Ch1_Modulation.pdf

1-1
McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004
Modulation
of Digital Data
This chapter contains some slides from chapter 5 of “Data Communications and networking”,
3rd ed. by B. Forouzan. The borrowed slides contain the copyright notice of McGraw-Hill at the
bottom of the slide. Most of these slides are either modified or supplemented by additional
information.
Part II : Wireless Communication
Chapter 1

1-2
In order to get a digital signal into the air (through an antenna) it has to be
modulated first. The modulation assumes a high frequency (radio frequency)
sinusoidal oscillation, called carrier, into which the data signal is somehow
“impressed” before being sent to antenna. Antenna transforms the electric
oscillations into electromagnetic waves (radio waves). Radio waves propagate
from transmitting antenna to all receiving antennas, where they induce electrical
signals equal (although attenuated and delayed) to the signal in transmitting
antenna. Each receiver then needs to take the impress off the carrier and thus
recover the original data signal. This process reversed to modulation is called
demodulation.
Carrier c(t) = sin(2pfct )
Signal in antenna s(t)
Data signal d(t)

1-3
Taken from: “Wireless Communications and Networks”, by W. Stallings, Prentica Hall 2002
Frequencies used in telecommunications

1-4
ASK = Amplitude Shift Keying
FSK = Frequency Shift Keying
PSK = Phase Shift Keying
QAM = Quadrature Amplitude Modulation (combination of ASK and PSK)
A carrier signal has three parameters which can be used for
impressing:
c(t) = A sin(2π fct + φ)
Amplitude Frequency Phase

1-5
Bit rate and baud rate
Bit rate: number of bits per second
Baud rate: number of signal units per second
If a signal unit is composed of n bits, then the bit rate is n times
higher than baud rate
Question:
An analog signal carries 4 bits in each signal unit. If 1000 signal
units are sent per second, find the baud rate and the bit rate,
Answer:
baud rate = 1000 bauds/sec
bit rate: 1000*4 = 4000 bps

1-6
ASK
s(t) = (Ao+ ∆A d(t)) sin(2π fc t)

1-7
Spectrum of ASK
ASK is not noise immune
NOTICE: The bandwidth shown above is the minimal required bandwidth for
the ASK. In practice the required bandwidth is much larger. These issues will
be discussed in slide “Spectral Analysis of Digital Signals”. (Similar remark
holds for FSK and PSK considered in the following slides.)

1-8
FSK
s(t) = Asin(2π(fc + ∆f d(t))t)

1-9
Spectrum of FSK
FSK is noise immune but requires large bandwidth

1-10
PSK
s(t) = Asin(2πfct+ ∆φd(t)) =
= Asin(2πfct+ πd(t)) =
= AdNRZ(t)sin(2πfct)

1-11
Spectrum of PSK
PSK is noise immune and requires small bandwidth
PSK with two phases is called 2-level PSK (2-PSK),
or binary PSK (B-PSK)
+
This is the minimal bandwidth
(the Nyquist bandwidth).
In practical applications is
used BW = 2 Nbaud (see later)

1-12
Constellation of two-level PSK

1-13
Four-level PSK
Excellent performance of 2-PSK encourages us to go with
4-PSK, also called quadrature PSK (Q-PSK)
90o 180o 270o
180o 0o

1-14
Constellation of 4-PSK
4-PSK has more efficient usage of bandwidth than 2-PSK,
because each signal unit has two bits.
For the same bandwidth, the data bit rate doubles.

1-15
Example
Find the bandwidth for a 4-PSK signal transmitting at
2000 bps. Transmission is in half-duplex mode.
For 4-PSK the baud rate is one half of the bit rate, 1000
bauds per second. The bandwidth for any-level PSK is
equal to the baud rate, therefore, the bandwidth for 2000
bps 4-PSK is 1000 Hz.
To be revised (this slide is using Nyquist BW)

1-16
The idea can be extended to 8-PSK, 16-PSK, 32-PSK,….
The limitation is the ability of equipment to distinguish small
differences in signal’s phase.
8-PSK

1-17
Example
Given a bandwidth of 5000 Hz for an 8-PSK signal, what
are the baud rate and bit rate?
For PSK the baud rate is the same as the bandwidth,
which means the baud rate is 5000. But in 8-PSK the bit
rate is 3 times the baud rate, so the bit rate is 15,000 bps.
To be revised (this slide is using Nyquist BW)

1-18
Quadrature Amplitude Modulation
(QAM)
Combination of ASK and PSK which helps making a contrast
between signal units. The number of amplitude shifts should
be lower than the number of phase shifts due to noise susceptibility
of ASK.

