Bài giảng Xử lý tín hiệu số - Chapter 1: Sampling and Reconstruction

Chapter 1

Sampling and Reconstruction

Ha Hoang Kha, Ph.D.

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Ho Chi Minh City University of Technology

Email: [email protected]

Content

Sampling

Sampling theorem

Spectrum of sampling signals

Antialiasing prefilter

Ideal prefilter

Practical prefilter

Analog reconstruction

Ideal reconstructor

Practical reconstructor

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Sampling and Reconstruction

Review of useful equations

Linear system

x(t)

y(t) = x(t)∗h(t)

Linear system

h(t)

H(f)

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

X ( f )

x(t) = Acos(2π f₀t +θ)

Especially,

y(t) = A| H( f₀) | cos(2π f₀t +θ + arg(H( f₀)))

1

FT

Fourier transform:

cos(2π f₀t)←⎯→ [δ ( f + f₀) + δ ( f − f₀)]

2

1

FT

sin(2π f₀t)←⎯→ j[δ ( f + f₀) −δ ( f − f₀)]

2

1

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

Trigonometric formulas:

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

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Sampling and Reconstruction

1. Introduction

A typical signal processing system includes 3 stages:

The analog signal is digitalized 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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Sampling and Reconstruction

2. Analog to digital conversion

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

x_Q(n)

x(nT)

≡

x(n)

11010

x(t)

Sampler

t=nT

Quantizer

Coder

A/D converter

x_Q(n)

x(t)

x(n)

111

110

101

100

011

010

001

000

n

t

n

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Sampling and Reconstruction

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

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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Sampling and Reconstruction

3. Sampling-example 1

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

Hz. Find the discrete-time signal 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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Sampling and Reconstruction

3. Sampling-example 2

Consider the two analog sinusoidal signals

7

x₁(t) = 2cos(2π t),

8

1

x₂(t) = 2cos(2π t); t(s)

8

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

Find the discrete-time signals ?

Solution:

1

7 1

7

x₁(n) ≡ x₁(nT) = x₁(n ) = 2cos(2π n) = 2cos( πn)

f_s8 1

= 2cos((2 − )πn) = 2cos( n)

4

1

π

4

1

1 1

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

f_s8 1

1

4

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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Sampling and Reconstruction

3. Sampling-example 2

f₂=1/8 Hz

f₁=7/8 Hz

f_s=1 Hz

Fig: Illustration of aliasing

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Sampling and Reconstruction

3. Sampling-Aliasing of Sinusoids

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

at a sampling rate f =1/T results in a discrete-

x

(

t

)

=

A

cos(2

π

f

₀

t

+

θ

)

s

time signal x(n).

x

_k

(

t

)

=

A

cos(2

π

f

_k

t

+

θ

)

The sinusoids

is sampled at f

_s

, resulting in a

discrete time signal x_k(n).

If f_k=f₀+kf_s, k=0, ±1, ±2, …., then x(n)=x_k(n) .

Proof: (in class)

Remarks: We can that the frequencies f_k=f₀+kf_sare indistinguishable

from the frequency f₀after sampling and hence they are aliases of f₀

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Sampling and Reconstruction

4. Sampling Theorem-Sinusoids

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

the frequency (rad/s) of the analog signal, and f=Ω/2π is the

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

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

Fig: Sinusoid sampled at different rates

f_s

samples / sec samples

Note that

=

f cycles / sec

cycle

f_s

f

samples

cycle

To sample a single sinusoid properly, we must require

≥ 2

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Sampling and Reconstruction

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 f_max.

Fig: Typical bandlimited spectrum

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

f

_s

≥

2

f

_max

frequency f_max.

f_s=2f_maxis called Nyquist rate; f_s/2 is called Nyquist frequency;

[-f_s/2, f_s/2] is Nyquist interval.

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Sampling and Reconstruction

4. Sampling Theorem

The values of f_maxand f_sdepend on the application

Application

Biomedical

Speech

fmax

fs

1 KHz

4 KHz

20 KHz

4 MHz

2 KHz

8 KHz

40 KHz

8 MHz

Audio

Video

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Sampling and Reconstruction

4. Sampling Theorem-Spectrum Replication

∞

ꢀ

x(nT) = x(t) = x(t) δ (t − nT) = x(t)s(t)

s(t) = δ (t − nT)

Let

where

∑

n=−∞

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

∞

1

s

(

t

)

=

S

e

^{j2π f nt}

S

_n

=

δ

(

t

)

e

^{− j2π f nt}

dt

=

δ

(

t

)

dt

=

s

where

∑

n

∫

T

n=−∞

T

∞

1

e^{j2π f nt}

s

s(t) =

Thus,

∑

T

n=−∞

∞

1

ꢀ

x(t) = x(t)s(t) =

x(t)e^{j2πnf t}

s

which results in

∑

T

ꢀ

n=−∞

∞

ꢀ

1

X ( f ) =

X ( f − nf )

x(t)

Taking the Fourier transform of

yields

∑

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

interval f_s.

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Sampling and Reconstruction

4. Sampling Theorem-Spectrum Replication

f_s/2 ≥ f_max

Fig: Spectrum replication caused by sampling

Fig: Typical badlimited spectrum

f_s/2 < f_max

Fig: Aliasing caused by overlapping spectral replicas

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Sampling and Reconstruction

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 f ∈[− f / 2, f / 2]

⎧

⎨

⎩

s

H( f ) =

An ideal reconstructor (lowpass filter)

0

otherwise

Then

ꢀ

X_a( f ) = X ( f )H( f ) = X ( f )

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Sampling and Reconstruction

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 the discrete time signal x(n) ?

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

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

reconstructed signal x_a(t) ?

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Sampling and Reconstruction

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 the discrete time signal x(n) ?

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

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

reconstructed signal x_a(t) ?

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Sampling and Reconstruction

5. Analog reconstruction

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

Nyquist interval (NI) [-f_s//2, f_s/2].

sampling at f_s

x(t), f₀∈ NI ------------------> x(n) ----------------------> x_a(t), f_a=f₀

ideal reconstructor

sampling at f_s

ideal reconstructor

x_k(t), f_k=f₀+kf_s------------------> x(n) ----------------------> x_a(t), f_a=f₀

The frequency f_aof reconstructed signal x_a(t) is obtained by adding

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

within the Nyquist interval [-f

_s

//2, f

_s

/2].. That is

f_a= f mod( f_s)

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Sampling and Reconstruction

5. Analog reconstruction-Example 3

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

Hz. Find the reconstructed signal x_a(t) ?

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Sampling and Reconstruction

Bài giảng Xử lý tín hiệu số - Chapter 1: Sampling and Reconstruction - Hà Hoàng Kha