1 transmisi data
TRANSCRIPT
TRANSMISI DATA
Sujoko
Objektif
• Konsep dan Terminologi
Di dlm pengiriman data
• Transmisi data analog dan digital
• Gangguan transmisi (Impairment)
• Analisis Fourier
• Kuat sinyal (Signal Strength) dan Decibels
Sources of signal impairment Signal carried on transmission medium affected by:
- Attenuation(kebalikan dari penguatan)
- Limited Bandwidth(keterbatasan jalan)
- Delay Distortion(tertunda)
- Noise(ganguan)
Sources of signal impairment
Sumber dari perusakan sinyal
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
Attenuation (lanjutan)
•Signal Amplification ( Gain ): Increase in signal amplitude
Contoh (Attenuation)
Contoh (Lanjutan)
Signaling Rate vs. Data Bit Rate
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 ?
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.
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
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
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
Aplikasi analisis Fourier (Untuk transmisi data)
Aplikasi analisis Fourier (Untuk transmisi data)
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
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
Aplikasi
analisis Fourier
(Untuk transmisi data)
Analysis Effect of Limited Bandwidth:
Aplikasi analisis Fourier (Untuk transmisi data) Effect of Limited Bandwidth
Aplikasi analisis Fourier (Untuk transmisi data) Effect of Limited Bandwidth:
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
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
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
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
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
Delay Distortion (Lanjutan)
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)
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)
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.
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
Noise (Lanjutan)