1-19
Example of 8-QAM signal

1-20
Bit and baud rate comparison
8N
7N
6N
5N
4N
3N
2N
N
Bit
Rate
N
5
Pentabit
32-QAM
N
6
Hexabit
64-QAM
N
7
Septabit
128-QAM
N
8
Octabit
256-QAM
N
4
Quadbit
16-QAM
Tribit
Dibit
Bit
Units
N
3
8-PSK, 8-QAM
N
2
4-PSK, 4-QAM
N
1
ASK, FSK, 2-PSK
Baud rate
Bits/Baud
Modulation

1-21
Example
A constellation diagram consists of eight equally spaced
points on a circle. If the bit rate is 4800 bps, what is the
baud rate?
The constellation indicates 8-PSK with the points 45
degrees apart. Since 23 = 8, 3 bits are transmitted with
each signal unit. Therefore, the baud rate is
4800 / 3 = 1600 baud

1-22
Example
Compute the bit rate for a 1000-baud 16-QAM signal.
A 16-QAM signal has 4 bits per signal unit since
log216 = 4. Thus, 1000*4 = 4000 bps
Compute the baud rate for a 72,000-bps 64-QAM signal.
Example
A 64-QAM signal has 6 bits per signal unit since
log2 64 = 6. Thus, 72000 / 6 = 12,000 baud

1-23
0 5 10 15 20 25 30 35 40 45 50
0
2
4
0 5 10 15 20 25 30 35 40 45 50
0
2
4
0 5 10 15 20 25 30 35 40 45 50
0
2
4
Hartley-Shannon Theorem
Data signal
Signal with noise
( )
s t
( )
n t
( ) ( )
s t n t
+
Noise
Can’t discriminate
The question is how many levels in amplitude, frequency or phase we can go.
If the difference between two adjacent levels is too small and comparable to
the level of noise in the communication channel, then we cannot discriminate
the levels. The noise puts the limit on the number of levels.
Discrimination possible here
SDSU © Marko Vuskovic, 2004

1-24
Hartley-Shannon Theorem (cont.)
The theoretical limit of number of levels is:
= 0 if signal and noise are
uncorrelated and noise has
zero mean
Root-mean-square of signal and noise:
Root-mean-square of signal with noise
Root-mean-square of noise
__________________________________
M =
2 2
0 0
1 1
( ) ( )
T T
RMS RMS
s S s t dt n N n t dt
T T
= = = =
∫ ∫
2 2 2
0 0 0 0
1 1 1 1
( ( ) ( )) ( ( ) ( )) ( ) 2 ( ) ( ) ( )
T T T T
RMS
s t n t s t n t dt s t dt s t n t dt n t dt S N
T T T T
+ = + = + + = +
∫ ∫ ∫ ∫
( ( ) ( ))
(1)
( )
RMS
RMS
s t n t S N S N
M
n t N
N
+ + +
= = =
Root-mean-square of signal with noise:
(S and N are average power of signal and noise respectively):

1-25
Hartley-Shannon Theorem (cont.)
On the other hand we know that the data bit rate C is:
where W is signal bandwidth. After we combine (1) and (2):
This is the famous Hartley_Shannon equation which defines the upper
theoretical limit of a communication channel capacity.
2 (2)
2 log
C W M
=
2
2 log
S N
C W
N
+
=
2
log 1
S
C W
N
 
= +
 
 

1-26
Spectral Analysis of Digital Signals
In wired media the required bandwidth was determined by the rule:
1
2 2
b
b
R
B
T
= =
B
b
R
b
T
- bandwidth in [Hz] = [sec-1]
- bit rate in [bps]
- bit interval in [sec]
This rule is based on the assumption that the “worse case” signal is a
periodic signal, i.e. a train of pulses (0101010101…). The first harmonic of such
signal contains 81% of energy, which satisfies the 50%-energy rule.
.
In case of wireless communication the media is more hostile (noise,
interferences, longer propagation distances, Doppler effect, fading,…).
In addition, the signal through the media is more complex due to modulation
and spectrum spreading. Therefore, we need to use larger bandwidth, which is
determined based on a more exact spectral analysis of the signal.
.
Since the energy of the signal in a given frequency range (band) [f1,f2] can be
determined by: 2
1
( )
f
f
E P f df
= ∫
where P(f) is the power spectral density (PSD) of the signal, we will characterize
signals by this important function.

