Fundamentals of Digital Signal Processing. Fourier Transform of continuous time signals with t in sec and F in Hz (1/sec). Examples:

Fundamentals of Digital Signal Processing

Fourier Transform of continuous time signals

dtetxtxFTFX Ftj 2)()()(

dFeFXFXIFTtx Ftj 2)()()(

with t in sec and F in Hz (1/sec).

Examples:

0 0 0sinct

TFT rect T FT

02 0 FFeFT tFj

021

021

02cos FFeFFetFFT jj

Discrete Time Fourier Transform of sampled signals

n

fnjenxnxDTFTfX 2][][)(

21

21

2)()(][ dfefXfXIDTFTnx fnj

with f the digital frequency (no dimensions).

Example:

020( )j f n

k

DTFT e f f k

since, using the Fourier Series,

2( ) j nt

k n

t k e

Property of DTFT

• f is the digital frequency and has no dimensions

• is periodic with period f = 1. )1()( fXfX

21

21 f

f11

2

1

21

)( fX

• we only define it on one period

f1

2

)( fX

1

2

tFjetx 02)( nj

ssF

F

enTxnx02

)(][

ss TF /1

Sampled Complex Exponential: no aliasing

0F F

)(FX

f

( )X f

2

1

1

2

1. No Aliasing20

sFF

2sF

2

sF0f

sF

Ff 0

0 digital frequency

tFjetx 02)( nj

ssF

F

enTxnx02

)(][

ss TF /1

Sampled Complex Exponential: aliasing

0FF

)(FX

f

( )X f

2

1

1

2

2. Aliasing 0 2sF

F

2sF

2

sF0f

0 00

s s

F Ff round

F F

digital frequency

tFjetx 02)( 02[ ] ( ) j f nsx n x nT e

ss TF /1

Mapping between Analog and Digital Frequency

ss F

Fround

F

Ff 00

0

Example

tjetx 10002)( kHzFs 3

Then:

• analog frequency

• FT:

• digital frequency

•DTFT: for

)1000()( FFX FT

HzF 10000

0 0 1 1 10 3 3 3s s

F FF Ff round round

31)( ffX DTFT 2

1|| f

Example

tjetx 20002)( kHzFs 3

Then:

• analog frequency

• FT:

• digital frequency

•DTFT: for

)2000()( FFX FT

HzF 20000

31)( ffX DTFT 2

1|| f

0 0 2 2 10 3 3 3s s

F FF Ff round round

Example

tjjtjj eeeettx 80001.02180001.0

21)1.08000cos()(

kHzFs 3

Then:

• analog frequencies

• FT:

• digital frequencies

•DTFT

)4000()4000()( 1.0211.0

21 FeFeFX jj

FT

0 14000 , 4000F Hz F Hz

21|| f 3

11.021

311.0

21)( fefefX jj

DTFT

4 4 4 10 3 3 3 31f round

4 4 4 11 3 3 3 3( 1)f round

Linear Time Invariant (LTI) Systems and z-Transform

][nx ][ny][nh

If the system is LTI we compute the output with the convolution:

m

mnxmhnxnhny ][][][*][][

If the impulse response has a finite duration, the system is called FIR (Finite Impulse Response):

][][...]1[]1[][]0[][ NnxNhnxhnxhny

n

nznxnxZzX ][][)(

Z-Transform

Facts:

)()()( zXzHzY

][nx ][ny)(zH

Frequency Response of a filter:

fjezzHfH 2)()(

Digital Filters

)(zH][nx ][ny

Ideal Low Pass Filter

f21

21

)( fH

f21

21

)( fH

Pf

Pf

passband

constant magnitude in passband…

… and linear phase

A

Impulse Response of Ideal LPF

P

P

f

f

fnjfnjideal dfAedfefHnh 222

1

21

)(][

Assume zero phase shift,

nfsincAfnh PPideal 22][

This has Infinite Impulse Response, non recursive and it is non-causal. Therefore it cannot be realized.

