DISCRETE-TIME SINUSOIDAL SIGNALS Copyright 2012 | Instructor: Dr. Gülden Köktürk |...

DISCRETE-TIME SINUSOIDAL SIGNALS

Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

x[n]=Acos(Ωn+Φ);

n is integer (discrete-time variable)

n is called the sample number.

A: amplitude

Ω: discrete frequency (radians/sample(

Φ: phase (radians)

Instead of Ω, we can use the discrete cyclic frequency variable, F.

F: discrete cyclic frequency (cycles/sample)


Example:

x[n]=Acos


Property 1- A discrete-time sinusoid is periodic ONLY IF ITS FREQUENCY F IS RATIONAL NUMBER I (Compare with continuous sinusoids: A continuous-time sinusoid is always periodic, for any value of its frequency, f0)

Definition: x[n] is periodic with period N (N>0 and integer) if and only if x[n]=x[n+N] for all n.

The smallest value of N for which above equation is true is called the fundamental period.

Properties of Discrete-Time Sinusoids:


If x[n] is periodic; thenThis equation is true if there exists an integer K such that which is a rational number.Thus, x[n] is periodic if and only if its frequency F can be expressed as the ration of two integers.

Proof of Periodicity Property:


To determine the fundamental period N of a periodic discrete-time sinusoid, we express its frequency F as a rational number, , and cancel common factors so that K and N are relatively prime. Then, the fundamental period is equal to N. For example,

If (sinusoid repeats itself every 60 samples)

If (sinusoid repeats itself every 2 samples)


Property 2- Discrete-time sinusoids whose frequencies are separated by an integer multiple of 2π are identical.

Thus, any two discrete-time sinusoids with frequencies in the range

are distinct (unique).

A sinusoid with is identical to a sinusoid with frequency


We call the sinusoid with an alias of a corresponding sinusoid with .

(Compare with continuous sinusoids: Continuous-time sinusoids are distinct for all frequency values

-)

Property 3- Highest rate of oscillation in a discrete-time sinusoid is attained when (or equivalently, for F=1/2 or -1/2)


To illustrate this property, let x[n]=cos (Ω1n)

Consider frequency Ω1n=0, π/8, π/4, π/2, π (or equivalently, F1=0, 1/16, 1/8, 1/4, 1/2) which result in periodic discrete-time sinusoids with periods N= ∞, 16, 8, 4, 2.


To see what happens for π<Ω<2π, we consider the sinusoids with Ω1 and 2π-Ω1. Note that as Ω1 varies from π to 2π, 2π-Ω1 varies from π to 0. Let

Thus, x2[n] is an alias of x1[n].

As with continuous-time sinusoids, complex exponentials and negative frequencies can be introduced for discrete-time sinusoids.

(Euler formula)


Since discrete-time sinusoids with frequencies that are seperated by an integer multiple of 2π are identical, frequencies in any interval constitute all the existing discrete-time sinusoids or complex exponentials. Hence, frequency range for discrete-time sinusoids is finite with duration 2π. Usually, we choose the range

which we call the fundamental range.


Sampling is described by the equation

x[n] is the discrete-time signal obtained by taking samples of the analog signal xa(t) every Ts seconds.

SAMPLING OR ANALOG SIGNALS


Relation between continuous-time, t and discrete-time, n:

As a result of this, there exists a relationship between continuous frequency f0 (ω0) and discrete frequency F (or Ω).

Consider analog sinusoid

xa(t)=Acos(2πf0nTs+Φ)


discrete frequency

Recall that for continuous-time

However, for discrete-time sinusoids


Since the highest frequency in a discrete-time signal is ,

It follows that, with sampling rate fS, the corresponding highest values for f0 and ω0 are

What happens to frequencies above ? Consider following example:


Ex: x1(t)=cos(2π10t)

x2(t)=cos(2π50t)

Let x1(t) and x2(t) be sampled with sampling frequency fs=40 Hz.

However, since


discrete sinusoidal signals are identical.

Thus, we say that the frequency f02=50 Hz is an alias of the frequency f01=10 Hz at the sampling rate of fS=40 samples/sec.

f02=50 Hz is not the only alias of f01=10 Hz. In fact, for fS=40 samples/sec, f03=90 Hz is also an alias. f04=130 Hz is also an alias. All of the continuous sinusoids

cos[2π(10+40k)t]; k=1, 2, 3, 4, ... when sampled at 40 samples/sec, give identical values.

they are all aliases of fo1=10 Hz.


In general, sampling continuous sinusoid

xa(t)=Acos(2πf0nTs+Φ)

with sampling rate

Results in the discrete sinusoid

x[t]=Acos(2πFn+Φ) where

If we assume , frequency F of x[n] is the range which is the range for discrete sinusoids


On the other hand, if the sinusoids

xa(t)=Acos(2πf0kt+Φ) where f0k=f0+kfS

Are sampled at fS; then the sampled (discrete) sinusoid

which is identical to the discrete-time sinusoid obtained by sampling Acos(2πf0t+Φ) with fS.


Thus, frquencies f0k=f0+kfS;

Are indistinguishable from the frequency f0 after sampling with fS; hence they are aliases of f0.

Example of aliasing:


Two continuous-time with ,

Note that for k=+1 f01=f02+fS=(-7/8+1)=1/8

The frequency f0=fS/2 is called the folding frequency.


Example: Consider the analog signal xa(t)=3cos(100πt)

(a) Determine the minimum sampling rate to avoid aliasing.

Answer: since f0=50 Hz, the min. sampling rate to avoid aliasing is fS=100 Hz.

(b) If fS=200 Hz, what is the discrete-time signal obtained after sampling?

Answer:


(c) If fS=75 Hz, what is the discrete-time signal obtained after sampling?

Answer:

(d) What is the frequency 0<f0<fS/2 of a sinusoid that gives samples identical to those obtained in part (c)?

Answer: for fS=75 Hz

we have f0=FfS=75F


in part (c), we found the discrete-frequency F=1/3. Hence,

the sinusoid

ya(t)=3cos(2πf0t)=3cos(50πt) sampled at fS=75 samples/sec gives identical samples.

f0=50 Hz is an alias of f0=25 Hz for fS=75 Hz.


Given an analog signal, how should we select the sampling period TS, or equivalently, the sampling rate, fS?

We know that the highest frequency in an analog signal that can be unambiguously reconstructed when the signal is sampled at a rate fS=1/Ts is fS/2. Any frequency above fS/2 or below –fS/2 results in samples that are identical with a corresponding frequency in the range . To avoid the ambiguities resulting from aliasing, we must select the sampling rate to be sufficiently high.

The Sampling Theorem


That is, we must select the fS/2˃f0max where f0max is the highest possible frequency of the analog signal. Thus, to avoid the problem of aliasing, fS is selected so that

With the sampling rate selected in this manner, any frequency component in the analog signal is mapped into a discrete-time frequency value

or equivalently

The sampling rate fS=2f0max is called the Nyquist rate.


Example: Consider the analog signal

xa(t)=3cos(50πt)+10sin(300πt)-cos(100πt)

What is the Nyquist rate for this signal?

Answer: frequencies present in xa(t) are f1=25 Hz, f2=150 Hz, f3=50 Hz

f0max=150 Hz Nyquist rate=300 Hz

We should satisfy to avoid aliasing.

DISCRETE-TIME SINUSOIDAL SIGNALS Copyright 2012 | Instructor: Dr. Gülden Köktürk |...

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