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Discrete Time Periodic Signals
Definition:
A discrete time signal x[n] is periodic with period N if and only if
x[ n ] x[ n N ]
for all n .
Meaning: a periodic signal keeps repeating itself forever!
x[n]
N
n
Example: a Sinusoid
Consider the Sinusoid:
x [ n ] 2 cos 0.2 n 0.9
It is periodic with period N 10 since
x [ n 10 ] 2 cos 0 . 2 ( n 10 ) 0 . 9
2 cos 0 . 2 n 0 . 9 2 x [ n ]
for all n.
General Periodic Sinusoid
Consider a Sinusoid of the form:
k
x [ n ] A cos 2
n
N
with k, N integers.
It is periodic with period N since
k
x [ n N ] A cos 2
(n N )
N
k
A cos 2
n k 2 x[ n ]
N
for all n.
Example of Periodic Sinusoid
Consider the sinusoid:
x [ n ] 5 cos 0 . 3 n 0 . 1
We can write it as:
3
x [ n ] 5 cos 2
n 0 . 1
20
It is periodic with period N 20 since
x [ n 20 ] 5 cos 0 . 3 ( n 20 ) 0 . 1
5 cos 0 . 3 n 0 . 1 3 2 x [ n ]
for all n.
Periodic Complex Exponentials
Consider a Complex Exponential of the form:
x [ n ] Ae
2
j k
n
N
It is periodic with period N since
x [ n N ] Ae
Ae
2
j k
(n N )
N
2
j k
n
N
1
e
jk 2
x[ n ]
for all n.
Example of a Periodic Complex Exponential
Consider the Complex Exponential:
x[ n ] (1 2 j ) e
j 0 . 1 n
We can write it as
x [ n ] (1 2 j ) e
1
j 2
n
20
and it is periodic with period N = 20.
Reference Frames
Goal:
We want to write all discrete time periodic signals in terms
of a common set of “reference signals”.
It is like the problem of representing a vector
reference frame defined by
x
• an origin “0”
• reference vectors e1 , e 2 ,...
x
e2
Reference
Frame
0
e1
in a
Reference Frames in the Plane and in Space
For example in the plane we need two reference vectors
x
e2
Reference
Frame
0
e1 , e 2
e1
… while in space we need three reference vectors e1 , e 2 , e 3
Reference
Frame
x
e3
e2
0
e1
A Reference Frame in the Plane
If the reference vectors have unit length and they are
perpendicular (orthogonal) to each other, then it is very simple:
a 2 e2
x a 1 e1 a 2 e 2
a 1 e1
0
Where
a1
a2
projection of x along e1
projection of x along e 2
The plane is a 2 dimensional space.
A Reference Frame in the Space
If the reference vectors have unit length and they are
perpendicular (orthogonal) to each other, then it is very simple:
a 3 e3
a 2 e2
Where
0
x a1e1 a 2 e 2 a 3 e 3
a 1 e1
a 1 projection of x along e1
a 2 projection of x along e
2
a 3 projection of x along e
3
The “space” is a 3 dimensional space.
Example: where am I ?
x
N
E
e2
0
200 m
e1 300 m
Point “x” is 300m East and 200m North of point “0”.
Reference Frames for Signals
We want to expand a generic signal into the sum of reference
signals.
The reference signals can be, for example, sinusoids or complex
exponentials
x[n ]
n
reference signals
Back to Periodic Signals
A periodic signal x[n] with period N can be expanded in terms of N
complex exponentials
ek [ n ] e
j 2
k
N
n
k 0 ,..., N 1
,
as
N 1
x[ n ]
a
k 0
k
ek [n ]
A Simple Example
Take the periodic signal x[n] shown below:
2
1
n
0
Notice that it is periodic with period N=2.
Then the reference signals are
e0 [ n ] e
e1 [ n ] e
j 2
j 2
x[ n ] 1 . 5 e 0 [ n ] 0 . 5 e1 [ n ]
n
n
n
2
We can easily verify that (try to believe!):
1 . 5 1 0 . 5 ( 1)
0
for all n.
1
2
n
1 1
n
( 1)
n
Another Simple Example
Take another periodic signal x[n] with the same period (N=2):
1 .3
0
n
0 .3
Then the reference signals are the same
e0 [ n ] e
e1 [ n ] e
We can easily verify that (again try to believe!):
x [ n ] 0 . 5 e 0 [ n ] 0 . 8 e1 [ n ]
0 . 5 1 0 . 8 ( 1)
n
n
for all n.