1-27
The statistical properties of the signal are as follows:
Random Binary Sequence
A good representation of digital signals are random binary sequences, which
are sequences of pulses of constant duration Tb and random amplitudes Ak,
which take values A or –A with equal probability. The sequence of pulses is
randomly positioned along the time axis.
2
1
[ ] [ ]
2
0
[ ]
0
k k
k k m
P A A P A A
A if m
E A A
otherwise
+
= = = − =
 =
= 

t
A
-A
Tb
2Tb
Tb
Tb
td
d(t)
- Equal probability of the pulse being A or -A
- Two pulses are uncorrelated
td is uniformly distributed over the interval [0,Tb]
(P[e] – probability of event e; E[x] – Mathematical expectation of random variable x)

1-28
1
b
T
2
b
T
3
b
T
1
b
T
−
2
b
T
−
3
b
T
− 0
f
2
b
A T
90% energy
95% 97%
PSD of Random Binary Sequences
It can be shown that the PSD of the random binary sequence is:
( )
2
2 sin
[ ( )] ( ) b
b
b
fT
PSD d t P f A T
fT
 
π
= =  
π
 
As seen the PSD of random binary sequences is a continuous function,
as opposed to PSD of periodic functions, which is discrete (see next slide).
SDSU
© Marko Vuskovic, 2004

1-29
1
b
T
2
b
T
1
b
T
−
2
b
T
−
0
f
2
b
A T
PSD of Random Binary Sequences (Cont.)
The PSD of a periodic sequence of pulses dPER(t) is a discrete function:
( )
2
2
1 3 5
[ ( )] ( ) , , , ,...
2 2 2
b
PER PER
b b b
b
A T
PSD d t P f f
T T T
fT
= = =
π
[ ( )]
PSD d t
[ ( )]
PER
PSD d t
1
2 b
T
−
3
2 b
T
−
5
2 b
T
−
1
2 b
T
3
2 b
T
5
2 b
T

1-30
It is convenient to represent the PSD in decibels (dB)
1
b
T
2
b
T
3
b
T
1
b
T
−
2
b
T
−
3
b
T
− 0
f
dB
10
( )
10 log
(0)
P f
P
 
 
 
0.3 0.3
3 10 1.995 2, 3 10 0.5012 1 2
10 1, 10 0.1, 20 100, 1 1.259
dB dB
dB dB dB dB
−
= = ≈ − = = ≈
= − = = =

1-31
1
b
T
2
b
T
3
b
T
1
b
T
−
2
b
T
−
3
b
T
− 0
f
dB
13 dB 18 dB
p1
p2
p3
Decibels are used to compare power. In diagram below the peak p1 is 13 dB higher than
peak p2 meaning that the corresponding peak is 20 times higher in original diagram.
Similarly 18 dB mans 63 times higher.
1 2 1 2
1 3 1 3
[ ] [ ] 13 20
[ ] [ ] 18 63
P dB P dB dB P P
P dB P dB dB P P
= + ⇒ =
= + ⇒ =
1
b
T
2
b
T
3
b
T
1
b
T
−
2
b
T
−
3
b
T
− 0
f
p1
p2 p3
20 times
higher 63 times
higher

1-32
BPSK Modulator
Binary PSK (BPSK) modulation can be accomplished by simply multiplying
the original signal d(t) (which is a binary random sequence) by the carrier
signal, which is an analog sinusoidal oscillation. After multiplication a band-
pass filter is required (see next slide).
NRZ
Encoding
Binary bit
stream
X
Crystal
Oscillator
( ) cos( )
c
c t C t
= ω
1
0
0
1
 
 
 
 
 
 
1
1
1
1
−
 
 
 
 
 
−
 
cos( )
cos( )
cos( )
cos( )
c
c
c
c
t
t
C
t
t
− ω
 
 
ω
 
 
ω
 
− ω
 
 
BPSK
output
( )
d t
( )
s t
Carrier signal
Band
Pass
Filter
'( ) ( ) ( )
s t d t c t
=
This is an idealized picture. In reality there are dynamic transitions between
cos(ω
ω
ω
ωt) and –cos(ω
ω
ω
ωt), and between -cos(ω
ω
ω
ωt) and cos(ω
ω
ω
ωt) etc., which make the
signal s’(t) much more complex. Therefore we need to use the concept of PSD,
and PSD shifting by the carrier (see next slide).
( ) cos( ),
{0, 180}
c
s t C t
= ω + ϕ
ϕ∈

