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# Analog Building Blocks

Apr 06, 2018

## Documents

Amit Mehta
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Analog Building Blocks

Sampling theorem

Undersampling, antialiasing

FIR digital filters

Quantization noise, oversampling

6.111 Fall 2009 1Lecture 10

Lab #3 report due on-line @ 5pm today.Lpset #8 due Tuesday 10/20

Digital Representations ofAnalog Waveforms

Continuous timeContinuous values

Discrete timeDiscrete values

6.111 Fall 2009 2Lecture 10

Discrete Time

=

=n

nTttp )()(

)(tx )(txp

Lets use an impulse train to sample a continuous-time function at aregular interval T:

time)(tx

time)(tp

T1

(x) is a narrow impulse at x=0,where

time)(txp

Time Domain

= )()()( afdtattf

6.111 Fall 2009 3Lecture 10

ReconstructionIs it possible to reconstruct theoriginal waveform using only thediscrete time samples?

)(txp Rp )(tx

Frequency Domain

)( jX

)( jP

MM

T

2

s s20s2 s

Ts

2=

1

)( jXp

MM

T

1

s

Ms

So, if m < s-m, we can recoverthe original waveform with a low-pass filter!

)( jRp

cc

T MscM

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Sampling TheoremLet x(t) be a band-limited signal, ie, X(j)=0 for || > M. Thenx(t) is uniquely determined by its samples x(nT), n = 0, 1, 2, ,if

s > 2Mwhere

s =T

2

Given these samples, we can reconstruct x(t) by generating aperiodic impulse train in which successive impulses haveamplitudes that are successive sample values, then passing thetrain through an ideal LPF with gain T and a cutoff frequencygreater than M and less than s-M.

2Mis called the

Nyquist rate ands/2 the Nyquistfrequency

6.111 Fall 2009 5Lecture 10

Undersampling Aliasing

If s 2M theres an overlapof frequencies between oneimage and its neighbors and we

discover that those overlapsintroduce additional frequencycontent in the sampled signal, aphenomenon called aliasing.

)( jX

2 5-2-5

)( jP

-6 60

6,5 == sM

)( jXp

-5 -2 2 5

1 4 8-6 6-1-4-8

There are now tones at 1(= 6 5) and 4 (= 6 2) in

addition to the originaltones at 2 and 5.

6.111 Fall 2009 6Lecture 10

Antialias Filters

The frequency response of human ears essentially drops to zero above 20kHz. Sothe Red Book standard for CD Audio chose a 44.1kHz sampling rate, yielding a

Nyquist frequency of 22.05kHz. The 2kHz of elbow room is needed becausepractical antialiasing filters have finite slopefs = (3 samples/line)(490 lines/frame)(30 frames/s) = 44.1 kHz

If we wish to create samples at some fixed frequency s, then toavoid aliasing we need to use a low-pass filter on the originalwaveform to remove any frequency content s/2.

2

sC

=

Discrete-Time

sampler

s

We need this antialiasing filter it hasto have a reasonably sharp cutoff

This is the symbol for alow-pass filter see the

little x marks on themiddle and highfrequecies?

6.111 Fall 2009 7Lecture 10

Digital Filters

y[n] = bkk= 0

N

x[n k]

Equation for an N-tap finite impulse response (FIR) filter:

What components are part of the tPD of this circuit?How does tPD grow as N gets larger?

x[n-1]x[n]

x[n-2] x[n-3] x[n-N-1] x[n-N]

Provide next xevery CLK

shift registerremembers lastN values

6.111 Fall 2009 8Lecture 10

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Lab 4 overview

ac97

ac97-

commands

audio

8

8

64K x 8 BRAM

ENTER button(push to record)

lab4.v

Assignment: build a voice recorder that recordsand plays back 8-bit PCM data @ 6KHz

About 11 seconds of speech @ 6KHz6.111 Fall 2009 13Lecture 10

BRAM Operation

Source: Xilinx App Note 463

BRAM

Single-port

Config.CLK

WE

Data_in Data_out

6.111 Fall 2009 14Lecture 7

AC97: PCM data

Slot 0 (16) Slot 1 (20) Slot 20 (20)

AC97_SYNCH

Slot 2 (20)

256 bits @ 12.288Mhz = 48kHz frame rate

Slot 3 (20)

Frame info commands LData

readyselects a particular clock_27mhzclock edge whenyou should store input data from the AC97(from_ac97_data) and provide new output to the AC97(to_ac97_data).

PCM = pulse code modulation

Sample waveform at 48kHz,encode results as an N-bit signed

number. For our AC97 chip, N =18.

Slot 4 (20)

RData

FPGA sends output frame to AC97 while AC97 sends input frame to FPGA

6.111 Fall 2009 15Lecture 10

Lab 4 w/ FIR filter

Since were down-sampling by a factor of 8, to avoid aliasing(makes the recording sound scratchy) we need to pass theincoming samples through a low-pass antialiasing filter to removeaudio signal above 3kHz (Nyquist frequency of a 6kHz samplerate).

