Orthogonal Frequency Division Multiplexing (OFDM)

Multi-Carrier Modulation
Dr. Ahmad Gomaa
Assistant Professor
Electronics and Communications department
Faculty of Engineering, Cairo University
http://scholar.cu.edu.eg/gomaa
Contact: aarg_2010@yahoo.com

Outline
• Single carrier (SC) modulation
• Shortcomings of SC systems
• Multi-Carrier modulation (MCM)
– Main idea
– Implementation using IFFT and FFT
– MCM in multi-path fading– MCM in multi-path fading
– OFDM parameters
– Bit-Interleaved Coded OFDM
– PAPR
– Subcarrier loading
– OFDM under RF impairments (CFO, PN, IQI, Doppler)
– Single-Carrier Frequency-Division Multiple-Access (SC-
FDMA)

Single Carrier (SC) Modulation
• Bits are modulated according to M-QAM or M-
PSK
• Modulated symbols are passed through a pulse
shaping filter (e.g., Square-Root Raised Cosineshaping filter (e.g., Square-Root Raised Cosine
(SRRC) filter)
• Output is up-converted to the carrier frequency
(fc)

Single Carrier Modulation
M-QAM
(or M-PSK)
bits
Pulse Shaping filter
π/2
Pulse Shaping filter
Pulse shaping filter specifies
 symbol rate & signal shape in frequency-domain

Single Carrier (SC) Modulation
• Transmitted signal passes through multi-path
channel
• Inter-Symbol Interference (ISI) occurs due to
the channelthe channel
• The receiver has to apply equalizers to extract
the modulated symbols
• Complex equalizers and not even optimal!

Multi-path channel
Transmitter Receiver
Reflecting object
(car, building, ….)
Reflecting object
(car, building, ….)
Channel impulse response

Multi-path channel
• Multi-path channel causes inter-symbol
interference (ISI)
• Need equalizers at the receiver to reverse the
channel effect
• Baseband model• Baseband model
][][][][
0
nzknxkhny
L
k
 
Channel
h
x[n]
z[n]
y[n]
Modulated
QAM symbols
Receiver noise
Received samples
Convolution between x[n] and h

Multi-path channel
• y[n] is not only function of x[n] but also
function of previous symbols {x[n-1], x[n-2], ….
x[n-L]}
• These previous symbols act as interference on• These previous symbols act as interference on
the current symbol x[n]
• Need to extract x[n] from y[n]
• This is done using equalizers

Equalizers
Channel
h
x[n]
z[n]
y[n]
Modulated
QAM symbols
Equalizer
w
][ˆ nx
M


Noise
M
k
nwnznwnhnx
knykwnwnynx
][][][][][
][][][][][ˆ
[n]beshould
0






If w[n] is designed to make h[n]*w[n] = δ[n], then we force the ISI = 0
 Zero-Forcing equalizer  effect on noise?

Equalizers
1)()(
][][][


fWfH
nnwnh 
Channel frequency response
Time-domain
n
f
Time-domain
To
Frequency-domain
As channel length (rms delay spread) increases, frequency-selectivity increases
One-tap channel has flat frequency response
h(n)
H(f)
Dip frequencies

Shortcomings of SC systems
• For W(f) H(f) = 1, W(f) will have large magnitude at dip
frequencies
• W(f) will enhance the noise at these dip frequencies
• When looking in time-domain, these enhanced noise will
impact the whole time-domain  all symbols are impacted byimpact the whole time-domain  all symbols are impacted by
noise enhancement
• ZF equalizer  Noise enhancement  Poor performance! 
Solution? (Multi-Carrier Modulation)

Multi-Carrier Modulation - Main Idea
• If we can divide H(f) into N subbands (subchannels or subcarriers), each of
small BW
• Then, channel is approx FLAT over each band
• We equalize every subcarrier alone by dividing by the channel gain at this
subcarrier
• We transmit 1 QAM symbol over every• We transmit 1 QAM symbol over every
subcarrier
• In receiver, we divide received signal
into N subcarriers, divide every subcarrier
by channel gain at this subcarrier to detect
QAM symbol at this subcarrier
• Equalizer design is much easier  No convolution
• Only QAM symbols at dip frequencies will suffer  No effect on other
QAM symbols as they are transmitted on other subcarriers
……
Subcarrier (channel approx flat)
f

• For QAM symbol to be transmitted over a small BW symbol
duration has to be large  small data rate?
• Every symbol will have small symbol rate but we transmit N symbols
in parallel  Effective rate is high!
5us
cos(w1t)
Example
1us
SerialtoParallel
Bit rate = 1 bit / 1us
= 1Mbps
.
.
. cos(w5t)
Example
Bit rate = 5 bits / T
= 5 bits / 5us
= 1 Mbps
T

