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Does Hz and bps mean same? Can a signal be transferred at rate of say Mbps on a channel bandwidth of few Khz?

sk1
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    No they are not the same. The relationship between the symbol rate, the bit rate and the bandwith is often complex and depends on the modulation, noise, and other things. – Gustavo Litovsky Jan 27 '13 at 16:43
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    Here's a relevant question: http://electronics.stackexchange.com/questions/21854/why-is-the-nyquist-data-rate-lower-than-the-shannon-data-rate – The Photon Jan 27 '13 at 16:44
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    Thanks for the accept, but there's still lots more to say about this topic. You will encourage more answers if you hold off accepting for a while. 24 hrs is a good minimum because it gives people all around the world a chance to see the question and respond. – The Photon Jan 27 '13 at 17:53
  • @ThePhoton I didn't knew accepting an answer closes further response. As suggested I have toggled the option. – sk1 Jan 27 '13 at 17:59
  • No problem. The accept doesn't block new answers in the software, but it does tend to psychologically discourage them. – The Photon Jan 27 '13 at 18:28

5 Answers5

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There are actually three terms you want to know about

Bandwidth

Bandwidth is measured in Hz. It describes the frequency band that a communication channel is able to transmit with low loss.

Typically we talk about a 3-dB bandwidth, meaning the range of frequencies a channel can transmit with less than 3 dB of loss. For a baseband system, the bandwidth extends from 0 Hz to a frequency B which we call the bandwidth. For a modulated system if the carrier is at f0, then the transmission band would be from \$f_0 - B/2\$ to \$f_0 + B/2\$.

Also, outside of information theory, the term bandwidth may be used more broadly as a synonym for bit rate, or for data processing capacity, but when the units are Hz, we know we're talking about the analog bandwidth of a signal path of some kind.

Baud

You didn't ask about this, but its also important to keep this separated in your mind from the other two terms. Baud is the number of symbols transferred per second on the channel.

Bit rate

Bit rate indicates the amount of information transferred on a channel, and is measured in bits per second or bps. Bit rate is different from baud if more than one bit is transferred per symbol. For example, in a 4-level amplitude modulation scheme, each symbol can encode 2 bits of information. Alternately, for example when an error-correcting code is used, the bit rate can be less than the baud rate, as a larger number of symbols are used to convey a smaller number of bits of independent information.

The Shannon Theorem shows how bit rate is limited by bandwidth and the signal-to-noise ratio of the channel:

\$C = B\ \log_2(1 + \mathrm{SNR})\$

where C is the capacity (maximum bit rate of the channel), B is the bandwidth of the channel, and SNR is the signal to noise ratio.

The Photon
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    Overall a good post. It would be good idea, however, to clarify that the term bandwidth means a frequency range. You implied this, but it could be clearer. For example, an FM radio station might transmit at a frequency of 100 MHz, but the bandwidth is about 200 KHz-- meaning that it transmits over a range of frequencies that is 200 KHz wide and approximately centered on 100 MHz. –  Jan 27 '13 at 19:06
  • @DavidKessner, good point, edited. – The Photon Jan 27 '13 at 19:47
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Hertz and Bits per Second do not mean the same thing. They do have a relationship, determined by the bit encoding used.

To illustrate:

  • Quadrature Phase Shift Keying: By encoding to one of 4 phase positions for each "wave" or symbol, 2 bits can be carried per symbol. Thus:
  • Thus a 100 KHz carrier can carry 200 kbps of data in ideal case, ignoring any protocol overhead.

To achieve Mbps transmission on a KHz channel, encoding would need to accomplish hundreds of unique values per symbol. While this is not conceptually impossible, it isn't trivial enough to be in practical use, to my knowledge.

For just 3 bits per symbol, one needs 8 possible values.

How would one conceivably encode 8 possible values per symbol?

By having 8 (or 9) different voltage values, for instance, imposed on a signal... for the 8 possible values each symbol (wave duration) carries. The ninth value if used would be for a "no-op" or "ignore this" value.

While this is simple in a lab experiment, it isn't so simple in real-world transmission media. The problem just gets worse with higher encode level requirements. 4 bits need 16 values, 8 bits per symbol need 256 values, which would just yield a bps rate 8 times the KHz rate.

Anindo Ghosh
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They are similar concepts in that they both measure a rate of a thing, but not the same. Hz, or hertz, means cycles per second, and is a measure of frequency. bps is "bits per second", or less frequently "bytes per second". The relationship between the two will depend on how a bit is encoded.

