Chapter

p 1

Sampling and Reconstruction

Ha Hoang Kha, Ph.D.Click to edit Master subtitle style

Ho Chi Minh City University of Technology

@

Email: hhkha@hcmut.edu.vn

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Content

Sampling

Sampling theorem

Spectrum of sampling signals

Antialiasing prefilter

Ideal

Id l prefilter

fil

Practical prefilter

Analog reconstruction

Ideal reconstructor

Practical reconstructor

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Review of useful equations

Linear system

Linear system

h(t)

H(f)

x(t )

X(f )

y (t ) = x(t ) ∗ h(t )

Y ( f ) = X ( f )H ( f )

Especially,

Especially x(t ) = A cos(2π f 0t + θ )

y (t ) = A | H ( f 0 ) | cos(2π f 0t + θ + arg( H ( f 0 )))

1

cos(2π f 0t ) ←⎯→ [δ ( f + f 0 ) + δ ( f − f 0 )]

2

1

sin(2π f 0t ) ←⎯→

← FT → j[δ ( f + f 0 ) − δ ( f − f 0 )]

2

1

Trigonometric formulas: cos(a) cos(b) = [cos(a + b) + cos(a − b)]

2

1

sin( a) sin(b) = − [cos( a + b) − cos( a − b)]

2

1

sin( a) cos(b) = [sin( a + b) + sin( a − b)]

2

Fourier transform:

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1. Introduction

A typical signal processing system includes 3 stages:

The

Th analog

l signal

i l is

i digitalized

di i li d b

by an A/D converter

The digitalized samples are processed by a digital signal processor.

The digital processor can be programmed to perform signal processing

operations such as filtering, spectrum estimation. Digital signal processor can be

a general purpose computer, DSP chip or other digital hardware.

The resulting output samples are converted back into analog by a

D/A converter.

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2. Analog to digital conversion

Analog to digital (A/D) conversion is a three-step process.

x(t)

Sampler

t=nT

x(t)

x(nT)≡x(n)

( ) ( ) Quantizer xQ(n) Coder

A/D converter

x(n)

t

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11010

111 xQ(n)

110

101

100

011

010

001

000

n

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3. Sampling

Sampling is to convert a continuous time signal into a discrete time

signal The analog signal is periodically measured at every T seconds

signal.

x(n)≡x(nT)=x(t=nT),

( ) ( ) (

), n=….-2,, -1,, 0,, 1,, 2,, 3……..

T: sampling interval or sampling period (second);

fs=1/T: sampling rate or sampling frequency (samples/second or

Hz)

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3. Sampling-example 1

The analog signal x(t)=2cos(2πt) with t(s) is sampled at the rate fs=4

Hz. Find the discrete-time signal

g x(n)

( )?

Solution:

x(n)≡x(nT)=x(n/fs)=2cos(2πn/fs)=2cos(2πn/4)=2cos(πn/2)

n

0

1

2

3

4

x(n)

2

0

‐2

0

2

Plot the signal

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3. Sampling-example 2

Consider the two analog sinusoidal signals

7

1

x1 (t ) = 2 cos(2π t ),

) x2 (t ) = 2cos(2π t ); t (s)

8

8

These signals are sampled at the sampling frequency fs=1 Hz.

Fi d the

Find

h discrete-time

di

i signals

i l ?

Solution:

1

71

7

) = 2 cos(2π

n) = 2 cos( π n)

fs

81

4

1

π

= 2 cos((2 − )π n) = 2 cos( n)

4

4

1

11

1

x2 (n) ≡ x2 (nT ) = x2 (n ) = 2 cos(2π

n) = 2 cos( π n)

fs

81

4

x1 (n) ≡ x1 (nT ) = x1 (n

Observation: x1(n)=x2(n) Æ based on the discrete-time signals, we

cannot tell which of two signals are sampled ? These signals are

called “alias”

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3. Sampling-example 2

f2=1/8 Hz

f1=7/8 Hz

fs=1 Hz

Fig: Illustration of aliasing

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3. Sampling-Aliasing of Sinusoids

In general, the sampling of a continuous-time sinusoidal signal

p g rate fs=1/T

/ results in a discretex(t ) = A cos(2π f 0t + θ ) at a sampling

time signal x(n).

