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Frequency Domain Representation of Sinusoids:
Continuous Time
Consider a sinusoid in continuous time:
x(t ) A cos(2F0t )
A j j 2F0t A j j 2F0t
e e
e e
2
2
Frequency Domain Representation:
magnitude
A
2
A
2
F0
radians
phase
F0
F (Hz)
F (Hz)
Example
Consider a sinusoid in continuous time:
x(t ) 10cos(2000 t 0.15)
5e j 0.15 e j 2000 t 5e j 0.15 e j 2000 t
Represent it graphically as:
5
5
magnitude
1,000
1,000
F (Hz)
radians
phase
0.15
0.15
F (Hz)
Continuous Time and Frequency Domain
In continuous time, there is a one to one correspondence between a
sinusoid and its frequency domain representation:
magnitude
x(t )
A
2
A
2
F0
A
F0
F (Hz)
phase
t
T0
radians
One-to-One correspondence
(no ambiguity!!)
F (Hz)
Example
Let
magnitude
x(t )
2
5
2
500
0
500
F (Hz)
phase
-5
0
5
10
msec
msec
15
Given this sinusoid, its
frequency, amplitude and
phase are unique
20
radians
0 .3
0 .3
F (Hz)
Example
Consider a sinusoid in discrete time:
x[n] 8 cos(0.45 n 0.35)
4e j 0.35 e j 0.45 n 4e j 0.35 e j 0.45 n
Represent it graphically as:
4
4
magnitude
0.45
0.45
(rad )
radians
phase
0.35
0.35
(rad )
Frequency Domain Representation of Sinusoids:
Discrete Time
Same for a sinusoid in discrete time:
x[n] A cos(0 n )
A j j0n A j j0n
e e
e e
2
2
Frequency Domain Representation:
magnitude
A
2
0
phase
A
2
0 (rad )
(rad )
Discrete Time and Frequency Domain
In discrete time there is ambiguity.
All these sinusoids have the same samples:
x[n] A cos(0 n )
A cos(1n )
A cos(2 n )
1 0 k 2
2 k 2 0
with k integer
Example
All these sinusoids have the same samples:
x[n] 5 cos(0.1 n 0.2)
5 cos((0.1 2 )n 0.2) 5 cos(2.1 n 0.2)
5 cos((0.1 4 )n 0.2) 5 cos(4.1 n 0.2)
5 cos((2 0.1 )n 0.2) 5 cos(1.9 n 0.2)
5 cos((4 0.1 )n 0.2) 5 cos(3.9 n 0.2)
… and many more!!!
Ambiguity in the Digital Frequency
A
2
A
2
0 0
(rad )
x[n]
n
The given sinusoid can come
from any of these
frequencies, and many more!
A
2
2 0
A
2
2 0
A
2
2 0
(rad )
A
2
2 0 (rad )
In Summary
A sinusoid with frequency 0
x[n] A cos(0n )
is indistinguishable from sinusoids with frequencies
0 2 , 0 4 ,...,0 k 2 ,...
2 0 , 4 0 ,...,k 2 0 ,...
These frequencies are called aliases.
Where are the Aliases?
Notice that, if the digital frequency is in the interval
0
all its aliases are outside this interval
0 k 2
0 k 2
…
…
0 0
0
…all aliases here…
(rad)
Discrete Time and Frequency Domains
If we restrict the digital frequencies within the interval
there is a one to one correspondence between sampled
sinusoids and frequency domain representation (no aliases)
A
2
x[n]
A
2
magnitude
0 0 0 (rad)
n
0
phase
(rad)
Continuous Time to Discrete Time
Now see what happens when you sample a sinusoid: how do we
relate analog and digital frequencies?
x(t )
x[n]
Fs
F0
F0
0
F (Hz)
0 2
F0
Fs
0
(rad)
Which Frequencies give Aliasing?
F0
F0 kFs
2
k 2 2
Fs
Fs
F0
kFs F0
2
k 2 2
Fs
Fs
Fs F0
…
…
…
F0 0 F0
F
S
2
Aliases:
Fs F0
F0 kFs
kFs F0
F (Hz)
FS
2
k integer
Example
Given: a sinusoid with frequency F0 2kHz
sampling frequency
Fs 10kHz
the aliases (ie sinusoids with the same samples as the one given)
have frequencies
F0 kFs 12kHz, 22kHz, 32kHz,...
kFs F0 8kHz, 18kHz, 28kHz,...
