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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 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
©James Buckwalter 35
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.
©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 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 ???
©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 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
©James Buckwalter 47