Slide Set Data Converters ————————— Background Elementsims.unipv.it/Courses/download/ACD/CAD_SlidesI.pdf · Slide Set Data Converters —————————

F. MalobertiDATA CONVERTERS Springer2007

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BACKGROUND ELEMENTSF. MalobertiDATA CONVERTERS Springer2007

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BACKGROUND ELEMENTS0

Slide Set

Data Converters

—————————

Background Elements


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Summary• Introduction

• The Ideal Data Converter

• Sampling

• Amplitude Quantization

• Quantization Noise

• kT/C Noise

• Discrete and Fast Fourier Transforms

• The D/A Converter

• The z-Transform


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The Ideal Data Converter

Transformation from continuous-amplitude, continuous-time into discrete-amplitude discrete-time and vice-versa.

(a)

(b)


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SamplingA sampler transforms a continuous-time signal into its sampled-data equiv-alent.

x∗(t) = x∗(nT ) =∑

x(t)δ(t− nT ) (1)

x(t)

t t

x*(t)

T 2T 3T 6T .....

4T 5T

Only the values at the sampling instant matter (independently on the realrepresentation).


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Sampling as a non-linear ProcessThe Laplace transform is:

X X*

T

X X*

S d(t-nT)-∞

∞

L

∞∑−∞

δ(t− nT )

=∞∑−∞

e−nsT ; (2)

L[x∗(nT )

]=∞∑−∞

(X(s− jnωs) =∞∑−∞

x(nT )e−nsT (3)


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Spectral Implicationsf1f2

f1f2

f1f2

-1 1 2 3-2

1 2 3 4 5-1-2-3

fs

fs

-fs

-fs

f

f

f

(a)

(b)

(c)

fB

is it possible to return back to the continuous-time?


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Nyquist Theorem

A band limited signal, x(t), whose Fourier spectrum, X(jω), vanishes forangular frequencies |ω| > ωs/2 is fully described by a uniform samplingx(nT ), where T = 2π/ωs. The band limited signal x(t) is reconstructedby

x(t) =∞∑−∞

x(nT )sin(ωs(t− nT )/2)

ωs(t− nT )/2(4)

Half the sampling frequency, fs/2 = 1/2T , is often named the Nyquist frequency. The

frequency interval 0 · · · fs/2 is referred to as the Nyquist band (or band-base) while fre-

quency intervals, fs/2 · · · fs, fs · · ·3fs/2, · · · are named the second and third Nyquist

zones, and so forth.


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Something that you need to ...

Remember!

The anti-aliasing filter pro-tects the information contentof the signal. Use an anti-aliasing filter in front of everyquantizer to reject unwantedinterferences outside of theband of the interest!


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Antialiasing Filter

B fS/2 fS fS

Rejectfrom here

Passuntil here

Stop-band Attenuation

Pass-Band Transition-Band Stop-Band

Ripple in the Pass-Band

(a)

(b)

H(s)

[dB]

ASB

f B-f

The complexity of the antialiasing filter depends on the band-pass ripplethe transition region, and the attenuation in the stop-band


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Undersampling

SignalSignal

SignalSignal ImageImage

Signal SignalImage Image Image Image Image

Image ImageImage

SignalSignal

(a)

(b)

(c)

(d)

fS

fS

fL fH

fL fH


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Aliasing is an important issue ...

Remember That

Under-sampling requires ananti-aliasing filter! This re-moves unwanted spurs, whichcan occur in the base-bandor be aliased back from anyother Nyquist zones. Theanti-aliasing filter for under-sampled systems is band-passaround the signal band.


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Sampling-time Jitter

The sampling time depends on the used clock generator that can be af-fected by jiitter.

x(nT)

T

DX(0)

d(0)

DX(T)

d(T)

DX(2T)≈0

2T 3T

d(2T) d(3T)

DX(3T)

t0

The error is large with a large signal slopes.


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Error caused by Jitter

For a sine wave Xin(t) = A · sin(ωint), the error ∆X(nT ) is given by

∆X(nT ) = A · ωin · δ(nT ) cos(ωinnT ) (5)

Assume that δ(nT ) is the sampling of a random variable δji(t)

< xji(t)2 >=< [Aωin cos(ωinnT )]2 >< δij(t)

2 > (6)

=A2ωin

2

2< δij(t)

2 > (7)

The SNR caused by jitter becomes

SNRji,DB = −20 · log< δij(t) > ωin (8)


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Jitter Requirements

106 107 10810−14

10−13

10−12

10−11

10−10

Input frequency [Hz]

r [s]

ettij kcolC

-66 dB

-78 dB

-90 dB

-102 dB


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Assume that the noise is a fixed part plus the one coming from jitter

v2n = 0.4 · 10−8 + 0.1 · 10−8

[f

20 · 106

]2(9)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 172

73

74

75

76

77

78

79

80

81

Normalized input frequency, f/fCK

SN

R

[d

B]


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Amplitude Quantization

Amplitude quantization changes a sampled-data signal from continuous-level to discrete-level. The amplitude of each quantization interval or quan-tization step, ∆, is

∆ =XFSM

(10)

An input level other thanXm,n the mid point of the n-th interval leads to thequantization error, εQ leading to a quantization output Y corresponding toan Xin input

Y = Xin + εQ = (n+ 1/2)∆; n∆ < Xin < (n+ 1)∆ (11)


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Quantization Error

The quantization error is an additive signal with amplitude limited by thequantization step amplitude

∑X(nT) Y(nT)

eQ(nT)

D/2

D/2

D

Xmin

Xmax

(a)

(b)

eQ

Xin


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Before going ahead it worths to ...

Remark

The quantization error is afundamental limit of the quan-tization process: εQ cannotbe avoided: it becomes zeroonly when the number of bitsgoes to infinity, which is un-feasible in practice.


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Quantization Noise

Remind the definition of the SNR

SNR|dB = 10 · logPsign

Pnoise(12)

Psign and Pnoise are the power of signal and noise in the band of interest.

It would be convenient to assimilate the quantization error to noise ...

That’s possible but only under certain conditions.


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Quantization Noise: Conditions

? all the quantization levels are exercised with equal probability;? a large number of quantization levels are used;? the quantization steps are uniform;? the quantization error is not correlated with the input.

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1x 10

−3

0 50 100 150 200 250 300−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1


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Quantization Noise: Properties

Time average power

p(εQ) =1

∆for εQ ∈ −∆/2 · · ·∆/2

p(εQ) = 0 otherwise (13)

The time average power of εQ is given by

PQ =∫ ∞−∞

ε2Q · p(εQ)dεQ =∫ ∆/2

−∆/2

ε2Q

∆dεQ =

∆2

12(14)


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Quantization Noise: Calculation of the SNR

The knowledge of the Noise power enables calculating the SNR.

The power of a sine wave with maximum amplitude is

Psin =1

T

∫ T0

X2FS

4sin2(2πft)dt =

X2FS

8=

(∆ · 2n)2

8(15)

The power of a triangular wave with maximum amplitude is

Psin =X2FS

12=

(∆ · 2n)2

12(16)

leading to

SNRsine|dB = (6.02 · n+ 1.78) dB (17)

SNRtrian|dB = (6.02 · n) dB (18)


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Equivalent number of bits (ENB or ENoB)

The effective SNR is the real measure of the data converter resolution.Measured in bits it is the ENB

ENBsin =SNRtot|dB − 1.78

6.02(19)

ENBtrinag =SNRtot|dB

6.02(20)

If, for example there is quantization an jitter δji

ENB =10 · log(π2f2

inδ2ji + 2−2N/12)− 1.78

6.02(21)


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Equivalent number of bits with jitter

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 18

9

10

11

12

13

14

15

16

Sampling uncertanty [psec]

Equi

vale

nt n

umbe

r of b

it

f= 40 MHz

f= 80 MHz

f= 120 MHz

f= 160 MHz

f= 200 MHz


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Quantization noise: Properties

Noise power spectrum

The power spectrum is the Laplace transform of the auto-correlation func-tion. For sampled data signals

Pε(f) =∫ ∞−∞

Rε(τ)e−j2πfτdτ =∞∑−∞

Rε(nT )e−j2πfnT (22)

Assume that the auto correlation function, Re(nT ), goes rapidly to zero for|n| > 0 or, for simplicity, use Re(0) only.

The auto correlation becomes a delta in the time domain and the Laplacetransform becomes frequency independent.

The power spectral density is white with power PQ = ∆2/12 spread uni-formly over the unilateral Nyquist interval 0 · · · fs/2.


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Quantization noise: PropertiesNoise power spectrum (ii)

The bilateral power spectrum is

pε(f) =∆2

12 · fs; meeting the condition

∫ ∞−∞

pε(f)df = ∆2/12

(23)

fs/2-fs/2 fs/2-fs/2

S

fs/2-fs/2

Signal

QuantizationNoise

Quantized Signal

area

D2

/12


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kT/C NoiseAn unavoidable limit is the kT/C noise due to the thermal noise associatedwith the sampling switch.

