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LOGO

iugaza2010.blogspot.commelasmer@gmail.com

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Signals, Systems and Signal Processing

Signal: is a function which is dependent of some variables

which are independent variables.

signal: f(x1,x2,…) , x :time,distance,temperature,……

In DSP the independent variable is number(small n)

: f(n)

System: defined as a physical device that performs an

operation on a signal.

Processed the signal: occurs when we pass the signal through

the system.

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Basic element of DSP system

A/D

converterDSP

Analog

input signal

Analog

output signalD/A

converter

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Advantages of Digital over analog signal processing

1) Flexibility: simply changing program.

2) Amenable to full integration

3) More accuracy

4) Less sensitivity to components(R,L,C) and environment changes

5) Dynamic range

6) Easy adjustment of processor characteristic by changing coefficients(numbers)

In analog we change value of R,L,C .

7) Easy storage in Taps,disks,…..

Disadvantages of Digital

1) Increase complexity (A/D,D/A)

2) It has certain limitations due to A/D (fs=2fo for no aliasing)

3) Power dissipation.

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(1) Mono channel versus Multichannel

multichannel generated by multisource

Ex: Electrocardiograms(ECG)

(2) One dimensional versus multidimensional

If the signal is a function of one independent variable then it is

one-dimensional signal otherwise it is multidimensional.

Ex: - Any picture is 2-dim signal, f(x,y)

- Black and white TV picture is a 3-dim signal, f(x,y,t)

- for color TV :the signal is 3-dim (x,y,t) and 3-channel (R,G,B)

Classification of Signals:

Ir(x,y,t)

I(x,y,t)= Ig(x,y,t)

Ib(x,y,t)

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(3) Real valued or comples value signalS1(t)=A sin πt real signal

S2(t)=A e(jπt) complex signal

(4) Continuous time versus discrete time signal(a) CT + CV analog

(b) DT + CV discrete

(c) DT + DV digital

(5) Deterministic versus randomdeterministic signal// can be described by mathematical equation.

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Continuous-time sinusoidal signal

( ) cos( )

A:amplitude , θ:phase

Ω:analog frequency in rad

1Ω=2πF , F= , T:period

x t A t

T

Discrete-time sinusoidal signal

( ) cos( )

n:sampler number , w=2πf

1f= , N:period

x n A wn

N

( ) x(n)

w

F f

T N

x t

The concept of frequency in CT and DT signals

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For Continuous-time sinusoidal signal:

1) xa(t) always periodic.

2) Continuous-time sinusoidal signals with different frequencies

are themselves distinct.

3) Increasing the frequency F results in an increase in the rate

of oscillation of the signal.

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(1) Periodic of discrete-sinusoidal signal

x(n) is periodic only if its frequency in Hertz is rational number.

integer k

integer Nf

:

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

( ) cos( ( ) ) cos( )

cos( 2 ) ................. (2)

for: (1)=(2)

2 2 k= 1, 2,......

kf=

N

proof

x n A wn

x n N A w n N A wn wN

A wn fN

fN k

Note: analog sinusoidal signal always periodic.

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Determine which of the following sinusoids are periodic and compute

their fundamental period:

(a) cos(0.01πn)

0.01 10.01 2 ( )

2 200

periodic with period N=200

kw f f rational

N

0 50 100 150 200 250 300 350 400-1

-0.5

0

0.5

1

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(b) cos(30πn/105)

30 30 12 ( )

105 2 (105) 7

periodic with period N=7

kw f f rational

N

0 5 10 15 20 25 30-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

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(c) sin(3n)

33 2 (not rational)

2

periodic

w f f

non

0 50 100 150 200 250 300-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

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x(n)=cos(πn/6) cos(πn/3)

1

2

1cos( )cos( ) cos( ) cos( )

2

1 1 3( ) cos( )cos( ) cos( ) cos( ) cos( ) cos( )

3 6 2 3 6 3 6 2 6 6

3 3 3 1cos( ) 2 = ( rational, N =4)

