lecture i introduction baseband pulse ... - soc: sochassan/cirf08_comm_theory.pdf · n2 t 0 2 2 ( )...

CIRFCircuit Intégré Radio Fréquence

Lecture I

•Introduction•Baseband Pulse Transmission•Digital Bandpass Transmission•Circuit Non-idealities Effect

Hassan AboushadyUniversity of Paris VI

Disciplines required in RF design

H. Aboushady University of Paris VI

• S. Haykin, “Communication Systems”, Wiley 1994.

• B. Razavi, “RF Microelectronics”, Prentice Hall, 1997.

• M. Perrott, “High Speed Communication Circuits andSystems”, M.I.T.OpenCourseWare, http://ocw.mit.edu/,Massachusetts Institute of Technology, 2003.

• D. Yee, “ A Design methodology for highly-integratedlow-power receivers for wireless communications”,http://bwrc.eecs.berkeley.edu/, Ph.D. University ofCalifornia at berkeley, 2001.

References



Lecture I




Matched Filter

Linear Receiver Model

• Filter Requirements, h(t) :• Make the instantaneous power in the output signal g0(t) ,measured at time t=T, as large as possible compared withthe average power of the output noise, n(t).

)()()( 0 tntgty +=Tttwtgtx ≤≤+= 0,)()()( )(th

• g(t) : transmitted pulse signal, binary symbol ‘1’ or ‘0’.• w(t) : channel noise, sample function of a white noise processof zero mean and power spectral density N0/2.


Maximize Signal-to-Noise Ratio

Linear Receiver Model

power noiseoutput averagesignaloutput in thepower ousinstantane=SNR

)(

)(2

20

tn

TgSNR =


Math Review

Fourier Transform

Inverse Fourier Transform

Power Spectral Density of a Random Process SY(f)applied to a Linear System H(f)

dfefGtg tfj∫∞

∞−

= π2)()(

dtetgfG tfj∫∞

∞−

−= π2)()(

)()()( 2 fSfHfS XY =


Compute Signal-to-Noise Ratio

dfefGfHtg tfj∫∞

∞−

= π20 )()()(

2

220 )()()( dfefGfHTg Tfj∫

∞

∞−

= π

20 )(2

)( fHNfSN = ∫∞

∞−

= dffStn N )()(2 ∫∞

∞−

= dffHNtn 202 )(2

)(

∫

∫∞

∞−

∞

∞−=dffHN

dfefGfHSNR

Tfj

20

2

2

)(2

)()( π

• Optimization Problem:• For a given G(f) , find H(f) in order to maximize SNR.


Schwarz’s Inequality

• If we have 2 complex functions φ1(x) and φ2(x) in the realvariable x, satisfying the conditions:

∞<∫∞

∞−

dxx 21 )(φ ∞<∫

∞

∞−

dxx 22 )(φ

then we may write that:

dxxdxxdxxx ∫∫∫∞

∞−

∞

∞−

∞

∞−

≤ 22

21

2

21 )()()()( φφφφ iff )()( *

21 xkx φφ =

where constantarbitrary :k

dffGdffHdfefGfH Tfj ∫∫∫∞

∞−

∞

∞−

∞

∞−

≤ 222

2 )()()()( π

setting:TfjefGx πφ 2

2 )()( =and)()(1 fHx =φ


Matched Filter

dffGN

SNR ∫∞

∞−

≤ 2

0

)(2 dffGN

SNR ∫∞

∞−

= 2

0max )(2

)()( *21 xkx φφ = Tfj

opt efGkfH π2* )()( −=

dfefGkth tTfjopt ∫

∞

∞−

−−= )(2* )()( π dfefGkth tTfjopt ∫

∞

∞−

−−−= )(2)()( π

)()( tTgkthopt −=• for a real signal g(t) we haveG*(f)=G(-f)

• The impulse response of the optimum filter, except for thescaling factor k, is a time-reversed and delayed version of theinput signal g(t)


Properties of Matched Filters

)()( tTgkthopt −=

Tfjopt efGkfH π2* )()( −=

Tfj

Tfj

opt

efGk

efGfGk

fGfHfG

π

π

22

2*

0

)(

)()(

)()()(

−

−

=

=

=

• Taking the inverse Fourier transform at t=T:

EkdffGkdfefGTg Tfj === ∫∫∞

∞−

∞

∞−

2200 )()()( π

Where E is theenergy of thepulse signal g(t)

∫∞

∞−

= dffHNtn 202 )(2

)( ENkdffGkNtn2

)(2

)( 022202 == ∫∞

∞−

002

22

max2

2N

E

ENk

EkSNR ==


Matched Filter for Rectangular Pulse

)()( tTgkthopt −=

g(t)

T

g(t)

tT

h(t) = g(T-t)

tT

g(-t)

t

h(t) for a rectangular Pulse:

Filter Output g(t)*h(t):

Implementation:


Error Rate due to Noise

In the interval , the received signal:bTt ≤≤0

⎩⎨⎧

+−++

=sent was'0' symbol,)(sent was'1' symbol,)(

)(twAtwA

tx

Tb is the bit duration, A is the transmitted pulse amplitude

• The receiver has prior knowledge of the pulse shape but notits polarity.