• In general, we have QAM symbol (which is complex a+jb) rather
than just BPSK symbols (which are real as in the previous example)
• To modulate Xk = ak+jbk over subcarrier k, we need exp(jwkt) and
rather than just cos(jwkt) 
 tjX kk exp
• So, we can write the total output as
 
durationSymbolTotal
0,exp)(
1

 
T
TttjXtx
N
k
kk 
k k kXk = ak+jbk  QAM symbol

• How to choose wk’s?
• To easily separate these N subcarriers at the receiver, we need
them to be orthogonal, i.e.,
     


T
t
llkk dttjXtjX
0
0expexp 
• To satisfy this, we can show that
Nkk
T
k ,.....,2,1,
2



Ttt
T
k
jXtx
N
k
k 





 
0,
2
exp)(
1


• In digital domain, set t = n Ts
MCM – Implementation with IFFT
s
s
N
k
sks
T
T
nTnTT
T
kn
jXnTx 





 
00,
2
exp)(
1

This equation can be effectively implemented usingThis equation can be effectively implemented using
Inverse Fast Fourier Transform (IFFT)
N-point
IFFT
{X1,X2,….XN} {x(0), x(1),…. x(N-1)}
Nn
N
kn
jXnx
N
k
k 





 
0,
2
exp)(
1


• For x(nTs) = x(n), we need
• We choose the sampling time (Ts) to be total symbol time (T ) over
number of subcarriers (N)
N
T
T
T
T
N s
s

number of subcarriers (N)
• Since we choose the subcarrier frequencies as:
• Then, subcarrier bandwidth is the difference between any two
subcarriers which is:
Nkk
T
k ,.....,2,1,
2



T
fBWSubcarrier kk 1
22
1




 





• Hence, the total BW of the system is
sTT
N
fN
BWsubcarrierssubcarrierofNumberBWT
1
otal



MCM – Waveform in time-domain
Nn
N
kn
jXnx
N
k
k 





 
0,
2
exp)(
1

Time-domain signal = weighted summation of sinusoidal signals of
frequencies {k/N, k=1,2,…,N}
Total symbol duration = T
nTs
0
N-1
Freq = 2/N
Freq= 1/N
Multiplied by X1
Multiplied
by X2
Every subcarrier carries
One QAM symbol
Total symbol duration = T
Ts

MCM – Waveform in frequency-domain
Signal in Frequency domain
Subcarrier spacing = 1/T
At the peak of any subcarrier,
contributions from all other subcarriers = zeros  Orthogonal

• From the previous figure, we clearly notice the subcarriers are
orthogonal to each other
• This system is called OFDM (Orthogonal Frequency Division
Multiplexing) where orthogonal frequency comes from the fact that
the subcarriers are orthogonal to each other
• In the receiver, how to recover Xk from x(n) ?
• Exploit that the subcarriers are orthogonal as follows:
MCM – Subcarriers orthogonality
• Exploit that the subcarriers are orthogonal as follows:
• We multiply x(n) by the conjugate of the kth subcarrier and run the
summation.
• This will null out all subcarriers except for the kth one, it will give NXk
• We divide by N to get Xk
Nk
N
kn
jnx
N
X
N
n
k 





 


1,
2
exp)(
1ˆ
1
0


• This previous equation is exactly what FFT
(Fast Fourier Transform) does
• So, in Tx we run IFFT and in Rx we run FFT
MCM – IFFT/FFT implementation
QAM
mod
bits
SerialtoParallel
N-pointIFFT
ParalleltoSerial
QAM
demod
 kXˆ kX
bits
ParalleltoSerial
N N
SerialtoParallel
N-pointFFT
N N

• This works great if channel is AWGN with no
multipath, i.e., channel is flat with no frequency
selectively
• Even the single-carrier case works great with flat
channel (no multi-path) without the headache of
MCM in multipath fading
channel (no multi-path) without the headache of
FFT and IFFT
• Now, what if there’s multi-path channel?

• Consider a multipath channel with only 2 paths:
A direct path with zero delay and another path
with some delay
• Focus on the subcarriers 1/N and 2/N
• In the next page, the solid curves are the two• In the next page, the solid curves are the two
subcarriers of the direct path (no delay)
• The dashed curve is the delayed version due to
the delayed path
• We plot only the delayed path of subcarrier 2/N
to avoid crowding

Now, to get X1, we multiply the whole received signal by the blue signal and integrate (sum)
This will cancel out the solid red signal BUT not the dashed red because it’s not a COMPLETE
Sinusoidal.
So, we will have inter-carrier interference (ICI) from X2 and other subcarriers on X1

 So, multi-path fading  ICI in OFDM
 How to solve that?
 Solution is to make the dashed sinusoidal a COMPLETE wave, i.e., Solution is to make the dashed sinusoidal a COMPLETE wave, i.e.,
have an integer number of periods in the integration (summation) time
 This is done using Cyclic prefix, i.e., copy the last portion of the total
OFDM symbol and put it before the beginning of the OFDM symbol

Now, the delayed version of subcarrier 2 will have an integer number of periods4
within the OFDM symbol duration (summation interval), so it will be cancelled out
when we multiply the received signal by subcarrier 1 and integrate  No ICI