The sinusoids xk (t ) = A cos(2π f k t + θ ) is sampled at fs , resulting in a

discrete time signal xk(n).

If fk=f0+kfs, k=0,

k=0 ±1,

±1 ±2,

±2 …., then x(n)=xk(n) .

Proof: ((in class))

Remarks: We can that the frequencies fk=f0+kfs are indistinguishable

from the frequency f0 after sampling and hence they are aliases of f0

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4. Sampling Theorem-Sinusoids

Consider the analog signal x(t ) = A cos(Ωt ) = A cos(2π ft ) where Ω is

the frequency

q

y (rad/s)

(

) of the analogg signal,

g , and f=Ω/2π is the

frequency in cycles/s or Hz. The signal is sampled at the three rate

fs=8f, fs=4f, and fs=2f.

Fi Sinusoid

Fig:

Si

id sampled

l d at diff

different rates

Note that

f s samples / sec samples

=

=

f

cycles

l / sec

cycle

l

To sample a single sinusoid properly, we must require f s ≥ 2 samples

f

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4. Sampling Theorem

For accurate representation of a signal x(t) by its time samples x(nT),

two conditions must be met:

1) The signal x(t) must be bandlimitted, i.e., its frequency spectrum must

be limited to fmax .

Fig:

g Typical

yp

bandlimited spectrum

p

2) The sampling rate fs must be chosen at least twice the maximum

f s ≥ 2 f max

frequency fmax.

fs=2fmax is called Nyquist rate; fs/2 is called Nyquist frequency;

[ fs/2,

[-f

/2 fs/2] is

i Nyquist

N i interval.

i

l

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4. Sampling Theorem

The values of fmax and fs depend on the application

Application

pp

fmax

fs

Biomedical

1 KHz

2 KHz

Speech

4 KHz

8 KHz

Audio

20 KHz

40 KHz

Video

4 MHz

8 MHz

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4. Sampling Theorem-Spectrum Replication

∞

Let x(nT ) = x (t ) = x(t ) ∑ δ (t − nT ) = x(t ) s(t ) where s(t ) =

n =−∞

∞

∑ δ (t − nT )

n =−∞

s(t) is periodic, thus, its Fourier series are given by

s (t ) =

∞

∑

n =−∞

S n e j 2π f s nt where S n =

1

1

1

− j 2π f s nt

δ (t )e

dt = ∫ δ (t )dt =

∫

TT

TT

T

1 ∞ j 2π f s nt

Thus, s(t ) = ∑ e

T n =−∞

∞

1

which results in

x (t ) = x(t ) s(t ) = ∑ x(t )e j 2π nf st

T n =−∞

1 ∞

Taking the Fourier transform of x (t ) yields X ( f ) = ∑ X ( f − nf s )

T n =−∞

Observation: The spectrum of discrete-time signal is a sum of the

original spectrum of analog signal and its periodic replication at the

i

interval

l fs.

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4. Sampling Theorem-Spectrum Replication

fs/2 ≥ fmax

Fig: Spectrum replication caused by sampling

Fi Typical

Fig:

T i l badlimited

b dli i d spectrum

fs/2 < fmax

Fig: Aliasing caused by overlapping spectral replicas

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5. Ideal Analog reconstruction

Fig: Ideal reconstructor as a lowpass filter

An ideal reconstructor acts as a lowpass filter with cutoff frequency

equal to the Nyquist frequency fs/2.

⎧T

H

(

f

)

=

An ideal reconstructor (lowpass filter)

⎨

⎩0

Then

f ∈ [− f s / 2, f s / 2]

otherwise

othe

wise

X a ( f ) = X ( f )H ( f ) = X ( f )

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5. Analog reconstruction-Example 1

The analog signal x(t)=cos(20πt) is sampled at the sampling

frequency fs=40 Hz.

a) Plot the spectrum of signal x(t) ?

b) Find

Fi d the

h discrete

di

time

i signal

i l x(n)

( )?

c) Plot the spectrum of signal x(n) ?

d) The signal x(n) is an input of the ideal reconstructor,

reconstructor find the

reconstructed signal xa(t) ?

