Transmission Media and Data Communication: Cables, Fibre, Channel Capacity and Multiplexing · ট্রান্সমিশন মিডিয়া ও ডেটা কমিউনিকেশন: ক্যাবল, ফাইবার, চ্যানেল ক্যাপাসিটি ও মাল্টিপ্লেক্সিং

1. Data communication basics

Data communication is the exchange of data between two devices over a transmission medium. Every system has five components: the message, the sender, the receiver, the transmission medium and the protocol (the agreed rules). A channel is the path that carries signals between a source and a destination.

Direction of flow (transmission modes):

Mode	Direction	Example
Simplex	One way only	Keyboard → computer, TV and radio broadcast
Half duplex	Both ways, but one at a time	Walkie-talkie, a hub-based (shared) Ethernet
Full duplex	Both ways at the same time	Telephone call, switched Ethernet

Serial vs parallel. Parallel transmission sends several bits at once over several wires (short distances, such as old printer cables); serial transmission sends one bit after another over one channel (all networks, USB).

Asynchronous vs synchronous serial transmission.

Point	Asynchronous	Synchronous
Unit	One character (byte) at a time	Long blocks (frames) of bytes
Framing	A start bit before and one or more stop bits after every byte	A few synchronisation (flag) bytes per frame; no start and stop bits per byte
Timing	No shared clock; gaps between bytes may vary	Sender and receiver kept in step by a clock (sent or recovered from the data)
Overhead and speed	High overhead (≈ 20 % or more), slower	Low overhead, faster
Used in	Keyboards, RS-232 serial ports, UART links	Ethernet, SONET/SDH, HDLC links

The problem with asynchronous transmission is that the start and stop bits waste capacity and each byte must be re-synchronised; synchronous transmission solves this by sending large blocks with timing carried by the clock or the line code, so only a small header and trailer are added per block.

Worked example 1: baud rate for a serial link. Each character has 8 data bits, 1 parity bit and 2 stop bits, sent at 300 characters per second. Asynchronous framing always adds 1 start bit, so each character takes 1 + 8 + 1 + 2 = 12 bits, and the line needs 300 × 12 = 3,600 bits per second (3,600 baud with two-level signalling).

2. Guided media: twisted pair, coaxial cable and optical fibre

Twisted pair. Pairs of insulated copper wires twisted together; the twisting makes noise affect both wires almost equally, so it cancels at the receiver. UTP (unshielded) is cheap, thin and easy to install, which is why it is used instead of STP in almost every office LAN; STP (shielded) adds a metal foil or braid against interference, at higher cost and with the need for grounding. LAN cables use RJ-45 connectors and a maximum channel length of 100 m for Ethernet.

Category	Rated bandwidth	Typical Ethernet use
Cat 3	16 MHz	10 Mbps (10BASE-T), telephone
Cat 5	100 MHz	100 Mbps (100BASE-TX)
Cat 5e	100 MHz	1 Gbps (1000BASE-T)
Cat 6	250 MHz	1 Gbps to 100 m; 10 Gbps only up to about 55 m
Cat 6A	500 MHz	10 Gbps to 100 m
Cat 8	2,000 MHz	25 and 40 Gbps up to 30 m (data-centre links)

So the difference between Cat 5 and Cat 6: Cat 6 is rated to 250 MHz instead of 100 MHz, has tighter twists and better crosstalk performance, and supports 10 Gbps over short runs.

Coaxial cable. A central copper conductor, insulation, a braided metal shield and an outer jacket. It has more bandwidth and better noise immunity than twisted pair and uses BNC or F connectors; 50 Ω cable was used for early Ethernet and 75 Ω for cable TV and cable internet.

Optical fibre. Data travels as light in a glass core surrounded by cladding of slightly lower refractive index (the core is the solid dielectric cylinder; the surrounding dielectric is the cladding), then a protective buffer and jacket. Light that hits the core–cladding boundary at more than the critical angle is reflected back completely (total internal reflection), so it stays in the core along the fibre.