• How to choose the length of cyclic extension?
• It should be chosen such that it’s greater than
effective channel length
• Cyclic extension converts the linear convolution
between Tx signal and channel into Cyclic Convolutionbetween Tx signal and channel into Cyclic Convolution
• Cyclic convolution in time-domain is equivalent to
multiplication of FFTs in Frequency-domain
• FFT of transmitted signal is simply the QAM symbols
(Xk)
• FFT of the channel is the channel frequency response

tzthtxty


getweFFTTaking
)()()()(
signalReceived
Circular convolution
Thanks to Cyclic extension!
Nk
H
Y
X
NkZHXY
k
k
k
kkkk


1,ˆ
knowledge,channelAssuming
1,
Here, equalizer is ONLY 1-tap and it’s Maximum Likelihood (ML)  Optimal
 ML minimizes error rate

MCM – Cyclic Extension
Cyc.
Exten
IFFT output
(N samples)
Cyc.
Exten
IFFT output
(N samples)
OFDM symbol OFDM symbol
…..
time
OFDM symbol
1
OFDM symbol
2
• Cyclic extension  Prevents multipath channel from
introducing ICI after taking FFT at the receiver
• Cyclic extension is also called Cyclic Prefix
• Does it have another function ?

• When OFDM signal passes through multi-path channel, every
symbol will leak into its successor (the symbol coming after it)
• To prevent inter-symbol interference (where symbol here
refers to OFDM symbol), we need to have Guard Interval
between OFDM symbols
• Thanks to its position lying between successive OFDM
symbols, Cyclic Extension acts also as Guard Interval
MCM – Cyclic Extension [Guard Interval]
symbols, Cyclic Extension acts also as Guard Interval
IFFT output
(N samples)
IFFT output
(N samples)
time
OFDM symbol 1 OFDM symbol 2
Leakage of
OFDM symbol 1
Does NOT reach
OFDM symbol 2

MCM – Pilot and Guard subcarriers
Nused  Number of used subcarriersZEROS ZEROS

OFDM parameters
Parameter Definition Notes
Ts Sampling time Ts = 1/B = 1/bandwidth (seconds)
Tcp
Guard interval length
(Cyclic Prefix length)
Tcp = NcpTs > Effective channel length
(seconds)
T IFFT output length T = NTs = 1/Δf (seconds)
Ttot OFDM symbol length Ttot = T+Tcp= (N+Ncp)Ts (seconds)Ttot OFDM symbol length Ttot = T+Tcp= (N+Ncp)Ts (seconds)
Δf
Subcarriers frequency spacing =
frequency of 1st subcarrier
Δf = 1/T = B/N (Hz)
B OFDM bandwidth B = 1/Ts = N Δf (Hz)
N
Number of subcarriers
(IFFT size) = # samples in IFFT output
N = B/Δf = T/Ts
Bused Occupied OFDM bandwidth Bused = Nused Δf(Hz)
Nused Number of used subcarriers

OFDM parameters (Cnt’d)
Parameter Definition Notes
Bguard Guard subcarriers bandwidth B = Bguard + Bused (Hz)
Nguard Number of guard subcarriers N = Nguard + Nused
Ncp # samples in cyclic prefix Ncp = Tcp/Ts
f Sampling rate f = 1/T = N Δf (Hz)fs Sampling rate fs = 1/Ts = N Δf (Hz)
How to choose T = 1 /Δf ?
lengthchannel
1
αbandwidthCoherence
subcarrieroverchannelFlatbandwidthCoherence
)efficiencyr(rate/poweoverheadCPReasonable4
al
CP
f
TT



4G LTE OFDM parameters
(N)
Licensed BW 3 MHz
= B (OFDM BW)
IFFT output
length (T)
= 1 /subcarrier spacing (Δf) = 1/ 15000 = 67 us
Nused = Bused/Δf 72+1(DC) 180+1 300+1 600+1 900+1 1200+1
Nguard = N-Nused 55 75 211 423 535 847
Bused = Nused Δf
(MHz)
1.08 2.7 4.5 9 13.5 18
BW efficiency =
Bused /licensed
BW x 100 %
86 % 90 %

4G LTE spectrum
Used subcarriers
Left Guard
Subcarriers
Right Guard
Subcarriers freq
Licensed BW  Purchased spectrum
Used BW (Bused)
ZEROSZEROS
Licensed BW  Purchased spectrum
We transmit zeros on part of licensed BW  In air, it accommodates side lopes
of used subcarriers  side lopes will not get outside licensed BW 
So we do not interfere with neighboring bands
We transmit outside licensed BW but we just transmit zeros  No
interference on neighboring bands
B = fs = sampling frequency

