Week 5 – Nyquist-Shannon ESE 250 – S’12 Kod & DeHon 1 ESE250: Digital Audio Basics Week 5 Feb....

Week 5 – Nyquist-ShannonESE 250 – S’12 Kod & DeHon 1

ESE250:Digital Audio Basics

Week 5 Feb. 9, 2012

Nyquist-Shannon Theorem

ESE 250 – S’12 Kod & DeHon

2

Course Map

Numbers correspond to course weeks

2,5 6

11

13

12

Today

Week 5 – Nyquist-Shannon

Where are we ?• Week 2 Received signal is sampled &

quantized q = PCM[ r ]

• Week 3 Quantized Signal is Coded c =code[ q ]

• Week 4 Sampled signal first

transformed into frequency domain

Q = DFT[ q ]• Week 5

signal oversampled & low pass filtered

Q = LPF[ DFT(q+n) ]• Week 6

Transformed signal analyzed Using human psychoaoustic

models• Week 7

Acoustically Interesting signal is “perceptually coded”

C = MP3[ Q]

OverSample

DFT LPF

DecodeProduce

r(t)

p(t)

q + n

CPerceptual

CodingStore /

Transmit

Q + N Q

Week 4

Week 6

Week 5 Week 3

[Painter & Spanias. Proc.IEEE, 88(4):451–512, 2000]



Reconstruction• Acquisition Side

we’ve convinced ourselves to sample, then process, then store and transmit discretely sampled values

• On the other end? how should the stored sound be produced? ? send strings of sampled spl levels

o off to the sound cardo at what rate?

? Interpolateo with what family of interpolants?

• Week 4: harmonic reconstruction Gets better with more terms But need more samples to compute them and error seems unpredictable

• This week: general reconstruction introduce further assumptions about signal

o to guarantee exact finite reconstructiono with appropriate basis functions

introduce another processing step to achieve those assumptions

Generic Digital Signal Processor

Sampler(t)q

Codec

Store/Transmit Decode Produce p(t)

Week 5 – Nyquist-Shannon 4


Shannon’s TheoremM. Unser. Proc. IEEE, 88(4):569–587, 2000

Questions:• are the real

functions countable after all?

• what is “frequency content”?

1 5 1 0 5 5 1 0 1 5

0 . 5

0 . 5

1 . 0

b0(t)

• Hypothesis given a (sufficiently “nice”) real function, r(t) whose frequency content does not exceed

M = 2 fM

• Conclusion There is a set of “sampling” basis functions (we’ll not discuss)

BS = { …, b-2(t), b-1(t), b0(t), b1(t), b2(t), … } which can exactly reconstruct the function

r(t) = … + r-2 b-2(t) + r-1 b-1(t) + r0 b0(t) + r1 b1(t) + r2 b2(t) + … from samples

rk = r(k TM) taken at sampling intervals

TM = /M = 1 / (2 fM ) called the “Nyquist” rate


Trial Application to Our Setting

• Psychoacoustic measurements introduced in next (week 6) lecture show human audition is bandlimited at frequency A = 2 fA for fA = 22 kHz

• Naïve Processing Strategy sample received signal, r(t) at intervals TA = /A = 1 /2 fA = [ 44 ¢ 103 ]-1 sec reconstruct exactly as needed

r(t) = … + r-2 b-2(t) + r-1 b-1(t) + r0 b0(t) + r1 b1(t) + r2 b2(t) + … from samples r = {… , r-2 , r-1 , r0 , r1 , r2 , … } where rk = r(k TA)

• Why bother with DFT at all??Week 5 – Nyquist-Shannon 6


Poking the Trial Balloon• In reality we receive a “mixed” signal

r(t) = q(t) + n(t) q(t)

o signal component of auditory interesto has bandwidth fA = 22 kHz

n(t)o noise (uninteresting information)o has (typically) high frequency content

impurities associated with transmission and recording background sounds

• We only get to sample the received signal, r(t), not the desired signal, q(t) !!



