Discrete Time Periodic Signals

Discrete Time Periodic Signals

A discrete time signal x[n] is periodic with period N if and only if

][][ Nnxnx for all n .

Definition:

N

][nx

n

Meaning: a periodic signal keeps repeating itself forever!

Example: a Sinusoid

[ ] 2cos 0.2 0.9x n n

Consider the Sinusoid:

It is periodic with period since 10N

][29.02.0cos2

9.0)10(2.0cos2]10[

nxn

nnx

for all n.

General Periodic Sinusoid

n

N

kAnx 2cos][

Consider a Sinusoid of the form:

It is periodic with period N since

][22cos

)(2cos][

nxknN

kA

NnN

kANnx

for all n.

with k, N integers.

1.03.0cos5][ nnx

Consider the sinusoid:

It is periodic with period since 20N

][231.03.0cos5

1.0)20(3.0cos5]20[

nxn

nnx

for all n.

We can write it as:

1.0

20

32cos5][ nnx

Example of Periodic Sinusoid

nN

kj

Aenx

2

][

Consider a Complex Exponential of the form:

for all n.

It is periodic with period N since

Periodic Complex Exponentials

][

][

22

)(2

nxeAe

AeNnx

jkn

Nkj

NnN

kj

1

njejnx 1.0)21(][

Consider the Complex Exponential:

We can write it as

Example of a Periodic Complex Exponential

nj

ejnx

20

12

)21(][

and it is periodic with period N = 20.

Goal:

We want to write all discrete time periodic signals in terms of a common set of “reference signals”.

Reference Frames

It is like the problem of representing a vector in a reference frame defined by

• an origin “0”

• reference vectors

x

,..., 21 ee

x

01e

2e

Reference Frame

Reference Frames in the Plane and in Space

For example in the plane we need two reference vectors

x

01e

2e

21,ee

Reference Frame

… while in space we need three reference vectors 321 ,, eee

0

1e

2e

Reference Frame

x

3e

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:

2211 eaeax

0

11ea

22ea

Where projection of along

projection of along

1a

2a 2e1e

x

x

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:

332211 eaeaeax

0

11ea

22ea

Where projection of along

projection of along

projection of along

1a

2a 2e1e

x

x

The “space” is a 3 dimensional space.

3a x

3e

33ea

Example: where am I ?

N

E0

1e

2e

x

m300

m200

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

n

][nx

reference signals

Back to Periodic Signals

A periodic signal x[n] with period N can be expanded in terms of N complex exponentials

1,...,0 ,][2

Nkenen

N

kj

k

as

1

0

][][N

kkk neanx

A Simple Example

Take the periodic signal x[n] shown below:

n0

1

2

Notice that it is periodic with period N=2.

Then the reference signals are

nnj

nnj

ene

ene

)1(][

11][

2

12

1

2

02

0

We can easily verify that (try to believe!):

nn

nenenx

)1(5.015.1

][5.0][5.1][ 10

for all n.

Another Simple Example

Take another periodic signal x[n] with the same period (N=2):

n0

3.0

3.1

Then the reference signals are the same0

22

0

12

21

[ ] 1 1

[ ] ( 1)

j n n

j n n

e n e

e n e

We can easily verify that (again try to believe!):

nn

nenenx

)1(8.015.0

][8.0][5.0][ 10

for all n.

Same reference signals, just different coefficients

Orthogonal Reference Signals

Notice that, given any N, the reference signals are all orthogonal to each other, in the sense

kmN

kmnene

N

nmk if

if 0][][

1

0

*

1

0 2

)(221

0

21

0

*

1

1][][

N

n N

kmj

kmjn

N

kmjN

n

nN

kmjN

nmk

e

eeenene

Since

by the geometric sum

… apply it to the signal representation …

k

N

m

N

nmkm

N

nk

nx

N

mmm

N

nk

Nanenea

neneanenx

1

0

1

0

*

1

0

*

][

1

0

1

0

*

][][

][][][][

and we can compute the coefficients. Call then kNakX ][

1,...,0 ,][][1

0

2

NkenxNakX

N

n

knNj

k

Discrete Fourier Series

1,...,0 ,][][1

0

2

NkenxkX

N

n

knNj

Given a periodic signal x[n] with period N we define the

Discrete Fourier Series (DFS) as

Since x[n] is periodic, we can sum over any period. The general definition of Discrete Fourier Series (DFS) is