-50 -40 -30 -20 -10 0 10 20 30 40 50-0.05

0

0.05

0.1

0.15

0.2

n

fp=0.1

[ ]h n

n

0.1

1Pf

A

Non Ideal Ideal LPF

The good news is that for the Ideal LPF

0][lim

nhidealn

n

][nh

L L

nL2L

][nh

Frequency Response of the Non Ideal LPF

Pf STOPf

passstop stop

f

transition region

attenuation

ripple

|)(| fH11

11

2

LPF specified by:

• passband frequency

• passband ripple or

• stopband frequency

• stopband attenuation or

Pf

1 dB RP 1

1

11

10log20

STOPf

2 dB R 2S 10log20

Best Design tool for FIR Filters: the Equiripple algorithm (or Remez). It minimizes the maximum error between the frequency responses of the ideal and actual filter.

1f 2f 21

attenuation

ripple

|)(| fH11

11

2

1 2 3 3 1 2, 0, , , / , 1,1,0,0 , ,h firpm N f f f f w w

impulse response ][],...,0[ Nhhh

1f 2f 21

3 f0

Linear Interpolation

11/ w

2/ w

The total impulse response length N+1 depends on:

• transition region

• attenuation in the stopband

1f 2f

|)(| fH

2

12 fff

N

f1

~ 22)(log20 210

Example:

we want

Passband: 3kHz

Stopband: 3.5kHz

Attenuation: 60dB

Sampling Freq: 15 kHz

Then: from the specs

We determine the order the filter

301

0.150.35.3 f

8230~ 2260 N

Frequency response

0 0.1 0.2 0.3 0.4 0.5-100

-80

-60

-40

-20

0

20magnitude

digital frequency

dB

0 0.1 0.2 0.3 0.4 0.5-120

-100

-80

-60

-40

-20

0

20magnitude

digital frequency

dB

N=82

N=98

Example: Low Pass Filter

0 0.1 0.2 0.3 0.4 0.5-120

-100

-80

-60

-40

-20

0

20magnitude

digital frequency

dB

Passband f = 0.2

Stopband f = 0.25 with attenuation 40dB

Choose order N=40/(22*(0.25-0.20))=37

| ( ) |H f

f

Almost 40dB!!!

Example: Low Pass Filter

Passband f = 0.2

Stopband f = 0.25 with attenuation 40dB

Choose order N=40 > 37

| ( ) |H f

f0 0.1 0.2 0.3 0.4 0.5

-80

-70

-60

-50

-40

-30

-20

-10

0

10magnitude

digital frequency

dB OK!!!

General FIR Filter of arbitrary Frequency Response

],...,,[

],...,,,0[

10

21

M

M

HHHH

ffff

01f 2f 3f 1Mf 2

1Mf

0H1H

2H

3H 1MH MH

Weights for Error:

1w2w

2/)1( Mw

],...,,[ 2/)1(21 Mwwww

Then apply:

, / , ,Mh firpm N f f H w

… and always check frequency response if it is what you expect!

Example: ( ) 1/ sinc( )H f f for 0 0.2f

( ) 0H f 0.25 0.5f

0 0.2 0.25 0.5 f

fp=0:0.01:0.2; % vector of passband frequencies

fs=[0.25,0.5]; % stopband frequencies

M=[1./sinc(fp), 0, 0]; % desired magnitudesDf=0.25-0.2; % transition regionN=ceil(A/(22*Df)); % first guess of orderh=firpm(N, [ fp, fs]/0.5,M); % impulse response

40A dB

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4magnitude

digital frequency

0 0.1 0.2 0.3 0.4 0.5-90

-80

-70

-60

-50

-40

-30

-20

-10

0

10

not very good here!

dB

37N

To improve it:

1. Increase order

2. Add weights

0 0.2 0.25 0.5 f

40A dB

1w 0.2w

w=[1*ones(1,length(fp)/2), 0.2*ones(1, length(fs)/2)];

h=firpm(N, [fp, fs]/0.5,M,w);

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4magnitude

digital frequency

0 0.1 0.2 0.3 0.4 0.5-160

-140

-120

-100

-80

-60

-40

-20

0

20

100N dB