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5. Analog reconstruction-Example 2

The analog signal x(t)=cos(100πt) is sampled at the sampling

frequency fs=40 Hz.

a) Plot the spectrum of signal x(t) ?

b) Find

Fi d the

h discrete

di

time

i signal

i l x(n)

( )?

c) Plot the spectrum of signal x(n) ?

d) The signal x(n) is an input of the ideal reconstructor,

reconstructor find the

reconstructed signal xa(t) ?

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5. Analog reconstruction

Remarks: xa(t) contains only the frequency components that lie in the

Nyquist interval (NI) [-f

[ fs//2,

//2 fs/2].

/2]

sampling

p g at fs

ideal reconstructor

x(t), f0 ∈ NI ------------------> x(n) ----------------------> xa(t), fa=ff0

sampling at fs

ideal reconstructor

xk(t), fk=f0+kfs------------------> x(n) ----------------------> xa(t), fa=f0

The

Th frequency

f

fa off reconstructedd signal

i l xa(t)

( ) is

i obtained

b i d by

b adding

ddi

to or substracting from f0 (fk) enough multiples of fs until it lies

within the Nyquist interval [[-ffs//2,

//2 fs/2]..

/2] That is

f a = f mod( f s )

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5. Analog reconstruction-Example 3

The analog signal x(t)=10sin(4πt)+6sin(16πt) is sampled at the rate 20

Hz. Findd the

h reconstructed

d signall xa(t) ?

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5. Analog reconstruction-Example 4

Let x(t) be the sum of sinusoidal signals

x(t)=4+3cos(πt)+2cos(2πt)+cos(3πt)

x(t)

4+3cos(πt)+2cos(2πt)+cos(3πt) where t is in milliseconds.

milliseconds

a) Determine the minimum sampling rate that will not cause any

aliasing effects ?

b) To observe aliasing effects, suppose this signal is sampled at half its

Nyquist rate. Determine the signal xa(t) that would be aliased with

x(t) ? Plot the spectrum of signal x(n) for this sampling rate?

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6. Ideal antialiasing prefilter

The signals in practice may not bandlimitted, thus they must be

f l d by

filtered

b a lowpass

l

fl

filter

Fi Ideal

Fig:

Id l antialiasing

ti li i prefilter

p filt

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6. Practical antialiasing prefilter

A lowpass filter: [-fpass, fpass] is the frequency range of interest for the

application

l

(ffmax=ffpass)

The Nyquist frequency fs/2 is in the middle of transition region.

The stopband frequency fstop and the minimum stopband attenuation

Astop dB must be chosen appropriately to minimize the aliasing

effects.

effects

f s = f pass + f stop

Fig: Practical antialiasing lowpass prefilter

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6. Practical antialiasing prefilter

The attenuation of the filter in decibels is defined as

A( f ) = −20 log10

H( f )

(dB)

H ( f0 )

where f0 is a convenient reference frequency, typically taken to be at

p filter.

DC for a lowpass

α10 =A(10f)-A(f) (dB/decade): the increase in attenuation when f is

g byy a factor of ten.

changed

α2 =A(2f)-A(f) (dB/octave): the increase in attenuation when f is

changed by a factor of two.

Analog filter with order N, |H(f)|~1/fN for large f, thus α10 =20N

(dB/decade) and α10 =6N (dB/octave)

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6. Antialiasing prefilter-Example

A sound wave has the form

x(t ) = 2 A cos(10π t ) + 2 B cos(30π t ) + 2C cos(50π t )

+ 2 D cos(60π t ) + 2 E cos(90π t ) + 2 F cos(125π t )

where t is in milliseconds. What is the frequency content of this

g ? Which parts

p

of it are audible and whyy ?

signal

This signal is prefilter by an anlog prefilter H(f). Then, the output y(t)

prefilter is sampled

p at a rate of 40KHz and immediatelyy

of the p

reconstructed by an ideal analog reconstructor, resulting into the final

analog output ya(t), as shown below:

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p 1

Sampling and Reconstruction

Ha Hoang Kha, Ph.D.Click to edit Master subtitle style

Ho Chi Minh City University of Technology

@

Email: hhkha@hcmut.edu.vn

CuuDuongThanCong.com

https://fb.com/tailieudientucntt

Content

Sampling

Sampling theorem

Spectrum of sampling signals

Antialiasing prefilter

Ideal

Id l prefilter

fil

Practical prefilter

Analog reconstruction

Ideal reconstructor

Practical reconstructor

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Review of useful equations

Linear system

Linear system

h(t)