6 6 12 4

1cos( ) 2 ( rational, N =12

6 6 12

x y x y x y

n n n n n n n nx n

nw f f

nw f f

1 2

)

x(n) periodic with period=LCM(N , N ) LCM(4,12) 12

0 10 20 30 40 50 60-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

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Determine which of the following sinusoids are periodic and compute

their fundamental period:

(a) x(t)=3cos(5t+π/6)

analog sinusoidal signal always periodic

5 1 25 2 ( ) 1.256

2 5F F T peiod

F

0 0.5 1 1.5 2 2.5 3-3

-2

-1

0

1

2

3

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(2) Discrete time sinusoid whose radian frequencies are

separated by integer multiples of 2π are identical

1 1

2 2

2 1

2 1 1

1

:

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

( ) cos( )

2

( ) cos(( 2 ) ) cos( 2 )

= cos( ) ....................(2)

(1)=(2)

2

2

1

2

o

o

proof

x n A w n

x n A w n

w w k

x n A w k n A w n kn

A w n

w w k

w f

1

2

example:

x(n)=cos( )o

f

for

w n

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-20 -10 0 10 200

0.5

1w=0

-20 -10 0 10 20-1

0

1w=30

-20 -10 0 10 20-1

0

1w=60

-20 -10 0 10 20-1

0

1w=90

-20 -10 0 10 20-1

0

1w=120

-20 -10 0 10 20-1

0

1w=150

-20 -10 0 10 20-1

0

1w=180

Note:

The period of the sinusoidal decreases as the frequency increases.

The rate of oscillation increase as the frequency increase.

The sequences of any two sinusoidal with frequencies in the range :

-π ≤ w ≤ π or -0.5 ≤ f ≤ 0.5 are different(unique).

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-20 -10 0 10 200

0.5

1w=360

-20 -10 0 10 20-1

0

1w=390

-20 -10 0 10 20-1

0

1w=420

-20 -10 0 10 20-1

0

1w=450

-20 -10 0 10 20-1

0

1w=480

-20 -10 0 10 20-1

0

1w=510

-20 -10 0 10 20-1

0

1w=540

Sequence that result from a sinusoid with a frequency |w| > π is identical to a

sequence obtained from a sinusoidal with frequency : -π ≤ w ≤ π

for continuous-time sinusoids result in distinct signals for Ω or F in the range

-∞< F <∞ or -∞< Ω <∞

w=360 aliased with w=0 , w=30 aliased with w=390 …………

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The frequency range for discrete-time sinusoids is finite with duration 2π

: choose,

0 ≤ w ≤ 2π (0 ≤ f ≤ 1) or -π ≤ w ≤ π (-0.5 ≤ f ≤ 0.5) fundamental range

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( ) cos( ) ( ) cos( ) ( )

( ) ( ) ( )

1( ) cos( ) cos(2 )

cos(2 ) cos(2 )

frequency=analog frequency*sampling time

for on

*

c

*

a s s

a s

s

s

s

s

s

s s

s

x t A t x nT A nT x n

x t x nT x n

x n A nT

nt nT

F

Ff

A FnF

FA

F T w T

n A nfF

dig a

F

it l

tinuous: ( , )

0.5 0.5for discrete sinsuoidal signals:

F

f

w

the infinit analog frequency is mapping into finite digital frequency.

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max

max

fold

2 rate

2 alias

1 1

2 2

2 2

ing

F

2

=2

s

s

s

s

s

s

F F N

f

F FF

FF

yquits

F F

F

To determine the mapping of any (alias) frequency above Fs/2

into the equivalent frequency below Fs/2 ,we can use Fs/2 to

reflect or fold the alias frequency to the fundamental range.