1-33
BPSK Modulator (Cont.)
The modulated signal s’(t) can be represented by a PSD that is shifted along
the frequency axis by fc. In order to restrict the frequencies which go into the
air (and may interfere with other channels), we need to apply a bandspass filter.
The filter can be designed to pass the first lobe of the PSD, or the first two
lobes, etc. The first lobe usually suffices as it carries 90% of energy.
.
The first lobe requires the bandwidth which twice the bit rate of the original
signal:
0
1
b
T
2
b
T
3
b
T
1
b
T
−
2
b
T
−
3
b
T
−
c
f 1
c
b
f
T
+
2
c
b
f
T
+
1
c
b
f
T
−
2
c
b
f
T
−
c
f
2Rb band
4Rb band
Pass band
[ ( )]
PSD d t
[ '( )]
PSD s t
1
2
2 b
b
B R
T
= =
PSD of BPSK signal
Base band

1-34
Serial to
Parallel
Converter
Binary bit
stream
x
Crystal
Oscillator
cos( )
ct
ω
1
0 90
1 180
1
 
 
 
  =  
   
 
 
cos( )
cos( )
c
c
t
t
− ω
 
 
− ω
 
Band
Pass
Filter
QPSK
output
( )
d t ( )
s t
NRZ
Encoding
NRZ
Encoding x
1
1
 
 
 
QPSK Modulator
0
1
 
 
 
1
1
 
 
−
 
90o
sin( )
ct
− ω
+
sin( )
sin( )
c
c
t
t
− ω
 
 
ω
 
cos( ) sin( )
cos( ) sin( )
cos( 135) cos( 90 45)
2 2
cos( 135) cos( 180 45)
c c
c c
c c
c c
t t
t t
t t
t t
− ω − ω
 
=
 
− ω + ω
 
ω + ω + +
   
=
   
ω − ω + +
   
Phase
Shift
1
1
−
 
 
−
 
( ) cos( ) sin( ), , { 1, 1}
k k c k c k k
s t A t B t A B
= ω − ω ∈ − +
k-th transmitted symbol:
0o
90o
180o
10
11
01
00
270o
I-channel
(in phase)
Q-channel
(quadrature
phase)

1-35
Determining the phase and amplitude
cos( ) cos( )cos( ) sin( )sin( )
cos( ) sin( )
c c c
c c
C t C t C t
A t B t
ω + ϕ = ω ϕ − ω ϕ =
= ω − ω
cos( ) sin( ) cos( )
c c c
A t B t C t
ω − ω = ω + ϕ
2 2
atan2( , )
C A B B A
= + ϕ =
cos( ) sin( )
tan( )
sin( ) cos( )
A C C B
B C C A
= ϕ ϕ
⇒ ⇒ = ϕ =
= ϕ ϕ

1-36
Serial to
Parallel
Converter
Binary bit
stream
x
Crystal
Oscillator
cos( )
ct
ω
0
0
1
0 135
1 135
1
 
   
  =  
  −
 
 
 
cos( )
cos( )
c
c
t
t
− ω
 
 
− ω
 
Band
Pass
Filter
QPSK
output
( )
d t ( )
s t
NRZ
Encoding
NRZ
Encoding x
1
1
 
 
 
π
π
π
π/4-Shifted QPSK Modulator
0
1
 
 
 
1
1
 
 
−
 
90o
sin( )
ct
− ω
+
even bits
odd bits
sin( )
sin( )
c
c
t
C
t
− ω
 
 
ω
 
Phase
Shift
1
1
−
 
 
−
 
( ) cos( ) sin( ), , { 1, 1}
k k c k c k k
s t A t B t A B
= ω − ω ∈ − +
k-th transmitted symbol:
45o
45o
−
135o
135o
−
10
11 01
00
0
cos( ) sin( ) cos( 135 )
2
cos( ) sin( ) cos( 135
c c c
c c c
t t t
t t t
− ω − ω  
ω +
 