Same reference signals, just different coefficients
j 2
0
n
2
j 2
1
2
n
1 1
n
( 1)
n
Orthogonal Reference Signals
Notice that, given any N, the reference signals are all orthogonal
to each other, in the sense
0 if m k
e [ n ]e m [ n ] N if m k
n0
N 1
*
k
Since
N 1
n0
N 1
e k [ n ]e m [ n ]
*
n0
j 2
e
mk
N
n
j 2
e
n0
N 1
mk
N
n
j 2 ( m k )
1 e
mk
j 2
N
1 e
by the geometric sum
… apply it to the signal representation …
N 1
n0
N 1
N 1
*
*
x [ n ]e k [ n ] a m e m [ n ] e k [ n ]
n0 m 0
x[ n ]
N 1
N 1 *
a m e k [ n ]e m [ n ] Na k
m 0
n0
and we can compute the coefficients. Call X [ k ] Na k then
N 1
X [ k ] Na k
x [ n ]e
n0
j
2
N
kn
,
k 0 ,..., N 1
Discrete Fourier Series
Given a periodic signal x[n] with period N we define the
Discrete Fourier Series (DFS) as
N 1
X [k ]
x [ n ]e
j
2
N
kn
,
k 0 ,..., N 1
n0
Since x[n] is periodic, we can sum over any period. The general
definition of Discrete Fourier Series (DFS) is
X [ k ] DFS x [ n ]
n 0 N 1
n n0
for any n
0
j
x [ n ]e
2
N
kn
,
k 0 ,..., N 1
Inverse Discrete Fourier Series
The inverse operation is called Inverse Discrete Fourier Series
(IDFS), defined as
x [ n ] IDFS
X [ k ]
1
N
N 1
k 0
j
X [ k ]e
2
N
kn
Revisit the Simple Example
Recall the periodic signal x[n] shown below, with period N=2:
2
1
n
0
1
X [k ]
j
x [ n ]e
2
2
nk
x [ 0 ] x [1]( 1) 1 2 ( 1) , k 0 ,1
k
k
n0
Then
X [ 0 ] 3,
X [1] 1
Therefore we can write the sequence as
x [ n ] IDFS
1
X [ k ]
1
2
k 0
1 . 5 0 . 5 ( 1)
n
j
X [ k ]e
2
2
kn
Example of Discrete Fourier Series
Consider this periodic signal
x[n ]
1
10
0
n
The period is N=10. We compute the Discrete Fourier Series
9
X [k ]
x [ n ]e
n0
j
2
10
kn
4
e
n0
j
2
10
kn
2
j
5k
10
1 e
2
j
k
10
1 e
5
if k 1, 2, ..., 9
if k 0
… now plot the values …
| X [k ] |
magnitude
5
0
0
X [k ]
2
4
6
8
10
k
8
10
k
phase (rad)
2
0
-2
0
2
4
6
Example of DFS
Compute the DFS of the periodic signal
x [ n ] 2 cos( 0 . 5 n )
Compute a few values of the sequence
x [ 0 ] 2 , x [1] 0 , x [ 2 ] 2 , x [ 3 ] 0 ,...
and we see the period is N=2. Then
1
X [k ]
j
x [ n ]e
2
2
kn
x [ 0 ] x [1] ( 1)
n0
which yields
X [ 0 ] X [1] 2
k
Signals of Finite Length
All signals we collect in experiments have finite length
x[ n ] x ( nT s )
x (t )
Fs
1
Ts
T MAX
N TM AX FS
Example: we have 30ms of data sampled at 20kHz (ie 20,000
samples/sec). Then we have
N 30 10
3
20 10 600
3
data points
Series Expansion of Finite Data
We want to determine a series expansion of a data set of length N.
Very easy: just look at the data as one period of a periodic sequence
with period N and use the DFS:
n
0
N 1
Discrete Fourier Transform (DFT)
Given a finite interval of a data set of length N, we define the Discrete
Fourier Transform (DFT) with the same expression as the Discrete
Fourier Series (DFS):
X [ k ] D F T x [ n ]
N 1
x [ n ]e
j
2
kn
N
k 0, ..., N 1
,
n0
And its inverse
x [ n ] ID F T X [ k ]
1
N
N 1
n0
j
X [ k ]e
2
N
kn
,
n 0, ..., N 1
Signals of Finite Length
All signals we collect in experiments have finite length in time
x[ n ] x ( nT s )
x (t )
Fs
1
Ts
T MAX
N TM AX FS
Example: we have 30ms of data sampled at 20kHz (ie 20,000
samples/sec). Then we have
N 30 10
3
20 10 600
3
data points
Series Expansion of Finite Data
We want to determine a series expansion of a data set of length N.