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signal signal

= 2 F = 2 f

rad Hz

sec

Continuous time Discrete time

w

rad cycles

sample sample

0.5 0.5

w

F f

, * s

s

Ff T

F

* ,s

s

F f FT

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Consider the following two analog sinusoidal signal signal:

x1(t)=cos(2π10t) , x2(t)=cos(2π50t) , Fs=40 Hz

Find the corresponding discrete sequence:

s 1 2F =40 Hz , F =10 Hz , F =50 Hz

11( ) ( ) cos(2 10 )

40

12( ) ( ) cos(2 50 ) cos(5 ) cos(5 2 )

40 2 2

= co

cos( )2

coss(5 2 )2

1(

( )2

) 2( )

: 1( ) 2( )

a s

a s

n

n

x n x nT n

x n x nT n n n k

n

x n x n

but x t x t

comment : The frequen2 1 s

cy F = 50 Hz is an alias of F = 10 Hz at F = 40 Hz

foldF = 20 Hz , 20

2

0.540s

sF Ff Hz

F

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Leads a unique signal if sampled at Fs

Ex:

If Fo=10 Hz find the alias frequencies of Fo if Fs=40 Hz

F1=10+1(40)=50 Hz

F2=10+2(40)=90Hz

F3=10+3(40)=130Hz

……

……

k o s

aliased base s

aliased base

F =F + k F ,k= 1, 2,.........

F =F + k F

f =f + k

foldF = 20 Hz , 20

2

0.540s

sF Ff Hz

F

Note: All sinusoidal with frequency

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f=F/Fs=50/40=1.25 , fbase=1.25-1=0.25Hz

f=F/Fs=10/40=0.25Hz

F1=10+1(40)=50 Hz

k o s

aliased base s

aliased base

F =F + k F ,k= 1, 2,. ........

F =F + k F

f =f + k

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ss

s

F1 1 sampling freq.t=nT = n F analog freq.=

F F n #of samples

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An analog signal contains frequencies up to 10 kHz

s

base fold

s 1

F4 kHz < 5kHz

2

or F 8 kHz < 2F 10 kHz

aliasing happe

F =F kF 5 (1)4 1

ns

al

fo

iase

ld

d kHz

F

(a) What range of sampling frequencies allow

exact reconstruction of this signal from its samples?

max2 2(10) 20 kHz

20 kHz

s

s

F F

F

(b) If Fs=8 kHz .Examine what happens to the

frequency F1=5kHz ?

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An analog signal contains frequencies up to 100 Hz

sfold

F125 Hz , f =0.5

2foldF

(a) What is the Nyquist rate of this signal ?

max2 2(100) 200 HzsF F

(b) If Fs=250 sample/sec ,what is the highest frequency that

can be represented uniquely at this sampling.

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An analog signal x(t)=sin(480*pi*t)+3sin(720*pi*t)

is sampled 600 times per second

sF 600300 Hz

2 2foldF

(a) What is the Nyquist sampling rate?

1 2

, max

480 720240 Hz , 360 Hz

2 2

2 2(360) 720 Hzs N

F F

F F

(b) Determine the folding frequency.

(c) Find x(n) .

( ) ( ) sin(480 ) 3sin(720 )

1 1 =sin(480 ) 3sin(720 ) =sin(0.8 ) 3sin(1.2 )

600 600

=sin(0.8 ) 3sin( 0.8 ) sin(0.8 ) 3sin(0.8 )

a s s sx n x nT nT nT

n n n n

n n n n

= 2sin(0.8 )n

0 500 1000 1500 2000-4

-3

-2

-1

0

1

2

3

4

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(d) If x(n) is pass through an ideal D/A converter ,

what is the reconstructed signal y(t).

