83
(Zamanda Öteleme veya Kaydırma) t -1 0 1 2 1 - - 1 0 1 0 1 0 1 2 2 1 0 ) ( t t t t t t x t 0 1 2 2 1 - - - - - - - - 2 1 1 0 2 1 1 1 0 1 1 0 0 1 1 2 2 ) 1 ( 2 0 1 1 0 ) 1 ( t t t t t t t t t t t x t -2 -1 0 2 1 - - - - - - 0 1 1 0 0 1 1 1 0 1 1 2 0 1 1 4 2 2 ) 1 ( 2 2 1 1 0 ) 1 ( t t t t t t t t t t t x Continuous Time (Sürekli Zaman)

Continuous Time (Sürekli Zaman) (Zamanda Öteleme veya ...akademikpersonel.kocaeli.edu.tr/yunusee/poster/yunusee30.09.2013_10.35... · Sayısal Haberleşme Alıcı-Verici Birimi

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(Zamanda Öteleme veya Kaydırma)

t -1 0 1

2

1

-

-

10

101

0122

10

)(

t

t

tt

t

tx

t 0 1 2

2

1

-

-

---

--

-

2110

211101

1001122)1(2

0110

)1(

tt

tt

tttt

tt

tx

t -2 -1 0

2

1

-

---

--

0110

011101

12011422)1(2

2110

)1(

tt

tt

tttt

tt

tx

Continuous Time (Sürekli Zaman)

(Zamanda Ölçekleme)

t -1 0 1

2

1

-

-

10

101

0122

10

)(

t

t

tt

t

tx

t -1/2 0 1/2

2

1

--

--

2/1120

2/101201

02/1021242)2(2

2/1120

)2(

tt

tt

tttt

tt

tx

t -2 -1 0 2

2

1

--

--

212

10

2112

101

0202

1122)

2

1(2

212

10

)2

1(

tt

tt

tttt

tt

tx

(Zamanda Tersine Çevirme)

t -1 0 1

2

1

-

-

10

101

0122

10

)(

t

t

tt

t

tx

t -1 0 1

2

1

-

-

----

--

-

10

011

1001222)(2

110

)(

t

t

tttt

tt

tx

Örnek:

))2/1(2(2)12(2 --- txtx

t -1 0 1

2

1

t -1 0 1

2

1

t - ½ 0 ½

2

1

t 0 ½ 1

2

4

x(t)

x1(t)=x(-t)

x2(t)=x1(2t)= x(-2t)

x3(t)=2x2(t- ½)= 2x(-2t+1)

Discrete Time (Ayrık Zaman)

n -3 -2 -1 0 1 3 4

1 1 1

. . .

2

-1

2 . . .

-

-

-

-

40

31

21

10

01

12

21

30

][

n

n

n

n

n

n

n

n

nx

-

-

--

-

-

--

---

---

-

5410

4311

3211

2110

1011

0112

1211

2310

]1[

nn

nn

nn

nn

nn

nn

nn

nn

nx n -3 -2 -1 0 1 2 4 5

1 1 1

. . .

2

-1

. . . 3

n -3 -2 -1 0 1 3 4

1 1 1

. . .

2

-1

2 . . .

-

-

-

-

40

31

21

10

01

12

21

30

][

n

n

n

n

n

n

n

n

nx

n -4 -3 -1 0 1 2 3

1 1 1

. . .

2

-1

2 . . .

--

--

---

--

-

--

--

--

-

440

331

221

110

001

112

221

330

][

nn

nn

nn

nn

nn

nn

nn

nn

nx

n -3 -2 -1 0 1 3 4

1 1 1

. . .

2

-1

2 . . .

-

-

-

-

40

31

21

10

01

12

21

30

][

n

n

n

n

n

n

n

n

nx

-

--

--

---

2420

2/3321

1221

2/1120

0021

2/1122

1221

22/3320

]2[

nn

nn

nn

nn

nn

nn

nn

nnn

nx

(n: tamsayı!)

(n: tamsayı!)

(n: tamsayı!)

(n: tamsayı!)

n -2 -1 0 2

1 1

. . .

-1

1 . . .

n -3 -2 -1 0 1 3 4

1 1 1

. . .

2

-1

2 . . .

-

-

-

-

40

31

21

10

01

12

21

30

][

n

n

n

n

n

n

n

n

nx

-

--

--

--

84210

63211

42211

21210

00211

21212

42211

63210

]2

1[

nn

nn

nn

nn

nn

nn

nn

nn

nx

n-5

n-3

n-1

n1

n3

n5

n7

Genlik değeri

tanımlanmamış

aradaki n’ler için

x[½n]0

n -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

1 1 1

. . .

2

-1

4 . . .

Örnek:

n -3 -2 -1 0 1 3 4

1 1 1

. . .

2

-1

2 . . .

-

-

-

-

40

31

21

10

01

12

21

30

][

n

n

n

n

n

n

n

n

nx

-

-

---

-

-

--

---

---

-

2413002

3/4313212

12132)1(2

3/2113002

3/1013 212

0113 422

3/1213 212

1313 002

]13[2

nn

nn

nn

nn

nn

nn

nn

nn

nx

n -1 0 2

4

. . .

-2

1 . . .

RoC

z=x+jy

x=Re{z}

y=Im{z}

|z|=(x2+y2)1/2 Re{z}

Complex z-plane

jIm{z}

z

x

jy

q

q =arctan(y/x)

x=|z|cos(q)

y=|z|sin(q)

z=|z|ejq=|z|{cos(q)+jsin(q)}

=

: RoC

x

y

)sin()cos(

1)sin(

)cos(22

je

yxzy

x

ezjyxz

j

jZero

(Sıfır)

Pole

(Kutup)

( )

Problem

k

Telekomünikasyon Trafiği Gelişimi

Haberleşme

Servislerindeki

Artış

Electronic Communications: Telephone, wireless phone, TV, radar, etc.

