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Modern Standards and Distortion

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

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Text of Modern Standards and Distortion - UCSB€¦ · •GSM 33 dBm transmit power (TDD) 30 dB path loss...

  • 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

    ©James Buckwalter 4

    Verizon uses Band 2, 4, 13

  • GSM

    ©James Buckwalter 5

  • 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

    ©James Buckwalter 6

  • 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.

    ©James Buckwalter 7

  • 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

    ©James Buckwalter 8

  • 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

    ©James Buckwalter 13

  • 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

    ©James Buckwalter 17

  • 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

    ©James Buckwalter 19

  • 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.

    ©James Buckwalter 20

  • Input-Intercept/Output-Intercept Points

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

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

    ©James Buckwalter 21

  • 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?

    ©James Buckwalter 26

  • 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.

    ©James Buckwalter 27

    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

    ©James Buckwalter 28

  • 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?

    ©James Buckwalter 31

    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 w1t( )

    • 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

    ©James Buckwalter 35

  • Review of Cross Modulation Distortion

    ©James Buckwalter

    XMD = 2 10log10

    VTX

    2

    Rs

    æ

    èçç

    ö

    ø÷÷-10log10

    VIP3

    2

    2Rs

    æ

    èçç

    ö

    ø÷÷

    æ

    è

    çç

    ö

    ø

    ÷÷

    XMD = 2 PTX ,TOTAL

    - PIIP3( ) ® PIIP3 = PTX ,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.

    ©James Buckwalter 38

    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?

    ©James Buckwalter 39

  • 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.

    ©James Buckwalter 40

    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 uVpJ: 1 mVpT: 10 mVp

    S:-110 dBm →1 uVpJ: -50 dBm →1 mVp

    S: 10 uVpJ: 10 mVpT: 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 ???

    ©James Buckwalter 43

  • 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 mVpJ: 1 mVpT: 10 mVp

    S:-110 dBm →1 mVpJ: -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

    ©James Buckwalter 47