DISCRETE-TIME
SINUSOIDAL SIGNALS
x[n]=Acos(Ωn+Φ); −∞ < n < ∞
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.
Ω = 2πF
⟹ x n = Acos 2πFn + ∅ ; −∞ < n < ∞
F: discrete cyclic frequency (cycles/sample)
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
Example:
π
n
6
π
π
+ ⟹ Ω = rad/samples
3
6
cycles
π
⟹F=⋯
and ∅ = radians
sample
3
x[n]=Acos
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
Properties of DiscreteTime Sinusoids:
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.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
Proof of Periodicity
Property:
If x[n] is periodic; then
cos 2πF N + n + ∅ = cos 2πFn + ∅
x[n+N]
x[n]
This equation is true if there exists an integer
K such that 2πFN = k2π
⟹F=
k
N
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.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
To determine the fundamental period N of a
periodic discrete-time sinusoid, we express
k
its frequency F as a rational number, F = ,
N
and cancel common factors so that K and N
are relatively prime. Then, the fundamental
period is equal to N. For example,
31
If F1 =
then N1 = 60 samples
60
repeats itself every 60 samples)
30
If F1 =
then N2 = 2 samples
60
repeats itself every 2 samples)
(sinusoid
(sinusoid
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
Property 2- Discrete-time sinusoids whose
frequencies are separated by an integer
multiple of 2π are identical.
cos Ω + 2π n + ∅ = cos Ωn + 2πn + ∅
= cos Ωn + ∅
Thus, any two discrete-time sinusoids with
frequencies in the range
1
1
−π ≤ Ω ≤ π or − ≤ F ≤
2
2
are distinct (unique).
1
A sinusoid with Ω > π (or F > )
is
2
identical to a sinusoid with frequency Ω <
1
π (or F < )
2
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
We call the sinusoid with Ω > π , an alias
of a corresponding sinusoid with Ω < π .
(Compare with continuous sinusoids:
Continuous-time sinusoids are distinct for all
frequency values
-∞ < ω0 < ∞ or − ∞ < f0 < ∞)
Property 3- Highest rate of oscillation in a
discrete-time sinusoid is attained when Ω =
π (or Ω = −π) (or equivalently, for F=1/2 or 1/2)
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
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.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
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
x1 n = Acos Ω1 n
x2 n = Acos 2π −Ω1 n = Acos Ω1 n = x1 n
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.
A j(Ωn+ϕ) A −j(Ωn+ϕ)
x n = Acos Ωn + ϕ = e
+ e
2
2
(Euler formula)
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
Since
discrete-time
sinusoids
with
frequencies that are seperated by an integer
multiple of 2π are identical, frequencies in
any interval Ω1 ≤ Ω ≤ Ω1 + 2π 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
0 ≤ Ω ≤ 2π or − π ≤ Ω ≤ π
1
1
(0 ≤ F ≤ 1 or − ≤ F ≤ )
2
2
which we call the fundamental range.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
SAMPLING OR ANALOG
SIGNALS
Sampling is described by the equation
x n = xa nTs ; −∞ < n < ∞
x[n] is the discrete-time signal obtained by
taking samples of the analog signal xa(t)
every Ts seconds.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
Relation between continuous-time, t and
discrete-time, n:
n
t = nTs =
fs
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)
sample with TS ⟹ xa nTS = x n
2πf0 n
= Acos 2πf0 nTS + ϕ = Acos(
+ ϕ)
fS
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
⟹discrete frequency
F=
f0
fS
or equivalently Ω = ω0 TS
Recall that for continuous-time
−∞ < f0 < ∞ and − ∞ < ω0 < ∞
However, for discrete-time sinusoids
−1
2
≤F≤
1
2
and
⇓
−1
2
−1
2TS
=
−fS
2
≤
f0
fS
≤
1
2
⇓
fS
< f0 < =
2
−π ≤ Ω ≤ π
⇓
−π ≤ ω0 TS ≤ π
⇓
−π
= −πfS ≤ ω0 ≤ πfS =
1 TS
TS
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
π
TS
Since the highest frequency in a discrete1
time signal is Ω = π (or F = ),
2
It follows that, with sampling rate fS, the
corresponding highest values for f0 and ω0
are
fS
1
π
f0 max = =
or ω0 max = πfS =
2 2TS
TS
What happens to frequencies above
Consider following example:
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
fS
2
?
Ex: x1(t)=cos(2π10t)
x2(t)=cos(2π50t)
Let x1(t) and x2(t) be sampled with sampling
frequency fs=40 Hz.
10
π
x1 n = cos 2π n = cos( n)
40
2
50
5π
x2 n = cos 2π n = cos( n)
40
2
However, since
5π
π
π
cos
n = cos 2πn + n = cos( n)
2
2
2
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
⟹ x1 n = x2 [n] ⟹ 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.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
In general, sampling continuous sinusoid
xa(t)=Acos(2πf0nTs+Φ)
with sampling rate fs =
1
Ts
Results in the discrete sinusoid
x[t]=Acos(2πFn+Φ) where F =
If we assume
−fS
2
< f0 <
−1
fS
,
2
f0
Ts
frequency F of
1
x[n] is the range
≤ F ≤ which is the
2
2
range for discrete sinusoids
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
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
f0 + kfS
x n = xa nTS = Acos(2π
n + ∅)
fS
f0
= Acos 2πn + ∅ + 2πkn
fS
F
= Acos(2πFn + ∅)
which is identical to the discrete-time
sinusoid obtained by sampling Acos(2πf0t+Φ)
with fS.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
Thus, frquencies f0k=f0+kfS; k = ∓1, ∓2, ⋯
Are indistinguishable from the frequency f0
after sampling with fS; hence they are
aliases of f0.
Example of aliasing:
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
1
Two continuous-time with f01 = Hz,
8
7
f02 = − Hz, sampling rate fS = 1 Hz (k = +1)
8
Note that for k=+1 f01=f02+fS=(-7/8+1)=1/8
The frequency f0=fS/2 is called the folding
frequency.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
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:
100πn
π
x n = 3 cos
= 3cos( n)
200
2
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
(c) If fS=75 Hz, what is the discrete-time
signal obtained after sampling?
Answer:
100πn
4π
x n = 3 cos
= 3 cos
n
75
3
2π
2π
= 3 cos 2π −
n = 3cos( n)
3
3
(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
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
in part (c), we found the discrete-frequency
1
F=1/3. Hence, f0 = 75. = 25 Hz.
3
⟹ 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.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
The Sampling Theorem
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
−fS
f
in the range
< f0 < S . To avoid the
2
2
ambiguities resulting from aliasing, we must
select the sampling rate to be sufficiently
high.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
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
fS > 2f0 max (sampling theorem)
With the sampling rate selected in this
manner, any frequency component f0 <
f0 max in the analog signal is mapped into a
discrete-time frequency value
−1
2
≤F=
f0
fS
1
2
≤ or equivalently
−π ≤ Ω = 2πF ≤ π
The sampling rate fS=2f0max is called the
Nyquist rate.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING
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 f0 > 300 samples/sec to
avoid aliasing.
Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING