each byte. There may be a gap
between each byte.
Asynchronous here means “asynchronous at the byte level,” but the bits
are still synchronized; their durations are the same.
Synchronous transmission
In synchronous transmission, we
send bits one after another without
start or stop bits or gaps. It is the
responsibility of the receiver to group
the bits.
Isochronous Transmission
• For realtime audio and video, uneven delays between
frames is not acceptable – so synchronous
transmission doesn’t work well
• The entire stream of bits must be synchronized – this
is isochronous transmission
• Isochronous transmission guarantees data at a fixed
rate
Chapter 5
Handout #4
Dr. Clincy
Professor of CS
Dr. Clincy
Lecture 2
Slide 16
Digital-to-analog conversion
Based on the digital data, the
Modulator changes characteristics of the “controllable” analog signal
(bandpass analog signal) on the transmitter side to represent the digital
data
Demodulator interprets the analog signal in re-creating the digital data on
the receiver side
Terminology: “modulating digital data into an analog signal”
The analog signal we can control ? Sine Wave, Carrier Signal, Periodic
Signal
Dr. Clincy
Lecture
17
Types of digital-to-analog conversion
Change amplitude to represent a bit
Change frequency to represent a bit
Change phase to represent a bit
Combination of changing both amplitude and phase
to represent a set of bits
Dr. Clincy
Lecture
18
Recall
• For digital transmission, bit rate (data rate)
and signal rate (baud rate) relationship was
»S = N x 1/r where r = # of data
elements per signal element and N is the
data rate in bps (and S is the signaling or
baud rate)
• For analog, r = log2L where L is the type of
signal (versus level)
Dr. Clincy
Lecture
19
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
Dr. Clincy
Lecture
20
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.
Dr. Clincy
Lecture
21
Binary amplitude shift keying
changing the
original amplitude
Explain this
not changing the original
amplitude
B = (1 + d) x S
Dr. Clincy
Bandwidth (B) is proportional to the signal
rate (S) and depending on the modulation and
filtering process, the required bandwidth can
range between S to 2S (where middle
bandwidth is fc). The value of d relates to the
modulation and filtering process
Lecture
22
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).
Dr. Clincy
S = N * 1/r
Lecture
23
Binary frequency shift keying
changing the
original frequency
Explain this
Dr. Clincy
not changing the original
frequency
Lecture
Use two different
carrier frequencies, f1
and f2, for 0 and 1
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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?
The
difference
(delta)
between
the two
frequencie
s
Solution
The midpoint of the band is at 250 kHz. We choose
2Δf to be 50 kHz; this means
Dr. Clincy
Lecture
25
Binary phase shift keying
changing the
original phase
Explain this
Dr. Clincy
not changing the original
phase
Lecture
26
QPSK and its implementation
QPSK –
Quadrature Phase
Shift Keying
Use 2 bits in each
signal element –
decreases baud
rate and
bandwidth
Uses 4 possible phases (versus 2)
2 composite signals are created
Because the 2 signals are using the same bandwidth – each
signal has ½ bandwidth
Dr. Clincy
Lecture
27
Example
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 = S = 6 MHz.
B = (1 + d) x S
Dr. Clincy
Lecture
28
Concept of a constellation diagram
Helps define the amplitude and phase
of a signal element
the amplitude of the 2nd
carrier
Peak Amplitude
Phase
This is the amplitude
given one carrier
Only use 1 carrier and
phase is static and 2
amplitude levels
Dr. Clincy
Only use 1 carrier and
1 amplitude and 2
phases (0o and 180o)
Lecture
Binary Phase Shift Keying – uses 2 different
phases and the same amplitude
Uses 2 carriers and 1
amplitude and 4 phases
(45o, 135o, -45o, -135o)
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Constellation diagrams for some QAMs
QAM – Quadrature Amplitude Modulation
For QPSK, we only changed the phase
For QAM, we change both the phase and amplitude
Has a 0 amplitude
and a positive
amplitude (with 2
carriers)
Dr. Clincy
Has a negative
amplitude and a
positive amplitude
(with 2 carriers)
Lecture
Has 2 positive
amplitudes (with 2
carriers)
Has 4 negative
levels and 4
positive levels
(with 2 carriers)
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