Example
x(t )
x[n]
15 .0kHz
4 .0
4 .0
F (kHz)
8
15
F0 8
0 2
rad
Fs 15
8
15
(rad)
Aliased Frequencies
F
s
2
F0
0 2
F0
Fs
Fs
2
Fs F0 Fs F0
aliases
0
F
Sampling Theorem for Sinusoids
If you sample a sinusoid with frequency F0 such that | F0 | Fs / 2 ,
there is no loss of information (ie you reconstruct the same sinusoid)
x(t )
x[n] x(nTs )
magnitude
Fs / 2
Fs / 2
F
Fs
1
Ts
y(t )
DAC
Digital to
Analog
Converter
Extension to General Signals: the Fourier Series
Any periodic signals with period T0 can be expanded in a sum of
complex exponentials (the Fourier Series) of the form
x(t )
j 2kF0t
a
e
k
k
with
F0
ak
1
T0
the fundamental frequency
The Fourier Coefficients
Example
A sinusoid with period T0 1.0m sec 103 sec
x(t ) 5 cos(2000 t 0.1)
We saw that we can write it in terms of complex exponentials as
x(t ) 2.5e j 0.1 e j 2000 t 2.5e j 0.1 e j 2000 t
Which is a Fourier Series with
F0 1000Hz
a1 2.5e j 0.1
a1 2.5e j 0.1
ak 0 if k 1
Computation of Fourier Coefficients
For general signals we need a way of determining an expression for the
Fourier Coefficients.
From the Fourier Series multiply both sides by a complex exponential and
integrate
x(t )
j 2kF0t
a
e
k
k
T0 / 2
j 2mF0t
j
2
k
m
F
t
0
a T
x
(
t
)
e
dt
a
e
dt
k
m 0
k
T0 / 2
T0 / 2
T0 / 2
T0 if k m
0 otherwise
Fourier Series and Fourier Coefficients
Fourier Series:
x(t )
j 2kF0t
a
e
k
k
Fourier Coefficients:
1
ak
T0
T0 / 2
j 2kF0t
x
(
t
)
e
dt
T0 / 2
Example of Fourier Series…
x(t )
2
1
Period T0 4 103 sec
4
t (m sec)
Fundamental Frequency: F0 1 / T0 250Hz
Fourier Coefficients:
1
ak
4 103
1
a0
4 103
10 3
2e
j 2k 250 t
10 3
10 3
2dt 1
10 3
sin k / 2
dt 2
if k 0
k
… Plot the Coefficients
| ak |
1250
750
250
1
0.636
250
0.212
750
0.1273 0.0909
1250
1750
Fourier Coefficients:
1
ak
4 103
1
a0
4 103
10 3
2e
j 2k 250 t
10 3
10 3
2dt 1
10 3
sin k / 2
dt 2
if k 0
k
F (Hz)
Parseval’s theorem
| ak |
1250
750
250
1
0.636
250
0.212
750
0.1273 0.0909
1250
1750
F (Hz)
The Fourier Series coefficients are related to the average
power as
1
T0
T0 / 2
2
|
x
(
t
)
|
dt
T0 / 2
2
|
a
|
k
k
Sampling Theorem
If a signal is a sum of sinusoids and B is the maximum frequency (the
Bandwidth) you can sample it at a sampling frequency Fs 2B without
loss of information (ie you get the same signal back)
x(t )
x[n] x(nTs )
t
magnitude
Fs / 2
B
F
Fs / 2
Fs
1
Ts
y (t ) x(t )
DAC
t
Digital to
Analog
Converter
Example
x(t ) 2 cos(2000 t 0.1 ) 3 cos(3000 t )
it has two frequencies
1 .5 1 .0
The bandwidth is
1 .0 1 .5
F (kHz)
B 1.5kHz
The sampling frequency has to be
Fs 2B 3.0kHz
so that we can sample it without loss of information
Example
The bandwidth of a Hi Fidelity audio signal is
approximately
B 22 kHz
22 .0
since we cannot hear above this frequency.
The music on the Compact
Disk is sampled at
Fs 44.1kHz
i.e. 44,100 samples for every second of music
F (kHz)
Example
For an audio signal of telephone quality we need only the
frequencies up to 4kHz.
The sampling frequency on digital phones is
Fs 8kHz