Time and frequency division multiplexing

Why it matters

A pipeline plant may have hundreds of sensors but only one radio link or one fibre back to the control room. Multiplexing lets many signals share one channel: frequency-division multiplexing (FDM) gives each signal its own slice of spectrum, and time-division multiplexing (TDM) gives each its own slice of time. Telephone trunks, FM telemetry with subcarriers, PCM data-acquisition frames and wavelength-division multiplexed fibres all apply these two ideas.

Key ideas

FDM. Each message modulates its own subcarrier, the modulated signals are added, and the composite occupies adjacent frequency bands. At the receiver, band-pass filters separate the channels and each is demodulated. Small guard bands between channels allow practical filters with finite roll-off and tolerate oscillator drift.

• Bandwidth: for N channels each occupying B_ch after modulation, with guard band B_g between adjacent channels, B_total = N·B_ch + (N − 1)·B_g (some systems also leave a guard at each edge — read the question).
• B_ch depends on the modulation: SSB uses W, DSB uses 2W.
• Weaknesses: non-linearity in the common amplifier creates intermodulation products that fall into other channels (crosstalk), and the filters are bulky analog parts.
• Examples: the analog telephone hierarchy (12 SSB voice channels of 4 kHz form a 48 kHz group), broadcast radio, FM/FM telemetry with IRIG subcarriers, and wavelength-division multiplexing (WDM) in fibres, which is FDM at optical frequencies.

TDM. Each signal is sampled in turn by a commutator (or multiplexer IC), and the samples are interleaved in a frame. One frame contains one sample (or one PCM word) from every channel, and frames repeat at the sampling rate. The receiver uses a decommutator synchronised to the frame.

• Frame timing: T_frame = 1/f_s. With N equal channels and no overhead, each slot lasts T_frame/N.
• Bit rate for PCM-TDM: R_b = (N·n + framing bits)·f_s.
• Synchronisation: framing bits or a sync word mark the start of each frame; losing frame sync scrambles which samples belong to which channel until the receiver re-acquires lock.
• Unequal bandwidths: a channel with twice the bandwidth is sampled twice per frame (super-commutation); slow channels can share one slot across successive frames (sub-commutation). All samples are kept at the Nyquist rate of their own channel or above.
• Crosstalk in TDM comes from pulse spreading (ISI) into the neighbouring slot when the channel bandwidth is too narrow, so a minimum bandwidth is needed: for a pulse train of rate r pulses per second, B_min = r/2.
• Examples: the T1 carrier (24 channels, 1.544 Mbit/s), the E1 carrier used in India and Europe (32 slots × 64 kbit/s = 2.048 Mbit/s, with 30 voice channels plus framing and signalling), and data-acquisition systems with an analog multiplexer feeding one ADC.

Comparison.

• FDM suits analog signals and continuous transmission; TDM suits digital signals and benefits from cheap digital logic.
• FDM needs many filters and is sensitive to amplifier non-linearity; TDM needs accurate timing and is sensitive to channel bandwidth (ISI).
• Both, ideally, need the same total bandwidth for the same number of channels: N·W for SSB-FDM and N·f_s/2 = N·W for ideal PAM-TDM at the Nyquist rate.

Statistical TDM assigns slots only to active sources, adding address headers; it is the basis of packet networks. CDMA, a third option, separates users by codes rather than time or frequency.

Formulas

B_total = N·B_ch + (N − 1)·B_g N number of channels, B_ch bandwidth per modulated channel (Hz), B_g guard band (Hz).

B_ch = W (SSB), B_ch = 2W (DSB) W message bandwidth (Hz).

T_frame = 1/f_s, T_slot = T_frame/N f_s sampling rate per channel (Hz), times in s.

R_b = (N·n + b_f)·f_s n bits per sample, b_f framing bits per frame, R_b bit rate (bit/s).

B_min = r/2 r total pulse (symbol) rate (pulses/s) of the TDM stream; minimum channel bandwidth (Hz) with ideal Nyquist pulses.

Worked examples

Example 1 (standard, T1 carrier). Twenty-four voice channels are sampled at 8 kHz and encoded with 8 bits each; one framing bit is added per frame. Find the frame length in bits, the frame period and the line bit rate.

1. Bits per frame = 24 × 8 + 1 = 193 bits.
2. T_frame = 1/f_s = 1/8000 = 125 µs.
3. R_b = 193 × 8000 = 1 544 000 bit/s. Answer: 193 bits, 125 µs, 1.544 Mbit/s.

Example 2 (GATE level). Four sensor signals have bandwidths 1 kHz, 1 kHz, 2 kHz and 4 kHz. Each is sampled at its own Nyquist rate, the samples are time-division multiplexed, and each sample is encoded with 8 bits. Find the total sample rate, the bit rate and the minimum transmission bandwidth.

1. Nyquist rates: 2, 2, 4 and 8 kHz.
2. Total sample rate = 2 + 2 + 4 + 8 = 16 ksample/s. (A commutator running at 2 kHz with 8 slots per frame — 1, 1, 2 and 4 slots for the four signals — achieves this.)
3. R_b = 16 000 × 8 = 128 kbit/s.
4. B_min = R_b/2 = 64 kHz. Answer: 16 ksample/s, 128 kbit/s, 64 kHz.

Example 3 (FDM). Twelve voice channels of 3.4 kHz are frequency multiplexed using DSB with 1 kHz guard bands between adjacent channels. Find the total bandwidth, and compare with SSB.

1. DSB: B_ch = 2 × 3.4 = 6.8 kHz; B_total = 12 × 6.8 + 11 × 1 = 81.6 + 11 = 92.6 kHz.
2. SSB: B_ch = 3.4 kHz; B_total = 12 × 3.4 + 11 × 1 = 40.8 + 11 = 51.8 kHz. Answer: 92.6 kHz with DSB, 51.8 kHz with SSB.

Common mistakes

• Counting N guard bands instead of N − 1 when they lie only between adjacent channels.
• Using the message bandwidth W instead of the modulated channel bandwidth (2W for DSB).
• Forgetting framing or signalling bits when computing the TDM bit rate.
• Sampling all channels at the rate of the slowest one; each must be sampled at least at its own Nyquist rate.
• Mixing frame rate and bit rate: a 125 µs frame does not mean 8 kbit/s.

For GATE IN

• TDM bit-rate and minimum-bandwidth numericals for PCM channels with different bandwidths.
• FDM total bandwidth with guard bands for AM, DSB or SSB channels.
• Frame structure questions (slot time, bits per frame, T1 and E1).
• Conceptual comparison of crosstalk mechanisms in FDM and TDM.

Quick check

1. Ten channels, each 4 kHz with SSB, no guard bands. What is the FDM bandwidth?
2. What is the frame period of a TDM system whose channels are sampled at 10 kHz?
3. An E1 frame has 32 slots of 8 bits at 8000 frames/s. What is its bit rate?
4. Name one source of crosstalk in FDM and one in TDM. Answers: 1. 40 kHz. 2. 100 µs. 3. 2.048 Mbit/s. 4. FDM: amplifier non-linearity (intermodulation) or poor filters; TDM: pulse spreading from limited bandwidth (ISI) or timing error.