# What is the minimum freqeuncy required to transmit 2400 bits per second? [closed]

Yes we can use diiferent symbols and modulation sechemes for example QAM ,DPSK,QPSK etc... you can use any modulation secheme. What will be or can be the minimum frequency to transmit information equals to 2400 bits per second over a channel.

## closed as unclear what you're asking by Marcus Müller, brhans, PeterJ, Dmitry Grigoryev, Voltage SpikeSep 11 '17 at 16:35

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• Is this a homework question? If so you need to show some work. – Transistor Sep 4 '17 at 18:13
• 1.2 kHz. You can figure the rest out. – Andy aka Sep 4 '17 at 18:37
• Your question does not really make sense. You cannot transmit any information with a single frequency. You need to modulate that frequency in some way-either by changing its frequency, phase or amplitude. Any of these modulation schemes will increase the bandwidth of the signal. How much bandwidth is needed to transmit a fixed data rate is given by Shannon's Equation (which you can google) and depends on both the data rate and the signal-to-noise ratio (SNR). Since you haven't specified SNR, the required bandwidth cannot be determined. – Barry Sep 4 '17 at 18:54
• No this is not home work. Yeah i know modulation and etc should be use. I have memtioned it in my description. Let say i choose 2400 hz to transmit 2400 bits per second. If i choose bpsk. I think I can ideally transmit 2400 bps (kindly correct me on this). But practically it is not possible ? Is it ? What if on real channel i want accuracy of 98% for detecting symbol. I can use qpsk which can help us to increase time frquency resolution. What other ideas can u share... – Hassan Mahmood Sep 4 '17 at 20:30
• What signal-noise ratio is the channel? What error probability is tolerable? – Brian Drummond Sep 4 '17 at 22:56

$C = B \log_2 (1+\frac{S}{N})$

• $B$ is the channel bandwidth
• $\frac SN$ is the signal to noise ratio or SNR

So, you can push 2400bps through a channel with a lower bandwidth than 2400bps, depending on coding, and of course depending on SNR.

For example, if B = 240Hz, then you will need a S/N ratio of 1023, or about 30dB, which is feasible on short copper cables.

A simplistic code to achieve this would be to assemble the bits into 10-bit symbols, translate this into a voltage between 0 and 1024V, and use a channel with noise lower than 1V.

Gigabit Ethernet does not use 1GHz bandwidth...

• Gigabit has 125 MBd IIRC, that makes 8b/S; compare to DVB-C2, which does up to 4096-QAM – that's a freaking 12b/S, and really pretty much. I know that people talked about $2^{16}$-QAMs, but I don't see them happening anytime soon; there's simply some noise and linearity constraints to be solved before you can throw a >16 bit ADC on a 10 MHz channel and really make use of all 16 bit on both I and Q simultaneously. – Marcus Müller Sep 4 '17 at 19:21
• The first time I read that a Gb-Ethernet PHY actually runs a highspeed ADC followed by a DSP which performs channel equalization and decoding, I was kinda floored, considering a switch with 8 of those costs like €20... – peufeu Sep 4 '17 at 20:07
• now, factor between BOM cost and selling point is usually 2.5, but since this is extreme commodity hardware, let's say it's 1.5; subtract (Germany) 19% taxes, leaves me with a 16€ switch, leaves us with 11€ component cost. Take away 4€ for the wall-wart supply, 2€ for case, and then think the price relation between magnetics and semiconductors... – Marcus Müller Sep 4 '17 at 20:11
• Yeah, a piece of wire bent into a loop is worth as much as silicon which needs a \$10Billion factory to manufacture. I love modern tech. – peufeu Sep 4 '17 at 21:13