When we are talking about "channel bandwidth", we are probably talking about RF modulation. RF signals are typically said to have a carrier frequency, which is a central frequency that is then modulated (by any number of means) to encode the data. Wi-Fi, for example, often has carrier frequencies around 2.4 GHz. Each Wi-Fi channel is a slightly different frequency.

To encode the signal of interest, we change this carrier somehow. We might vary its frequency (frequency modulation, FM) or it's amplitude (amplitude modulation, AM). Or we might switch it on and off (carrier wave modulation, CW). These are all simple modulation schemes. Something like Wi-Fi uses a much more complex scheme.

If we take the fourier transform of the resulting carrier + modulation, we can see the range of frequencies used by this signal. Other signals that use the same range will interfere. The difference between the lowest and the highest frequencies is the channel bandwidth.

Again, how much data (bits per second) you can fit in a given channel bandwidth is highly dependant on your modulation scheme.

Phil Frost
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Hz means cycles per second. Only, cycles is something which is understood. For reasons of convenience, this cycles does not appear in the units, so the units are just \$1/\$s. This is because many kinds of formulas whose result is frequency will not produce this cycles unit. Frequency will come out as \$1/s\$. For instance, ine the formula for the -3dB corner frequency of an RC filter, \$f = \frac{1}{2\pi RC}\$, the right hand side does not have cycles in it anywhere. The units cancel down to \$1/s\$. Capacitance is coulombs per volt, \$C/V\$. Resistance is volts per current, \$V/I\$, so the voltage cancels out and RC becomes \$C/I\$, making its reciprocal \$I/C\$. But current is coulombs per second, and so the \$C\$'s cancel out leaving \$1/s\$.

But not any occurrence of \$1/s\$ in a formula can be replaced by Hertz! For example, velocity is meters per second. An object moving at a constant velocity in a straight line has meters-per-second speed, but its motion does not exhibit anything connected with frequency (quantum connections between energy and frequency aside). Hertz is usually for frequencies of signals, oscillations, and periodic events resembling oscillations. It's for situations where we can identify a \$1/s\$ in the formula, and when it makes sense to pretend that the \$1\$ can be replaced by cycles, because of some repetitive process or signal exhibiting the phenomenon of frequency.

Is bits per second Hertz? First of all, the communication of bits doesn't have to be periodic. If you receive 3600 bits in one hour, that doesn't mean there was a 1 Hz signaling involved. The bits could have arrived at sporadic intervals. For instance, 3599 bits could have arrived in the first 5 minutes, and then you waited 55 minutes more for the last one.

Even if the data rate is perfectly smooth, that doesn't mean that bits per second is Hertz. Suppose that bits are neatly clocked over eight parallel lines. Then 800 bits per second actually means that the frequency of arrival of any one bit is 100 Hz, same as that of the eight bit word which contains it.

Re: Can a signal be transferred at rate of say Mbps on a channel bandwidth of few Khz?

Yes, if the channel is completely noise-free. The analog bandwidth alone does not restrict the digital bandwidth. Bandwidth together with noise, however, limit the upper bound on the channel capacity. See the Shannon-Hartley theorem Wikipedia page. Why doesn't bandwidth limit capacity? Intuitively, we can look at it like this: consider functions over an \$[a,b]\$ interval on the real number line. Even if we limit our imagination to just those functions which are continuous, smooth, differentiable everywhere in \$[a,b]\$ and which have no components above a certain frequency (are bandwidth limited), there is still an uncountable infinity of all possible such functions. Thus the functions correspond to the real numbers. I.e. this signal of duration \$[a,b]\$ can represent any real number by mapping it to some function shape within the allowed bandwidth. It's only up to the resolution of the sender and receiver to decide how much advantage they take of the noise-free channel's theoretically unlimited capacity.

Kaz
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Lets assume that you have two frequencies f1 and f2 and f1 represents 0 and f2 represents 1. Further assume that you need at least a separation of delta between the two frequencies so that they do not interfere. Lastly each of the frequencies has to be transmitted for T seconds for it to be reliably transmitted and detected. So the bit rate is (1/T) bits/second.

Now you want to increase the bit rate. One way to do that is to use 4 frequencies instead of 2. So the mapping may be something like.

f1: 00, f2: 01, f3: 10, f4: 11

So now you can transmit 2 bits in the same duration T. So the bit rate is (2/T) bits/second. The bandwidth requirement has increased from 2*delta to 4*delta (3 deltas between the 4 frequencies and delta/2 at the two ends). So this example in very simple terms shows you the relationship between bandwidth and data rate. Increasing the bandwidth does increase the data rate.

Yasir Ahmed
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