4G LTE spectrum
Side lopes of modulated subcarriers

WiFi OFDM parameters
IFFT size (N) = 64 (52 used subcarriers + 1 DC + 11 guard subaccreirs)
IFFT output duration (T) = 1 / subcarrier spacing = 1/(312.5x103)= 3.2 us
Cyclic prefix length = Tg = 0.8us = T/4
Sample time (Ts) = 1/(20x106) = 0.05 us
Used BW (Bused)= 53 (52 + 1 DC) subcarriers x 312.5 kHz = 16.6 MHz = OBW = Occupied BW

Comments on OFDM parameters
• Cyclic prefix (CP) length in LTE > CP in WiFi, why?
• Because WiFi works in indoor environment while LTE works outdoor.
• In outdoor, channel paths can be reflected from far objects so will
come to Rx after long delay  Need longer cyclic prefix
• What if we use too long CP, much longer than channel length?
• Then, we need IFFT output length (T > 4Tcp) to be very long as well
– Channel can vary during OFDM symbol (because it’s too long in
duration)  Doppler effect  ICI See slide
– Channel can vary during OFDM symbol (because it’s too long in
duration)  Doppler effect  ICI See slide
– Large latency  receiver will need to wait too long before it can receive
OFDM symbol  OFDM detection cannot start before the whole OFDM
symbol is received  Large latency not suitable for real-time
applications (voice/video chatting, gaming, …)
• What if we use too short CP, shorter than channel length?
• Interference between successive OFDM symbols (every OFDM
symbol will leak into its successor)  ISI
• Subcarriers orthogonality will be destroyed leading to Inter-Carrier
interference (ICI)

• What about subcarriers with DEEP fading?
• What if a Narrow-Band Interference (NBI) hits
OFDM signal? (e.g., BLUETOOTH into WLAN)
MCM – Deep fading and NBI
……
Deep Fading
Narrow-band Interference (NBI)
OFDM signal

• In case of deep fading, Hk is very small, so dividing by it will amplify
the noise  Gives wrong QAM symbol after dividing by Hk
• In case of NBI, noise power seen at impacted subcarriers is high 
Gives wrong QAM symbol after dividing by Hk
MCM – Deep fading and NBI
NkZHXY kkkk  1,
Large in case of NBISmall in case of Deep fading
k
k
k
k
k
k
H
Z
X
H
Y
X
ˆˆ
ˆ  Large in deep fading
and NBI
WRONG detection
kk XX ˆ
Channel estimate

• How to solve this problem?
• Answer : Channel coding
• Channel coding: Introduce redundancy of information bits
(e.g., Repetition code, Convolutional code, Turbo code)
• Then, distribute the resulting bits over subcarriers
• If a subcarrier is hit by deep fading or NBI, we lose its bits but
Bit-Interleaved Coded OFDM
• If a subcarrier is hit by deep fading or NBI, we lose its bits but
we use the redundancy bits distributed to other subcarriers to
correctly decode information bits
Channel
Coding
Interleaver
QAM
modulation
SerialtoParallel
N-pointIFFT
N
ParalleltoSerial
Nbits
AddCyclicPrefix

• Example, Use repetition code, every information bit is
repeated 3 times
1 0 1 1 1 1 0 0 0 1 1 1
Channel
Coding
Interleaver
1 0 1 1 0 1 1 0 1
BPSK
modulation
NBI
1 0 1 1 0 1 1 0 1
freq
NBI
Subcarrier
1
Subcarrier
4
Subcarrier
6

• Subcarriers 4-6 are hit by NBI  their data is wrongly detected
• We can still detect the original information bits after decoding
1 0 1 1 0 1 1 0 1
NBINBI
freqAfter detection
1 0 1 00 11 00 1 0 1
1 00 1 0 11 0 1 00 1
After De-Interleaving
After Decoding (Majority Rule)
1 0 1
WRONG!
CORRECT!

• Note the importance of the Interleaver!
• It distribute code bits among subcarriers, so code bits
corresponding to any information bit are assigned to NON-
ADJACENT subcarriers, so if one is lost, others are not lost
• One possible Interleaver design: Matrix Interleaver
• Write code bits row by row in a matrix and read them out
Interleaver design
• Write code bits row by row in a matrix and read them out
column by column
c1 c2 c3 ……. c10
c11 c12 c13 ……. c20
c21 c22 c23 ……. c30
.
.
c91 c92 c93 ……. c100
{c1 c11 c21 …… c91
c2 c12 c22 …… c92
c3 c13 c23 …… c93
….….…. .…. c100}
{c1 c2 c3 …… c10
c11 c12 c13 …… c20
c21 c22 c23 …… c30
….….…. .…. c100}
c1 And c2 separated
by 10 bits!