H(f)

x(t )

X(f )

y (t ) = x(t ) ∗ h(t )

Y ( f ) = X ( f )H ( f )

Especially,

Especially x(t ) = A cos(2π f 0t + θ )

y (t ) = A | H ( f 0 ) | cos(2π f 0t + θ + arg( H ( f 0 )))

1

cos(2π f 0t ) ←⎯→ [δ ( f + f 0 ) + δ ( f − f 0 )]

2

1

sin(2π f 0t ) ←⎯→

← FT → j[δ ( f + f 0 ) − δ ( f − f 0 )]

2

1

Trigonometric formulas: cos(a) cos(b) = [cos(a + b) + cos(a − b)]

2

1

sin( a) sin(b) = − [cos( a + b) − cos( a − b)]

2

1

sin( a) cos(b) = [sin( a + b) + sin( a − b)]

2

Fourier transform:

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1. Introduction

A typical signal processing system includes 3 stages:

The

Th analog

l signal

i l is

i digitalized

di i li d b

by an A/D converter

The digitalized samples are processed by a digital signal processor.

The digital processor can be programmed to perform signal processing

operations such as filtering, spectrum estimation. Digital signal processor can be

a general purpose computer, DSP chip or other digital hardware.

The resulting output samples are converted back into analog by a

D/A converter.

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2. Analog to digital conversion

Analog to digital (A/D) conversion is a three-step process.

x(t)

Sampler

t=nT

x(t)

x(nT)≡x(n)

( ) ( ) Quantizer xQ(n) Coder

A/D converter

x(n)

t

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11010

111 xQ(n)

110

101

100

011

010

001

000

n

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3. Sampling

Sampling is to convert a continuous time signal into a discrete time

signal The analog signal is periodically measured at every T seconds

signal.

x(n)≡x(nT)=x(t=nT),

( ) ( ) (

), n=….-2,, -1,, 0,, 1,, 2,, 3……..

T: sampling interval or sampling period (second);

fs=1/T: sampling rate or sampling frequency (samples/second or

Hz)

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3. Sampling-example 1

The analog signal x(t)=2cos(2πt) with t(s) is sampled at the rate fs=4

Hz. Find the discrete-time signal

g x(n)

( )?

Solution:

x(n)≡x(nT)=x(n/fs)=2cos(2πn/fs)=2cos(2πn/4)=2cos(πn/2)

n

0

1

2

3

4

x(n)

2

0

‐2

0

2

Plot the signal

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3. Sampling-example 2

Consider the two analog sinusoidal signals

7

1

x1 (t ) = 2 cos(2π t ),

) x2 (t ) = 2cos(2π t ); t (s)

8

8

These signals are sampled at the sampling frequency fs=1 Hz.

Fi d the

Find

h discrete-time

di

i signals

i l ?

Solution:

1

71

7

) = 2 cos(2π

n) = 2 cos( π n)

fs

81

4

1

π

= 2 cos((2 − )π n) = 2 cos( n)

4

4

1

11

1

x2 (n) ≡ x2 (nT ) = x2 (n ) = 2 cos(2π

n) = 2 cos( π n)

fs

81

4

x1 (n) ≡ x1 (nT ) = x1 (n

Observation: x1(n)=x2(n) Æ based on the discrete-time signals, we

cannot tell which of two signals are sampled ? These signals are

called “alias”

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3. Sampling-example 2

f2=1/8 Hz

f1=7/8 Hz

fs=1 Hz

Fig: Illustration of aliasing

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3. Sampling-Aliasing of Sinusoids

In general, the sampling of a continuous-time sinusoidal signal

p g rate fs=1/T

/ results in a discretex(t ) = A cos(2π f 0t + θ ) at a sampling

time signal x(n).

The sinusoids xk (t ) = A cos(2π f k t + θ ) is sampled at fs , resulting in a

discrete time signal xk(n).

If fk=f0+kfs, k=0,

k=0 ±1,

±1 ±2,

±2 …., then x(n)=xk(n) .