30-tap low-passFIR filter

30-tap low-passFIR filter

Down-sampleby 8

Down-sampleby 8

48kHz samples 6kHz samples

6.111 Fall 2009 16Lecture 10

Up-sampleby 8

Up-sampleby 8

30-tap low-passFIR filter

30-tap low-passFIR filter

6kHz samples 48kHz samples

We need a low-pass reconstruction filter (the same filter as forantialiasing!) when playing back the 6kHz samples. Actually wellrun it at 48kHz and achieve a 6kHz playback rate by feeding it asample, 7 zeros, the next sample, 7 more zeros, etc.

,Xi,0,0,0,0,0,0,0,Xi+1,0,0,0,0,0,0,0,Xi+2,

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Discrete ValuesIf we use N bits to encode the magnitude of one of thediscrete-time samples, we can capture 2N possible values.

So well divide up the range of possible sample values into 2N

intervals and choose the index of the enclosing interval as theencoding for the sample value.

sample voltage

quantized value 11-bit

32-bit

63-bit

134-bit

0

00

0123456789101112131415

1

2

3

4

5

6

7

1

2

3

1

VMAX

VMIN

6.111 Fall 2009 17Lecture 10

Quantization Error

53

54

55

56

57

Note that when we quantize the scaled sample values we may be offby up to step from the true sampled values.

54 55 56 55 55

The red shaded region showsthe error weve introduced

6.111 Fall 2009 18Lecture 10

Quantization Noise

samplescale &

quantize+

-

+)(tx

QuantizationNoise

N

scale

Time Domain Freq. Domain

Max signal

Noise

N2

1

)( jNOISE

2

s

s

2

s

In most cases its white noise with auniform frequency distribution

6.111 Fall 2009 19Lecture 10

SNR: Signal-to-Noise Ratio

SNR = 10log10P

SIGNAL

PNOISE

= 10log10

ASIGNAL

2

ANOISE2

= 20log10

ASIGNAL

ANOISE

SNR is measured in decibels (dB). Note that its a logarithmicscale: if SNR increases by 3dB the ratio has increased by a factor2. When applied to audible sounds: the ratio of normal speechlevels to the faintest audible sound is 60-70 dB.

Max signal

Noise

N2

1

SNR = 20log10Asignal

Anoise

20log10(2

N)

N 6.02dB

RMS amplitude

6.111 Fall 2009 20Lecture 10

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Oversampling

Oversampling+LPF reduces noise by 3dB/octave

To avoid aliasing we know that s must be at least 2M. Is thereany advantage to oversampling, i.e., s = K2M?

Suppose we look at the

frequency spectrum ofquantized samples of a sinewave: (sample freq. = s) 2/s

1

Lets double the samplefrequency to 2s.

1

2/

)2/(2 s

Total signal+noise power remains thesame, so SNR is unchanged. But noiseis spread over twice the freq. range soits relative level has dropped.

Now lets use a low pass filterto eliminate half the noise!Note that were not affectingthe signal at all

2/s

=

NOISE

SIGNAL

P

PSNR s 10log10

dBSNRP

PSNR

ss

NOISE

SIGNAL 32/

log10102 +=

=

1

2/

6.111 Fall 2009 21Lecture 10

Our Analog Building Block: OpAmp

+

-

i+ ~ 0

i- ~ 0 +-

+

-

vid

vout

vout

vid

VCC = 10V

-VCC = -10V

e = 100mV

-100mV

Reasonableapproximation

+

-

vid +- avid+

-

vout

Linear Mode

If -VCC < vout < VCC

+

-

vid -VCC+

-

vout

Negative Saturation

vid< - e

-++

-

vid+

-

vout

Positive Saturation

vid> e

-+ +VCC

-VCC

VCC

Very small input range for open loop configuration6.111 Fall 2009 22Lecture 10

The Power of (Negative) Feedback

invoutv

1R2R

-+ -

+vid +- avid

+

-voutinv

R2

-+

R1

021

=+

++

R

vv

R

vvidoutidin

a

vv outid =

++=

2211

11

RR

a

Ra

v

R

v outin

( )( )1

1 1

2

21

2 >>++

= aifR

R

RRa

aR

v

v

in

out

Overall (closed loop) gain does not depend on open loop gain Trade gain for robustness Easier analysis approach: virtual short circuit approach

v+ = v- = 0 if OpAmp is linear

+-

6.111 Fall 2009 23Lecture 10

Basic OpAmp CircuitsVoltage Follower (buffer)

invoutv

2R

1R

inR

RR

out vv 121+

+

Non-inverting

invoutv

1inv

outv2inv

1R

1R

2R

2R

( )121

2

ininR

R

outvvv

Differential Input

inout vv

inv

outv

R C

t

inRCoutdtvv 1

Integrator

+

-

6.111 Fall 2009 24Lecture 10

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OpAmp as a Comparator

Analog Comparator:Is V+ > V- ? The Output is a DIGITAL signal

LM311 uses asingle supplyvoltage

6.111 Fall 2009 25Lecture 10

Digital to Analog

Common metrics: Conversion rate DC to ~500 MHz (video) # bits up to ~24

Voltage reference source (internal / external; stability) Output drive (unipolar / bipolar / current) & settling time Interface parallel / serial Power dissipation

