Bandwidth and throughput are
fundamentally different
concepts: the
difference is much deeper than that between
theory and
practice. I
shall take the meaning of 'throughput' as
data-transfer
capacity,
measured in
information per unit time, which I believe others will
agree on.
Unfortunately, the original meaning of bandwidth has been
deformed, possibly during the 1990s Internet marketing craze (see
broadband). According
to The Jargon File, the use of 'bandwidth' as a synonym for data
throughput originated as an inside joke of hackers; it has since
leaked into common usage as a supposedly proper synonym.
I acknowledge that languages evolve, and a word can come to mean something different over time. However, in the case of bandwidth the original
meaning is alive and well, at least in science and engineering. For the sake of
clarity we should remember the distinction.
Examples
Consider CD audio data. It is sampled with 16 bits at
44.1 kHz. This means a data rate of 705.6 kbps per one
channel. On the other hand, Nyquist's theorem tells that the maximum bandwidth of CD
audio is half of the sampling rate, i.e. 22.05 kHz.
The numbers 22.05 and 705.6 are different, but that does not explain
much; bandwidth and throughput are measured in different units, so
they cannot be meaningfully compared.
Turning this idea around, if we have an analogue channel with the
bandwidth of 22.05 kHz, we could imagine using it to transfer data at
705.6 kbps. Just divide the data into 16-bit chunks and perform
DAC.
That 16 bits converted into analogue means
216 = 65536 different signal
levels. Each change in the signal level
conveys 16 bits of information. This is the crux of the
misunderstanding. Bandwidth basically tells how often you can change the signal level, while throughput also depends on the number of possible levels (Baud is not bps). Even a modest bandwidth can have
a huge data throughput, given enough signal levels.
Which brings us to another example: Telephone lines have a bandwidth
of only a few kHz. Yet it is possible to transfer data at 56 kbps. In
this case, 256 signal levels are used. The number of signal levels is
a compromise, because finer level separations mean more problems with
noise. For example, a range of 0 to 5 volts divided into 256 levels
means a separation of about 0.02 V, so the noise voltage must be less
than this.
For a general mathematical relation between bandwidth and throughput, see Shannon's Law.
Conclusions
From the examples you may infer that throughput is proportional to
bandwidth, other things being equal. It is because of these other
things that bandwidth is not the same as throughput; for example, a
different number of signal levels will change the constant of
proportionality.
Another kind of change comes from the number of channels used. In our example of CD audio, one channel has a data rate of 705.6 kbps, so the typical setup of two channels (left and right) has 1411.2 kbps. Yet the bandwidth remains at 22.05 kHz, because the individual bandwidths do not add up in any well-defined way.
Interestingly enough, the term broadband does bear some relevance
to home Internet services such as DSL and cable. It has to do with
the details of the technology, AFAIK the kind of modulation
used (please correct me if I'm wrong). However, the broadness does not
directly refer to high data
capacity, and it is physically quite possible to have a fat pipe
(i.e. high data rate) without using broadband technology.