Point	Single-mode fibre (SMF)	Multimode fibre (MMF)
Core diameter	About 8 – 10 µm (cladding 125 µm)	50 or 62.5 µm (cladding 125 µm)
Light paths	One mode, straight down the core	Many modes (step-index or graded-index)
Light source	Laser	LED or VCSEL laser
Dispersion	Very low (no modal dispersion)	Modal dispersion limits the distance
Distance	Tens of kilometres and more (long haul, submarine)	Up to a few hundred metres at high speed
Typical wavelengths	1310 nm and 1550 nm	850 nm and 1300 nm
Cost	Cheaper fibre, dearer optics	Cheaper optics; the usual choice inside buildings and LANs

Silica fibre has low-loss "windows" around 850, 1310 and 1550 nm; the 1550 nm window has the lowest loss (about 0.2 dB/km) and is used for long-haul and WDM systems. Advantages of fibre: very high bandwidth, very low attenuation, immunity to electromagnetic interference, hard to tap, thin and light, no electrical hazard. Disadvantages: higher installation cost, fragile glass, skilled splicing and special connectors (SC, LC, ST), and a fibre link normally carries light one way, so a duplex link usually needs two fibres (BiDi optics use two wavelengths to share one fibre). A media converter joins copper and fibre, for example to connect a switch with no fibre ports to a fibre link. SONET/SDH backbones run over fibre.

Point	Twisted pair	Coaxial cable	Optical fibre
Signal	Electrical	Electrical	Light
Bandwidth	Lowest	Medium	Highest
Attenuation	Highest	Medium	Lowest
Noise immunity	Low (better with STP)	Good	Immune to EMI
Cost and installation	Cheapest, easiest	Moderate	Most expensive, needs skill
Typical use	LANs, telephone lines	Cable TV, older Ethernet	Backbones, long haul, submarine links

Ethernet cable names. In 10BASE5, "10" is the speed in Mbps, "BASE" means baseband, and "5" means about 500 m per segment (thick coax). 10BASE2 is thin coax (185 m), 10BASE-T is twisted pair (100 m), and an F in the name, as in 100BASE-FX, means fibre. A straight-through cable connects unlike devices (PC to switch), a crossover cable connects like devices (switch to switch, PC to PC, router to router, and also PC to router, because a PC's and a router's ports are wired the same way; modern ports switch automatically with auto-MDIX), and a rollover (console) cable connects a PC's serial port to a router's or switch's console port.

3. Unguided media: radio, microwave, infrared and satellite

Medium	Frequency (common textbook ranges)	Propagation	Typical use
Radio waves	3 kHz – 1 GHz	Omnidirectional, pass through walls, long range	AM/FM radio, TV, mobile and paging
Microwaves	1 GHz – 300 GHz	Directional, need line of sight; affected by rain	Point-to-point towers, satellite, mobile backhaul, Wi-Fi (2.4/5 GHz)
Infrared	300 GHz – 400 THz	Line of sight, short range, blocked by walls	TV remotes, short indoor links

The band edges are conventions: the ITU names bands VLF, LF … SHF, EHF by decades of frequency, and some texts start microwaves at 300 MHz.

Satellite links are microwave links that use a satellite as a relay. A geostationary satellite sits 35,786 km above the equator, so a signal going up and back down travels about 71,600 km: roughly 0.24 s one way and about 0.5 s for a round trip, which is noticeable on calls. This is one reason submarine fibre cables carry most international traffic instead of satellites: much higher capacity (many terabits per second), far lower delay, lower cost per bit, and no weather effects. Satellites remain useful for remote areas, ships, aircraft and broadcasting.