• There are two possible kinds of error to be considered: (1) Symbol ‘1’ is chosen when a ‘0’ was transmitted. (2) Symbol ‘0’ is chosen when a ‘1’ was transmitted.

x(t)


Error Rate due to Noise

Suppose that symbol ‘0’ was sent: bTttwAtx ≤≤+−= 0,)()(

The matched filter output is: ∫∫ +−==bb T

b

T

dttwT

AdttxY00

)(1)(

Y is a random variable with Gaussian distribution and a mean of -A.

The variance of Y: ∫ ∫=+=b bT T

Wb

Y dudtutRT

AY0 02

22 ),(1)(σ

Where RW(t,u) is the autocorrelation function of the white noise w(t) .Since w(t) is white with a PSD of N0/2 :

)(2

),( 0 utNutRW −= δ

bY T

N2

02 =σ


PDF: Probability Density Function

Normalized PDFµY= 0 and σY=1

⎥⎦

⎤⎢⎣

⎡ −−= 2

2

2)(exp

21)(

Y

Y

Y

Yyyf

σµ

πσ

• Gaussian Distribution:

• Symbol ‘0’ was sent:

dyyf

yPP

Y

e

∫∞

=

>=

λ

λ

)0(

sent) was'0' symbol(0

bYY T

NA 02, =−= σµ

• Symbol ‘1’ was sent:

dyyf

yPP

Y

e

∫∞−

=

<=λ

λ

)1(

sent) was'1' symbol(1

bYY T

NA 02, =+= σµ


BER in a PCM receiver

dyTNAy

TNP

bb

e ∫∞

⎥⎦

⎤⎢⎣

⎡ +−=λπ /

)(exp/

1

0

2

0

0

• let λ=0 and the probabilities of binary symbols: p0 = p1 = 1/2 .

bTNAyz/0

+= [ ]dzzPNE

eb

∫∞

−=0/

20 exp1

π

• where Eb =A2Tb , is the transmitted signal energy per bit.

[ ]dzzuerfcu∫∞

−= 2exp1)(π

• the complementary error function:

⎟⎟⎠

⎞⎜⎜⎝

⎛==

001 2

1NEerfcPP b

ee


BER in a PCM receiver

1100 eee PpPpP +=

10 ee PP =

21

10 == pp

10 eee PPP +=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

021

NEerfcP b

e


Lecture I



• In wired systems, coaxial lines exhibit superior shieldingat higher frequencies

• In wireless systems, the antenna size should be asignificant fraction of the wavelength to achieve areasonable gain.

• Communication must occur in a certain part of thespectrum because of FCC regulations.

• Modulation allows simpler detection at the receive end inthe presence of non-idealities in the communicationchannel.

Why Modulation?


• mi : one symbol every T seconds• Symbols belong to an alphabet of M symbols: m1, m2, …, mM

MmPp

mPmPmP

ii

M

1)(

)(...)()( 21

==

===• Message output probability:

• Example: Quaternary PCM, 4 symbols: 00, 01, 10, 11


Message Source

MidttsET

ii ...,,2,1,)(0

2 == ∫• Energy of si(t) :


Transmitter

• Signal Transmission Encoder: produces a vector si made up of Nreal elements, where N M .• Modulator: constructs a distinct signal si(t) representing mi ofduration T .

≤

• si(t) is real valued and transmitted every T seconds.

Examples of Transmitted signals: si(t)


• The modulator performs a step change in the amplitude, phaseor frequency of the sinusoidal carrier

• ASK:Amplitude Shift Keying

• PSK:Phase Shift Keying

• FSK:Frequency Shift Keying

Special case: Symbol Duration T = Bit Duration, Tb

⎩⎨⎧

=≤≤

+=Mi

Tttwtstx i ,...,2,1

0),()()(

• Received signal x(t) :


Communication Channel

• Two Assumptions:•The channel is linear (no distortion).• si(t) is perturbed by an Additive, zero-mean, stationnary,White, Gaussian Noise process (AWGN).

≤


Receiver

m̂

•TASK: observe received signal, x(t), for a duration T and make abest estimate of transmitted symbol, mi .

•Detector: produces observation vector x .•Signal Transmission Decoder: estimates using x, themodulation format and P(mi) .

• The requirement is to design a receiver so as to minimizethe average probablilty of symbol error:

∑=

==M

iiie mPmmPP

1

)()ˆ(


Coherent and Non-Coherent Detection

• Coherent Detection:- The receiver is time synchronized with the transmitter.- The receiver knows the instants of time when themodulator changes state.- The receiver is phase-locked to the transmitter.

