DSP-Lec 01-Sampling and Reconstruction

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 pro

Review of useful equations

™ Linear system x t( ) Linear systemh(t) y t( ) = x t( )∗h t( )

x t = A π f t + θ

h(t) H(f)

2

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

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

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

2 Analog to digital conversion

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

x(t) Sampler x(nT)≡x(n) Quantizer Coder 11010

t=nT

A/D converter

xQ(n) 111

011 100 101 110

n

000001

010011

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)

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 x(n) ? g ( )

3 Sampling-example 2

™ Consider the two analog sinusoidal signals

7 ( ) 2 cos(2 )

3 Sampling-Aliasing of Sinusoids

at a sampling rate fs=1/T results in a

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

™ The sinusoids x t k( ) = Acos(2 π f t k + θ ) is sampled at f resulting in a

™ The sinusoids is sampled at fs , resulting in a discrete time signal xk(n)

Fig: Typical bandlimited spectrum

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

frequency fmax f s ≥ 2 fmax

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

[ f /2 f /2] i N i i l

[-fs/2, fs/2] is Nyquist interval

4 Sampling Theorem-Spectrum Replication

™ Taking the Fourier transform of yields x t ( )

™ Observation: The spectrum of discrete-time signal is a sum of the original spectrum of analog signal and its periodic replication at the

interval fs.

4 Sampling Theorem-Spectrum Replication

™ fs/2 ≥ fmax

Fi T i l b dli i d

Fig: Spectrum replication caused by sampling

Fig: Typical badlimited spectrum

™ fs/2 < fmax

Fig: Aliasing caused by overlapping spectral replicasFig: Aliasing caused by overlapping spectral replicas

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

™ An ideal reconstructor (lowpass filter) ( ) [ / 2, / 2]

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

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

reconstructed signal xa(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

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

reconstructed signal xa(t) ?

5 Analog reconstruction

™ Remarks: xa(t) contains only the frequency components that lie in the Nyquist interval (NI) [ f //2 f /2]

Nyquist interval (NI) [-fs//2, fs/2]

™ x(t), f0 ∈ NI -> x(n) -> xp g s a(t), fa=f0

™ xk(t), fk=f0+kfssampling at f -> x(n) -> xs ideal reconstructor a(t), fa=f0

™ The frequency fa of reconstructed signal xa(t) is obtained by adding

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

within the Nyquist interval [-f //2 f /2] That is

5 Analog reconstruction-Example 4

™ Let x(t) be the sum of sinusoidal signals

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

x(t) 4+3cos(πt)+2cos(2πt)+cos(3πt) where t is in 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?

6 Ideal antialiasing prefilter

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

filtered by a lowpass filter

Fi Id l ti li i p filtFig: Ideal antialiasing prefilter

6 Practical antialiasing prefilter

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

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

application (fmax=fpass)

™

™ The stopband frequency fstop and the minimum stopband attenuation

Astop dB must be chosen appropriately to minimize the aliasing

6 Practical antialiasing prefilter

™ The attenuation of the filter in decibels is defined as

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

DC for a lowpass filter.p

™ α10 =A(10f)-A(f) (dB/decade): the increase in attenuation when f is changed by a factor of ten.g y

™ 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)

6 Antialiasing prefilter-Example

™ A sound wave has the form

( ) 2 cos(10 ) 2 cos(30 ) 2 cos(50 )

2 cos(60 ) 2 cos(90 ) 2 cos(125 )

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

signal ? Which parts of it are audible and why ?g p y

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

of the prefilter is sampled at a rate of 40KHz and immediately p p y

reconstructed by an ideal analog reconstructor, resulting into the final analog output ya(t), as shown below:

6 Antialiasing prefilter-Example

Determine the output signal y(t) and ya(t) in the following cases:

a)When there is no prefilter, that is, H(f)=1 for all f

b)When H(f) is the ideal prefilter with cutoff fs/2=20 KHz

c)When H(f) is a practical prefilter with specifications as shown

c)When H(f) is a practical prefilter with specifications as shown

below:

The filter’s phase response is assumed to be ignored in this example

7 Ideal and practical analog reconstructors

™ An ideal reconstructor is an ideal lowpass filter with cutoff Nyquist

frequency fs/2

7 Ideal and practical analog reconstructors

™ The ideal reconstructor has the impulse response:

=

which is not realizable since its impulse response is not casual π f ts

™ It is practical to use a

staircase reconstructor

7 Ideal and practical analog reconstructors

Fig: Frequency response of staircase recontructor

7 Practical reconstructors-antiimage postfilter

™ An analog lowpass postfilter whose cutoff is Nyquist frequency fs/2

is used to remove the surviving spectral replicas

Fig: Analog anti-image postfilter

Fig: Spectrum after postfilter

8 Homework

™ Problems: 1.2, 1.3, 1.4, 1.5, 1.9