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Analog Building Blocks
Sampling theorem
Undersampling, antialiasing
FIR digital filters
Quantization noise, oversampling
OpAmps, DACs, ADCs
6.111 Fall 2009 1Lecture 10
Lab #3 report due online @ 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 continuoustime 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 < sm, we can recoverthe original waveform with a lowpass filter!
)( jRp
cc
T MscM
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Sampling TheoremLet x(t) be a bandlimited 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 sM.
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
2 525
)( jP
6 60
6,5 == sM
)( jXp
5 2 2 5
1 4 86 6148
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 lowpass filter on the originalwaveform to remove any frequency content s/2.
2
sC
=
DiscreteTime
sampler
s
We need this antialiasing filter it hasto have a reasonably sharp cutoff
More info: http://www.cs.columbia.edu/~hgs/audio/44.1.html
This is the symbol for alowpass 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 Ntap finite impulse response (FIR) filter:
What components are part of the tPD of this circuit?How does tPD grow as N gets larger?
x[n1]x[n]
x[n2] x[n3] x[nN1] x[nN]
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
Serial linksto/from AC97chip
recorder(your job!)
ready
8
8
64K x 8 BRAM
we 88Addr 16
ENTER button(push to record)
lab4.v
Assignment: build a voice recorder that recordsand plays back 8bit PCM data @ 6KHz
About 11 seconds of speech @ 6KHz6.111 Fall 2009 13Lecture 10
BRAM Operation
Source: Xilinx App Note 463
BRAM
Singleport
Config.CLK
WE
Address
Data_in Data_out
6.111 Fall 2009 14Lecture 7
AC97: PCM data
Slot 0 (16) Slot 1 (20) Slot 20 (20)
AC97_SYNCH
READY
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 Nbit 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 downsampling by a factor of 8, to avoid aliasing(makes the recording sound scratchy) we need to pass theincoming samples through a lowpass antialiasing filter to removeaudio signal above 3kHz (Nyquist frequency of a 6kHz samplerate).
30tap lowpassFIR filter
30tap lowpassFIR filter
Downsampleby 8
Downsampleby 8
48kHz samples 6kHz samples
6.111 Fall 2009 16Lecture 10
Upsampleby 8
Upsampleby 8
30tap lowpassFIR filter
30tap lowpassFIR filter
6kHz samples 48kHz samples
We need a lowpass 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 thediscretetime 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 11bit
32bit
63bit
134bit
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: SignaltoNoise 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 6070 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+
+
Noninverting
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 Nbit 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
R2R Ladder DAC Architecture
R2R 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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Nonidealities 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 1bit change
Binary code
Ana
log Ideal
Nonmonoticity
6.111 Fall 2009 29Lecture 10
Labkit: ADV7125 Triple Out Video DAC
Three 8bit DACs
Single Supply Op.: 3.3 to 5V
Internal bandgap voltage ref
Output: 226 mA 330 MSPS (million samples per
second)
Simple edgetriggered registerbased 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
SuccessiveApproximation A/D
Example: 3bit 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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SuccessiveApproximation 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
Bruteforce 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
Sigma Delta ADC
integrator
1bit DAC
+ + Bit stream
+
Analoginput
1bit ADC
DecimatorBit stream samples
REFV
REFINREF VVV
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Sigma Delta ADC
A simple ADC:
R
C
Vref Controller(FPGA)
Vin
RVin
Controller(FPGA)
C
Poor Mans ADC:
6.111 Fall 2009 37Lecture 10
AD Supply Voltages Consideration
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, ..
3axisaccelerometernow used in cellphones, games,iPods, laptops,6.111 projects
406.111 Fall 2009
3 Axis 5G accelerometer