=  
 
− ω + ω ω −
   

1-37
π
π
π
π/4-Shifted QPSK Demodulator
( ) cos( ) sin( )
c c
s t A t B t
= ω + ϕ − ω + ϕ
! !
!
( ) 2cos( )
c
c t t
= ω + ϕ
( ) 2sin( )
c
c t t
= − ω + ϕ
Incoming
PSL
Signal
X
VCO
Data
Signal
X
90o
X
LPF
LPF
Controller
Compa-
rator
Compa-
rator
( )
s t
!
( )
c t
( )
c t
cos( ) sin( )
A B A
ϕ − ϕ − ϕ − ϕ ≈
! !
sin( ) cos( )
A B B
− ϕ − ϕ + ϕ − ϕ ≈
! !
( )
{ 1,1}
A∈ −
cos(2( ))
AB ϕ − ϕ
!
{ }
0,1
a ∈
( )
{ 1,1}
B ∈ −
[ ]
,
a b
ϕ
!
{ }
0,1
b ∈
VCO – Voltage Controlled Oscillator
LPF – Low Pass Filter
This is simplified diagram, carrier tracking and noise suppression omitted
ϕ
! - Unknown phase shift of the signal during propagation
ϕ - Estimation of ϕ
!
Parallel
to serial
converter

1-38
Trigonometric identities used in previous slide
2 2
sin( )cos( ) 1 2[sin( ) sin( )]
sin( )sin( ) 1 2[ cos( ) cos( )]
cos( ) cos( ) 1 2[cos( ) cos( )]
cos ( ) sin ( ) cos(2 )
x y x y x y
x y x y x y
x y x y x y
x x x
= + + −
= − + + −
= + + −
− =
{ }
2 2
0 , 1,1
A B if A B
− = ∈ −
In addition:

1-39
Simplified Representation of
QPSK Transmitter/Receiver
c
f
( )
s t
P/S COMP LPF DEMOD
( )
d t
( )
d t ( )
s t
NRZ BPF
[ , ]
k k
A B
MOD
c
f
S/P
I/Q
I/Q

1-40
Differential π
π
π
π/4-Shifted QPSK (DQPSK)
BPSK, QPSK and p/4-shifted QPSK require coherent demodulation, which
assumes exact knowledge of the frequency and phase of the incoming carrier
wave. The synchronization of the local (receiver’s) oscillator can be done with
an additional pilot carrier signal transmitted along with the data signal, or with a
very complex circuitry that increase the space and energy requirements for the
wireless device.
Alternative is differential PSK which uses the phase changes instead of the
absolute phase values for each symbol. This allows for noncoherent
demodulation. The phase changes act as synchronous reference.
Differential π/4-Shifted QPSK (DQPSK) has slightly higher BER but requires
simpler receiver circuitry. It is used in wireless LANs (802.11) and in second
generation cellular telephony (IS-95).

1-41
Multi Carrier Modulation
This is a modulation technique which is recently used to fight
multipath interference at very high data rates. The modulation is performed
with a multitude of carriers (subcarriers) instead of with only a single carrier.
The modulators in each carrier can be any of single carrier modulators
considered so far (BPSK, QPSK, DQPSK, QAM).
One of the important MCM methods is Orthogonal Frequency Division
Multiplexing (OFDM).

1-42
Multipath Transmission and
Intersymbol Interference (ISI)
It is a common situation (specially in urban areas) that receiver picks several
signals from the same transmitter, which have propagated through different
paths.

1-43
ISI (Cont.)
Since delays for different paths can be different, it is possible that the
previous symbol interferes with the current symbol, specially if the difference
in delay between two paths (τ
τ
τ
τ) is in the same order of magnitude as the symbol
interval (Ts).
Symbol
k
t
t
t
Symbol
k-1
τ
τ
τ
τ
Delay
Path 1
Path 2
Symbol
k
Ts
Received
signal
Typical interpath delays are
100 ns, therefore the symbol
rates of several Mbps and
greater can be affected by ISI.