( ) ( ) ( ) 2sin(0.8 600) 2sin(480 )s

s

ty t x tF x t t

T

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Sampling Theorem

Sampling x(t) by multiplying x(t) by train of impulses

x(n)=x(t)P(t)

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1 ( )

1 1 1 1 1....... ( 2 ) ( ) ( ) ( ) ( 2 ) .......

s s

n

s s s s s

X (f) = X f nfT

X f f X f f X f X f f X f fT T T T T

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Anti-aliasing Filter

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Suppose that an analog signal is given as:

X(t)=5cos(2*pi*1000t)

and is sampled at the rate of 8000 Hz

(a) Sketch the spectrum for the original signal

(b) Sketch the spectrum for the sampled signal from 0 to 20 kHz

1cos(2 ) ( ) ( )

2

5cos(2 1000 ) 2.5 ( 1000) ( 1000)

o o of t f f f f

t f f

1( )

1 1 1 1 1....... ( 2 ) ( ) ( ) ( ) ( 2 ) .......

s s

n

s s s s s

X (f) = X f nfT

X f f X f f X f X f f X f fT T T T T

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Notice that the spectrum of the sampled signal contains the images of the original

spectrum that the images repeat at multiplies of the sampling frequency

fs,2fs,3fs,….. (8 kHz,16 kHz,24 kHz,…

All images must be removed since they convey no additional information.

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- Sampling converts the analogue signal into discrete value of samples.

- We need to encode each sample value in order to store it in b bits

memory location.

Quantization

After quantizationBefore quantization

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x(t)=(0.9)t , Fs=1 sample/sec

x(n)=(0.9)n

If b=4 bits L=16 levels( choose L=11){0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1} , step=0.1

0000

0001

0010

0011

0011

0100

0101

0101

0110

0110

max min 1 0Q. step= 0.1

1 11 1

x x

L

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Q ua ntization error : ( ) ( ) ( )q qe n x n x n

2 2error

max minQuantization step =1

x x

L

0.1 0.1

2 2error

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Quantization of sinusoidal signal

2 2

2 2

0 0

2 2

2

2

2

1 1P ( ) ( sin )

2 2

( ) for ( T t T )2

1 1P ( ( )) ( )

2 2

1

2 2

2

2

sig

q

T T

q q

T T

T

T

S t dt A wt dt

e t tT

e t dt t dtT T T

t dtT T

A

Average power of sinusoidal signal :

Average power of quantized signal :

2

12

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Signal to quantization noise ratio

2

12

2

max min

2

2

2

2

2

the signal toquantization noise ratio & #13;

P

P

( )

2

12

3

2

P

1

2

P 2

2

2

sig

q

sig

q

A

SQNR

x x A A A

L L L

A

SQNRA

L

L

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X(t)= 3cos(600*pi*t) + 2cos(1800*pi*t)

Operated at 10,000 bits/sec , quantized into 1024 levels

(a) What is the sampling frequency and folding frequency?

2 2

s

sfold

log10242 1024 log 1024 log 2 10 bits/sample

log 2

1F ( )(10,000) 1000 sample/sec

sec sec 10

FF 500 Hz

2

b bL b

sample sample bit

bit

(b) What is the Nyquist rate .

1 2

, max

600 1800300 Hz , 900 Hz

2 2

2 2(900) 1800 Hzs N

F F

F F

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(c) Find x(n) .

1 1( ) ( ) 3cos(600 ) 2cos(1800 )

1000 1000

3cos(0.6 ) 2cos(1.8 )

3cos(0.6 ) 2cos( 0.2 )

3cos(0.6 ) 2cos(0.2 )

a sx n x nT n n

n n

n n

n n

(d) What is the resolution.

max min 5 ( 5) 10

1 1024 1 1023

x x

L

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X(n)=6.35 cos(pi*n/10) is quantized with a resolution=0.02

How many bits are required in the A/D converter ?

max min

10

10

6.35 ( 6.35)0.02 636 2

1 1

log 6369.3 10

log 2

bx xL

L L

b bits

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Determine the bit rate and the resolution in the sampling of a seismic

signal with dynamic range of 1 volt if the sampling rate is

Fs=20 samples/s and we use an 8-bit A/D converter ? What is the

maximum frequency that can be present in the resulting digital seismic

signal ?

max min

8

fold

(a) bit rate= * (8)(20) 160 bits/secsec sec

1 1(b)

1 2 1 255

20(c) F = =10 Hz

2

s

bit bit sampleb F

sample

x x

L

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Homework#1 Deadline: Tue: 08/03/2011