Concept and Model of Communications

General Communication Model

Source Transmitter Transmission

System Receiver Destination

Microphone Telephone Computer Scanner

Transformer Encoder Compress Modulator

Line/Cable Fiber/Air Satellite Network

Transformer Decoder Uncompress Demodulator

Speaker Earphone Computer Printer

Basic Communication Criteria: Performance, Reliability, Security

Important Reasons for Modulation:

• Ease of Radiation: c=l×f

• Simultaneous Transmission of Several Signals

FDM/TDM

• Effecting the Exchange of SNR with B

Channel Capacity Channel Bandwidth

Signal-to-Noise Ratio: S_

N

Modulation

Carrier: Acos(2πfct+φ) where fc is called carrier frequency

Modulation: change or modify values of A, fc, φ according to input signal m(t) - modify A A[m(t)]: Amplitude Modulation (AM)

- modify fc fc[m(t)]: Frequency Modulation (FM) - modify φ φ[m(t)]: Phase Modulation (PM)

Modulator m(t)

Acos(2πfct+φ)

modulated signal: s(t)

• Sayısal sinyaller analog sinyallere göre gürültü ve parazit sinyallerinden

daha az etkilenirler.

• Sayısal sinyallerdeki bozulmalar tekrar ediciler (regenerative repeaters)

tarafından giderilebilir.

• Hata sezme (error detection) ve düzeltme (correction) teknikleri sayesinde az hata

oranlı sinyal iletimi yapılabilir.

• Sayısal sinyallere parazit ve karıştırıcı sinyal etkilerinden korunabilmek için

güvenlik ve kriptolama gibi sinyal işleme teknikleri uygulanabilir.

• Sayısal devreler analog devrelere göre daha esnek, daha dayanıklı, ve daha az

maliyetli olarak tasarlanabilir.

Neden

Sayısal

Haberleşme?

Sayısal Haberleşme Alıcı-Verici Birimi

Sayısal Haberleşme Çoklu-Atlama Kanalı

Sayısal Tekrarlayıcı

Modulator, Demodulator & Modem

Modulator accepts bit sequence and modulates a carrier.

Demodulator accepts a modulated signal and regenerates bit sequence.

Modem is a single device which includes both modulator and demodulator.

Multiplexing, Multiplexer & Demultiplexer

Multiplexing is a technique that allows simultaneous transmissions of multiple

signals across a single data link.

CompA1

CompB1

CompC1

Da

Db

Dc D ≥ Da+Db+Dc

D E M U X

CompA2

CompB2

CompC2

1 shared link: rate D

Multiplexer Demultiplexer

FDM – Frequency Division Multiplexing

- A set of signals are put in different frequency positions of a link/medium. - Bandwidth of the link must be larger than the sum of signal bandwidths. - Each signal is modulated using its own carrier frequency. - Examples: radio, TV, satellite, etc.

A1

B1

C1

Mod

Mod

Mod

1

2

3

+

f

Dem

Dem

Dem

1

2

3

A2

B2

C2

1

2

3

1

2

3

f1

f2

f3

TDM – Time Division Multiplexing

- Multiple data streams are sent in different time in single data link/medium. - Data rate of the link must be larger than a sum of the multiple streams. - Data streams take turn to transmit in a short interval. - Widely used in digital communication networks.

CompA1

CompB1

CompC1

CompA2

CompB2

CompC2

D E M U X

… C1 B1 A1 C1 B1 A1 …

For no aliasing:

Bit rate:

Bandwidth of PCM waveform:

R=n.fs=(bit sayısı/örnek veri)x(örnek veri/sn )=bit/s=bps

m 255 Quantizer

6-dB Law:

Depends on:

input waveshapes

quantification characteristics

A-tipi sıkıştırma eğrisinin parçalı gösterimi

Amaç; giriş genliğinin herhangi bir değeri için belirli sınırlar içinde kalan bir kuantalama

hatası elde etmektir.

Lokal kuantalama

seviye (adım) sayısı: M

Encoder Transmission

System/Channel Bandwidth=B

Decoder

t

0 1 0 0 1 0

Maximum Signal Rate

Channel Capacity

Shannon Theorem (1948):

For a system/channel bandwidth B and signal-to-noise ratio S/N, its channel capacity is,

C = Blog2(1+S/N) bits/sec (bps, bit rate)

C is the maximum number of bits that can be transmitted per second with a Pe=0.

To transmit data in bit rate D, the channel capacity of a system/channel must be

C ≥ D

+

Noise n(t)

s(t)

t

Relationship between Transmission Speed and Noise

Information / Hz: 1 Hertz can transmit a maximum of 2 pieces of information per second.

2 bits / sec / Hz

Shannon theorem C = Blog2(1+S/N) shows that the maximum rate or channel Capacity of a system/channel depends on bandwidth, signal energy and noise intensity. Thus, to increase the capacity, three possible ways are 1) increase bandwidth; 2) raise signal energy; 3) reduce noise. Shannon theorem tell us that we cannot send data faster than the channel capacity, but we can send data through a channel at the rate near its capacity.

Examples

1. For an extremely noise channel S/N 0, C 0, cannot send any data regardless of bandwidth

2. If S/N=1 (signal and noise in a same level), C=B

3. The theoretical highest bit rate of a regular telephone line where B=3000Hz and S/N=35dB. 10log10(S/N)=35 log2(S/N)= 3.5x log210

C= Blog2(1+S/N) =~ Blog2(S/N) =3000x3.5x log210=34.86 Kbps If B is fixed, we have to increase signal-to-noise ratio for increasing transmission rate.

Channel Capacity