Transmission Fundamentals¶

Exam-ready answer

A telecommunication network is an organized set of terminals, access links, switching or routing nodes, transmission systems, signaling, and operations support used to convey voice, data, image, or video between users. Its bearer path is: source terminal converts information to an electrical, optical, or radio signal; the access network carries it to a local exchange/router; metro and core nodes select a route and aggregate traffic; the destination access network and terminal recover the information. A separate control plane performs address analysis, call/session setup and release through systems such as SS7 or SIP, while OSS/NMS performs fault, configuration, accounting, performance, and security management.

The evolution was telegraph -> manual telephone exchange -> automatic Strowger/crossbar exchange -> analog FDM coaxial/microwave trunks -> PCM/TDM digital exchanges and PDH/SDH fiber -> packet data, cellular systems, and NGN. Early calls occupied a dedicated analog circuit and required an operator; automatic dialing introduced electromechanical switching. PCM then sampled a nominal voice channel at \(8\,\text{kHz}\) with 8 bits/sample, producing \(64\,\text{kbit/s}\), and digital multiplexing improved quality and capacity. Modern NGN uses shared IP/MPLS transport, software session control, media gateways, and service platforms so voice, video, and data use one core.

Thus the architectural trend is from service-specific, hardware-controlled circuit networks to software-controlled, packet-switched convergence. Circuit switching still offers deterministic delay for traditional voice, whereas packet sharing is more efficient for bursty traffic but needs QoS, synchronization, security, and legacy interworking. Examples are PSTN local-loop-to-exchange calling and a VoIP call carried through broadband access and an IP core.

Practice target: 8 minutes; draw and label the three planes, then give the evolution stages and one circuit-versus-packet caveat.

Exam-ready answer

Guided media confine electromagnetic energy to a physical path. Twisted pair consists of two insulated copper conductors twisted to cancel induced noise. It is inexpensive, easy to terminate, and already reaches most subscribers; balanced telephone loops, Ethernet, and DSL are typical applications. Its resistance, dielectric loss, crosstalk, and external interference limit distance and bandwidth, although category and DSL technology determine the actual rate.

Coaxial cable has a central conductor, dielectric, and concentric shield, giving controlled characteristic impedance, commonly \(50\,\Omega\) for RF/data or \(75\,\Omega\) for video systems. Its shielding and bandwidth exceed ordinary pair cable, so it suits cable television, RF feeders, and legacy digital trunks, but it is bulkier and harder to install. Optical fiber guides light in a dielectric core and cladding by total internal reflection. Single-mode fiber offers extremely high bandwidth and long repeater spacing; multimode fiber is cheaper for shorter links. Fiber is light, secure, immune to EMI and has low attenuation, but lasers, receivers, connectors, and precision splicing add cost and bending must be controlled.

Selection depends on required bit rate and distance, installed cost, attenuation, EMI, security, mechanical environment, and maintenance skill. For example, copper pair is economical for the final subscriber drop, while fiber is preferred for an aggregation or core link. Published bandwidth and reach are technology- and installation-dependent rather than universal values.

Practice target: 8 minutes; define guided media, compare all three under four common headings, and finish with one selection example and caveat.

Exam-ready answer

The characteristic impedance \(Z_0\) is the voltage-to-current ratio of one travelling wave on a uniform line; equivalently, it is the input impedance of an infinitely long line or of a finite line terminated in \(Z_0\). For distributed primary constants \(R,L,G,C\) per unit length,

At a load \(Z_L\), the incident wave is partly reflected unless \(Z_L=Z_0\). The voltage reflection coefficient and voltage standing-wave ratio are

Return loss measures the incident-to-reflected power ratio:

Therefore a high positive return loss means a small reflection and good matching; \(RL=20\,\text{dB}\) means \(P_r/P_i=10^{-2}\), so only 1% of incident power is reflected. For example, a \(50\,\Omega\) line terminated in \(75\,\Omega\) has \(\Gamma=(75-50)/(75+50)=0.2\), \(RL\approx13.98\,\text{dB}\), and VSWR \(=1.5\). Matching improves delivered power, suppresses standing waves and waveform distortion, and reduces hybrid echo or transmitter stress. An open or short has \(|\Gamma|=1\) and \(RL=0\,\text{dB}\); an ideal match has \(\Gamma=0\) and theoretically infinite return loss. In practice \(Z_0\) and return loss vary with frequency, connectors, and line tolerances, so they must be specified over the operating band.

Practice target: 8 minutes; write all three boxed relations and solve one numerical mismatch without confusing high return loss with high reflected power.