OFDM Transmitter and Receiver
Channel
Coding
Interleaver
QAM
modulation
SerialtoParallel
N-pointIFFT
N
ParalleltoSerial
Nbits
AddCyclicPrefix
Channel
decoding
De-
Interleaver
QAM
demod
ParalleltoSerial
N-pointFFT
SerialtoParallel
Nbits
RemoveCyclicPrefix
.
.
.
.
Freq domain Channel equalization (1 tap/subcarrier)
Channel is estimated
using pilots
1
ˆ/1 H
kHˆ/1
NHˆ/1

Peak-to-Average Power Ratio
andand
Subcarrier Loading

Subcarrier Loading
• If the channel has slow variation over time (e.g., in xDSL)
• We can adapt the modulation of each subcarrier
• Subcarriers with deep fades  loaded with few bits (e.g., BPSK
or QPSK)  Low poweror QPSK)  Low power
• Subcarriers with good channel  loaded with many bits (e.g.,
64-QAM)  High power
• We can design the loading of every subcarrier to maximize
overall throughput (bit rate)  Use waterfilling technique

Subcarrier Loading
Subcarrier
With bad
Channel 
Not assigned
Any bits (not used)
Subcarrier
With good
Channel 
Assigned
many bits
(high QAM)
 Large
power

Subcarrier Loading
• Can we use adaptive loading in Single-Carrier systems?
• No because by definition adaptive loading is used in multi-
carrier systems
• This is one advantage of OFDM over Single-Carrier• This is one advantage of OFDM over Single-Carrier

Peak-to-Average Power Ratio (PAPR)
• One disadvantage of OFDM is that it has high PAPR
• This means that the peak power is much higher than the
average power
Peak power
Average power
time
OFDM symbol power

• OFDM has high PAPR, so what?
• After generating OFDM signal, we feed it into Power Amplifier (PA) to
amplify it
• Any PA has a linear region and a saturation (non-linear) region
• To avoid saturation and generation of non-linear components, we
need to work in the linear region
• Hence, we need to push the OFDM signal such that both its peak and
average are within the linear region
• This means that most of the time (where average power lies), we will
work in low gain  Low efficiency of PA

Output
signal
power
Non-linear (Saturation) region
Linear region
Input signal power
Most
Of time,
We work
In low
Gain region
High gain region is rarely utilized!

• Why does OFDM have high PAPR?
• OFDM signal = Summation of N random variables
Nn
N
kn
jXnx
N
k
k 





 
0,
2
exp)(
1

Random variable
• OFDM signal = Summation of N random variables
• Central Limit Theory  Summation of many random variables has
a Gaussian PDF regardless of PDF of every random variable
• x(n) is Gaussian distributed for large N (number of subcarriers)
• Gaussian random variables have high PAPR, why?

Peak Power (rare!)
Average power (Most of time)
Peak power is much higher than average power for large variance
Peak power is less likely to happen than average power

• What about PAPR of Single-Carrier systems?
• It’s actually lower than OFDM! Why?
• In single-carrier (SC), transmitted signal = modulated
• In single-carrier (SC), transmitted signal = modulated
QAM signal directly  No IFFT
• Hence, PAPR of SC = PAPR of QAM symbol
• Consider QPSK  What’s PAPR?

PAPR of QPSK
QPSK Constellation
Inphase
Quadrature
A
A s1
s2
s s4
d1d2
d3 d4
PAPR of QPSK s3
s4
M  Number of
QAM symbols
 
dB0
PowerAverage
PowerPeak
log10dBinPAPR
1
2
2
PowerAverage
PowerPeak
PAPR
2maxmaxPowerMaximumPowerPeak
22
4
111
PowerAverage
10
2
2
22
2
4
1
2
1
2
1



  
A
A
AdP
AAd
M
P
M
i
i
i
i
i
M
i
i
M
i
i

PAPR of SC with QPSK = 0 dB  Very good for PA efficiency
If we used OFDM with QPSK, PAPR will be much higher
In terms of PAPR, Single Carrier is better than OFDM!
Exercise: Compute PAPR of Single Carrier with 16-QAM

OFDM under RF impairments
• Effect of Carrier Frequency Offset (CFO)
• Effect of Phase Noise (PN)
• Effect of Doppler [Not RF impairment but has
similar impact]similar impact]
• Effect of I/Q imbalance (IQI)

OFDM under CFO
• CFO means that carrier frequency of receiver local
oscillator is different from that of the transmitter
fc,Tx
fc,Rx
π/2
π/2
fc,Tx
fc,Rx
CFO,,  fff TxcRxc
CFO ≠0 due to inaccuracies
of crystal oscillators
used at Tx and Rx

OFDM under CFO
• Frequency of Crystal oscillator (XO) has manufacturing error
measured as part per million (ppm)
• ppm = error (in Hz) in every 1 million Hz (1 MHz)
• Specifications of XO tells its frequency and expected error in ppm
• Example: XO has frequency = 2 GHz with 0.1 ppm
This means we have 0.1 Hz maximum error in every 1 MHz
XO frequency = 2 GHz = 2000 MHz• XO frequency = 2 GHz = 2000 MHz
• Maximum error will be 2000*0.1 = 200 Hz
• XO frequency = 2 GHz ± 200 Hz
• We expect XO frequency to be any where in the range
2 GHz -200 Hz < XO frequency < 2 GHz + 200 Hz
• If both Tx and Rx XOs have 0.1 ppm, then maximum carrier
frequency offset (CFO) between them is when one of them is
2 GHz -200 Hz while the other is 2 GHz + 200 Hz  Δfmax= 400 Hz

