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DIGITAL SIGNAL PROCESSING

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General information

Professor: Mihnea UDREA– B209– mihnea@comm.pub.ro

Grading:– Laboratory: 15%– Project: 15%– Tests: 20%– Final exam : 50%– Course quiz: ±10%

Web: www.electronica.pub.ro MOODLE

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1. Introduction

Discrete Time Signals and Systems

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Analog signal processing systems

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Digital signal processing systems

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periodically sampling a continous time signal at time intervals nTs.

1.1 Discrete-Time Signals

tA

-A

( )ax t

nA

-A

0 1 2 3 4 5 10

( )x n

Ts 2Ts…………nTs

a Sx n x nT

Ts sampling period1

ss

FT

sampling frequency

s a sn

x nT x t t nT

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Basic signals

1 , 00 , 0

nu n

n

n

1

-4 -3 -2 -1 0 1 2 3 4

( )u nThe discrete unit step

n

1

-4 -3 -2 -1 0 1 2 3 4

( )n

1 , 00 , 0

nn

n

The discrete impulse

Periodical signals

A discrete signal is N samples periodical if:N Z x n x n N n Z

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The discrete-time sinusoid

f0 – normalised frequency

ω0 – angular normalised frequency

00 0 S

S

Ff F TF

0 0 02ST f

tA

-AT0

0 0sinax t A t 0 02 F

0 0sina Sx n x nT A n

nA

-A

0 1 2 3 4 5 10

00 0 2S

S

Ft nT nF

The discrete-time sinusoid

x(n) is N samples periodical if there exist integers N and k:

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sin 0.2x n A n

nA

-A

0 1 2 3 4 5 10

N=10, k=1

F0= 1 kHz; FS= 10 kHz 00 2 0.2

S

FF

0 022N k kN

The discrete-time sinusoid

x(n) is N samples periodical if there exist integers N and k:

10N=20, k=3

F0= 1.5 kHz; FS= 10 kHz 01.5 32 210 20

0 022N k kN

sin 0.3x n A n

nA

-A

0 1 2 3 4 20

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The discrete-time signal energy

2def

nE x n

2a S a S

nE T x nT

21lim2 1

N

N n NP x n

N

1a

S

E ET

The average power

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Frequency analysis of discrete-time signals

0

x(t)

Ts 2Ts 3Ts nTs

0 t

d(t)

0 t

d(t - nTs)

nTs

s s sn n

x nT x t t nT x t t nT

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Frequency analysis of discrete-time signals

12s s

nx nT x t t nT

1 12 a s s a s

sk kX X k X k

T

Denote:

sx nT X

ax t X

22s ss

FT

s sn

x nT x t t nT

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1

Ω

Xa(Ω)

0 ΩM-ΩM Ωs 2Ωs-Ωs-2Ωs

X(Ω)1/Ts

Ω0 ΩM-ΩM Ωs 2Ωs-Ωs-2Ωs

1/Ts

Ωs

-Ωs/2 Ωs/2

aliasing

Conditions:1) |Xa(Ω)|=0, for |Ω|>ΩM

2) Ωs 2ΩM

Sampling Theorem (Nyquist)

1a s

s kX X k

T

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F

X(F)

0 FM-FM Fs 2Fs-Fs-2Fs

The minimum sampling rate is achieved forΩs = 2ΩM Fs = 2FM = Nyquist rate (Nyquist frequency)

f

X(f)

0 0.5-0.5 1 2-1-2

- normalized frequencys

FfF

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ω

X(ω)

0 π-π 2π 4π-2π-4π

f

X(f)

0 0.5-0.5 1 2-1-2

2 2s

FfF

2π(Ωs)

1(Fs)

[ ; ) , [ 0.5;0.5) f (basic interval)

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Frequency analysis of discrete-time signals

n

nX z Z x n x n z

Z transform:

1 112

n

C

x n Z X z X z z dzj

,D z z R R R R

Im{z}

Re{z}R- R+

Most important properties:

kx n X z x n k z X z

1 2 1 2x n x n X z X z 18

Frequency analysis of discrete-time signals

0

n

n nX z Z x n x n z

Z transform:a) signal with left limited support

particular case: 00,x n n n

D z z R

Im{z}

Re{z}R-

0supp ,x n n

0 0n

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Frequency analysis of discrete-time signals