4. Bandwidth, data rate, throughput and latency

• Analogue vs digital signals. An analogue signal varies smoothly and can take any value in a range (voice on an old telephone line); a digital signal uses a small, fixed set of levels (the line codes in section 7). For a periodic signal, period T = 1/f and wavelength λ = c/f: a 50 Hz mains wave has T = 0.02 s, and a 2.4 GHz Wi-Fi wave in air has λ = 3 × 10⁸ ÷ 2.4 × 10⁹ = 0.125 m (12.5 cm).
• Bandwidth has two meanings: in hertz, the range of frequencies a channel passes (a telephone line passes about 300 – 3,300 Hz, so 3,000 Hz); in bits per second, the maximum data rate of a link.
• Data (bit) rate is bits per second; baud (signal) rate is signal elements per second. bit rate = baud × log₂L when each signal element carries log₂L bits (multilevel signalling and modulation), so there baud ≤ bit rate. Codes with a mid-bit transition (RZ, Manchester) can use two signal elements per bit, so 10 Mbps Manchester Ethernet runs at up to 20 Mbaud.
• Throughput is the rate actually achieved, always at most the bandwidth (a 100 Mbps link might deliver 70 Mbps).
• Latency (delay) = propagation time + transmission time + queuing time + processing time, where propagation time = distance ÷ propagation speed and transmission time = message size ÷ bandwidth.

Baseband vs passband (broadband). A baseband signal occupies frequencies from near 0 Hz upward and uses the whole medium for one signal (Ethernet over twisted pair). A passband signal has been modulated onto a carrier, so it occupies a band around the carrier frequency; broadband systems place many such channels side by side (FDM), as in cable TV and DSL. The main benefit of broadband is carrying many signals and services on one medium at once.

Worked example 2: propagation vs transmission time. A 2.5 KB email (2,500 bytes) is sent over a 1 Gbps link to a receiver 12,000 km away; signals travel at 2.4 × 10⁸ m/s.

1. Propagation time = 12,000,000 m ÷ (2.4 × 10⁸ m/s) = 0.05 s = 50 ms.
2. Transmission time = 2,500 × 8 bits ÷ 10⁹ bps = 0.00002 s = 0.02 ms.
3. On a long, fast link the propagation time dominates; for a large file on a slow link the transmission time does.

5. Channel capacity: Nyquist and Shannon

	Nyquist bit rate	Shannon capacity
Channel	Noiseless (ideal)	Noisy (real)
Formula	C = 2B log₂L	C = B log₂(1 + SNR)
Depends on	Bandwidth and number of signal levels L	Bandwidth and signal-to-noise ratio
Tells you	The rate for a chosen signalling scheme	The absolute upper limit no scheme can beat

Decibels. SNR(dB) = 10 log₁₀(SNR), so SNR = 10 raised to the power (dB ÷ 10): 10 dB = 10, 20 dB = 100, 30 dB = 1,000, and 3 dB ≈ 2 (exactly 1.995). Always convert dB to a plain ratio before using Shannon's formula. Example: a signal of 10 mW over noise of 1 µW gives SNR = 0.01 ÷ 0.000001 = 10,000, which is 10 log₁₀(10,000) = 40 dB.

Worked example 3: Nyquist.

1. A noiseless 3,000 Hz channel with 2 levels: C = 2 × 3,000 × log₂2 = 6,000 bps.
2. A 200 kHz channel with 4 levels: C = 2 × 200,000 × log₂4 = 400,000 × 2 = 800 kbps.
3. We need 230 kbps over a noiseless 25 kHz channel. 230,000 = 2 × 25,000 × log₂L, so log₂L = 4.6 and L = 2⁴·⁶ ≈ 24.3. The number of levels should be a power of 2, so use 32 levels (then the channel can carry 250 kbps), or use 16 levels and accept 200 kbps.

Worked example 4: Shannon.

