CM · Digital modulation

Bandwidth and Symbol Rate

Every symbol costs both time and bandwidth. Within a given bandwidth B, no matter how hurriedly symbols are sent, no more than 2B of them fit into 1 second. This unit traces where that Nyquist ceiling comes from, and why sending more bits means making symbols heavier rather than faster.

How tightly can symbols be packed?

Each faint bump is one symbol pulse, and the bold curve is the received waveform, the sum of all the pulses. Raising the symbol rate shrinks the spacing between pulses, but the pulse width is set by the bandwidth and stays the same. So once Rs passes 2B, the pulses overlap, the received waveform smears into a single lump, and it turns red.

Symbol rate RsRs = 1.50 × 2B

Rs = 2B is the Nyquist ceiling. The value on the right gives the symbol rate as a multiple of that ceiling, and the starting value sits above it, in red.

Symbol rate versus the Nyquist ceiling

Rs > 2B → ISI

ISI · overlap

Symbols take time

Sending one symbol takes one slot of time Ts, and the symbol rate, the number of symbols sent per 1 second, is Rs = 1/Ts. To send faster, it seems enough to narrow the slots and pack the pulses closer together. But pulses cannot be packed without limit, because the shorter a pulse is, the wider the band of frequencies it occupies.

Nyquist — bandwidth sets the ceiling

The Nyquist signaling theorem fixes the height of that ceiling. Over a channel of bandwidth B, at most 2B symbols per 1 second can be sent without interfering with one another. Push the symbol rate above that, and neighboring pulses spill into each other's time slots, so the receiver picks up the next symbol while reading the current one. This spillover is called intersymbol interference (ISI), and once the pulses have smeared into one lump the bits cannot be recovered.

ObserveRs ≤ 2B

Within bandwidth B, the symbol rate cannot exceed 2B.

ChooseRb = Rs · log₂ ?

Multiply the symbol rate by log₂ M bits per symbol.

More bits make heavier symbols

With the symbol rate stuck at 2B, the only way to carry more bits in the same band is to load more bits onto each symbol. An M-ary scheme carries log₂ M bits per symbol, so the bit rate becomes Rb = Rs log₂ M. The efficiency per 1 hertz, η = Rb/B, can rise as high as 2 log₂ M. The catch is that a larger M packs the signal points tighter and demands a higher signal-to-noise ratio, and beyond that lies the ceiling Shannon set, which nothing can cross.

Fill inη = Rb / ? (bit/s/Hz)

Efficiency is the bit rate divided by the bandwidth.

On your ownC = B log₂(1 + ?)

The higher ceiling is the Shannon capacity, covered in a later unit.

Back to the first screen

As you raised the symbol rate, the pulses drew closer, and the moment Rs passed 2B they overlapped and the received waveform smeared into a red lump of ISI. Just below 2B the pulses stayed clearly apart while filling the band with no room to spare, which showed gold; push the rate too low and gaps opened between pulses, wasting the band in orange. One thing to remember: bandwidth sets the symbol-rate ceiling 2B, so more bits come from making symbols heavier, not from sending them faster.

By the Nyquist signaling rate, the symbol rate of a channel with bandwidth B cannot exceed Rs ≤ 2B. Go faster and the pulses overlap, causing intersymbol interference (ISI). The bit rate multiplies in the bits per symbol, Rb = Rs log₂ M, and the spectral efficiency is η = Rb/B ≤ 2 log₂ M. To carry more bits in the same band, M must grow so symbols become heavier; the price is a higher SNR, and the absolute limit is the Shannon capacity.

In the next unit

Whether symbols were made heavier or ISI was avoided, one quantity kept showing up: the signal-to-noise ratio, SNR. The next unit starts with what that noise actually is. It follows where thermal noise comes from, what the ratio of signal power to noise power means, and why that ratio sets the limits of communication.