+CsVin

SW

Rs

Csvn=4kTRs

Vout

vn,C

2 2

(a) (b)

v2n,Cs(ω) =

4kTRs1 + (ωRsCs)2

(24)

Pn,Cs =∫ ∞

0vn,out(f)df = 4kTRs

∫ ∞0

df

1 + (2πfRsCs)2=kT

Cs(25)


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kT/C Noise (ii)

10−1

100

101

102

10−6

10−5

10−4

10−3

Sampling capacitance [pF]

kT

/C n

ois

e v

olta

ge

10-bit

14-bit

11-bit

12-bit

13-bit

15-bit

kT/C noise voltage versus the capacitance value and quantization step for 1 VFS.


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In addition to quantization ...

Remember

The kT/C noise is a funda-mental limit caused by sam-pling. Sampling any signalusing 1 pF leads to 64.5 µVnoise voltage. If the samplingcapacitance increases by k

the noise voltage diminishesby√k.


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Example

A pipeline data-converter uses a cascade of two sample-and-hold circuitsin its first stage. The clock jitter is 1 psec. Determine the minimum sam-pling capacitance that enables 12 bit resolution. The full scale voltage is 1V; the input frequency is 5 MHz.

Solution

The quantization noise power is ∆2/12. We assume that an extra 50%noise is acceptable (the system would lose 1.76 dB, 0.29-bit). Thus thenoise budget for KT/C and jitter is

v2n,budget =

V 2FS

24 · 224= 2.48 · 10−9V 2 (26)


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The noise jitter only affects the signal in the first stage: the second stagesamples the held signal from the first stage. The jitter noise is

v2n,ji =

VFS2· 2πfδji2 = 2.47 · 10−10V 2 (27)

Thus, the total noise power that the sampler can generate is 2.23·10−9V 2.Assuming equal capacitance in both S&H circuits the noise for each isvn,C = 1.12 · 10−9V 2, leading to a sampling capacitance

CS =kT

v2n,C

=4.14 · 10−21

1.12 · 10−9= 3.7pF (28)

Observe that the noise jitter establishes a maximum achievable resolutionthat can not be exceeded even with very large sampling capacitances. Thislimit using the same 50% margin used above is 15.3 bit.


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Discrete and Fast Fourier TransformsThe spectrum of a sampled data signal is estimated using

L[x∗(nT )

]=∞∑−∞

x(nT )e−nsT (29)

F[x∗(nT )

]= X∗(jω) =

∞∑−∞

x(nT )e−jωnT (30)

Unfortunately (30) requires an infinite number of samples that, of course,are not available in practical cases.

A convenient approximation is the Discrete Fourier Transform (DFT ).

X(fk) =N−1∑n=0

x(nT )e−j2πkn/(N−1) (31)


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Discrete and Fast Fourier Transforms (ii)

The DFT is a complex function; the real and the imaginary parts are

|X(fk)| =√Real [X(fk)]2 + Im [X(fk)]2 (32)

PhX(fk) = arctan

[ImX(fk)RealX(fk)

](33)

Equation (31) requires N2 computations. For long series the Fast FourierTransform algorithm (FFT ) is more effective.

The FFT reduces the number of computations fromN2 down toN ·log2(N).

Use power of 2 elements in the series


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WindowingThe DFT and the FFT assume that the input is N-periodic.

Real signals are never periodic and the N-periodic assumption lead to discontinuity be-

tween the last and first samples of successive sequences.

0 100 200 300 400 500 600 700 800 900 1000−1.5

−1

−0.5

0

0.5

1

1.5


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Windowing (ii)Windowing tackles the missed N-periodicity by tapering the endings of the series.

xw(kT ) = x(kT ) ·W (k) (34)

10 20 30 40 50 60

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Hamming

Gaussian

Blackman-Harris

Flattop


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It is suggested to follow some ...

Useful Tips

When using FFT or DTFmake sure that the sequenceof samples is N-periodic. Oth-erwise use windowing.With sine wave inputs avoidrepetitive patterns in the se-quence: the ratio betweenthe sine wave period and thesampling period should be aprime number.


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Windowing Example

200 400 600 800 1000−1

−0.5

0

0.5

1Input

200 400 600 800 1000−1

−0.5

0

0.5

1Windowed Input

(a) (b)

Time domain input of the example before (a) and after windowing (b).


