Frequency Domain Representation of Sinusoids:
Continuous Time
Consider a sinusoid in continuous time:
x(t )  A cos(2F0t   )
 A j  j 2F0t  A  j   j 2F0t
  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  j0n  A  j   j0n
  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 2kF0t
a
e
 k
k  
with
F0 
ak
1
T0
the fundamental frequency
The Fourier Coefficients
Example
A sinusoid with period T0  1.0m sec  103 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
a1  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 2kF0t
a
e
 k
k  
T0 / 2


 j 2mF0t
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 2kF0t
a
e
 k
k  
Fourier Coefficients:
1
ak 
T0
T0 / 2
 j 2kF0t
x
(
t
)
e
dt

T0 / 2
Example of Fourier Series…
x(t )
2
1
Period T0  4 103 sec
4
t (m sec)
Fundamental Frequency: F0  1 / T0  250Hz
Fourier Coefficients:
1
ak 
4 103
1
a0 
4 103
10 3
 2e
 j 2k 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 103
1
a0 
4 103
10 3
 2e
 j 2k 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