0n

n

nX z Z x n x n z

Z transform:b) signal with right limited support

00,x n n n

D z z R

Im{z}

Re{z}R+

0supp ,x n n

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Frequency analysis of discrete-time signals

Z transform:A bilateral sequence can be decomposed:

x n x n x n

( ), 00 else

x n nx n

( ), 00 else

x n nx n

D z z R D z z R

D D D z R z R

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Frequency analysis of discrete-time signals

Z transform:Let:and:Then:and:

D z R z R

1' 'X z Z x n X z

'x n x n

X z Z x n

1 1'D z zR R

1' n n

n nX z x n z x n z X z

1 1z R zR

1 1z R zR

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Frequency analysis of discrete-time signals

Discrete time Fourier transform (DTFT):

DTFTj jn

nX e x n e x n

1 IDTFT2

j jn jx n X e e d X e n

2 2s

FfF

Discrete Fourier transform (DFT): sup 0, 1x n N

1

0DFT

Nnk

Nn

X k x n k x n W

1

0

1IDFTN

nkN

kx n X k n X k W

N

2jN

NW e

2k k

N

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Frequency analysis of discrete-time signals

X z jX e jz e X k

2k k

N

n

nX z x n z

j jn

nX e x n e

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0

N j nkN

nX k x n e

Z transform

DTFT

DFT

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Frequency analysis of discrete-time signals

012…k………N-1

X k

k

jX e X k

2k k

N

-3π -2π -π 0 π 2π 3π

jX e

ω

Discrete time Fourier transform (DTFT):

Discrete Fourier transform (DFT):

2 , 0, , 1k k NN

Example: The rectangular window

Is defined by: 1, 0, 1( )

0, in restD

n Nw n

1

-2 -1 0 1 2 ...... N-1 N N+1 n

( )Dw n

1

10

1( )1

NNn

Dn

zW z zz

The spectrum is:

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sin1 2( ) 11 sin2

NjN jjD j

NeW e ee

0 ( )DW

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The rectangular window frequency characteristic

jDW e

2 20 2 2N N

N

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Frequency analysis of discrete-time signals

N

0sinx n A n

n

-π -ω0 0 ω0 π 2π-ω0 2π 2π+ω0

jX e

ω

jDW e

0 π 2π2N

ω

-π -ω0 0 ω0 π 2π-ω0 2π 2π+ω0

j jDX e W e

ω

( )u n u n N

DTFT

Windowing effect

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Frequency analysis of discrete-time signals

N

0sinx n A n

n

0 1 2 3 4 5 6 7 8 9 10 11

k

( )u n u n N

X k

02 4N

jX e X k2

k kN

jDW e

0 π 2π2N

ω

DTFT DFT

0 ω0 π 2π-ω0 2π2N 2

k kN

ω

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Frequency analysis of discrete-time signals

0 1 2 3 4 5 6 7 8 9 10 11

k

X k

0 ω0 π 2π-ω0 2π2N

02 4.8N

jX e X k2

k kN

2k k

N

jDW e

0 π 2π2N

ω

N

0sinx n A n

n

( )u n u n N

DTFT DFT

Spectral leakage

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1.2 Discrete-time Systems

Linear systems satisfy the superpositionprinciple:

y n T x ny(n)x(n)

T { }

1 1 2 2 1 1 2 2 1 1 2 2T a x n a x n a T x n a T x n a y n a y n

The impulse response: h n T n

k

x n x k n k

k

x k T n k

kT n k h n

y n T x n k

T x k n k

k

kx k h n

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Discrete-time systems

Time invariant systems have the following property:

y n T x n y n k T x n k k Z

kh n T n k h n k k

y n x k h n k

Discrete time linear convolution:

1 2 1 2 1 2k

x x n x n x n x k x n k

y n h x n x h n x(n)h(n)

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Discrete-time systemsCauzal systems.

LTI cauzal systems: 0, 0h n n

0n n 0pentrua bx n x n n n

0pentrua by n y n n n

Stable systems – the impulse response has to be absolutely sumable.

k

h k

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LTIS difference equation: 0 0

N M

k kk k

y n k x n k

0 1

M N

k kk k

y n b x n k a y n k

If denote:0 0

0

kka

0

kkb

FIR finite impulse response system (N = 0):

0

M

kk

y n b x n k

x n n 1

, [0, ]0 , in rest

Mn

kk

b n Mh n b n k

IIR infinite impulse response (N > 0).