Very easy: just look at the data as one period of a periodic sequence
with period N and use the DFS:
n
0
N 1
Discrete Fourier Transform (DFT)
Given a finite of a data set of length N we define the Discrete Fourier
Transform (DFT) with the same expression as the Discrete Fourier
Series (DFS):
X [ k ] D F T x [ n ]
N 1
x [ n ]e
j
2
kn
N
k 0, ..., N 1
,
n0
and its inverse
x [ n ] ID F T X [ k ]
1
N
N 1
n0
j
X [ k ]e
2
N
kn
,
n 0, ..., N 1
Example of Discrete Fourier Transform
Consider this signal
x[n ]
1
9
0
n
The length is N=10. We compute the Discrete Fourier Transform
9
X [k ]
x [ n ]e
n0
j
2
10
kn
4
e
n0
j
2
10
kn
2
j
5k
10
1 e
2
j
k
10
1 e
5
if k 1, 2, ..., 9
if k 0
… now plot the values …
| X [k ] |
magnitude
5
0
0
X [k ]
2
4
6
8
10
k
8
10
k
phase (rad)
2
0
-2
0
2
4
6
DFT of a Complex Exponential
Consider a complex exponential of frequency
x[ n ] Ae
j 0 n
,
0 rad.
n
We take a finite data length
x[ n ] Ae
j 0 n
,
0 n N 1
… and its DFT
X [ k ] DFT x [ n ]
N 1
x [ n ]e
j
2
N
kn
, k 0 ,..., N 1
n0
How does it look like?
Recall Magnitude, Frequency and Phase
Recall the following:
1. We assume the frequency to be in the interval
0
2. We represented it in terms of magnitude and phase:
magnitude
0
phase
|A|
( rad )
A
0
( rad )
Compute the DFT…
X [ k ] D F T x [ n ]
N 1
x [ n ]e
j
2
kn
N
n0
N 1
Ae
j 0 n
j
e
2
kn
N
n0
N 1
Ae
2
j k
0
N
n0
Notice that it has a general form:
2
X [k ] A W N k
0
N
where (use the geometric series)
N 1
W N ( )
e
n0
j n
1 e
j N
1 e
j
n
,
k 0, ..., N 1
See its general form:
N 1
W N ( )
e
n0
j n
e
j ( N 1) / 2
N
sin
2
sin
2
… since:
N 1
W N ( )
e
j n
n0
e
j N / 2
e
e
j / 2
1 e
j N
1 e
j
j N / 2
e
j N / 2
j / 2
e
j / 2
e
e
e
j N / 2
j / 2
e j N / 2 e j N / 2 e j N / 2
j ( N 1) / 2
j / 2
e
j / 2
j / 2
e
e
e
N
sin
2
sin
2
… and plot the magnitude
W N ( )
12
N
10
8
6
4
2
0
-3
-2
-1
0
1
2
2
N
N
2
3
Example
Consider the sequence
x[ n ] e
In this case
j 0.3 n
, n 0, ..., 31
0 0 . 3 , N 32
Then its DFT becomes
X [ k ] W 32 0.3
k 2
32
Let’s plot its magnitude:
, k 0, ..., 31
... first plot this …
W 32 0 . 3
40
30
N 32
20
10
0
0
1
0 0 . 3
2
3
4
5
6
2
… and then see the plot of its DFT
X [ k ] W N 0.3
k 2
,
k 0, ..., N 1
N
35
30
25
20
15
10
5
0
0
5
10
The max corresponds to frequency
15
20
25
30
5 2 / 32 0 . 312 0 . 3
k
Same Example in Matlab
Generate the data:
>> n=0:31;
>>x=exp(j*0.3*pi*n);
Compute the DFT (use the “Fast” Fourier Transform, FFT):
>> X=fft(x);
Plot its magnitude:
>> plot(abs(X))
… and obtain the plot we saw in the previous slide.
Same Example in Matlab
Generate the data:
>> n=0:31;
>>x=exp(j*0.3*pi*n);
Compute the DFT (use the “Fast” Fourier Transform, FFT):
>> X=fft(x);
Plot its magnitude:
>> plot(abs(X))
… and obtain the plot we saw in the previous slide.
Same Example (more data points)
Consider the sequence
x[ n ] e
In this case
j 0.3 n
, n 0, ..., 255
0 0.3 , N 256
>> n=0:255;
>>x=exp(j*0.3*pi*n);
>> X=fft(x);
>> plot(abs(X))
See the plot …
… and its magnitude plot
| X [k ] |
200
150
100
50
0
0
50
100
150
200
250
300
k
What does it mean?