1,...,0 ,][][][1 20

0

NkenxnxDFSkX

Nn

nn

knNj

for any0n

Inverse Discrete Fourier Series

1

0

2

][1

][][N

k

knNjekX

NkXIDFSnx

The inverse operation is called Inverse Discrete Fourier Series (IDFS), defined as

Revisit the Simple Example

Recall the periodic signal x[n] shown below, with period N=2:

n0

1

2

1,0,)1(21)1](1[]0[][][1

0

2

2

kxxenxkX kk

n

nkj

Then 1]1[,3]0[ XX

Therefore we can write the sequence as

n

k

knjekXkXIDFSnx

)1(5.05.1

][2

1][][

1

0

2

2

Example of Discrete Fourier Series

Consider this periodic signal

The period is N=10. We compute the Discrete Fourier Series

25

102 29 4

210 10

100 0

1 if 1, 2,...,9

[ ] [ ]15 if 0

j k

j kn j kn

j k

n n

ek

X k x n e ee

k

][nx

n010

1

… now plot the values …

0 2 4 6 8 100

5magnitude

0 2 4 6 8 10-2

0

2phase (rad)

k

k

|][| kX

][kX

Example of DFS

Compute the DFS of the periodic signal

)5.0cos(2][ nnx

Compute a few values of the sequence

,...0]3[,2]2[,0]1[,2]0[ xxxx

and we see the period is N=2. Then

k

n

knjxxenxkX )1(]1[]0[][][

1

0

2

2

which yields

2]1[]0[ XX

Signals of Finite Length

All signals we collect in experiments have finite length

)(tx )(][ snTxnx

ss TF

1

MAXTMAX SN T F

Example: we have 30ms of data sampled at 20kHz (ie 20,000 samples/sec). Then we have

points data 60010201030 33 N

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

1N0

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):

21

0

[ ] [ ] [ ] , 0,..., 1N j kn

N

n

X k DFT x n x n e k N

And its inverse

21

0

1[ ] [ ] [ ] , 0,..., 1

N j knN

n

x n IDFT X k X k e n NN

Signals of Finite Length

All signals we collect in experiments have finite length in time

)(tx )(][ snTxnx

ss TF

1

MAXTMAX SN T F

Example: we have 30ms of data sampled at 20kHz (ie 20,000 samples/sec). Then we have

points data 60010201030 33 N

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

1N0

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):

21

0

[ ] [ ] [ ] , 0,..., 1N j kn

N

n

X k DFT x n x n e k N

and its inverse

21

0

1[ ] [ ] [ ] , 0,..., 1

N j knN

n

x n IDFT X k X k e n NN

Example of Discrete Fourier Transform

Consider this signal

The length is N=10. We compute the Discrete Fourier Transform

25

102 29 4

210 10

100 0

1 if 1, 2,...,9

[ ] [ ]15 if 0

j k

j kn j kn

j k

n n

ek

X k x n e ee

k

][nx

n0

9

1

… now plot the values …

0 2 4 6 8 100

5magnitude

0 2 4 6 8 10-2

0

2phase (rad)

k

k

|][| kX

][kX

DFT of a Complex Exponential

Consider a complex exponential of frequency rad. 0

,][ 0njAenx n

We take a finite data length

,][ 0njAenx 0 1n N

… and its DFT

1,...,0,][][][1

0

2

NkenxnxDFTkX

N

n

knNj

How does it look like?

Recall Magnitude, Frequency and Phase

0

2. We represented it in terms of magnitude and phase:

( )rad0

0

magnitude

phase

|| A

A

Recall the following:

1. We assume the frequency to be in the interval

( )rad

Compute the DFT…

00

21

0

221 1

0 0

[ ] [ ] [ ]

, 0,..., 1

N j knN

n

N N j k nj knj n NN

n n

X k DFT x n x n e

Ae e Ae k N

Notice that it has a general form:

0

2[ ] NX k A W k

N

1

0

1( )

1

j NNj n

N jn

eW e

e

where (use the geometric series)

See its general form:

1( 1)/2

0

sin2

( )sin

2

Nj n j N

Nn

N

W e e

… since:

1

0

/2 /2 /2 /2

/2 /2 /2 /2

/2 /2 /2( 1)/2

/2 /2 /2

1( )

1

sin2

sin

2

j NNj n

N jn

j N j N j N j N

j j j j

j N j N j Nj N

j j j

eW e

e

e e e e

e e e eN

e e ee

e e e

… and plot the magnitude

-3 -2 -1 0 1 2 30

2

4

6

8

10

12

( )NW

N

2

N

2

N

Example

Consider the sequence

0.3[ ] , 0,...,31j nx n e n

In this case 32,3.00 NThen its DFT becomes

23232

[ ] 0.3 , 0,...,31k

X k W k

Let’s plot its magnitude:

... first plot this …

0 1 2 3 4 5 60

10

20

30

40

3.032 W

2

32N

3.00

… and then see the plot of its DFT

0 5 10 15 20 25 300

5

10

15

20

25

30

35

2[ ] 0.3 , 0,..., 1N kN

X k W k N

kThe max corresponds to frequency 3.0312.032/25

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

0.3[ ] , 0,..., 255j nx n e n

In this case 0 0.3 , 256N >> n=0:255;

>>x=exp(j*0.3*pi*n);

>> X=fft(x);

>> plot(abs(X))

See the plot …

… and its magnitude plot

0 50 100 150 200 250 3000

50

100

150

200

k

| [ ] |X k

What does it mean?

The max corresponds to frequency

38 2 / 256 0.2969 0.3

A peak at index means that you have a frequency 0k

0 50 100 150 200 250 3000

50

100

150

200

k

| [ ] |X k

0 38k

0 0 2 /k N

Example

You take the FFT of a signal and you get this magnitude:

0 50 100 150 200 250 3000

200

400

600

800

1000

1200

|][| kX

k271 k 2 81k

There are two peaks corresponding

to two frequencies:

6328.0256

281

2

2109.0256

227

2

22

11

Nk

Nk

DFT of a Sinusoid

Consider a sinusoid with frequency rad. 0

0[ ] cos( ),x n A n n

We take a finite data length

0 1n N

… and its DFT

1,...,0,][][][1

0

2

NkenxnxDFTkX

N

n

knNj

How does it look like?

0[ ] cos( ),x n A n

Sinusoid = sum of two exponentials

Recall that a sinusoid is the sum of two complex exponentials

njjnjj eeA

eeA

nx 00

22][

( )rad0

0

magnitude

phase

/ 2A

( )rad

0

0

/ 2A

Use of positive frequencies

0 0[ ]2 2

j n j nj jA AX k DFT e e DFT e e

Then the DFT of a sinusoid has two components

… but we have seen that the frequencies we compute are positive. Therefore we replace the last exponential as follows:

0 0(2 )[ ]2 2

j n j nj jA AX k DFT e e DFT e e

Represent a sinusoid with positive freq.

Then the DFT of a sinusoid has two components

( )rad0

0

magnitude

phase

/ 2A

( )rad2

202

/ 2A

02

0 0(2 )[ ]2 2

j n j nj jA AX k DFT e e DFT e e

Example

Consider the sequence

[ ] 2cos(0.3 ), 0,...,31x n n n

In this case 32,3.00 NThen its DFT becomes

232 3232

[ ] 0.3 1.7 , 0,...,31k

X k W W k

Let’s plot its magnitude:

3.02

... first plot this …

0 1 2 3 4 5 60

5

10

15

20

32 32

10.3 1.7

2W W

/ 2 32 / 2N

3.00 0 1.7

/ 2 32 / 2N

2

… and then see the plot of its DFT

kThe first max corresponds to frequency 3.032/25

32 322

1[ ] 0.3 1.7 , 0,..., 1

2 kN

X k W W k N

0 5 10 15 20 25 30 350

5

10

15

20

This is NOT a frequency

Symmetry

If the signal is real, then its DFT has a symmetry: ][nx

*][][ kNXkX

In other words:

][][

|][||][|

kNXkX

kNXkX

Then the second half of the spectrum is redundant (it does not contain new information)

Back to the Example:

0 5 10 15 20 25 30 350

5

10

15

20

If the signal is real we just need the first half of the spectrum, since the second half is redundant.

Plot half the spectrum

If the signal is real we just need the first half of the spectrum, since the second half is redundant.

0 5 10 150

5

10

15

20

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

[ ] cos(0.3 ), 0,..., 255x n n n

In this case 0 0.3 , 256N >> n=0:255;

>>x=cos(0.3*pi*n);

>> X=fft(x);

>> plot(abs(X))

See the plot …

… and its magnitude plot

k

| [ ] |X k

0 50 100 150 200 2500

20

40

60

80

100

0 38k 0 218N k

The first max corresponds to frequency 0 38 2 / 256 0.3

Example

You take the FFT of a signal and you get this magnitude:|][| kX

k

There are two peaks corresponding

to two frequencies:

6328.0256

281

2

2109.0256

227

2

22

11

Nk

Nk

0 50 100 150 200 250 3000

50

100

150

271 k 2 81k