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10 20 30 40 50 60 70 80 90 1000

50

100

150

200

250

300

350Input Spectrum

10 20 30 40 50 60 70 80 90 1000

20

40

60

80

100

120

140Windowed Input Spectrum

Spectra with and without windowing.


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Is windowing avoidable?? Windowing is essentially an amplitude modulation of the input that produces unde-

sired effects.? A spike at the beginning or at the end of the sequence is completely masked being

windowed away almost completely? Windowing gives rise to spectral leakage.? The SNR measurement can be accurate but an input sine wave does not give a pure

tone.

With input sine waves use coherent sampling for which an integernumber of clock cycles,k, fits into the sampling window. In addition,k must be a prime number.

fin =k

2N − 1fs (35)


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What the FFT Represents?The FFT of an N sample sequence is made by N discrete lines equally spaced in thefrequency interval 0− fs.

Each line gives the power falling within fs/N centered around the line itself.

The FFT operates like a spectrum analyzer with N channels whose bandwidth is fs/N .

If the number of points of the series increases then the channelbandwidth of the equivalent spectrum analyzer decreases and eachchannel will contain less noise power.


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Processing Gain

? An M-bit quantization gives V 2FS/12 · 22M noise power.

? The power of the full scale sine wave is V 2FS/8.

? The FFT of the quantization noise is, on average, 3/2 ·22M/N belowthe full scale

x2noise|dB = Psign − 1.78− 6.02 ·M − 10 · log(N/2) (36)

The term 10 · log(N/2) is called the processing gain of the FFT .


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For distinguishing tones from noise ...

Notice

If the length of the input se-ries increases by a factor 2the floor of the FFT noisespectrum diminish by 3 dB.Tones caused by harmonicdistortion do not change.Only long input series revealsmall tones.


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Example of FFT Spectra

0 50 100 150 200 250 300 350 400 450 500−120

−100

−80

−60

−40

−20

0Input Spectrum in DB

SNR=62 dB

Processing gain 33.12 dB

0 50 100 150 200 250 300 350 400 450 500−120

−100

−80

−60

−40

−20

0Input Spectrum in DB

SNR=62 dB

Processing gain 42.14 dB

(a) (b)

4096 points 32768 points


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Coding Schemes

• USB - Unipolar Straight Binary: is the simplest binary scheme. It isused for unipolar signals. The USB represents the first quantizationlevel, −Vref + 1/2VLSB with all zero’s (· · ·0000) . As the digitalcode increases, the analog input increases by one LSB at a time, andwhen the digital code is at the full scale (· · ·1111) the analog input isabove the last quantization level Vref − 1/2VLSB. The quantizationrange is −Vref · · ·+ Vref .

• CSB - Complementary Straight Binary: the opposite of the USB.CSB coding is also used for unipolar systems. The digital code (· · ·0000)

represents the full scale while the code (· · ·1111) corresponds to thefirst quantization level.


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• BOB - Bipolar Offset Binary: is a scheme suitable for bipolar systems(where the quantized inputs can be positive and negative). The mostsignificant bit denotes the sign of the input: 1 for positive signals and 0for negative signals. Therefore, (· · ·0000) represents the full negativescale. The zero crossing occurs at (01 · · ·111) and the digital code(1 · · ·1111) gives the full positive scale.

• COB - Complementary Offset Binary: this coding scheme is comple-mentary to the BOB scheme. All the bits are complemented and themeaning remains the same. Therefore, since (01 · · ·111) denotesthe zero crossing in the BOB scheme the zero crossing of COB isbecomes (10 · · ·000).

• BTC - Binary Two’s Complement: is one of the most used codingschemes. The bit in the MSB position indicates the sign in a comple-mented way: it is 0 for positive inputs and 1 for negative inputs. The


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zero crossing occurs at (· · ·0000)). For positive signals the digitalcode increases normally for an increasing analog input. Thus, thepositive full scale is (0 · · ·1111). For negative signals, the digitalcode is the two’s complement of the positive counterpart. This leads(1 · · ·0000) to represent the negative full scale. The BTC codingsystem is suitable for microprocessor based systems or for the imple-mentation of mathematical algorithms. It is also the standard for digitalaudio.

• CTC - Complementary Two’s Complement: is the complementarycode of BTC. All the bits are complemented and codes have the samemeaning. The negative full scale is (0 · · ·1111); the positive full scaleis (1 · · ·0000).


