Download pdf - Modern Standards and Distortion - UCSB€¦ · •GSM 33 dBm transmit power (TDD) 30 dB path loss so 1m away the power is 0 dBm at 800 MHz. This means that you have a 0 dBm blocker

Modern Standardsand Distortion

Transmit/Receive Schemes

• Time Division Duplex (TDD)

– Example: GSM

– TX/RX in the same/separate bands. Antenna is switched between TX and RX on millisecond scales.

• Frequency Division Duplex (FDD)

– Example: CDMA/LTE

– TX and RX are simultaneously operating at separate bands. Require a duplexer to isolate the bands.

©James Buckwalter 2

Basic FDD System

LTE-A Frequency Bands


Verizon uses Band 2, 4, 13

GSM


Implications

• GSM 33 dBm transmit power (TDD) 30 dB path loss so 1m away the power is 0 dBm at 800 MHz. This means that you have a 0 dBmblocker

• LTE with TDD/FDD still uses 24 dBm max power. LTE 2.7 GHz and higher path loss is higher. In TD LTE systems, -15 dBm.

• Duplex offset . 40/50 , 80/100 MHz. center to center


Distortion in Cellular Systems

• Signals are represented as “CW” signals. This isn’t really true. We can interpret two ways:

1) Each CW signal is a wideband modulated signal.

2) Each wideband modulated signal is modelledwith a multitone (two tones) within 1 BW.


In-band (IB) and Out-of-band (OOB) Distortion

• TX blockers are typically considered to be around 0 dBm (for GSM) or 10 dBm (for 3GPP/4G)

• RX blockers are typically around -40 dBm


Effect of Device Nonlinearity:Harmonic Distortion

©James Buckwalter

cosin pk ov V t 3

,

1

cosk

out m k pk o

k

i g V t

9

Solving for Basic Harmonic Distortion

• Features??

©James Buckwalter

,2 ,32 3

,1

,2 ,32 3

3cos

2 4

cos 2 cos 32 4

m m

out pk m pk pk o

m m

pk o pk o

g gi V g V V t

g gV t V t

10

Defining Harmonic Distortion

• Total Harmonic Distortion

©James Buckwalter

,2

,1

,3 2

,1

Signal at 2 12

Signal at 2

Signal at 3 13

Signal at 4

mo

pk

o m

mo

pk

o m

gHD V

g

gHD V

g

2 22 3 ...THD HD HD

11

Gain Compression

©James Buckwalter

3

,1 ,3

10

,1

3420log 1

m pk m pk

m pk

g V g V

g V

,1

1

,3

40.11

3

m

dB

m

gV

g

12

Intermodulation Distortion


Effect of Device Nonlinearity:Intermodulation Distortion

©James Buckwalter

1 2cos cosin pkv V t t

3

, 1 2

1

cos coskk

out m k pk

k

i g V t t

14

Solve for Intermodulation Distortion (I)

©James Buckwalter

,1 1 2

22

,2 1 2

33

,3 1 2

cos cos

cos cos

cos cos

out m pk

m pk

m pk

i g V t t

g V t t

g V t t

,1 1 2

1 22

,2

2

,2 1 2 1 2

33

,3 1 2

cos cos

1 cos 2 1 cos 2

2 2

cos cos

cos cos

out m pk

m pk

m pk

m pk

i g V t t

t tg V

g V t t

g V t t

15

Solve for Intermodulation Distortion (II)

©James Buckwalter

,1 1 2

22

,2 1 2

33

,3 1 2

cos cos

cos cos

cos cos

out m pk

m pk

m pk

i g V t t

g V t t

g V t t

3

,3 1 1 2 2

3

,3 1 1 2 1 2

3

,3 2 2 1 2 1

1cos 3 3cos cos 3 3cos

4

3 1 1cos cos 2 cos 2

2 2 2

3 1 1cos cos 2 cos 2

2 2 2

m pk

m pk

m pk

g V t t t t

g V t t t

g V t t t

16

Picture of distortion generated from two tones


Definition of IntermodulationDistortion

©James Buckwalter

,21 2

1 ,1

,3 21 2

1 ,1

Signal at 2

Signal at

Signal at 2 33

Signal at 4

m

pk

m

m

pk

m

gIM V

g

gIM V

g

18

Graph of IM3 Products


IM3 Issues with FDMA systems

• IM3 tone generated by 2 jammers in adjacent channels will fall into desired signal channel.

• Typically, we desire to keep the IM3 tone at least 20 dB below the desired signal.


Input-Intercept/Output-Intercept Points

• 3rd-order Input-intercept Point (IIP3)

• 3rd-order Output-intercept Point (OIP3)


Definition of the Input Intercept Point

©James Buckwalter

Signal at 2w1-w

2= Signal at w

1

3

4g

m,3V

pk

3 = gm,1

Vpk

IIP3 =4

3

gm,1

gm,3

22

Note the Relationship between IIP3 and P1dB

• Factor in front of the 1dB voltage suggests that the 1dB compression occurs 9.6 dB below IIP3.

• A cruder rule of thumb is that IIP3 is 10 dB higher than P1dB.

©James Buckwalter

,1 ,1

1

,3 ,3

4 40.11 3

3 3

m m

dB

m m

g gV IIP

g g

23

Jammer Linearity Requirement

• Non-linearity in gain

• In terms of power

©James Buckwalter

2 3

1 2 3

1

3

43

3

o i i iv g v g v g v

gIIP

g

IM3 =

3

4g

3V

J

3

g1V

J

=4

3

g3

g1

VJ

2

®P

J

PIIP3

= IM3

®10log10

PJ

-10log10

PIIP3

=1

220log

10IM3

PJ

= PIIP3

+IM3

2Factor of two becauseIM3 is based on ratio of voltage

Jammer Rejection

• Typically we want to find the distortion at a desired band

©James Buckwalter

P

IIP3= P

J-

IM3

2 or IM3 = 2(P

J- P

IIP3)

25

Since IM3 = PIM 3

- PJ

PIIP3

= PJ

-P

IM 3- P

J

2=

3

2P

J-

1

2P

IM 3

P

IM 3= 3P

J-2P

IIP3

Example

• We assumed that IB blockers are -40 dBm.

• What IIP3 is required to keep distortion 10 dB below noise floor?


Example Cont.

• We assumed that IB blockers are -40 dBm.

• For sensitivity, assume 10 MHz channel for 4G.

• You can assume NF = 0dB and SNR = 0dB.

• Not too bad for CMOS LNA.


PIIP3

= 3

2P

J-

1

2P

IM 3

PIIP3

= 3

2-40dBm( ) -

1

2-111dBm( ) =-4.5dBm

P

sens= -174dBm+10log

1020MHz = -101dBm

Spur-Free Dynamic Range


Minimum Detectable Signal

©James Buckwalter

i

o

MDS FkT f

MDS FGkT f

29

Solve for the SFDR

• Minimum detectable signal is the weakest signal that can be resolved in a given bandwidth

©James Buckwalter

13 3

3

23

3

i i

i

SFDR IIP MDS IIP MDS

SFDR IIP MDS

Units are dBc Hz2/3

30

Noise Power Ratio Test

• Are two tones sufficient to interrogate the amplifier linearity?


White space available

Cross-Modulation Distortion (XMD)

©James Buckwalter

1 2cos 1 cos cosin D m Jv V t m t V t

• Occurs with high power transmitters. AM modulated signal copies itself onto neighboring signal.