0ka

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LTIS transfer function

h(n)y(n)x(n)

impulse response

y n x n h n

H(z)Y(z)X(z)

Y z X z H z

system function

Similar analysis with Fourier transform:

H z jH e jz e j j jY e X e H e

H z Z h n

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LTIS transfer function

0 1

M N

k kk k

y n b x n k a y n k

Assume a LTIS having the difference equation:

0 1

M Nk k

k kk k

Y z b X z z a Y z z

0

1

( )( ) ( )1

Mk

kk

Nk

kk

b zY z B zH zX z A za z

Applying the Z transform:

( )

( )

k

k

Zx n k X z zZy n k Y z z

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LTIS transfer function

zk – the system zeros:

pk – the system poles:

N – the system order (number of poles).

1

0 11

1

...( )( ) 1 ...

MM

NN

b b z b zB zH zA z a z a z

1 1 1

0 0 11 1 1

0 1

1 1 ... 1( )( ) 1 1 ... 1

M

N

b z z z z z zB zH zA z p z p z p z

0kH z

kH p

Stable systems – the poles must be less than 1 (in absolute value).

1kp

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LTIS transfer function

IIR systems:

FIR system: 1

0 10

...M

k Mk M

kH z b z b b z b z

0ka

0

1

( )( ) 1

Mk

kk

Nk

kk

b zB zH zA z a z

0( ) ( )

M

kk

h n b n k

1( )k

k kZb z b n k

0

( )M

k

kH z h k z

( ) kh k b

-1 0 1 2 3 4 5 6 7 n

h(n) M=6

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Example: The moving average filter

The mean of a length N signal:What if signal length is infinite?Use frames (windows) of last N samples

1

0

1 ( )N

nm x n

N

x(n–N+1) x(n–N+2) x(n–N+3) - - - x(n) x(n+1) x(n+2) - - -

1

0

1 ( )N

nk

m x n kN

nm1nm

2nm

0 50 100 150 200 250 300 350 400-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8Noisy signal

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The moving average filter

Denoising a signal:1

0

1( ) ( )M

km n x n k

N

0 50 100 150 200 250 300 350 400-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8N=4

original signaldenoised signal

0 50 100 150 200 250 300 350 400-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8N=20

original signaldenoised signal

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The moving average filter

The MA is a FIR filter of length Nthe impulse response:

1

0

1( ) ( )N

MAk

h n n kN

0 1 2 3 4 5 6 7 8 n

hMA(n) N=8

1/N

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The moving average filter

The MA is a FIR filter of length Nthe impulse response:

1

0

1 / , 0, 11( ) ( )0, in rest

N

MAk

N n Nh n n k

N

0 1 2 3 4 5 6 7 8 n

hMA(n) N=8

1/N

1( ) ( )MA Dh n w nN

1

10

1 1 1( )1

NNn

MAn

zH z zN N z

The moving average filter

jMAH e

2 20 2 2N N

1

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The moving average filter

The moving average filter output

1( ) ( ) ( 1) ... ( 1)MAy n x n x n x n NN

1 1( ) ( 1) ... ( 1)x n x n x n NN N

1( 1) ( )MAy n x n NN

1 1( ) ( 1) ( ) ( )MA MAy n y n x n x n NN N

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The moving average filter

The recursive equation

Corresponds to Z transform

1 1( ) ( 1) ( ) ( )y n y n x n x n NN N

1 1( ) 1 ( ) 1 NY z z X z zN

1( ) 1 1( )( ) 1

NY z zH zX z N z

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LTIS analysis in the frequency domain

The system transfer function:

j jk

kH e h k e

arg jj H e jj j jH e H e e H e e

j jH e H e

Systems having a real h(n):

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cos1 12 2

j jn j jn

x n A n

A e e A e e

( )

( )

2

2

j j j jn

j j j jn

Ay n e H e e e

A e H e e e

2

j n j njAy n H e e e

cosjy n A H e n

h(n)y(n)x(n)

LTIS analysis in the frequency domain

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jH e

cosjy n A H e n

is the filter gain and represents the magnitude–frequency characteristic;

is the phase delay of the system and represents the phase–frequency characteristic;

The time group delay: dd

LTIS analysis in the frequency domain

h(n)y(n)x(n)