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TRANSMISI DATA Sujoko

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TRANSMISI DATA

Sujoko

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Objektif

• Konsep dan Terminologi

Di dlm pengiriman data

• Transmisi data analog dan digital

• Gangguan transmisi (Impairment)

• Analisis Fourier

• Kuat sinyal (Signal Strength) dan Decibels

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Sources of signal impairment Signal carried on transmission medium affected by:

- Attenuation(kebalikan dari penguatan)

- Limited Bandwidth(keterbatasan jalan)

- Delay Distortion(tertunda)

- Noise(ganguan)

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Sources of signal impairment

Sumber dari perusakan sinyal

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Attenuation

•Signal Attenuation: Decrease in signal amplitude as it propagates along transmission medium- Consequence: Limit length of cable to be used

- Solution : Amplifiers (Repeaters) used to restore signal to original level.

- Property : Signal attenuation increases as a function of frequency.

Possible remedies : - Non Linear amplifiers or Equalizers

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Attenuation (lanjutan)

•Signal Amplification ( Gain ): Increase in signal amplitude

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Contoh (Attenuation)

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Contoh (Lanjutan)

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Signaling Rate vs. Data Bit Rate

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Limited Bandwidth

• Bandwidth of a communication / transmission medium:– band of sinusoidal frequency components (f1to

f2) that will– be transmitted by the channel unattenuated

• Question: What is the effect of channel bandwidth ontransmitted signal ?

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Fourier analysis

• A periodic signal is made of an infinite series of sinusoidal frequency components

- Fundamental frequency component : same frequency as initial periodic signal, cycles per sec (Hz)

- Harmonics : other frequency components, multiples of fundamental frequency.

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Fourier analysis

• V(t) = tegangan sinyal (periodis) sebagai fungsi waktu

• W0 = komponen frekuensi fondamental (rad/sec)

• T= 2π/W0 atau W0 = 2 πf0

1 1

00 )(sin)(cos)(n n

nn tnbtnaatv

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Aplikasi analisis Fourier (Untuk transmisi data)

Possible binary sequences (periodic) :(1) 1 0 1 0 1 0 …… period = 2 bit cell intervals(2) 1 1 0 1 1 0 ….. period = 3 bit cell intervals(3) 1 1 1 0 1 1 1 0 …. period = 4 bit cell intervalsNote: (1) has shortest period, highest fundamental frequencycomponent: worst-case sequence

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Aplikasi analisis Fourier (Untuk transmisi data)

Basic Binary Signal Types:

• Unipolar Signal (Return to zero, RZ) Amplitudes : +V, 0;

Mean Signal Level = V /2

• Bipolar Signal (Non Return–To–Zero , NRZ)

Amplitudes : +V, -V;

Mean Signal Level = 0

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Aplikasi analisis Fourier (Untuk transmisi data)

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Aplikasi analisis Fourier (Untuk transmisi data)

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Aplikasi analisis Fourier (Untuk transmisi data)

General Observations:

• A periodic binary sequence is made of an infinite series of

sinusoidal signals made of-A fundamental frequency component, f0

-A third harmonic component , 3f0

-A fifth harmonic component , 5f0

Note: Odd harmonics only

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Aplikasi analisis Fourier (Untuk transmisi data)/Lanjutan

• The amplitude of the harmonics diminishes with increasing frequency

• T (Signal Period) = 2*Tb (Bit Period)Tb = T/2; 1/ Tb = 2/T = 2 * 1/TR (bit rate) = 2 * f0 (fundamental frequency)

• When binary data signal transmitted on a channel, only those frequency components that are within channel bandwidth will be received

• A channel with a bandwidth :From 0 to fundamental frequency (half the bit rate)can often give satisfactory performance

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Aplikasi

analisis Fourier

(Untuk transmisi data)

Analysis Effect of Limited Bandwidth:

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Aplikasi analisis Fourier (Untuk transmisi data) Effect of Limited Bandwidth

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Aplikasi analisis Fourier (Untuk transmisi data) Effect of Limited Bandwidth:

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Contoh :

• A binary signal of R (bit rate) = 500 bps is transmitted on a communication channel

• What is the minimum bandwidth required, assuming(a) Fundamental frequency only(b) Fundamental and third harmonic(c) Fundamental,third and fifth harmonics

• Solution :For R = 500bps, f0 = R/2 = 250 Hz(a) 0 – 250 Hz(b) 0 – 750 Hz(c ) 0 – 1250 Hz

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Laju transfer informasi maksimum

Untuk kanal transmisi (noiseless) (the Nyquist Formula)

C = 2W log2M

whereW : bandwidth of the channel (Hz)M: number of levels per signaling element

log2M : number of bits/signaling element

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Laju transfer informasi maksimum(Contoh)

Data is sent over a PSTN with :

M = 8 levels /signaling element

W = 3,000 Hz

Q: What is the Nyquist maximum data transfer rate ?

A : C = 2W log2M

= 2*3000*log28

= 2*3000*3

= 18,000 bps

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Bandwidth Efficiency of a Transmission Channel

B = R/W = 1/(W* Tb) bps Hz-1

Observations:• The higher the bit rate relative to the available bandwidth, the higher

the bandwidth efficiency• Typical values for B: 0.25 - 3.0bps Hz-1, (B=3.0 requires a high signal

rate)•Another expression for bandwidth efficiency:

R = RS. m = RS log2 MB = R/W = (R S log2 M) /W

WhereM is number of levels per signaling elementlog2 M is number of bits per signaling element

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Delay Distortion

• Rate of propagation of a signal over a channel is function of frequency of signal

• Consequence: Delay Distortion – different frequency components arrive with different delays

• Delay distortion increases with bit rate• Inter symbol interference : frequency components

of a bit start to interfere with a later bit• Use of an eye diagram : oscilloscope displays all

possible signals superimposed

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Delay Distortion (Lanjutan)

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Noise

• Line Noise level :Random perturbations(gangguan) on the line, even when no signal present.Consequence: Interference between (attenuated) transmitted signal and line (background) noise

• Signal-to noise ratio: (SNR)SNR = 10log10(S/N) dBS : average power in received signal (watts)N : noise power (watts)

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Noise (Lanjutan)

Theoretical data rate of a transmission channel(Shannon – Hartley law)

C = W log2(1 + S/N) bps

Where

W = Bandwidth (Hz).

S = Average signal power (watts)

N = Random noise power (watts)

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Noise (Lanjutan)

Theoretical data rate of a transmission channel(Shannon – Hartley law)

Given a PSTN with: W = 3000 Hz, andSNR = 20 dBdetermine maximum theoretical data rateA : SNR = 10log10(S/N)20 = 10log10 (S/N) , untuk S/N = 102 = 100.C = W log2(1 + S/N)

= 3000 log2(1+100) = 19,963 bps.

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Noise (Lanjutan)

• Cross Talk: Noise caused by unwanted electrical coupling between adjacent lines

• Near – End Cross Talk (NEXT) or Self – Cross Talk: Strong signal output from transmitter circuit interferes with weak signal at receiver circuit Solution: Adaptive NEXT cancellers

• Impulse Noise: Caused by external electrical activity– impulses (lightning, impulses from old switching systems)Observation : Both cross talk and impulse noise arecaused by electrical activity external to transmission line

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Noise (Lanjutan)