• In class experiment: assume a nominal band limit at frequency N = 2 fN for fN = 5/4 Hz [samples/sec] Question: what is the Nyquist sample rate?

o TN = 1/(2 fN) = 2/5 sec [sample/sec]-1

• In this experiment, assume that q(t) ´ 0

o signal component of auditory interesto satisfies bandwidth requirement of fN = 5/4 Hz

n(t) = Cos[9/5 N t]o noise (uninteresting information)o has “high” frequency contento i.e., it is above (by 9/5) the presumed Nyquist rate

• Class will sample r(t) = q(t) + n(t) over “window” -T0/2 < t < T0/2 for T0 = 4 [sec] at the presumed Nyquist rate Questions:

o what is the number of Nyquist samples? nS ¢ TN = T0 Number of samples: nS = 2 fN T0 = 10

o what are the Nyquist sample times? t 2 {-2,-(8/5),-(6/5),-(4/5),-(2/5),0,2/5,4/5,6/5,8/5,2}

2 1 1 2

1 .0

0 .5

0 .5

1 .0

q(t)

2 1 1 2

1 .0

0 .5

0 .5

1 .0n(t)

2 1 1 2

1 .0

0 .5

0 .5

1 .0

r(t) = q(t) + n(t)

Bursting the Trial Balloon



2 1 1 2

1 .0

0 .5

0 .5

1 .0

1 .5

2 .0

Bursting the Trial Balloon• Class will sample r(t) = q(t) + n(t)

over “window”

-T0/2 < t < T0/2 for T0 = 4 at the presumed Nyquist rate

o Number of samples:

nS = 2 fN T0 = 10o Sample instants:

t 2 {-2,-(8/5),-(6/5),-(4/5),-(2/5),0,2/5,4/5,6/5,8/5,2}

• and get an “aliased” version of the noise! r(t) looks like a sampled version of n(t) = Cos[1/5 N t]

(D ebug) Out[730]=

2. 1.6 1.2 0.8 0.4 0 0.4 0.8 1.2 1.6 2. 1. 0.81 0.31 0.31 0.81 1. 0.81 0.31 0.31 0.81 1.

2 1 1 2

1 .0

0 .5

0 .5

1 .0

1 .5

2 .0



Aliasing• Widely familiar phenomenon

Wikipedia aliasing article “wagon wheel” effect Wikipedia Nyquist-Shannon Theorem

• “Folding:” another general manifestation of aliasing premise: r(t) = q(t) + n(t)

o we expect q(t) with bandwidth less than N = 2 fN o but turns out that n(t) is some “ superharmonic residue”

n(t) = cos[ mN(1+ ) t ] where 0 < <1

outcome: sample at “Nyquist rate” TN = 1/(2 fN)o rk = r( k TN) = q(k TN) + n(k TN)o where the noise

n( k TN) = cos[mN(1+ ) k TN ] = cos[mk(1+ ) ] = cos[ mk] ¢ cos[mk ] - sin[ mk] ¢ sin[mk ] = 1 ¢ cos[mk ] - 0 ¢ sin[mk ] = cos[ N k TN]

o “folds over” to act as if it were a low-band tone of frequency fN


http://en.wikipedia.org/wiki/Aliasing

http://scratch.mit.edu/projects/Giantfishy/408423

http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem


Anti-Aliasing• Given the “mixed” signal

r(t) = q(t) + n(t) q(t) – auditory signal with bandwidth fA = 22 kHz n(t) – high frequency noise

• We require some “anti-aliasing” pre-process that “smooths away” the noise which will otherwise appear in the samples

o r = {… , r-2 , r-1 , r0 , r1 , r2 , … }o where rk = r(k TA)

o and TA = /A = 1 /2 fA = [ 44 ¢ 103 ]-1 sec

• Question: how to do this?Week 5 – Nyquist-Shannon 11


Interlude: Visual Aliasing • Wolfram: Drawing a Line on Digital Display• Berkeley Course Demo• Typography

http://ptolemy.eecs.berkeley.edu/eecs20/week13/moire.html

http://ptolemy.eecs.berkeley.edu/eecs20/week13/fonts.html


Back to Shannon (new today)• Hypothesis

given a (sufficiently “nice”) real function, r(t) whose frequency content does not exceed

M = 2 fM

• Conclusion There is a set of “sampling” basis functions (we’ll not show)