OFDM under CFO
• At Rx, taking FFT of received signal is equivalent to
sampling the received signal in frequency-domain at the
locations of subcarriers as follows
Gives QAM symbol carried by red subcarrier
with ZERO contributions of other subcarriers

OFDM under CFO
• If we have CFO, then this sampling is SHIFTED
Gives QAM symbol carried by red subcarrier
with NON-ZERO contributions of other subcarriers
Inter-Carrier Interference (ICI)

OFDM under CFO
• CFO results in  Inter-Carrier Interference
• Every carrier sees interference from NEIGHBOURS
• Considering any subcarrier, it will see more interference
from close subcarriers than from far subcarriers. This isfrom close subcarriers than from far subcarriers. This is
clear from previous figure
• How to solve this problem?
• Use pilots to estimate CFO and compensate it in time-
domain BEFORE taking FFT

OFDM under CFO
 
 
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

kn
j
n
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fNB
T
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jnxnTjnTxnTx
fftjtxtx
N
f
s
f
sfss
RxcRxcff
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exp
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exp)(
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exp)(2exp)()(
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• Effect of CFO:
1. Desired symbol is rotated by phase = πα
2. ICI from neighboring subcarriers
OFDM under CFO
Hzinspacingsubcarrier
HzinCFO



f
f

Rotation effectICI effect

• For α to be small, we need to choose XO specs
such that CFO is small compared to subcarrier
spacing
• This means that 4G LTE (subcarrier spacing =
OFDM under CFO
• This means that 4G LTE (subcarrier spacing =
15 kHz) will have tougher spec on XO than
WLAN (subcarrier spacing = 312.5 kHz)
• To compensate for CFO, we estimate α (and
hence δf) using pilots and multiply received
signal by exp(jδfTsn), n=0:N-1 BEFORE taking
FFT

OFDM under Phase Noise
• Phase Noise (PN) means that oscillator has a random
phase that changes with time
π/2
π/2
π/2
  ttf txc  2cos   ttf rxc  2cos
Tx PN
Rx PN
fc
f
Ideal XO
No PN
fc f
Practical XO
With PN
Linewidth

• Phase Noise (PN) causes energy to leak around carrier
frequency
• Good oscillators have their energy leaking in a narrow
bandwidth (they have small linewidth)bandwidth (they have small linewidth)
• PLL is much better than free-running oscillators in terms of PN
• Effect of PN on OFDM (similar to CFO):
– Inter-Carrier Interference from neighboring subcarriers
– Phase rotation of every subcarrier

   
 
  symbolOFDMoverPNAverage)(exp
1
)(exp
2
exp)(
1ˆ
)(exp)()(,)(exp)()(
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ssubcarrierallforsamethesit'becauseCommon
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• How to solve PN problem?
– Estimate PN samples and multiply received signal by
exp(jφ(n)), n=0:N-1 BEFORE taking FFT
– Difficult solution because PN change from sample to
sample and difficult to track
• Practical solution• Practical solution
– Use pilots to estimate average PN φavg and multiply all
subcarriers by conj(g0) = exp(jφavg) to reverse rotation
induced by PN
– This removes CPE and is called CPE compensation
– Here, we ignore ICI caused by PN
– Practical oscillators have small PN linewidth  Less
leakage  Light ICI that can be ignored

OFDM under Doppler
• High Doppler means that channel rapidly changes over
time due to Transmitter and/or Receiver movement
Received
Signal
2fd
Transmitted
signal
Time-varying
channel
fc
f fc f
Doppler frequency = fd = v fc/c
c  Speed of light = 3x108 m/s
v  Relative speed between Tx and Rx
If Tx and Rx are not moving (e.g., WiFi), No Doppler spread
Doppler increases as carrier frequency and/or velocity increases
Doppler spread

OFDM under Doppler
• 4G LTE will have mobile phones moving in high
speed
• LTE has higher Doppler than WiFi
• Doppler effect is similar to PN effect  Both• Doppler effect is similar to PN effect  Both
have energy leakage in frequency domain
• Effect of Doppler on OFDM:
–ICI and rotation of every subcarrier

OFDM under Doppler
• Coherence time = 1/Doppler frequency
• Coherence time  Time over which channel can be
considered static
• To avoid Doppler effect, make sure that OFDM
symbol duration is smaller than Coherence time
• What’s the Coherence time if fc = 2 GHz and v = 300
Km/hr?
• Coherence time = 1/fd = c/ (v fc) = 1.8 ms
• OFDM symbol duration in LTE = 67 us << 1.8 ms
• So, Doppler spread can be safely ignored in LTE