1. A telephone line: B = 3,000 Hz, SNR = 3,162. C = 3,000 × log₂(3,163) ≈ 3,000 × 11.627 ≈ 34,881 bps (about 34.9 kbps). This is why fully analogue modems (V.34) topped out at 33.6 kbps; 56 kbps (V.90) modems got past it only because the ISP's end of the line was digital.
2. A teleprinter channel: B = 300 Hz, SNR = 3 dB. First convert: 10⁰·³ ≈ 1.995. Then C = 300 × log₂(2.995) ≈ 300 × 1.583 ≈ 475 bps (474.8 bps; taking SNR as exactly 2 gives 300 × log₂3 ≈ 475.5 bps, the same rounded answer).
3. B = 2 MHz, SNR = 255: C = 2 × 10⁶ × log₂256 = 2 × 10⁶ × 8 = 16 Mbps. To actually reach it, Nyquist gives 16 × 10⁶ = 2 × 2 × 10⁶ × log₂L, so log₂L = 4 and L = 16 levels.
4. B = 16 kHz, SNR = 15: C = 16,000 × log₂16 = 64 kbps.
5. If SNR is almost 0, log₂(1 + 0) = 0, so the capacity is 0, whatever the bandwidth.

Sampling theorem (Nyquist rate). To digitise an analogue signal without losing information, sample at least twice its highest frequency: fs ≥ 2 fmax. A signal whose highest frequency is 600 Hz needs at least 1,200 samples/s. Voice limited to 4 kHz is sampled 8,000 times per second; with 8 bits per sample that is the standard 64 kbps PCM voice channel.

6. Transmission impairments

Signals arrive weaker and less clean than they left, for three main reasons:

Impairment	What happens	Remedy
Attenuation	The signal loses energy with distance (measured in dB, a ratio, so losses add)	Amplifiers (analogue) or repeaters/regenerators (digital); shorter runs; lower-loss media
Distortion	The signal changes shape because its frequency components travel at different speeds; in fibre this is dispersion (modal in multimode, chromatic in all fibre)	Equalisers, single-mode fibre, dispersion compensation
Noise	Unwanted energy is added: thermal (random electron motion), induced (from motors and power lines), crosstalk (one wire picking up its neighbour) and impulse (spikes from lightning or switching)	Shielding, twisting, filtering, error-control coding

An amplifier boosts the noise together with the signal; a digital repeater (regenerator) detects the bits and sends a fresh, clean signal, which is why digital links can cross long distances without accumulating noise.

What happens if a cable is longer than the specified length (for example more than 100 m of UTP Ethernet)? Attenuation makes the signal too weak and distorted to read, so bit errors, retransmissions and dropped links increase; in shared CSMA/CD Ethernet a too-long segment can also cause collisions that are detected too late (late collisions) or not at all. The fix is to stay within the limit and use a repeater, a switch or fibre for longer distances.

Worked example 5: a fibre loss budget. A 40 km single-mode link has fibre loss 0.25 dB/km and two connectors of 0.5 dB each.

1. Fibre loss = 40 × 0.25 = 10 dB; connector loss = 2 × 0.5 = 1 dB; total = 11 dB.
2. If 1 mW (0 dBm) is launched, the received power is 0 − 11 = −11 dBm, that is 10⁻¹·¹ mW ≈ 0.079 mW.

7. Line coding: putting bits on the wire

Line coding converts a sequence of bits into a digital signal. A good line code keeps the receiver's clock in step (enough transitions), avoids a DC component, and helps detect errors.

Worked example 6: line codes for 10110001. Levels are written as + (high), − (low) and 0 (zero); two symbols in one cell are the first and second half of the bit. Conventions used here: NRZ-L uses + for 0 and − for 1; NRZ-I starts low and inverts at every 1; polar RZ sends + for 1 and − for 0; Manchester follows IEEE 802.3 (0 = high-to-low, 1 = low-to-high in mid-bit); differential Manchester starts low, always changes in mid-bit and also changes at the start of a 0; AMI starts its 1s with +. Other conventions exist (some courses draw NRZ-L with + for 1, and G. E. Thomas's original Manchester uses 0 = low-to-high, 1 = high-to-low, the reverse of IEEE 802.3), so in an exam state the convention you use: the shape of each code is what is marked.