BS = { …, b-2(t), b-1(t), b0(t), b1(t), b2(t), … } which can exactly reconstruct the function

r(t) = … + r-2 b-2(t) + r-1 b-1(t) + r0 b0(t) + r1 b1(t) + r2 b2(t) + … from samples

rk = r(k TM) taken at sampling intervals

TM = /M = 1 /(2 fM ) called the “Nyquist” rate

M. Unser. Proc. IEEE, 88(4):569–587, 2000

Questions:• are the real functions

countable after all?• what is “frequency content”? • what are the “basis

functions” ?Week 5 – Nyquist-Shannon 13


Fourier’s Theorem (review)• Hypothesis

given a (sufficiently “nice”) real function, r(t) which is periodic with period T0

r(t) = r(t + T0)

• Conclusion the set of “harmonic” basis functions

BH = { …, h-2(t), h-1(t), h0(t), h1(t), h2(t), … }

at frequency 0 = 2 f0 for f0 = 1/T0

defined by hk(t) = cos k 0 t (k>0)

hk(t) = 1 (k>0)

hk(t) = sin k 0 t (k<0) can exactly reconstruct the function

r(t) = … + R-2 h-2(t) + R-1 h-1(t) + R0 h0(t) + R1 h1(t) + R2 h2(t) + …

M. Unser. Proc. IEEE, 88(4):569–587, 2000

Question:• are the real functions

countable after all?


Clipping a Non-Periodic Function

• If we are willing to restrict attention to a specific

observation window

-T0/2 · t · T0/2• Then

Fourier’s Theorem yields

exact reconstruction of the periodic

extension of any function

Answer:• apparently, “periodic” (or, more practically, finite time)• real functions • are countable after all!

- T0/2 T0/2 T0-T0

- T0/2 T0/2 T0-T0



Fourier + Shannon• Fourier: period T0 ) exact

harmonic reconstruction at frequency 0 = 2 f0 for f0 = 1/T0

r(t) = … + R-2 h-2(t) + R-1 h-1(t) + R0 h0(t) + R1 h1(t) + R2 h2(t) + …

• Shannon: bandwidth M ) exact “sampled” reconstruction at frequency M = 2 fM for fM = 1/(2TM

) r(t) = … + r-2 b-2(t) + r-1 b-1(t) + r-0 b0(t) +

r1 b1(t) + r2 b2(t) + …

clipped frequencyforces finite harmonic series

clipped timeforces finite samples



• Fourier + Shannon: period T0 & bandwidth M = 2 fM ) finite exact harmonic reconstruction with nM = Round[T0 / TM] samples at “Maximal Rate” TM = 1 /(2 fM )

• Algebra: k 0 > M

) R-k sin k 0 t ´ Rk cos k 0 t ´ 0 (otherwise high frequency)

) R-k = Rk = 0 (sin & cos never identically zero)

k > nM ) r (t) = R-nM

h-nM (t) + … + R-2 h-2(t) + R-1 h-1(t) + R0 h0(t) + R1 h1(t) + R2 h2(t) + … + RnM hnM (t)

• Now we’re ready to solve the anti-aliasing problem

Fourier + Shannon ) Finite Series



• Now postulate: New signal, q(t), satisfying

o period T0o psychoacoustic bandwidth A = 2 fA for fA = 22 kHzo bounded by maximal bandwidth A < M

sampled with o nA = Round[T0 / TA] sampleso at “Auditory Nyquist Rate” TA = 1 /(2 fA )

• Fourier + Shannon: period T0 & bandwidth A

) finite exact harmonic reconstructionq(t) = Q-nA

h-nA (t) + … + Q-2 h-2(t) + Q-1 h-1(t) + Q0 h0(t) + Q1 h1(t) + Q2 h2(t) + … + QnA hnA (t)

vector representation:o Q = (Q-nA

, … , Q-2 , Q-1 , Q0 , Q1 , Q2 , … , QnA)

• Shannon + Fourier bandwidth A & period T0

) finite exact “sampled” reconstructiono q(t) = q-nA

b-nA (t) + … + q-2 b-2(t) + q-1 b-1(t) + q-0 b0(t) + q1 b1(t) + q2 b2(t) + … + qnA bnA (t)

vector representation:o q = (q-nA

, … , q-2 , q-1 , q0 , q1 , q2 , … , qnA)

Question: • what is the relationship • between the

representations frequency domain, Q time domain, q

Answer: Q = DFT(q)