OFDM under I/Q imbalance
• I/Q imbalance (IQI) means  Inphase and Quadrature paths
have different gains and non-90 degrees phase differences
ε
1
ε
hI,tx(t) hI,rx(t)
1
π/2+θt
π/2+θr
εt εr
εt ≠ 1  Gain imbalance @ Tx
θt ≠ 0  Phase imbalance @ Tx
hI,tx(t) ≠ hQ,tx(t)  Filter imbal @ Tx
εr ≠ 1  Gain imbalance @ Rx
θr ≠ 0  Phase imbalance @ Rx
hI,rx(t) ≠ hQ,rx(t)  Filter imbal @ Rx
hQ,tx(t) hQ,rx(t)

• Gain and Phase imbalance  Frequency-flat (Frequency-
Independent) I/Q imbalance (FI-IQI)
• Filter imbalance  Frequency-selective (Frequency-
dependent) I/Q imbalance (FD-IQI)
• Effect of FI-IQI on baseband signal:
y(t)tytyty
x(t)txtxtx
rrIQIFI
ttIQIFI
signalRxonIQIRxofEffect)()()(
signalTxonIQITxofEffect)()()(
*
*






 
  1,
2
exp1
1,
2
exp1
*
*
rr
rr
r
tt
tt
t
j
j











 If no IQI,
ε = 1, θ = 0
α =1, β = 0

• FD-IQI has a similar effect except that α and β are filters
convolved with y(t) and y*(t), respectively
• Assume y(t) = exp(j2πf1t)  1 Complex sinsoid
• What’s yFI-IQI (t) ?  α exp(j2πf1t) + β exp(-j2πf1t)
Y( f ) 1
ff1
Y( f )
With IQI α
ff1
FI-IQIYFI-IQI( f )
0 0-f1
β
IQI creates an image of every subcarrier

OFDM under I/Q imbalance - Example
π/2
εt
1
cos(2πf1t) Let f1 = 1 MHz, fc = 700 MHz
Without I/Q imbalance, RF signal x(t)
Should be cos(2π(fc+f1)t), i.e.,
@ 701 MHz
x(t)
fc =700 MHz
sin (2πf1t)
x(t) = cos(2πf1t) cos(2πfct)- εt sin(2πf1t) sin(2πfct)
= (1+ εt)/2 cos(2π(fc+f1)t) + (1- εt)/2 cos(2π(fc-f1)t)
Due to IQI, we get frequency component @ fc-f1=699 MHz in
addition to the expected 701 MHz. This 699 MHz is the image
of 701 MHz around the carrier frequency fc

OFDM under I/Q imbalance - Example
If εt = 1, i.e., no I/Q imbalance, we see only 701 MHz component
and there will be no frequency component @ 699 MHz
    
ComponentImageComponentOriginal
)MHz6992cos(
2
1
)MHz7012cos(
2
1
)( 



 tttx tt




and there will be no frequency component @ 699 MHz
1
f(MHz)701
Signal analyzer Output
(no IQI)
fc=700
2
1 t
2
1 t
f(MHz)701fc=700699
Signal analyzer Output
(With IQI)
image

• With IQI, subcarrier k will create an image for itself
• This image interferes with subcarrier –k
• IQI causes Inter-Carrier Interference (ICI) from image
subcarriers not neighboring subcarriers as CFO, PN and
Doppler
• Signal-to-image ratio is called Image Rejection Ratio (IRR) and
is defined for a mixer as:is defined for a mixer as:
• Good mixers have high IRRs (IRR > 30 dB). IRR = ∞ if no IQI
exists
• Baseband processing is used to estimate and compensate IQI
to increase IRR to +45 dB
frequencyimage@Power
frequencyoriginal@Power
log10IRR
2
10 



• Phase imbalance is given in degrees, e.g., 2o
• Gain imbalance is given in dBs or % as follows:
dB
1
20Log
GainQuad
GainInphase
20LogImbalanceGain 1010 













• If ε=0.9  Gain imbalance = 0.9 dB or 5%
• If ε=1 (No imbalance)  Gain imbalance = 0 dB or 0%
• Good mixers have Gain imbalance close to 0 dB or 0%
%100
1
1
%100
GainQuadGainInphase
GainQuadGainInphase
ImbalanceGain 






-

• An RF mixer has Gain imbalance = 3 % and phase
imbalance = 2 degrees, what’s IRR?
• ε = 0.94 and θ = 2 *180/pi  Convert from deg to rad
 
dB29log10IRR
0.01603.01
0.0160.97
2
180
2exp94.01
2
10
*











r
r
rr
r
j
j
j





IRR is the same if we used eqns of αt and βt instead of αr and βr

OFDM under Rx I/Q imbalance
Baseband received signal
Before Mixer
Baseband received signal
After IQI-impaired Mixer
1
0.5 0.5α
α
f0
f0
f0
β
β/2
Add both waveforms
To get baseband signal under IQI
Image conjugated

Baseband compensation of I/Q Imbalance
Use Calibration to estimate Gain and Phase imbalance
Calibration means sending single tone through the mixer and
using received samples to estimate ε and θ
Baseband I/Q compensation circuit
π/2+θr
1
εr
hI,rx(t)
hQ,rx(t) 1/εr 1/cos(θr) +

