ESE250 – Week 5 Nyquist-Shannon Theoremese250/week5/day5-S13.pdfESE250 S'13: DeHon, Kadric, Kod,...

ESE250 S'13: DeHon, Kadric, Kod, Wilson-Shah Week 5 – Nyquist-Shannon theorem

ESE250 – Week 5

Nyquist-Shannon Theorem

ESE250 S'13: DeHon, Kadric, Kod, Wilson-Shah Week 5 – Nyquist-Shannon theorem2

Course Map

Numbers correspond to course weeks

Today


Where are we ?

Week 2: SamplingReceived signal is sampled & quantized q = PCM[ r ] Week 3 : CompressionQuantized Signal is Coded c =code[ q ] Week 4 : Frequency DomainSampled signal first transformed into frequency domain Q = DFT[ q ]Week 5signal oversampled & low pass filteredQ = LPF[ DFT(q+n) ]Week 6Transformed signal analyzed Using human psychoaoustic modelsWeek 7Acoustically 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 5 Week 3

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

3


ReconstructionGeneric Digital Signal Processor

Sampler(t)q

Codec

Store/Transmit Decode Produce p(t)

4

● Acquisition side: we need to:● Sample● Then process● Then store discretely● Then transmit.

● On the other end:● How should the signal be reproduced?● How fast do we send information to sound

card?● What functions do we use to interpolate

samples?● Last week: Harmonic Reconstruction

● Gets better with more terms● But need more samples to compute them● And error seems unpredictable

● This week: General reconstruction● Make some assumptions about signal● That allow us to exactly reconstruct from

samples


Bohemian Rhapsody

● Versions sampled at different sampling rates

● 400Hz, then 1kHz, 2kHz, 3kHz.... 44kHz.● How can we describe the distortion?


Shannon's Theorem

● Given a signal whose frequency does not exceed f

max

● We can perfectly reconstruct the signal

● From samples taken once every Ts seconds

● If 2 x fmax

≤ 1/(Ts)

the “Nyquist Rate”

● More simply: 2 x fmax

≤ fN

- the “Nyquist Frequency”


Example 1● Consider a pure sine wave, sin(2πft), where f = 5Hz● Only one frequency component, so

fmax

= f

● Let's sample at 25Hz.


Example 1● Let's sample at 25Hz.

– 25Hz is greater than 2 x 5Hz, so should be able to perfectly reconstruct signal

– Connect the dots – looks good!

– Now let's sample at 10Hz.


Example 1● Let's sample at 10Hz.

– 10Hz is equal to 2 x 5Hz, so should still be able to perfectly reconstruct signal

– Connect the dots – looks good!

– Now let's sample at 6.25Hz.


Example 1● Let's sample at 6.25Hz.

– 6.25Hz is less than 2 x 5Hz, so we should expect some problems.

– Connect the dots – what do we see?

– We'll return to this later in the lecture


“Connect-The-Dots?”● We're actually doing “pulse shaping”

– Formally: use Δ – functions or sinc functions

– Practically: use pulses of height (sample) and width Ts.


Shannon's Theorem


max



● If 2 x fmax

≤ 1/(Ts)



≤ fN



Question● Imagine we have a signal with many harmonics● DFT will yield a large number of frequencies● For perfect reconstruction, we need to store

– the amplitude

– of each frequency

– at each sample point

● OR we could just sample at 2fmax

and store

– ONE amplitude

– per sample point

● Why do we need DFT at all?


Noise

● In reality we receive a mixed signal r(t):

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

q(t) = signal of auditory interest

n(t) = noise (no information)

– Typically high-frequency– From recording, transmission, background

sounds● We only get to sample r(t), not q(t)!


Example 2● Consider signal r(t):

r(t) = q(t) + n(t)– = 0 (silence)– = noise (33kHz and above)– Sample r @ 22kHz

● As in Example 1,– Higher-frequency noise

– Can alias down into the recorded signal


Aliasing● Widely known problem● “Folding” - a manifestation

of aliasing:● “Wagon wheel” effect

● Take a signal s = sin(2π(f+δ)t)● If we sample at f● The aliased noise will be sin(2πδt)● “Folds over” as if was actually a

lower frequency.