Common applications: Real world control (motors, lights)

Video signal generation

Audio / RF direct digital synthesis Telecommunications (light modulation) Scientific & Medical (ultrasound, )

6.111 Fall 2009 26Lecture 10

DAC: digital to analog converter

R

2R

4R

8R

RF

VOUT

B3

B2

B1

B0

Vi = 0 volts if Bi = 0Vi = V volts if Bi = 1

OPAMP will vary VOUTto maintainthis node at 0V, i.e., the sum ofthe currents flowing into thisnode will be zero.

0

842

0123 =++++

R

VB

R

VB

R

VB

R

VB

R

V

F

OUT

+++=

842

0123

BBBBV

R

RV FOUT

How can we convert a N-bit binary number to a voltage?

OKAY, thisll work, but thevoltages produced by thedrivers and various Rsmust be carefully matchedin order to get equal steps.

6.111 Fall 2009 27Lecture 10

R-2R Ladder achieves large current division ratios

with only two resistor values

VOUT = 2R

RV

B2

2+

B1

4+

B0

8

= V B2 +

B1

2+

B0

4

6.111 Fall 2009 28Lecture 10

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Non-idealities in Data Conversion

Binary code

Analog

Ideal

Offseterror

Offset a constant voltage offset that appears atthe output when the digital input is 0

Binary code

Ana

log

Ideal

Gain

error

Gain error deviation of slope from ideal value of 1

Binary code

Ana

log

Ideal

Integralnonlinearity

Integral Nonlinearity maximum deviation fromthe ideal analog output voltage

Differential nonlinearity the largest increment inanalog output for a 1-bit change

Binary code

Ana

log Ideal

Non-monoticity

6.111 Fall 2009 29Lecture 10

Labkit: ADV7125 Triple Out Video DAC

Three 8-bit DACs

Single Supply Op.: 3.3 to 5V

Internal bandgap voltage ref

Output: 2-26 mA 330 MSPS (million samples per

second)

Simple edge-triggered register-based interface

6.111 Fall 2009 30Lecture 10

Glitching and Thermometer D/A

Glitching is caused when switchingtimes in a D/A are notsynchronized

Example: Output changes from 011to 100 MSB switch is delayed

100011ou tv

t

0T

I

R

outv

( )210 TTTIRvout ++=

I I

1T 2T

Filtering reduces glitch butincreases the D/A settling time

One solution is a thermometercode D/A requires 2N 1switches but no ratioed currents

ThermometerBinary

11001

11111

10010

00000

ThermometerBinary

11001

11111

10010

00000

6.111 Fall 2009 31Lecture 10

Successive-Approximation A/D

Example: 3-bit A/D conversion, 2 LSB < Vin< 3 LSB

D/A converters are typically compact and easier to design. Why not A/Dconvert using a D/A converter and a comparator? DAC generates analog voltage which is compared to the input voltage If DAC voltage > input voltage then set that bit; otherwise, reset that bit This type of ADC takes a fixed amount of time proportional to the bit length

Vin code

D/A

Comparator

out

C+

6.111 Fall 2009 32Lecture 10

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Successive-Approximation A/D

Serial conversion takes a time equal to N(tD/A+ tcomp)

Successive

ApproximationGenerator

Control

Done

Go

-

+Sample/Hold

D/AConverter

vin

N

Data

6.111 Fall 2009 33Lecture 10

Flash A/D Converter

Brute-force A/D conversion Simultaneously compare the analog

value with every possible referencevalue

Fastest method of A/D conversion

Size scales exponentially withprecision(requires 2N comparators)

+

+

+

R

R

refV inv

0b

1b

Thermometertobinary

ComparatorsR

R

6.111 Fall 2009 Lecture 10 34

integrator

1-bit DAC

+ +- Bit stream

-+

DecimatorBit stream samples

REFV

REFINREF VVV

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R

C

Vref Controller(FPGA)

Vin

RVin

Controller(FPGA)

C

6.111 Fall 2009 37Lecture 10

AVDD Positive AnalogSupply Voltage

AVSS Analog Ground

DVDD Positive Digital

Supply Voltage

DVSS Digital Ground

dt

dv

c ci =

386.111 Fall 2009

Noise caused by currentspikes in fast switchingdigital circuits:

Digital/Analog Grounds

analog

circuit

digital

logic

AVdd DVdd

LM 4550

+

Vin

-

digital ground

DVss

analog ground

AVss

n signals

Connect the

grounds at a single

place396.111 Fall 2009

digital

logicanalogcircuit

Sensors

Many sensors havenative analogoutputs:

thermocouples,accelerometers,pressure gauge, ..

3-axisaccelerometernow used in cellphones, games,iPods, laptops,6.111 projects

406.111 Fall 2009

3 Axis 5G accelerometer

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