1-44
Multi Carrier Modulation
In order to diminish the impact of ISI the baseband signal is split into N
parallel signals, which are then modulated with different subcarriers.
While the baseband signal has symbol rate of Rs = 1/Ts
the subcarrier signals have N times smaller rate, R = Rs/N = 1/(NTs).
1010011110011110
1 0 1 1
0 1 0 1
1 1 0 1
0 1 1 0
10 10
10 01
01 11
11 10
1010011110011110
16 symbols in time τ
τ
τ
τ
8 symbols in time τ
τ
τ
τ
4 symbols in
time τ
τ
τ
τ
2 symbols in
time τ
τ
τ
τ
BPSK:
QPSK:

1-45
Multi Carrier Modulator
1
0
( ) ( )cos(2 ) ( )sin(2 )
N
k k k k
k
s t A t f t B t f t
−
=
= π − π
∑
Symbol
rate
Rs=1/Ts
Symbol
rate
R = Rs/N = 1/(NTs)
f
0 1/Ts
-1/Ts
( )
d t
0 ( )
d t
.
.
.
.
.
.
1( )
d t
2 ( )
d t
1( )
N
d t
−
+
( )
s t
BPF
BPF
BPF
BPF
f
f0
. . . . . .
f1 fN-1
2Rs/N
S/P
con-
verter
MOD
MOD
MOD
MOD
0
f
1
f
2
f
1
N
f −
2Rs
The data rate of this
signal is N times
smaller, so is the
effect of ISI

1-46
Multi Carrier Demodulator
( )
d t
.
.
.
.
.
( )
s t
LPF
LPF
LPF
LPF
P/S
con-
verter
0 ( )
d t
1( )
d t
2 ( )
d t
1( )
N
d t
−
Symbol
rate
Rs=1/Ts
Symbol
rate
R = Rs/N = 1/(NTs)
DEM
DEM
DEM
DEM
0
f
1
f
2
f
1
N
f −

1-47
OFDM
The multicarrier modulation solves the problem of ISI if there are enough
parallel carriers (typically few thousand). However this method suffers from
two serious problems:
(1) High bandwidth demand
(2) To complex circuitry (mainly because of bandpass filters)
OFDM solves both problems by using orthogonal subcarriers, which allow
spectrum overlapping and don’t require bandpass filters for each
subcarrier. The circuitry is still complex but with VLSI technology, the
approach became feasible, specially with the mass production. As will be
shown, the implementation makes use of highly efficient FFT (Fast Fourier
Transform) and IFFT (Inverse FFT) algorithms.
OFDM is used in IEEE 802.11a at bit rate 54 Mbps with 64 subcarriers.

1-48
OFDM (Cont.)
OFDM MCM uses subcarrier frequencies which satisfy the following condition:
n
s
n
f
NT
=
This has two consequences:
(1) The shifted spectra overlap 50% (which help saving bandwidth)
(2) The subcarrier waves are orthogonal
f
1
n
f − n
f
1
s
NT
1
s
NT
Orthogonality is defined as:
0 0
1
2 2 2 2 2 2
sin( )sin( ) cos( )cos( )
0
s s
NT NT
s s s s s s
if i j
i j i j
t t dt t t dt
if i j
NT NT NT NT NT NT
=

π π π π
= = 
≠

∫ ∫
0
2 2
sin( )cos( ) 0
s
NT
s s
i j
t t dt
NT NT
π π
=
∫

1-49
OFDM (Cont.)
The separation of a particular component from the signal mix can be simply
obtained by multiply-and-integrate procedure, no low pass filters needed:
1
0
0
2
cos(2 ) sin(2 ) cos(2 )
s
NT N
n n n n m m
n
s
A f t B f t f t dt A
NT
−
=
 
π + π π =
 
 
∑
∫
1
0
0
2
cos(2 ) sin(2 ) sin(2 )
s
NT N
n n n n m m
n
s
A f t B f t f t dt B
NT
−
=
 
π + π π =
 
 
∑
∫
x
( )
s t 0
2
(...)
s
NT
s
dt
NT ∫
2
cos( )
s
m t
NT
π
x
2
sin( )
s
m t
NT
π
−
m
A
m
B
0
2
(...)
s
NT
s
dt
NT ∫

1-50
f
f
MCM with FDM
MCM with OFDM
Saving of the bandwidth
Based on: “Introduction to OFDM”, TUD-TVS, by Dusan Matic

1-51
OFDM (Cont.)
After examining the modulation and demodulation equations it can be seen
that they resemble discrete inverse Fourier transform and discrete Fourier
transform. Therefore the implementation can use DSP chips for inverse FFT
(IFFT) and FFT. Since these chips work completely in digital domain, digital
to analog conversion is needed before outputting the signal to RF stage.
Similarly analog to digital conversion needed after the signal is received
from the RF input.
( )
s t
S/P IFFT P/S D/A LPF
( )
d t
( )
s t
P/S FFT S/P A/D LPF
( )
d t
Serial digital data
Parallel digital data Analog signal

Ch1_Modulation.pdf

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