Telecommunications System Components. Computer to process information. Terminals or input/output devices (source/destination)
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We can reduce the effect of cross talk & noise by using twisted wire.
Insulating outer cover
In its simplest form, coaxial consists of a core made of solid copper surrounded by insulation, a braided metal shielding, and an outer cover.
It minimizes both effect:
The low-loss regions of an optical fiber
Further improvements can be obtained by reducing the core
diameter to that of a single wavelength (3-10 Mm).
The emitted light propagates along a single (dispersionless) path.
Small core allows propagation in only one mode
The bit rate will be very high.
A formula derived by Nyquist can be used to find the capacity of the channel (cable) as a function of the bandwidth is:
C = 2B log2 M
where C = maximum capacity in bits per second
B = bandwidth of the cable
M = signaling level ( 8 - bit byte)
Data is to be transmitted over the PSTN using a transmission scheme with eight levels per signaling element ( 8-bit byte). If the bandwidth of the cable is 2600 Hz, deduce the Nyquist maximum data transfer rate.
C = 2 x 2600 x log2 8
= 2 x 2600 x 3
= 15 600 pbs
This formula is for noiseless cable, in practice, there are noises on the cable and we should use Shannon's formula.
C = B log2(1+S/N)
In the absence of a signal, a transmission line ideally has zero
electrical signal present.
In practice, however, there are random perturbations on the
line even when no signal is being transmitted.
This is called line noise level.
The ration of the average power in a received signal S, to the
power in noise level, N is called signal-to-noise ratio (SNR).
SNR = 10 log 10 ( S / N) dB
High SNR ratio indicates good-quality signal
Low SNR ratio indicates low-quality signal.
In practice, there are noises on the cable and we should use Shannon's formula to calculate the theoretical maximum information rate (C) of a transmission channel.
C = B log2(1+S/N)
where C = maximum capacity in bps
B = bandwidth of the channel in Hz
S/N = ratio of signal power (S) to Noise power (N)
expressed in decibels or dB.
Let us suppose that a phone line has a signal-to-noise power ratio of 20 dB.
SNR = 10 log10 (S/N)
20 = 10 log10 (S/N)
20/10 = log10 S/N
2 = log10 S/N
10 2 = S/N
Therefore S/N = 100.
If this line has a bandwidth of 2600Hz. Find C (maximum theoretical information rate that can be achieved).
C = 2600 x log2 (1 + 100)
C = 2600 x log2 (101)
= 2600 x log10 101 / log10 2
= 2600 x 2 / 0.3
= 2600 x 6.643
= 17, 270 bps