• Non-Coherent Detection:- No phase synchronism between transmitter and receiver.


Gram-Schmidt Orthogonalization Procedure

⎩⎨⎧

=≤≤

=∑= Mi

Tttsts

N

jjiji ,...,2,1

0,)()(

1

φ

⎩⎨⎧

==

= ∫ NjMi

dtttss j

T

iij ,...,2,1,...,2,1

,)()(0

φ

⎩⎨⎧

≠=

=∫ jiifjiif

dttt j

T

i 01

)()(0

φφ

• we represent the given set of real-valued energy signals s1(t), s2(t), … ,sM(t), each of duration T:

• where the coefficients of theexpansion are defined by:

• the real-valued basis functions φ1(t),φ2(t), … , φN(t) are orthonormal:

Modulator

Detector


Coherent Detection of Signals in Noise

• w(t) is a sample function ofan AWGN with power spectraldensity N0 /2.

⎩⎨⎧

=≤≤

+=Mi

Tttwtstx i ,...,2,1

0),()()(

Mi

s

ss

s

iN

i

i

i ,...,2,1,2

1

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=M

Miwsx i ,...,2,1, =+=• Observation vector x :

• Signal Vector si :

where w is the noise vector.


Coherent Binary PSK:

b

T

Edtttssb

== ∫ )()( 10

111 φ

• M=2, N=1

• One basis function:

• Signal constellation consists oftwo message points:

)2cos(2)(1 tfTEts cb

b π= )2cos(2)(2 tfTEts cb

b π−=bTt ≤≤0

• To ensure that each transmitted bit contains an integral number ofcycles of the carrier wave, fc =nc/Tb, for some fixed integer nc.

bcb

TttfT

t ≤≤= 0,)2cos(2)(1 πφ

b

T

Edtttssb

−== ∫ )()( 10

221 φ


Generation and Detection of Coherent Binary PSK

• Assuming white Gaussian Noise withPSD = N0/2 ,The Bit Error Rate for coherent binary PSK is: ⎟

⎟⎠

⎞⎜⎜⎝

⎛=

021

NEerfcP b

e

Binary PSK Transmitter

Coherent Binary PSK Receiver


Coherent QPSK:

• M=4, N=2:

• Two basis function:

⎪⎩

⎪⎨

⎧≤≤⎥⎦

⎤⎢⎣

⎡ −+=elsewhere , 0

0,4

)12(2cos2)( Ttitf

TE

ts ci

ππ

)4

2cos(2)(1ππ += tf

TEts c

TttfT

t c ≤≤= 0,)2cos(2)(1 πφ

TttfT

t c ≤≤= 0,)2sin(2)(2 πφ

)4

32cos(2)(2ππ += tf

TEts c

)4

52cos(2)(3ππ += tf

TEts c

)4

72cos(2)(4ππ += tf

TEts c


Constellation Diagram of Coherent QPSK System

⎟⎟⎠

⎞⎜⎜⎝

⎛=

021

NEerfcP b

e


QPSK waveform: 01101000


Generation and Detection of Coherent QPSK Signals

QPSK Transmitter

Coherent QPSK Receiver


Power Spectra of BPSK ,QPSK and M-ary PSK

QPSK

BPSK

8-PSK

• Symbol Duration: MTT b 2log=

• Power Spectral Density ofan M-ary PSK signal: )log(sinclog2

)(sinc2)(

22

2

2

MfTMETfEfS

bb

B

=

=

)(sinc2)( 2 fTEfS bbB =

)2(sinc4)( 2 fTEfS bbB =

)3(sinc6)( 2 fTEfS bbB =


Lecture I



Binary Dataantenna

011011Multiplexer

DecisionDevice

Threshold = 0

Threshold = 0

DecisionDevice

T

T

path I

path Q

∫T

dt0

∫T

dt0

)2cos(2)(1 tfT

t cπφ =

)2sin(2)(1 tfT

t cπφ =

I

Q

QPSK Constellation Diagram

00 10

1101


QPSK Receiver

• Circuit Noise (Thermal, 1/f)

• Gain Mismatch

• Phase Mismatch

• DC Offset

• Frequency Offset

• Local Oscillator phase noise


Receiver Circuit Non-Idealities

Circuit Noise:

• Thermal Noise• Resistors• Transistors

• Flicker (1/f) Noise• MOS transistors

I

Q


Circuit Noise

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Q

I

A(1-α/2)

A(1+α/2)


Gain Mismatch

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Q

I


Phase Mismatch

offset

offset

I

Q

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Q

I


DC Offset

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Q

I


Frequency Offset


Local Oscillator Phase Noise


Reciprocal Mixing

lecture i introduction baseband pulse ... - soc: sochassan/cirf08_comm_theory.pdf · n2 t 0 2 2 ( )...

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