Proof: ((in class))

Remarks: We can that the frequencies fk=f0+kfs are indistinguishable

from the frequency f0 after sampling and hence they are aliases of f0

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4. Sampling Theorem-Sinusoids

Consider the analog signal x(t ) = A cos(Ωt ) = A cos(2π ft ) where Ω is

the frequency

q

y (rad/s)

(

) of the analogg signal,

g , and f=Ω/2π is the

frequency in cycles/s or Hz. The signal is sampled at the three rate

fs=8f, fs=4f, and fs=2f.

Fi Sinusoid

Fig:

Si

id sampled

l d at diff

different rates

Note that

f s samples / sec samples

=

=

f

cycles

l / sec

cycle

l

To sample a single sinusoid properly, we must require f s ≥ 2 samples

f

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4. Sampling Theorem

For accurate representation of a signal x(t) by its time samples x(nT),

two conditions must be met:

1) The signal x(t) must be bandlimitted, i.e., its frequency spectrum must

be limited to fmax .

Fig:

g Typical

yp

bandlimited spectrum

p

2) The sampling rate fs must be chosen at least twice the maximum

f s ≥ 2 f max

frequency fmax.

fs=2fmax is called Nyquist rate; fs/2 is called Nyquist frequency;

[ fs/2,

[-f

/2 fs/2] is

i Nyquist

N i interval.

i

l

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4. Sampling Theorem

The values of fmax and fs depend on the application

Application

pp

fmax

fs

Biomedical

1 KHz

2 KHz

Speech

4 KHz

8 KHz

Audio

20 KHz

40 KHz

Video

4 MHz

8 MHz

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4. Sampling Theorem-Spectrum Replication

∞

Let x(nT ) = x (t ) = x(t ) ∑ δ (t − nT ) = x(t ) s(t ) where s(t ) =

n =−∞

∞

∑ δ (t − nT )

n =−∞

s(t) is periodic, thus, its Fourier series are given by

s (t ) =

∞

∑

n =−∞

S n e j 2π f s nt where S n =

1

1

1

− j 2π f s nt

δ (t )e

dt = ∫ δ (t )dt =

∫

TT

TT

T

1 ∞ j 2π f s nt

Thus, s(t ) = ∑ e

T n =−∞

∞

1

which results in

x (t ) = x(t ) s(t ) = ∑ x(t )e j 2π nf st

T n =−∞

1 ∞

Taking the Fourier transform of x (t ) yields X ( f ) = ∑ X ( f − nf s )

T n =−∞

Observation: The spectrum of discrete-time signal is a sum of the

original spectrum of analog signal and its periodic replication at the

i

interval

l fs.

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4. Sampling Theorem-Spectrum Replication

fs/2 ≥ fmax

Fig: Spectrum replication caused by sampling

Fi Typical

Fig:

T i l badlimited

b dli i d spectrum

fs/2 < fmax

Fig: Aliasing caused by overlapping spectral replicas

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5. Ideal Analog reconstruction

Fig: Ideal reconstructor as a lowpass filter

An ideal reconstructor acts as a lowpass filter with cutoff frequency

equal to the Nyquist frequency fs/2.

⎧T

H

(

f

)

=

An ideal reconstructor (lowpass filter)

⎨

⎩0

Then

f ∈ [− f s / 2, f s / 2]

otherwise

othe

wise

X a ( f ) = X ( f )H ( f ) = X ( f )

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5. Analog reconstruction-Example 1

The analog signal x(t)=cos(20πt) is sampled at the sampling

frequency fs=40 Hz.

a) Plot the spectrum of signal x(t) ?

b) Find

Fi d the

h discrete

di

time

i signal

i l x(n)

( )?

c) Plot the spectrum of signal x(n) ?

d) The signal x(n) is an input of the ideal reconstructor,

reconstructor find the

reconstructed signal xa(t) ?

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5. Analog reconstruction-Example 2

The analog signal x(t)=cos(100πt) is sampled at the sampling

frequency fs=40 Hz.

a) Plot the spectrum of signal x(t) ?

b) Find

Fi d the

h discrete

di

time

i signal

i l x(n)

( )?

c) Plot the spectrum of signal x(n) ?

d) The signal x(n) is an input of the ideal reconstructor,

reconstructor find the

reconstructed signal xa(t) ?

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5. Analog reconstruction

Remarks: xa(t) contains only the frequency components that lie in the

Nyquist interval (NI) [-f

[ fs//2,

//2 fs/2].