Exam-ready answer

A subscriber local loop is normally 2-wire: the same balanced pair carries speech in both directions. A trunk, repeater, codec, or digital switch uses 4-wire working, with separate transmit and receive pairs so each direction can be amplified without direct feedback. A hybrid is the bidirectional bridge that converts between these interfaces: speech from the 4-wire transmit port is delivered to the 2-wire line, incoming line speech is delivered to the 4-wire receive port, and ideally the local transmit signal is cancelled at that receive port.

The transformer windings and bridge arms divide the transmitted signal between the actual line impedance \(Z_L\) and a balance network \(Z_B\). When \(Z_B\) accurately models \(Z_L\) over the voice band, equal opposing voltages cancel at the receive arm, giving high trans-hybrid loss. If line length, handset, frequency, or weather makes \(Z_B\ne Z_L\), transmit energy leaks into receive and returns after network delay as echo; two imperfect hybrids plus trunk gain can also form a singing loop.

Telephone transformers provide galvanic isolation, DC blocking, balanced/unbalanced conversion, common-mode rejection, coupling, and impedance matching. For an ideal transformer with turns ratio \(a=N_p/N_s\),

Thus matching a \(150\,\Omega\) secondary load to a \(600\,\Omega\) primary line needs \(a=\sqrt{600/150}=2\). Real hybrids have insertion loss and finite bandwidth, and no fixed balance network exactly represents every subscriber loop; adaptive echo cancellation is therefore used on long-delay digital and packet paths.

Practice target: 8 minutes; label all four hybrid ports and the balance network, then state the transformer ratio and mismatch caveat.

5-mark answer and 10-mark extension

For 5 marks - write this¶

Echo is an intelligible delayed return of speech to the talker or listener. The usual electrical cause is impedance mismatch at a 2-wire/4-wire hybrid: part of the outgoing signal leaks into the opposite 4-wire path and returns after propagation delay. Acoustic coupling from a loudspeaker to a microphone causes acoustic echo. Singing is different: it is a sustained tone or howl produced when leakage through two hybrids and the gains and delays of the 4-wire path form positive feedback. At some frequency it satisfies the oscillation condition

Echo is controlled by accurate line termination and hybrid balance, high return/trans-hybrid loss, reduced delay, echo suppressors, or preferably an adaptive echo canceller. Singing is prevented by improving hybrid isolation and keeping loop gain below unity with a suitable stability margin. Echo is repeated speech and becomes more objectionable as delay rises; singing is regenerative oscillation and may persist without new speech.

Add for a 10-mark variant¶

In the signal path, near-end speech enters hybrid \(H_1\), travels on the four-wire send path through codecs, amplifiers or packet transport, and reaches hybrid \(H_2\). A fraction reflected by the mismatch at \(H_2\) returns on the receive path. The speaker hears talker echo after the round-trip delay; a second reflection may produce listener echo. Satellite, transcoder, and packet/jitter-buffer delays do not create the reflection, but they make an existing echo perceptually worse. The often-quoted delay at which control becomes important, about \(25\,\text{ms}\), is an engineering guideline and depends on echo level and the applicable standard.

An echo canceller models the echo path with an adaptive filter. If far-end/reference signal \(x(n)\) produces actual echo \(d(n)\), the filter forms \(\hat d(n)=\mathbf{w}^T(n)\mathbf{x}(n)\) and sends residual

An LMS-type update is \(\mathbf{w}(n+1)=\mathbf{w}(n)+\mu e(n)\mathbf{x}(n)\), with normalization and practical double-talk detection. Unlike an echo suppressor, which inserts loss in one direction and can clip simultaneous speech, cancellation permits full-duplex conversation; nonlinear loudspeakers, rapid path change, insufficient filter tail, and double talk limit cancellation.

For a two-hybrid singing loop, let the one-way gains be \(G_1,G_2\) and leakage factors \(k_1,k_2\). The loop magnitude is \(|G_1G_2k_1k_2|\); oscillation occurs only if magnitude and phase conditions coincide. Gain reduction, frequency shaping, improved balance networks, and stable automatic level control restore margin. As an example, a poorly balanced long VoIP/PSTN gateway call may have audible delayed self-speech but remain stable; raising trunk gain can turn the same feedback path into a continuous tone. Hence echo cancellation addresses delayed replicas, while singing control primarily addresses closed-loop stability.

Practice target: 9 minutes for the 5-mark core or 18 minutes for the full answer; distinguish reflection level, propagation delay, and closed-loop gain explicitly.