Psychoacoustics meets Nyquist



Q+N = DFT(q) + DFT(n) = DFT(q+n)• Assume we receive mixed signal

r(t) = q(t) + n(t) q(t)

o signal component of interesto has period T0 & bandwidth A

n(t)o noise (uninteresting information)o has (for now) only high frequency content, , i.e., satisfies: A < · M

• Problem: how to remove noise?• Frequency Domain approach

Take the received time-sampled representation, r = q + n r = (r-nM

, … , r-2 , r-1 , r0 , r1 , r2 , … , rnM)

= (q-nM + n-nM

, … , q-2 + n-2, q-1 + n-1 , q0 + n0, q1 + n1, q2 + n2, … , qnM + n-nM

) Compute frequency domain representation

o R = DFT(r) = DFT(q+n) = DFT(q)+ DFT(n) = Q + No R = (R-nM

, … , R-2 , R-1 , R0 , R1 , R2 , … , RnM)

= (Q-nM + N-nM

, … , Q-2 + N-2, Q-1 + N-1 , Q0 + N0, Q1 + N1, Q2 + N2, … , QnM + N-nM

) Recall the meaning of this representation: for hk(t) = trig[k 0 t] and trig 2 {cos, sin} we have

r(t) = (Q-nM + N-nM

) h-nM (t) + … + (Q-nA + N-nA

) h-nA (t) + … + (Q-nM

+ N-nM ) h-2(t) + (Q-nM

+ N-nM ) h-1(t) + (Q-nM

+ N-nM ) h0(t) + (Q-nM

+ N-nM ) h1(t) + (Q-nM

+ N-nM ) h2(t) + …

+ (Q-nA + N-nA

) hnA (t) + … + (Q-nM + N-nM

) hnM (t) Now apply assumptions about frequency content!


R = DFT(r) , r = DFT-1(R) • Compute frequency domain representation

R = (R-nM , … , R-2 , R-1 , R0 , R1 , R2 , … , RnM

)

= (Q-nM + N-nM


)

• Now introduce assumptions about frequency content: k > nA = A / 0 ) Qk = 0

k < nA = A / 0 ) Nk = 0

• Interpret in entries of R

= (Q-nM + N-nM


)

= (0 + N-nM, … , QnA

+ N-nA, … , Q-2 + 0, Q-1 + 0 , Q0 + 0, Q1 + 0, Q2 + 0, … ,

QnA + N-nM

, … , 0 + N-nM

)

= (0,…, 0, QnA … , Q-2 , Q-1 , Q0 , Q1 , Q2, …, QnA,,

, 0,…, 0 )

+ (N-nM,… , N-nA

, 0 , … , 0 , 0 , 0 , 0 , 0 , … , 0 , NnA

,…, NnM )

• Implement Algorithmically Zero out the “higher integer” entries of S = ZEROnA (R)

Obtain the inverse transform, s = DFT-1(S)

• Conclude that s = qWeek 5 – Nyquist-Shannon 20ESE 250 – S’12 Kod & DeHon

Visualizing a Low Frequency Subspace• Low Frequency Signal

Corrupted by High frequency noise


3 2 1 1 2 3

2

1

1

2

t 6 .28t 0 .69

t 4 .9

h0(t)

h1(t)

h-1(t)


Oversampled Pre-Filter• Anti-aliasing

requires“oversampling”o nM = Round[T0 / TM] samples o at “Maximal Rate” TM = 1 /(2 fM )o Where fM represents our best

estimate ofo the highest noise frequency

but how should we modelo “noise”o and its likely bandwidth?

• Analog pre-processing physical electro-mechanical

world o traditionally piezo-electric

microphoneo increasingly MEMS

Has its own limited bandwidth• Electronics need merely

sample up to the microphone “cut off”

frequency

Low-Pass Filter

Filter Transformr Q q

ADXL001 Accelerometer

Pre-Samplingr Low-Pass


http://www.analog.com/library/analogdialogue/archives/43-02/mems_microphones.html


ESE250:Digital Audio Basics

Week 5 Feb. 9, 2012

Nyquist-Shannon Theorem

Week 5 – Nyquist-Shannon ESE 250 – S’12 Kod & DeHon 1 ESE250: Digital Audio Basics Week 5 Feb....

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