/2]

sampling

p g at fs

ideal reconstructor

x(t), f0 ∈ NI ------------------> x(n) ----------------------> xa(t), fa=ff0

sampling at fs

ideal reconstructor

xk(t), fk=f0+kfs------------------> x(n) ----------------------> xa(t), fa=f0

The

Th frequency

f

fa off reconstructedd signal

i l xa(t)

( ) is

i obtained

b i d by

b adding

ddi

to or substracting from f0 (fk) enough multiples of fs until it lies

within the Nyquist interval [[-ffs//2,

//2 fs/2]..

/2] That is

f a = f mod( f s )

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5. Analog reconstruction-Example 3

The analog signal x(t)=10sin(4πt)+6sin(16πt) is sampled at the rate 20

Hz. Findd the

h reconstructed

d signall xa(t) ?

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5. Analog reconstruction-Example 4

Let x(t) be the sum of sinusoidal signals

x(t)=4+3cos(πt)+2cos(2πt)+cos(3πt)

x(t)

4+3cos(πt)+2cos(2πt)+cos(3πt) where t is in milliseconds.

milliseconds

a) Determine the minimum sampling rate that will not cause any

aliasing effects ?

b) To observe aliasing effects, suppose this signal is sampled at half its

Nyquist rate. Determine the signal xa(t) that would be aliased with

x(t) ? Plot the spectrum of signal x(n) for this sampling rate?

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6. Ideal antialiasing prefilter

The signals in practice may not bandlimitted, thus they must be

f l d by

filtered

b a lowpass

l

fl

filter

Fi Ideal

Fig:

Id l antialiasing

ti li i prefilter

p filt

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6. Practical antialiasing prefilter

A lowpass filter: [-fpass, fpass] is the frequency range of interest for the

application

l

(ffmax=ffpass)

The Nyquist frequency fs/2 is in the middle of transition region.

The stopband frequency fstop and the minimum stopband attenuation

Astop dB must be chosen appropriately to minimize the aliasing

effects.

effects

f s = f pass + f stop

Fig: Practical antialiasing lowpass prefilter

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6. Practical antialiasing prefilter

The attenuation of the filter in decibels is defined as

A( f ) = −20 log10

H( f )

(dB)

H ( f0 )

where f0 is a convenient reference frequency, typically taken to be at

p filter.

DC for a lowpass

α10 =A(10f)-A(f) (dB/decade): the increase in attenuation when f is

g byy a factor of ten.

changed

α2 =A(2f)-A(f) (dB/octave): the increase in attenuation when f is

changed by a factor of two.

Analog filter with order N, |H(f)|~1/fN for large f, thus α10 =20N

(dB/decade) and α10 =6N (dB/octave)

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6. Antialiasing prefilter-Example

A sound wave has the form

x(t ) = 2 A cos(10π t ) + 2 B cos(30π t ) + 2C cos(50π t )

+ 2 D cos(60π t ) + 2 E cos(90π t ) + 2 F cos(125π t )

where t is in milliseconds. What is the frequency content of this

g ? Which parts

p

of it are audible and whyy ?

signal

This signal is prefilter by an anlog prefilter H(f). Then, the output y(t)

prefilter is sampled

p at a rate of 40KHz and immediatelyy

of the p

reconstructed by an ideal analog reconstructor, resulting into the final

analog output ya(t), as shown below:

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## Tài liệu Bài giảng: Xử lý tín hiệu số ppt

## Bai giang Xu ly tin hieu so

## Bài giảng xử lý tín hiệu số đh kỹ thuật công nghệ thái nguyên

## Bài giảng xử lý tín hiệu số

## Bài giảng Xử lý tín hiệu số: Chapter 0 - Hà Hoàng Kha

## Bài giảng Xử lý tín hiệu số: Chapter 7 - Hà Hoàng Kha

## Bài giảng Xử lý tín hiệu số: Chapter 5 - Hà Hoàng Kha

## Bài giảng Xử lý tín hiệu số: Giới thiệu - Lê Vũ Hà

## Bài giảng Xử lý tín hiệu số: Chương 5 - Lê Vũ Hà

## Bài giảng Xử lý tín hiệu số: Chương 1 - Lê Vũ Hà

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