Chapter 3
Baseband Modulation
-
eeter
ware
Mah!r
w!rs
The output of the formatting blocks is a sequence of bits. However, there is nothing
physical about the digits resulting from this process. Digits are just abstractions-away to
describe the message information. We need a physical entity (electrical field,
electromagnetic wave,…). In other words, the objective of modulation is to associate
digital bits with analog signals. In the case of baseband communications, the information
will be carried through modulated pluses.
be,be
...
...
[01)
neöner!m annes!neneden
1. Baseband vs Passband
-fre
The
samples so s!g
-
--
en
b!sosa
!tmodusk!en
Figure 1: Passband and baseband communication chain
Baseband transmission (is also called line coding): transmit signals over passband
channel (medium) including the serial cables, local area networks and chip-to-chip line
(PCB). No filter is required.
Passband transmissions: transmit signal over radio frequency channels.
1|Page
Bandwidth:
hat the
mose
I
Rü
ht
+
D
↳
4) cos(2mfct)
Figure 2: bandwidth
-
𝟏𝟏
sampl!ng
-
𝟏
Aura
The minimum require baseband bandwidth: 𝑩 = 𝟐 𝑻 = 𝟐 𝑹𝒔 = 𝒇𝐦𝐚𝐱 ,
↓
E
where T is the symbol/pulse period
rate
𝟏
The minimum require passband bandwidth: 𝑩 = 𝑹𝒔 = 𝑻, where T is the
symbol/pulse period
To avoid inter-symbol interference the channel bandwidth has to be ≥ 𝑩 . The symbol
rate is 1/T symbol/s (baud)
2. Baseband modulation: Pulse code modulation (PCM)
The transmitter send a sequence (Serie) of modulated pulse (wave) according to the bit
sequnence
1- b!
!s
esarphe
N
𝑏1 , 𝑏2 , 𝑏3 … − −> ∑ 𝑎𝑘 𝑔(𝑡 − 𝑘𝑇)
𝑘
The questions that naturally raise are
2.1.
-
What form of pulse to adopt (PCM waveform)?
-
How to define the {𝑎1 , 𝑎2 , … } form the bit sequence {𝑏1 , 𝑏2 , … }?
Modulation/mapping: from bits to symbols
Considering PCM each symbol will correspond to one bit.
Example: 𝑏𝑖 = 1 −→ 𝑎𝑖 = 1 + 𝑉 and 𝑏𝑖 = 0−→ 𝑎𝑖 = 1 − 𝑉
𝑏𝑖 = 1 −→ 𝑎𝑖 = 1 + 𝑉 and 𝑏𝑖 = 0−→ 𝑎𝑖 = 0 0 volts
2|Page
and
In this case, the symbol rate 𝑅𝑠 = 1/𝑇 is equal 𝑅𝑏 = 1/𝑇𝑏 .
Figure 3: symbol rate
Example 1:
0
1110
↑ra 24Hz
->
=
↑s 24r
!ne
=
2
B 22Hz
->
=
44Hz
=
b!tsample
b!t
rate
2
=
b!t/cm4k
Rs
2.2.
=>
2330: Rb
22 bands
22/syd,
=
=
PCM Waveforms
One may use different pulse shape to encode the bits ‘0’ or ‘1:
-
Nonreturn-to-zero (NRZ)
-
Return-to-zero (RZ)
Phase encoded
&
-
ay
!
(91, 13
as
-
soslanf)
XLos
Hompes
3|Page
be
ca
d!stes!n
-
4|Page
Example 2: build on example 1 and encode the information using different
waveform.
100
111011
!nte
Discussions (on the difference between the waveforms):
5|Page
3. M-ary Pulse Modulation
3.1.
Performance Criterion
-
Energy efficiency:
- Spectral efficiency:
E
-
w!re
allow
tequenc!es
3.2.
come
↳o
-
be
deng!td
-M
Cost/complexity and use case:
Pulse-Amplitude Modulation (PAM)
Different from PCM where one bit is encoded into one symbol (two levels voltage),
PAM allows to encode multiple bits in one symbol by considering pulses with more
than two levels.
00
01
3
10
1
00-1
11
-
2
10
1000
10
-
-
I
6|Page
I
-
e
Example 3:
-
The information in an analog waveform, with maximum frequency 𝑓𝑚 = 3KHz, is to
be transmitted over an M-ary PAM system, where the number of pulse levels is
M=16. The quantization distortion is specified not to exceed ±1% of the peak-topeak analog signal.
6 b!/sample
↑ 1212 Br 7s 64Hz
-
D
n
=
=
=
1) What is the number of bits/sample that will be used indigitizing the analog
form.
6b!ssample
2) What is the minimum required sampling rate and what is the resulting bits
transmission rate?
62 Hz
6b1 x 64: 162b!/s
Rb
=
3) What is the symbol transmission rate? Derive the bandwidth efficiency
4 bas per scubd
considering 12KHz of available bandwidth M 16
de bands
=
-
-
~
ERs-He
2
4) Now, consider baseband channel with bandwidth equal to M5KHz. What should be theminimum number
of PAM levels (M) as such there will be no
o
inter-symbol interference?
D
e
-
-
Explain graphically the inter-symbol interference
⑰Rs
B losa
m
=
=
2 401.L.y
⑭
=
7|Page
3.3.
Pulse-Width Modulation (PWM)
on e
3.4.
r
Pulse-Position Modulation (PPM)
o
er
-mertebes
8|Page