Bit	1	0	1	1	0	0	0	1
NRZ-L	−	+	−	−	+	+	+	−
NRZ-I	+	+	−	+	+	+	+	−
Unipolar RZ	+ 0	0 0	+ 0	+ 0	0 0	0 0	0 0	+ 0
Polar RZ	+ 0	− 0	+ 0	+ 0	− 0	− 0	− 0	+ 0
Manchester	− +	+ −	− +	− +	+ −	+ −	+ −	− +
Differential Manchester	− +	− +	+ −	− +	− +	− +	− +	+ −
AMI (bipolar)	+	0	−	+	0	0	0	−

Notice the run of three 0s: NRZ-L and NRZ-I stay at one level (no transitions for the clock), unipolar RZ and AMI send nothing at all, while both Manchester codes still change in the middle of every bit.

• NRZ codes are simple and need the least bandwidth but lose synchronisation on long runs of equal bits (NRZ-I fixes long runs of 1s only).
• RZ returns to zero in the middle of each bit, giving more transitions at the cost of twice the bandwidth.
• Manchester and differential Manchester have a transition in every bit, so they carry their own clock and have no DC component (classic 10 Mbps Ethernet used Manchester; Token Ring used differential Manchester).
• AMI sends 0 as zero volts and alternates + and − for each 1, which removes the DC component and makes a single error visible as a broken alternation.

Block coding and scrambling fix the weak spots of simple line codes. Block coding adds redundancy before line coding: 4B/5B maps every 4 data bits to a 5-bit code word chosen so that the coded stream never has more than three 0s in a row, so NRZ-I stays synchronised (100BASE-FX, FDDI), and 8B/10B does the same job with 8-bit groups for gigabit links (1000BASE-X). Scrambling keeps AMI's bandwidth but replaces a long run of zeros with a pattern containing deliberate bipolar violations: B8ZS (8 zeros, North American T1) and HDB3 (4 zeros, European E1), so the receiver never loses its clock.

8. Modulation and digitisation

Modulation varies a property of a high-frequency carrier wave according to the message. It is necessary because low-frequency signals cannot be radiated efficiently (the antenna would need to be impossibly long), because different carriers let many signals share the medium (FDM), and because shifting a signal to a suitable band lets it pass through channels such as telephone lines and radio links.

Message	Changes amplitude	Changes frequency	Changes phase	Both amplitude and phase
Analogue (AM radio, FM radio)	AM	FM	PM	–
Digital (data)	ASK	FSK	PSK (BPSK, QPSK)	QAM

• Bits per symbol = log₂M: BPSK 1, QPSK 2, 8-PSK 3, 16-QAM 4, 64-QAM 6. More levels give more bits per hertz (16-QAM is more bandwidth-efficient than QPSK or BPSK) but need a cleaner channel. Example: 16-QAM at 1,000 baud carries 1,000 × 4 = 4,000 bps.
• AM bandwidth is 2 × the highest message frequency: a 10 kHz message on a 1 MHz carrier needs 20 kHz. FM needs more bandwidth but resists noise better; ASK is the most noise-sensitive digital scheme, PSK the most robust of the three basic ones. GSM phones use GMSK, a form of frequency-shift keying.
• FM example: a 5 V, 2 kHz tone into a modulator with sensitivity 40 Hz/V gives frequency deviation Δf = 5 × 40 = 200 Hz, modulation index β = 200 ÷ 2,000 = 0.1, and Carson's-rule bandwidth 2 × (200 + 2,000) = 4,400 Hz.
• QPSK bandwidth: 12 Mbps with 2 bits per symbol is 6 Mbaud; with no extra filtering factor (d = 0) the bandwidth is 6 MHz.

A modem (modulator–demodulator) converts digital data to an analogue signal and back. A codec does the opposite job: it digitises analogue signals such as voice using PCM (pulse code modulation): sample the signal (giving PAM, pulse amplitude modulation, a train of pulses whose heights follow the signal), quantise each sample to one of a fixed set of levels, and encode each level as bits. Delta modulation sends only 1 bit per sample (up or down by one step); it is simple but suffers slope overload when the signal changes faster than the step allows.

9. Multiplexing: FDM, TDM, WDM and CDM

Multiplexing lets many signals share one link: a multiplexer (MUX) combines them and a demultiplexer (DEMUX) separates them.