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


Example 1 Revisited● 5Hz Signal sampled at 6.25Hz.

– Folding predicts an aliased frequency of (10-6.25) = 3.75Hz

– Is this what we observe?

● Bottom line: Aliasing is undesirable.

● How do we prevent it?


Anti-Aliasing● Given the “mixed” signal

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

– q(t) = Auditory signal, fmax

= 22kHz

– n(t) = High-frequency noise

● We require a process to filter out the noise:– Some sort of “anti-aliasing” technique

– To “smooth away” the noise

– (which would otherwise appear in our samples)

– With some determined cut-off frequency between noise and signal

● How can we do this?


Interlude: Visual Aliasing

● Strobe light: making water run uphill

http://youtu.be/ZRlNOyxWWf8?t=4s


Shannon's Theorem


max



● If 2 x fmax

≤ 1/(Ts)



≤ fN



Fourier's Theorem (review)

● Given a “nice” periodic signal,● We can perfectly re-construct the signal● From just its harmonics

– Lecture 4

– Lab 4 (Synthesizer)


Clipping a non-periodic signal

● Consider a non-periodic signal

● Look at just a finite time window

● The signal is “clipped.”

t + Ts

t

t


Clipping a non-periodic signal● Fourier Analysis gives us

the frequency content of our clipped signal

● Can use this to – perfectly reconstruct– the periodic extension

– Of the clipped signal

t

t + Ts

t


Fourier + Shannon

● Fourier:– Over a finite time window

– We can construct an exact harmonic representation

● Shannon:– Over a finite range of

frequencies

– We can construct an exact representation in time

Clipped frequency forces finite harmonic series

Clipped time forces finite samples


Fourier + Shannon

● Summary:

– If we have a finite sampling window T0

– And finite bandwidth fmax

– We are guaranteed a finite, exact harmonic reconstruction

● We can now solve the anti-aliasing problem


Anti-Aliasing Revisited

● Given the “mixed” signal

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

– q(t) = Auditory signal, fmax

= 22kHz

– n(t) = High-frequency noise

● We need to remove n(t) to be left with just the audible signal q(t).

● If we remove n(t), then no high-frequency noise to alias into our sample!


Q + N = DFT(q) + DFT(n) = DFT(q+n)

● We can filter out the noise thus:

– Take the received time-sampled signal r(t)– Fourier: DFT(r):

● Finite sample period fmax

● So can find frequency-domain representation

– Assume q ranges from 0-fmax

Hz, n from fmax

-∞ Hz

● Can split DFT into terms < fmax

, and terms ≥ fmax

● i.e. DFT(q+n) = DFT(q) + DFT(n)

– Throw out DFT(n) terms

– Perform inverse DFT on DFT(q)

– Shannon: can now perfectly reconstruct q(t)!


Oversampled Pre-Filter● Anti-Aliasing requires

“oversampling”

– fs > 2 x maximum noise frequency

● How do we know this maximum noise frequency ?– Microphone is band-limited; has

its own cut-off frequency– Might also add a “low-pass filter”

Low-Pass Filter

Filter Transformr Q q

Pre-Samplingr Low-Pass

ADXL001 MEMS accelerometer


Open Questions

● How do we decide on fmax

?

– At what frequency do we distinguish between signal and noise?

● Why do CDs sample at 44.1kHz?– What is the significance of this number?


Big Ideas

● In order to perfectly sample a signal,

2 x fmax

≤ fs

● Sampling at too low a frequency will cause aliasing– which is detrimental to signal quality

● In order to prevent high-frequency noise from aliasing down into our signal– we must remove it with a low-pass filter


ESE250 – Week 5

Nyquist-Shannon Theorem

ESE250 – Week 5 Nyquist-Shannon Theoremese250/week5/day5-S13.pdfESE250 S'13: DeHon, Kadric, Kod,...

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