frequency domain representation of sinusoids: continuous time consider a sinusoid in continuous...
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Frequency Domain Representation of Sinusoids: Continuous Time
Consider a sinusoid in continuous time:
tFjjtFjj eeA
eeA
tFAtx
00 22
0
22
)2cos()(
Frequency Domain Representation:
0F
2
A
0F
2
A
magnitude
phaseradians
)(HzF
)(HzF
Example
Consider a sinusoid in continuous time:
tjjtjj eeee
ttx
200015.0200015.0 55
)15.02000cos(10)(
Represent it graphically as:
000,1
5
15.0
magnitude
phase
15.0
radians
5
000,1 )(HzF
)(HzF
Continuous Time and Frequency Domain
In continuous time, there is a one to one correspondence between a sinusoid and its frequency domain representation:
0F
2
A
0F
2
A
magnitude
phase
radians
)(HzF
)(HzF
t
)(tx
A
0T
One-to-One correspondence (no ambiguity!!)
Example
Let
Given this sinusoid, its frequency, amplitude and
phase are unique
500
2
3.0
magnitude
phase
radians
)(HzF
)(HzF
2
500
3.0
0 5 10 15 20-5
0
5
msecmsec
)(tx
Example
Consider a sinusoid in discrete time:
njjnjj eeee
nnx
45.035.045.035.0 44
)35.045.0cos(8][
Represent it graphically as:
45.0
4
35.0
magnitude
phase 35.0radians
4
45.0 )(rad
)(rad
Frequency Domain Representation of Sinusoids: Discrete Time
Same for a sinusoid in discrete time:
njjnjj eeA
eeA
nAnx
00
22
)cos(][ 0
2
A
magnitude
phase
0
2
A
0 )(rad
)(rad
Frequency Domain Representation:
Discrete Time and Frequency Domain
In discrete time there is ambiguity.
All these sinusoids have the same samples:
)cos(
)cos(
)cos(][
2
1
0
nA
nA
nAnx
201 k
02 2 k
with k integer
Example
All these sinusoids have the same samples:
)2.09.3cos(5)2.0)1.04cos((5
)2.09.1cos(5)2.0)1.02cos((5
)2.01.4cos(5)2.0)41.0cos((5
)2.01.2cos(5)2.0)21.0cos((5
)2.01.0cos(5][
nn
nn
nn
nn
nnx
… and many more!!!
Ambiguity in the Digital Frequency
][nx
n
02
A
2
A
0 )(rad
02
2
A
2
A
)(rad02
02
2
A
2
A
)(rad02
The given sinusoid can come from any of these
frequencies, and many more!
In Summary
A sinusoid with frequency 0
)cos(][ 0 nAnx
is indistinguishable from sinusoids with frequencies
,...2,...,4,2
,...2,...,4,2
000
000
k
k
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
2
2
0
0
k
k
)(rad0 00
…all aliases here…
……
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)
)(rad0 00
][nx
n
)(rad0
2
A
2
A
magnitude
phase
Continuous Time to Discrete Time
Now see what happens when you sample a sinusoid: how do we relate analog and digital frequencies?
)(tx ][nx
sF
)(HzF0F0F )(rad00
sF
F00 2
Which Frequencies give Aliasing?
s
s
s
s
s
s
F
FkFk
F
F
F
kFFk
F
F
00
00
222
222
Aliases:
0
0
FkF
kFF
s
s
k integer
0 0F0F
……
)(HzF
0FFs 0FFs …
2SF
2SF
Example
Given: a sinusoid with frequency
sampling frequency
kHzF 20
kHzFs 10
the aliases (ie sinusoids with the same samples as the one given) have frequencies
,...28,18,8
,...32,22,12
0
0
kHzkHzkHzFkF
kHzkHzkHzkFF
s
s
Example
)(tx ][nx
kHz0.15
)(kHzF0.4 )(rad
15
8
radF
F
s 15
82 0
0
0.415
8
Aliased Frequencies
0F 0FFs 2sF
2sF
F
0
0FFs
sF
F00 2
aliases
Sampling Theorem for Sinusoids
)(tx
)(][ snTxnx
ss TF
1
DAC
Digital to Analog
Converter
2/|| 0 sFF 0F
)(ty
If you sample a sinusoid with frequency such that , there is no loss of information (ie you reconstruct the same sinusoid)
F2/sF2/sF
magnitude
Extension to General Signals: the Fourier Series
Any periodic signals with period can be expanded in a sum of complex exponentials (the Fourier Series) of the form
0T
k
tkFjkeatx 02)(
00
1
TF
with
the fundamental frequency
ka The Fourier Coefficients
Example
A sinusoid with period sec10sec0.1 30
mT
We saw that we can write it in terms of complex exponentials as
)1.02000cos(5)( ttx
tjjtjj eeeetx 20001.020001.0 5.25.2)(
Which is a Fourier Series with
HzF 10000
1 if 0
5.2
5.21.0
1
1.01
ka
ea
ea
k
j
j
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
k
tkFjkeatx 02)(
0
2/
2/
22/
2/
20
0
0
0
0
0)( Tadteadtetx mk
T
T
tFmkjk
T
T
tmFj
otherwise 0
if 0 mkT
Fourier Series and Fourier Coefficients
2/
2/
2
0
0
0
0)(1
T
T
tkFjk dtetxT
a
k
tkFjkeatx 02)(
Fourier Series:
Fourier Coefficients:
Example of Fourier Series…
sec)(mt1 4
)(tx
2
Period sec104 30
T HzTF 250/1 00 Fundamental Frequency:
Fourier Coefficients:
12104
1
0 if 2/sin
22104
1
3
3
3
3
10
10
30
10
10
25023
dta
kk
kdtea tkj
k
… Plot the Coefficients
Fourier Coefficients:
12104
1
0 if 2/sin
22104
1
3
3
3
3
10
10
30
10
10
25023
dta
kk
kdtea tkj
k
)(HzF
1636.0
212.01273.0
|| ka
250 750 12502507501250
0909.01750
Parseval’s theorem
)(HzF
1636.0
212.01273.0
|| ka
250 750 12502507501250
0909.01750
k
k
T
T
adttxT
22/
2/
2
0
|||)(|1 0
0
The Fourier Series coefficients are related to the average power as
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 without loss of information (ie you get the same signal back)
BFs 2
)(tx
)(][ snTxnx
ss TF
1
DAC
Digital to Analog
Converter
)()( txty
F2/sF2/sF
magnitude
B
t
t
Example
)3000cos(3)1.02000cos(2)( tttx
it has two frequencies
5.10.15.1 0.1)(kHzF
The bandwidth is kHzB 5.1
The sampling frequency has to be kHzBFs 0.32
so that we can sample it without loss of information
Example
The bandwidth of a Hi Fidelity audio signal is approximately
since we cannot hear above this frequency.
kHzB 22
The music on the Compact Disk is sampled at
0.22)(kHzF
kHzFs 1.44
i.e. 44,100 samples for every second of music
Example
For an audio signal of telephone quality we need only the frequencies up to 4kHz.
The sampling frequency on digital phones is
kHzFs 8