Scheme	Shares by	Signal type	Examples
FDM	Frequency bands separated by guard bands	Analogue	Radio and TV broadcasting, first-generation mobile
TDM	Time slots in repeating frames	Digital	T1/E1 telephone lines (GSM's radio time slots are TDMA)
WDM	Wavelengths (colours) of light; DWDM packs them closely	Optical (FDM for light)	Fibre backbones and submarine cables
CDM (CDMA)	Orthogonal codes	Digital (spread spectrum)	3G CDMA phones, GPS

Worked example 7: FDM bandwidth. Six channels of 80 kHz each need a 4 kHz guard band between neighbours. Six channels have only 5 gaps, so the link needs 6 × 80 + 5 × 4 = 500 kHz. (Counting six guard bands by mistake gives 504 kHz.)

Worked example 8: synchronous TDM.

1. A TDM link sends 4,000 frames per second and each input has one 8-bit slot per frame. Each input therefore runs at 4,000 × 8 = 32 kbps.
2. Five sources each produce 200 characters per second; the slot is one 8-bit character and 1 synchronising bit is added to each frame. A frame holds 5 × 8 + 1 = 41 bits; the link sends 200 frames per second, so its rate is 41 × 200 = 8,200 bps, and each frame lasts 1 ÷ 200 s = 5 ms.

TDM and TDMA differ in where the sharing happens: TDM combines signals at one multiplexer onto a single link, while TDMA (time division multiple access) lets separate, distant stations take turns on a shared medium such as a radio channel.

10. Switching: circuit, packet and message

Point	Circuit switching	Packet switching	Message switching
Path	Dedicated path set up before data flows, held for the whole call	No reserved path; each packet is forwarded on its own (datagram) or along a virtual circuit	No reserved path; the whole message hops from node to node
Phases	Set-up, data transfer, tear-down	Data transfer only (datagram)	Store and forward
Resource use	Wasted when the line is silent	Shared on demand: efficient for bursty data	Needs large storage at nodes
Delay	Set-up delay, then constant	Variable (queuing)	High
Example	Traditional telephone network (PSTN)	The Internet (IP)	Telegraph, early email relays

The Internet uses packet switching because computer traffic is bursty: links are shared among many users instead of sitting idle, a failed link is simply routed around, and no set-up delay is needed. Circuit switching suits constant-rate traffic such as classic voice.

Worked example 9: a circuit-switched file transfer. A link of 1.536 Mbps is divided by TDM into 32 slots; one circuit uses one slot. A 960,000-bit file is sent after an 800 ms circuit set-up.

1. Rate of one circuit = 1,536,000 ÷ 32 = 48,000 bps.
2. Transmission time = 960,000 ÷ 48,000 = 20 s.
3. Total time = 20 + 0.8 = 20.8 s.

11. Tricky exam cases

• Shannon with dB: never put "3 dB" or "30 dB" straight into the formula; convert first (3 dB ≈ 2, 30 dB = 1,000).
• Nyquist vs Shannon: Nyquist has the factor 2 and the number of levels; Shannon has neither. For a noisy channel the Shannon limit is the true ceiling; the Nyquist formula then gives the levels needed to approach it.
• Guard bands: n channels need n − 1 guard bands.
• Sampling: use twice the highest frequency, not twice the bandwidth, unless the question is about band-pass sampling.
• Which device converts digital to analogue for a phone line? A modem. Which extends a cable segment? A repeater. Which has the highest bandwidth? Optical fibre.
• Media access: wired (shared) Ethernet uses CSMA/CD; Wi-Fi uses CSMA/CA, because a radio cannot hear a collision while it is transmitting and some stations cannot hear each other (the hidden-terminal problem).
• Single-mode or multimode for a LAN? Multimode is the usual choice inside buildings and campuses (cheaper optics, short distances); single-mode is for long distances.
• Baud vs bit rate: a 16-QAM signal at 1,000 baud carries 4,000 bps; plain 2-level NRZ has baud = bit rate, but Manchester needs up to twice as many baud as bits per second.