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D/A Conversion

Ideal Reconstruction

HR,id(f) = 1 for −fs

2< f <

fs

2

HR,id(f) = 0 otherwhise (37)

r(t) =sin(ωst/2)

(ωst/2)(38)


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Real ReconstructionS&H followed by the reconstruction filter.

HS&H(s) =1− e−sT

sτ(39)

HS&H(jω) = jT

τe−jωT/2 sin(ωT/2)

ωT/2(40)

fs 2fs 3fs 4fs 5fs

1

f

Ideal Reconstruction Filter

sin(wT/2)

wT/2

0.636

fs=1/T

fNyq


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Spectrum after the S&H

fS

2fS

3fS

1

f

Signal spectrum

S&H response

Residual images

fB

fS-f

B

Rule of thumb

The reconstruction mustuse an in-band x/sin(x)

compensation if the bandof the signal occupiesabout a quarter of theNyquist interval or more.


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Study of the S&H effect

100 200 300 400 500 600 700 800 900 1000−0.2

0

0.2

0.4

0.6

0.8

1

Input Signal in the Time−domain

0 10 20 30 40 50 60 700

50

100

150

200

250Spectrum of the Input

Input signal. Left: Time domain. Right: Spectrum.


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Study of the S&H effect (ii)

0 10 20 30 40 50 60 700

50

100

150

200

250Spectrum of the Sampled Signal

0 10 20 30 40 50 60 700

5000

10000

15000Spectrum of the Sampled−and−Held Signal

Spectrum of the sampled signal and its sampled-and-held version.


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The z-Transform

Is the time-discrete counterpart of the Laplace transform

Zx(nT ) =∞∑−∞

x(nT )z−n (41)

Za1 ·x1(nT )+a2 ·x2(nT ) = a1 ·Zx1(nT )+a2 ·Zx2(nT ) (42)

Moreover, the Z-transform of a delayed signal is

Zx1(nT − kT ) =∞∑−∞

x(nT − kT )z−(n−k)z−k = X(z)z−k (43)


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Mapping between s and z

z → esT (44)

|z| → eσT ; ω → Ω (45)

wT

s

p

p

p

p

p


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Mapping is good because of the following

Features

A sampled-data system isstable if all the poles of itstransfer function are insidethe unity circle.The frequency response ofthe system is the z-transferfunction calculated on theunity circle.


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Other Mappings

Points of the s-plane mapping the z-plane points: z = −0.5; z = −0.5 +

j√

3/2; z = −0.8j; z = 1.2

σ =1

Tslog |z|; ω =

1

Tsphase(z)± 2nπ (46)

wT

s

p

p

p

-p


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Approximate integrator in the z-domain

HI(s) =1

sτ. (47)

the use of the following discrete-time equations

y(nT + T ) = y(nT ) + x(nT + T )T, (48)

y(nT + T ) = y(nT ) + x(nT )T, (49)

y(nT + T ) = y(nT ) +x(nT + T )x(nT )

2T. (50)


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Approximate integrator in the z-domain (ii)

The use of the Z-transform obtains

zY = Y + zXT, (51)

zY = Y +XT, (52)

zY = Y +XTz + 1

2(53)

giving rise to three different approximate expressions of the integral trans-fer function

HI,F = Tz

z − 1; HI,B = T

1

z − 1; HI,Bil = T

z + 1

2(z − 1), (54)

where the indexes ”F ”, ”B” and ”Bil” indicate forward, backward and bilin-ear.


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Approximate integrator in the z-domain (iii)

The three above expression are equivalent to the continuous-time integra-tion through approximate mappings given by

sT →z − 1

Tz(Forward transformation), (55)

sT →z − 1

T(Backward transformation), (56)

sT →z − 1

2T (z + 1)(Bilinear transformation), (57)

Notice that the above mappings moves the poles differently than the idealmapping s→ ln(z)/T .


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Wrap-up

The background knowledge studied in this first part is essential to properlyunderstand and design data converters.

We have seen how a data converter performs the transformation fromcontinuous-time and continuous-amplitude to discrete-time and quantized-amplitude (and vice-versa).

We have studies that data conversion affects the spectrum of the signaland can sometimes modify its information content. It is therefore importantto know the theoretical implications and to be aware of the limits of theapproximations used for studying a data converter.

We also have studied the mathematical tools used for analysis and char-acterization of sampled-data systems.


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Documents

Slide Set Data Converters ————————— Background Elementsims.unipv.it/Courses/download/ACD/CAD_SlidesI.pdf · Slide Set Data Converters —————————