Single-Carrier Frequency-DivisionSingle-Carrier Frequency-Division
Multiple-Access (SC-FDMA)

Shortcomings of OFDM
• Large PAPR
• Vulnerable to RF impairments (CFO, PN, IQI)
– These impairments cause ICI which reduces effective
SNR of each subcarrier
– ICI can cause strong carriers (with good channels) to
interfere on weak carriers (with deep fading)interfere on weak carriers (with deep fading)
==causing==> low carrier to noise ratio
– Effect of these impairments on Single-Carrier does
not include inter-symbol interference (ISI) in AWGN
channel. Only symbol rotation and/or scaling but No
ISI in AWGN channel
– Hence, Single-Carrier is more robust to these
impairments than OFDM!

Shortcomings of OFDM
• RF impairments can be compensated in baseband using fairly
low-complexity methods
• Hence, we can live with RF impairments
• However, large PAPR of OFDM is a persistent problem that
impacts PA efficiency
• High PA efficiency is a key factor for long-lasting battery in• High PA efficiency is a key factor for long-lasting battery in
cell phones
• 4G LTE standard adopted Single-Carrier for Uplink where cell
phone is transmitting
• OFDM is chosen for 4G LTE downlink where eNodeB (Base
station) is transmitting
• There’s no battery constraints in base stations

Single-Carrier
• Single-carrier can be implemented using OFDM platform
but with FFT introduced before IFFT to cancel its effect
QAM
modulation
SerialtoParallel
IFFTFFT

Single-Carrier Frequency-Division
Multiple-Access (SC-FDMA)
• In 4G LTE uplink, multiple user equipments (UEs) (cell phones) can
transmit at the same time but need to be frequency-multiplexed
• This is achieved by making FFT and IFFT sizes different
• FFT size = M and IFFT size = N where M < N• FFT size = M and IFFT size = N where M < N
• User 1  Take FFT output (M entries) and assign to a part of IFFT
input and pad remaining part (N-M entries) by zeros
• User 2  Assign your FFT output to the part where User 1 put
zeros and put zeros in the part where User 1 put its data
• See next slides for Two-Users example

SC-FDMA Transmitters @ Cell phones
QAM
modulation
SerialtoParallel
N-pointIFFT
M-point
FFTM QAM
symbols
M
M
N-M zerosN-M zeros
User 1
Transmitter
AddCyclicprefix
ParalleltoSerial
N N
QAM
modulation
M-point
FFTM QAM
symbols
M
M
N-M zerosN-M zeros
User 2
Transmitter
N-pointIFFT
N
SerialtoParallel
AddCyclicprefix
ParalleltoSerial
N

SC-FDMA Receiver at eNodeB (Base station)
pointFFT
RemoveCyclicprefix
SerialtoParallel
Select U1
subcarriers
M-pointIFFT
ParalleltoSerialQAM
demod
1,
ˆ
1
UkH MM
User 1 Channel
estimate
To User 1
decoder
N-pointFFT
RemoveCyclicprefix
SerialtoParallel
N N
Select U2
subcarriers
M-pointIFFT
ParalleltoSerial
QAM
demod
Channel estimation
MM 2,
ˆ
1
UkH
User 2 Channel
estimateTo User 2
decoder

SC-FDMA
• Different users can be assigned different number of subcarrier
(M1, M2, ….)
• Subcarrier allocation  localized (adjacent) or distributed
(Interleaved) 
• eNodeB tells every user its allocation size M and location (indices)
of these subcarriers
• Difference between OFDM and SC-FDMA receivers:• Difference between OFDM and SC-FDMA receivers:
– In OFDM, we get (QAM symbol x Channel after FFT), so
division by channel is Maximum-Likelihood (ML) optimal
– In SC-FDMA, we get (FFT of QAM symbols x Channel after FFT),
so division by channel is NOT ML-optimal
• Subcarriers in OFDM carriers QAM symbols, while in SC-FDMA,
they carry FFT of QAM symbols
• Cyclic prefix is used in both OFDM and SC-FDMA to prevent ICI
and maintain orthogonality among subcarriers

SC-FDMA
• SC-FDMA PAPR depends on QAM order & subcarrier
allocation  Interleaved PAPR < Localized PAPR
• In 4G LTE, no pilots are multiplexed with data subcarriers in
order to maintain low PAPR as pilots are transmitted in
higher power than data
• Instead, pilots are transmitted alone in a separate symbol
every 6 data symbolsevery 6 data symbols
• In WiFi, OFDMA is used at both Tx and Rx. No high PA
efficiency is needed because
– Tx-Rx distance in WiFi < Tx-Rx distance in LTE
– More restrictions on output power in WiFi than in LTE because WiFi
transmits in unlicensed bands  To limit interference to others
– Hence, WiFi PA gain < LTE PA gain  No high efficiency needed in
WiFi

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