# 9bits/signal element, what's the bandwidth?

If I encode 9bits/signal element, what is the minimum required bandwidth of the channel in Hertz?

With the information that 9bits/signal element, is it possible to find its bit rate or any other things so as to find the minimum required bandwidth?

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"A bus seats 40 people. How fast is it driving?" See the problem? –  Chris Stratton Nov 17 '11 at 6:43
The minimum bandwidth can be calculated with $B_{min}=\frac{1}{\alpha\times\pi+2^{\beta}}\times\frac{\Delta}{\iota}$ where $\alpha$ is the weight of the element in ounces, $\beta$ is the number of bits in the element, $\Delta$ is the number of hits it takes to break the element into those bits with a hammer, and $\iota$ is the price of cheese at the time of the extinction of the dinosaurs. In shekels. Per hour. In Spain. –  Majenko Nov 17 '11 at 12:06

The bits/signal ratio is irrelevant, and can be anything. What matters is the (signals or symbols)/second. As such, the actual required bandwidth is completely independent of the modulation scheme. It is only a function of the data rate.

From your description, I assume 9bits/signal basically means something like encoding a 9 bit value as an analog value, where 512 discrete steps represent 512 possible 9 bit values.

Therefore, if you had a required data rate of 9 bits/second, your required signal bandwidth would be 1 Hz (or 1 symbol per second).

18 bits/second would be 2 Hz, 27 would be 3 Hz, etc...

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ohh..Thank you so much!! –  xenon Nov 17 '11 at 7:17
You may also be interested in the wikipedia page on Ampliture Shift Keying, which is effective this modulation scheme: en.wikipedia.org/wiki/Amplitude-shift_keying –  Connor Wolf Nov 17 '11 at 9:43
Does SNR come into it somewhere? en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem –  Martin Thompson Nov 17 '11 at 13:41

As shown many years ago by Claude Shannon, the capacity of a signal channel to transmit data is a function of the bandwidth of the channel (which determines how many symbols/second can be transmitted) and the number of bits/symbol that can be encoded (which is determined by the signal-to-noise ratio of the channel). The equation is:

$C = W \cdot log_2(1+\dfrac{S}{N})$

where
$C$ = channel capacity in bits/second
$W$ = channel bandwidth in Hz
$S$ = signal power
$N$ = noise power.

If the signal-to-noise ratio were infinite then one could send an infinite number of bits per symbol since an infinite number of amplitude levels could be identified. Also if the bandwidth were infinite one could send an infinite number of bits per second. One must know the required channel capacity, C, and the signal-to-noise ratio (S/N) in order to determine the required bandwidth, W. Knowing how many bits/symbol can be encoded is a step in that direction but more information is needed.

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If I encode 9bits/signal element, what is the minimum required bandwidth of the channel in Hertz?

You must choose ;

1. the minimum bits/error or bit error rate (BER)
2. the signal/noise ratio (SNR)

If one assumes the noise is Gaussian, then from Shannon-Hartley's Channel Capacity theorem, you can answer this question.

There may be other solutions for other noise types.

With the information that 9bits/signal element, is it possible to find its bit rate or any other things so as to find the minimum required bandwidth?

A flow of 28800 bps is obtained by 3200 baud at 9 bits/baud. ( or 9 bits per symbol element)

Some pattern recognition from signal content in amplitude/phase/frequency domains may indicate this. Training for symbol rate and assumed modulation schemes used in sequence for sync assist in this answer.

For telephony modem specific questions be sure to specify. For other types of characters, in good humour, specify if talking about cheese or language.

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