1

DIGITAL SIGNAL PROCESSING

Lecture 1 - Chapter 1

Classification of Signals: Continuous-Time verses Discrete-Time Signals

Continuous time or analog signals are signals that are defined for every value of a < t <

b, where (a, b) can be ( ∞+∞− , ), i.e., x (t) = e-|t| or ).cos()( ttx π= .

Discrete-time signals are defined at discrete-time instants and between the two discrete

time instants are undefined but are not zero. They can be obtained either by sampling

analog signals or they can be discrete in nature like discrete measurement signals.

A discrete-time signal having a set of discrete values is called a digital signal. Note that

sampling an analog signal produces a discrete-time signal. Then quantization of its

values produces a digital signal.

Deterministic versus Random Signals

Any signal that can be uniquely described by an explicit mathematical expression or a

well-defined rule is called “deterministic”. The past, present and future of a deterministic

signal are known with certainty. Otherwise, it is called “Random” and its properties is

explained by statistical techniques.

Review of Sinusoids in Continuous and Discrete Time

( ) ( )

( ) F2Ft2A

ttAtxa

πΩθπ

θΩ

=+=∞<<∞−+=

cos

cos

( ) ( )F

TtxTtx papa

1, ==+ : fundamental period. Increasing F means increasing

oscillation in time domain. F = 0 corresponds to Tp = ∞ .

2

Also, for complex exponential signals, ( ) ( )θΩ += tja Aetx . A sinusoidal signal then can

also be expressed as

( ) ( ) ( ) ( )θΩθΩθΩ +−+ +=+= tjtja Ae

21

Ae21

tAtx cos

Discrete-Time Sinusoid Signals

( ) ( )( ) ( ) ∞<<∞−+=

=+=nnfAnx

fnAnx

θπ

πωθω

0

0

2cos

2cos

A few important differences between continuous sinusoid and discrete sinusoids:

1) A discrete-time sinusoid is periodic only if its frequency is a rational number.

By default, ( ) ( )nxNnx =+ for all n if x(n) is periodic. The smallest N is called

Fundamental Period.

( ) ( )( ) ( )θπθπ +=++=+ nfNnfANnx 00 2cos2cos

This relationship is true if and only if ππ KNf 22 0 =

NK

f =⇒ 0 : a rational number.

To determine the period N of a periodic discrete time sinusoid, we express f as two

relatively prime numbers. Observe that a small change in frequency can result in a

large change in period. For example, ( ) ( )θπθπ +=

+

= nAnAnx cos

21

2cos1 its

period is 6030

21

1 ===fK

N .

Now consider ( ) 60517.06031

,6031

2cos 222 =→==

+

= NfnAnx θπ

2. An analog ( )+∞∞− ,F maps to 21

f21

≤≤− or equivalently to πωπ ≤≤− or in

other words, the highest rate of oscillation occurs at 21

f += or πω += . To see

what happens for πωπ 2≤≤ consider ω1 = ω0 and ω2 = 2π - ω0 . When ω1 varies

between π to 2π, then ω2 varies between π and 0. Now

( ) nAnAnx 011 coscos ωω ==

3

( ) ( ) ( ) ( )nxnAnAnAnAnx oo 1*

022 coscos2coscos)( ==−=−== ωωωπω

* this is only true because n is an integer value, i.e., x(n) is a discrete signal.

Hence, ( ) ( )nxnx 12 = is an alias because +∞<<∞− F maps only to 21

f21

≤≤−

by sampling.

Analog to Digital Conversion (A/D)

- sampling (sampling rate)

- quantization

Sampling: ( ) ( )

( ) nTsta

sa

tx

nnTxnx

==

∞<<∞−= where,

ss F

nnTt == , Fs = Sampling rate (Frequency) (Hz)

Consider an analog sinusoid: ( ) ( )φπ += Ft2Atxa cos

( ) ( ) ( )

+=+== φπφπ n

FF

2AnTF2AnTxnxs

ssa coscos

Now recall that ∞<<∞− F maps to 21

f21

≤≤−

→21

FF

f21

s≤=≤

− → max2FFs ≥ Nyquist rate

Example 1: ( )( ) Hz50Ft100tx

Hz10Ft20tx

22a

11a

=⇒=

=⇒=

π

π

cos

cos

If we digitize both of these signals with Fs = 40Hz, then

( ) ( ) n2

n40

2040n

xnTxnx 1as1a1ππ

coscos ==

==

( )

( ) !!cos

coscoscos

nxn2

n2

2n2

5n

40100

40n

xnx

1

2a2

==

+===

=

π

ππ

ππ

Therefore, by this sampling rate x2(n) has become same as x1(n), which is an aliasing

error. Equivalently, in this case, the frequency of 50 Hz is an alias of 10 Hz by sampling

max2FFs ≥

4

rate of 40 Hz. Furthermore, all frequencies (F1 + 40K) are aliases of F1. Hence, do not

use the Nyquist rate blindly.

Example 2: ( )43421443442143421

321 xxx

a t100t30010t503tx πππ cossincos −+=

Nyquist rate? f1 = 25, f2 = 150, f3 = 50 Hz

→ Fmax = 150 Hz → Fs = 300 Hz?!

Problem: with Fx = 300 Hz, ( ) 0n10Fn

xnxs

2a2 ==

= πsin all the time!

If it had a phase shift no <<θ , then it would have been fine, but it is best to choose a

higher sampling rate.

Sampling

( ) ( ) ( )tptxtx ap ⋅=

( ) ( )∑+∞

∞−−= snTttp δ

( ) ( ) ( ) ( )

( ) ( ) ( )[ ]

( ) ( )

( ) ( )sas

p

Kss

s

ap

ssnTtp

kXT

X

kT

P

PXX

nTtnTxtxnxx

ωωω

ωωδπ

ω

ωωπ

ω

δ

−=

−=

=

−==

∑

∑

∑

∞+

∞−

∞+

∞−

+∞

∞−=

1

2

*21

|

Therefore, Xp(ω) is a periodic function of shifted the Xa(ω).

x

P(t)

xa(t) x(n)

Ts

x(n)

( )txa

5

( )ωX

Obviously, if Ms ωω 2≥ there is no

aliasing and the signal can be

reconstructed accurately.

In practice however, generating very narrow impulse is very difficult. Therefore, the

practical way for sampling is zero-order hold. Such a system samples xa(t) at a given

sampling instant and holds that value until the succeeding sampling instant.

Reconstruction of xr(t) from the output of this system requires a cascade of low-pass

filters or a non-constant gain of LPF.

• • •

Xp(ω)

ωM ωs

xp(t)It is as if ( )→txa x

P(t) H0(t)( )txo

( )( )ωr

r

H

thxr(t)

|Hr(ω)|

-ωs /2 ωs /2

wωs /2

-ωs /2 2π

<Hr(ω)