ECEN689: Special Topics in High-Speed Links Circuits and ...

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Sam Palermo Analog & Mixed-Signal Center Texas A&M University ECEN689: Special Topics in High-Speed Links Circuits and Systems Spring 2010 Lecture 8: Channel Pulse Model

Transcript of ECEN689: Special Topics in High-Speed Links Circuits and ...

Page 1: ECEN689: Special Topics in High-Speed Links Circuits and ...

Sam PalermoAnalog & Mixed-Signal Center

Texas A&M University

ECEN689: Special Topics in High-Speed Links Circuits and Systems

Spring 2010

Lecture 8: Channel Pulse Model

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Announcements

• Reading• Will post some papers of link examples with

different modulation techniques

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Agenda

• ISI

• Channel Pulse Model

• Peak Distortion Analysis

• Modulation Schemes

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Inter-Symbol Interference (ISI)

• Previous bits residual state can distort the current bit, resulting in inter-symbol interference (ISI)

• ISI is caused by• Reflections, Channel resonances, Channel loss (dispersion)

• Pulse Response

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( )( )tc 1 ( )th ( )( )tc 1( )( )ty 1

( )( ) ( )( ) ( )thtcty ∗= 11

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NRZ Data Modeling

• An NRZ data stream can be modeled as a superposition of isolated “1”s and “0”s

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Data = “1000101”

“1” Symbol

“0” Symbol

( )( ) ( ) ( )( )TktukTtutck 11 +−−−≡

( )( ) ( )( )tctc kk10 −=

( )

<≥

≡0001

wherett

tu

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NRZ Data Modeling

• An NRZ data stream can be modeled as a superposition of isolated “1”s and “0”s

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( ) ( )( )∑∞

−∞=

=k

dki tctV k

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Channel Response to NRZ Data

• Channel response to NRZ data stream is equivalent to superposition of isolated pulse responses

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( ) ( )( ) ( )( )( ) ( )( )∑∑∞

−∞=

−∞=

−===k

d

k

dkio kTtytcHtVHtV kk

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Channel Pulse Response

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( )( ) ( )( ) ( )thtcty kk dd ∗=

y(1)(t) sampled relative to pulse peak:

[… 0.003 0.036 0.540 0.165 0.065 0.033 0.020 0.012 0.009 …]

k =[ … -2 1 0 1 2 3 4 5 6 …]

By Linearity: y(0)(t) =-1*y(1)(t)

cursor

post-cursor ISI

pre-cursor ISI

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Channel Data Stream Response

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Input Data Stream

Pulse Responses

Channel Response

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Channel “FIR” Model

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( )( )tc 10

( )( )tc 10

( )( )( ) ( )( )tytcH 10

10 =

( )( )( ) ( )( )tytcH 10

10 =

a-1

a0

a1

a2 a3

y(1)(t) sampled relative to pulse peak:

[… 0.003 0.036 0.540 0.165 0.065 0.033 0.020 0.012 0.009 …]

a =[ … a-2 a-1 a0 a1 a2 a3 a4 a5 a6 …]

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Peak Distortion Analysis

• Can estimate worst-case eye height and data pattern from pulse response

• Worst-case “1” is summation of a “1” pulse with all negative non k=0 pulse responses

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( ) ( ) ( )( )( )∑

≠−∞= <−

−+=

0

0

)1(01

kk kTty

d kTtytyts k

• Worst-case “0” is summation of a “0” pulse with all positive non k=0 pulse responses

( ) ( ) ( )( )( )∑

≠−∞= >−

−+=

0

0

)0(00

kk kTty

d kTtytyts k

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Peak Distortion Analysis

• Worst-case eye height is s1(t)-s0(t)

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( ) ( ) ( ) ( ) ( )( ) ( )( )( )

( )( )( )

−−−+−=−= ∑∑∞

≠−∞= >−

≠−∞= <−

0

0

0

0

)0(0

)1(001

kk kTty

d

kk kTty

d kTtykTtytytytststs kk

• If symmetric “1” and “0” pulses (linearity), then only positive pulse response is needed

( ) ( ) ( )( )( )

( )( )( )

−−−+= ∑∑∞

≠−∞= >−

≠−∞= <−

0

0

1

0

0

1)1(02

kk kTty

kk kTty

kTtykTtytyts

( ) ( )( )tyty )1(0

)0(0 1 Because −=

“1” pulse worst-case “1” edge

“1” pulse worst-case “0” edge

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Peak Distortion Analysis Example 1

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( )( )( )

( )

( )( )( )

( ) ( ) 288.0389.0007.0540.02

389.0

007.0

540.0

0

0

1

0

0

1

)1(0

=−−=

=−

−=−

=

≠−∞= >−

≠−∞= <−

ts

kTty

kTty

ty

kk kTty

kk kTty

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Worst-Case Bit Pattern• Pulse response can be used to find the worst-case bit pattern

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[ ]... ...a Pulse 654321012 aaaaaaaaa −−=

( )[ ]... )()(1)()()()()(- ... 21123456 −− −−−−−−− asignasignasignasignasignasignasignasign

• Flip pulse matrix about cursor a0 and the bits are the inverted sign of the pulse ISI

Worst-Case Bit Pattern Eye 10kbits Eye

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Peak Distortion Analysis Example 2

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( )( )( )

( )

( )( )( )

( ) ( ) 338.0542.0053.0426.02

542.0

053.0

426.0

0

0

1

0

0

1

)1(0

−=−−=

=−

−=−

=

≠−∞= >−

≠−∞= <−

ts

kTty

kTty

ty

kk kTty

kk kTty

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Modulation Schemes

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• Binary, NRZ, PAM-2• Simplest, most common modulation format

• PAM-4• Transmit 2 bits/symbol• Less channel equalization and circuits run ½ speed

• Duo-binary• Allows for controlled ISI, symbol at RX is current bit plus preceding bit• Results in less channel equalization

[ ] [ ] [ ]1−+= nxnxnw

0

1

00

01

11

10

0, if x[n-1]=0

1, if x[n-1]=0 OR

0, if x[n-1]=1

1, if x[n-1]=1

No Pre-Coding Case

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Modulation Frequency Spectrum

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Majority of signal power in 1GHz bandwidth

Majority of signal power in 0.5GHz bandwidth

Majority of signal power in 0.5GHz bandwidth

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Nyquist Frequency

• Nyquist bandwidth constraint:• The theoretical minimum required system bandwidth

to detect RS (symbols/s) without ISI is RS/2 (Hz)• Thus, a system with bandwidth W=1/2T=RS/2 (Hz)

can support a maximum transmission rate of 2W=1/T=RS (symbols/s) without ISI

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/Hz)(symbols/s 222

1≤⇒≤=

WRWR

TSS

• For ideal Nyquist pulses (sinc), the required bandwidth is only RS/2 to support an RS symbol rate

Modulation Bits/Symbol Nyquist Frequency

NRZ 1 Rs/2=1/2Tb

PAM-4 2 Rs/2=1/4Tb

Duobinary 1 (or 1.5) ?? Rs/2=1/2Tb (or 1/3Tb) ??

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Next Time

• Modulation scheme comparison

• Link Circuits• Termination structures• Drivers• Receivers

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