A peak at index k 0 means that you have a frequency
0
| X [k ] |
k 0 2 / N
200
150
100
50
0
0
50
k 0 38
100
150
200
250
300
k
The max corresponds to frequency
3 8 2 / 2 5 6 0 .2 9 6 9 0 .3
Example
You take the FFT of a signal and you get this magnitude:
| X [k ] |
1200
1000
800
600
400
200
0
0
50
k 1 27
100
k 2 81
There are two peaks corresponding
to two frequencies:
150
1 k1
2 k2
200
2
27
N
2
N
250
2
0 . 2109
256
81
2
256
0 . 6328
300
k
DFT of a Sinusoid
Consider a sinusoid with frequency
0
rad.
x[ n ] A cos( 0 n ), n
We take a finite data length
x[ n ] A cos( 0 n ),
0 n N 1
… and its DFT
X [ k ] DFT x [ n ]
N 1
x [ n ]e
j
2
N
kn
, k 0 ,..., N 1
n0
How does it look like?
Sinusoid = sum of two exponentials
Recall that a sinusoid is the sum of two complex
exponentials
x[ n ]
A
e
2
magnitude
e
j 0 n
A
A/2
0
0
0
e
j
e
j 0 n
2
A/2
phase
j
( rad )
0
( rad )
Use of positive frequencies
Then the DFT of a sinusoid has two components
A j j 0 n
A j j 0 n
X [k ] D FT e e
DFT e e
2
2
… but we have seen that the frequencies we compute
are positive. Therefore we replace the last exponential
as follows:
A j j 0 n
A j j ( 2 0 ) n
X [k ] D FT e e
DFT e e
2
2
Represent a sinusoid with positive freq.
Then the DFT of a sinusoid has two components
A j j 0 n
A j j ( 2 0 ) n
X [k ] D FT e e
DFT e e
2
2
magnitude
A/2
0
phase
A/2
0
2 0
2
( rad )
2 0
2
( rad )
Example
Consider the sequence
x [ n ] 2 cos(0.3 n ), n 0, ..., 31
In this case
0 0 . 3 , N 32
Then its DFT becomes
X [ k ] W 32 0.3 W 32 1.7
k 2
32
2 0 . 3
Let’s plot its magnitude:
, k 0, ..., 31
... first plot this …
1
2
W 32 0.3 W 32 1.7
20
N / 2 32 / 2
N / 2 32 / 2
15
10
5
0
0
1
0 0 . 3
2
3
4
5
0 1.7
6
2
… and then see the plot of its DFT
X [k ]
1
2
W 32 0.3 W 32 1.7
k
2
,
k 0, ..., N 1
N
20
15
This is NOT a frequency
10
5
0
0
5
10
15
20
The first max corresponds to frequency
25
30
5 2 / 32 0 . 3
35
k
Symmetry
If the signal x[n ]
is real, then its DFT has a symmetry:
X [k ] X [ N k ]
In other words:
*
| X [ k ] | | X [ N k ] |
X [ k ] X [ N k ]
Then the second half of the spectrum is redundant (it does not
contain new information)
Back to the Example:
20
15
10
5
0
0
5
10
15
20
25
30
35
If the signal is real we just need the first half of the spectrum,
since the second half is redundant.
Plot half the spectrum
20
15
10
5
0
0
5
10
If the signal is real we just need the first half of the spectrum,
since the second half is redundant.
15
Same Example in Matlab
Generate the data:
>> n=0:31;
>>x=cos(0.3*pi*n);
Compute the DFT (use the “Fast” Fourier Transform, FFT):
>> X=fft(x);
Plot its magnitude:
>> plot(abs(X))
… and obtain the plot we saw in the previous slide.
Same Example (more data points)
Consider the sequence
x [ n ] cos(0.3 n ), n 0, ..., 255
In this case
0 0.3 , N 256
>> n=0:255;
>>x=cos(0.3*pi*n);
>> X=fft(x);
>> plot(abs(X))
See the plot …
… and its magnitude plot
| X [k ] |
100
80
60
40
20
0
0
50
100
k 0 38
The first max corresponds to frequency
150
200
250
N k 0 218
k
0 38 2 / 256 0.3
Example
You take the FFT of a signal and you get this magnitude:
| X [k ] |
150
100
50
0
0
50
100
k 1 27 k 2 81
There are two peaks corresponding
to two frequencies:
150
1 k1
2 k2
200
2
27
N
2
N
250
2
0 . 2109
256
81
2
256
0 . 6328
300
k