32

Cross-Modulation Distortion (XMD)

©James Buckwalter

v

out»V

Da

1+ 3a

3V

J

2mcos wmt( )( )cos w

1t( )

• Modulation of undesired signal becomes impressed on desired signal

33

Cross Modulation Distortion

• Similar to intermodulation distortion however CMI results from only one modulated signal

©James Buckwalter

XMD =3 a

3V

J

2VD

2a1V

D

=3

2

a3

a1

VJ

2

34

XMD in Full Duplex Systems


Review of Cross Modulation Distortion

©James Buckwalter

XMD = 2 10log10

VTX

2

Rs

æ

èçç

ö

ø÷÷-10log

10

VIP3

2

2Rs

æ

èçç

ö

ø÷÷

æ

è

çç

ö

ø

÷÷

XMD = 2 PTX ,TOTAL

- PIIP3( ) ® P

IIP3= P

TX ,TOTAL-

XMD

2

SJ

= g1V

Jcos w

Jt( )

SXMD

=3

2g

3V

JV

TX

2 cos wJ

+ wTX ,1

-wTX ,2( )t( )

XMD =S

XMD

SJ

=

3

2g

3V

JV

TX

2

g1V

J

=3g

3

2g1

VTX

2

,1 ,2XMD J TX TX

TX,1 = 800 MHzTX,2 = 801 MHz

RX = 900 MHz1 = 901 MHz

36

XMD Requirement

• This equation was based on CW signals for the TX and blockers. Therefore this equation is “approximate”.

©James Buckwalter

P

IIP3= P

TX ,TOTAL-

XMD

2

37

Since XMD = PXMD

- PJ

PIIP3

=2P

TX ,TOTAL+ P

J- P

XMD

2

XMD Requirement

• Factor of 5 is added to account for the modulated nature of the TX, CW.

• Other factors are derived based on narrowband/wideband modulation.


P

IIP3=

PCW

+ 2PTX ,TOTAL

- PXMD

-5( )2Larson and Aparin, TMTT 2005

Example

• We assumed that IB blockers are -40 dBm and OOB blockers are 0 dBm.

• What XMD is required to keep distortion 10 dB below noise floor?


Example Cont.

• We assumed that IB blockers are -40 dBm and OOB blockers are 0 dBm.

• For sensitivity, assume 10 MHz channel for 4G.

• You can assume NF = 0dB and SNR = 0dB.

• This pretty tough for CMOS LNA.


P

sens= -174dBm+10log

1020MHz = -101dBm

P

IIP3=

PCW

+ 2PTX ,TOTAL

- PXMD

-5( )2

=-40dBm+ 2 ×0dBm- -111dBm( ) -5( )

2= 33dBm

Duplexer

©James Buckwalter

• RF to RX: 3 dB in band

• TX to RF: 3 dB in band

• TX to RX: 60 dB

41

Receiver Signal Levels

©James Buckwalter

S:-107 dBmJ: -47 dBm

T: 30 dBm30 dBm →10 Vp

S: 1 uVp

J: 1 mVp

T: 10 mVp

S:-110 dBm →1 uVp

J: -50 dBm →1 mVp

S: 10 uVp

J: 10 mVp

T: 100 mVp

20 dB

42

How much linearity is needed?

• Transmit signal is 10 mVp

• Jammer is 1mVp

• Signal is 1uVp

• Amplifier

– Power Gain: 20 dB

– Noise Figure: 2dB

– IIP3 ???


Linearity Requirement Exercise (I)

• What IIP3 (power) is required to keep the IM3 product at the minimum detectable signal for the example receiver described below?

©James Buckwalter

S:-107 dBmJ: -47 dBm

S: 1 mVp

J: 1 mVp

T: 10 mVp

S:-110 dBm →1 mVp

J: -50 dBm →1 mVp

44

Linearity Requirement for IM3

©James Buckwalter

PIIP3

= PJ

-IM3

2

PIIP3

= -50dBm --110dBm - -50dBm( )( )

2

PIIP3

= -20dBm

45

Linearity Requirement for XMD

• Cross modulation distortion imposes a tougher linearity requirement on the receiver than the two-tone intermodulation.

©James Buckwalter

IIP3 = PTX ,TOTAL

-XMD

2

XMD = -110dBm - -50dBm( ) = -60dB

IIP3 = -30dBm --60dB

2= 0dBm

46

Summary of Metrics

• One-dB Compression – One tone

• Intermodulation Distortion – Two tone

• Intercept Point – Two Tone

• Spur-Free Dynamic Range – Two tone

• Carrier-to-Interference Ratio – Multi-tone

• Cross-Modulation Index – Modulated one/two tone
