Estimate the bandwidth required to transmit 1Mbits/s of data using a modulation scheme which comprises 32 different frequencies and 2 amplitude levels

Yes, this is a homework question that ive been trying to do. What i have so far is the following:

I think 32 frequencies implies that i have 32 symbols, and therefore i have 64 bits, i saw somewhere that the number of bits is always 2*(number of symbols).

bps = 1Mbps = baud*bpb from here

=>baud = 1Mbps/64 = 15625

and i also saw somewhere that Band Width = 2*baud = 31250Hz

Is this correct?


  • 1
    \$\begingroup\$ number of bits is always 2*(number of symbols). Not true. In BPSK, there is 1 bit per symbol. In QPSK there are indeed 2 bits per symbol - because you at each frequency you have two carriers, an in-phase one (e.g. cos) and a quadrature-phase one (e.g. sine). \$\endgroup\$ Commented Jun 11, 2015 at 21:48
  • 2
    \$\begingroup\$ Key point : the number of frequencies is an irrelevant distraction. If you have 32 of them, each requires 1/32 of the BW required by a single carrier carrying all the information. \$\endgroup\$
    – user16324
    Commented Jun 11, 2015 at 22:26

1 Answer 1


There are two roads: one that lets you understand what's happening, and another that is fast. Let's start with the firts.

please note: I'm assuming 1Mbitps = \$2^{20}\$bps and not \$10^6\$bps.

Your transmitter would divide the bitrate across 32 modulators: each modulator would see a reduced bitrate, i.e. 1Mbps/32 = 32768bps = 32kbps. Since you have two amplitude levels the baud rate and the bit rate correspond: each of your modulators works with a 32kbaud rate needing 65.536kHz of bandwidth (of noiseless channel, Shannon would add). Your total bandwidth is then your number of channels times each channel's bandwidth, i.e. 64kHz*32 = 2.1MHz (approx.).

Now for the fast way: you have a bit rate of 1Mbps, all your modulator's symbols are the same, i.e. 1 bit per symbol, so you can just double your bit rate and get the bandwidth:

2*1Mbps = 2*1.048M = 2.1MHz

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    \$\begingroup\$ Based on the calculation in OPs the question (and typically with communications as far as I have seen) k,M,G etc are powers of 10 -> so 1Mbps is 10^6bps. \$\endgroup\$ Commented Jun 11, 2015 at 21:46
  • \$\begingroup\$ Apparently the IEC agrees with you: en.wikipedia.org/wiki/Mebibyte I will leave the answer as is because it's easy to punch in different numbers and the fact that he has 32 modulators makes me thing that his professor agrees with my first interpretation. \$\endgroup\$ Commented Jun 11, 2015 at 21:54
  • \$\begingroup\$ 32kbaud requires 16kHz not 64 kHz (assuming no redundancy or clock components) so the bandwidth required is >= 0.5 MHz. \$\endgroup\$
    – user16324
    Commented Jun 11, 2015 at 22:24
  • \$\begingroup\$ @BrianDrummond, so does this mean that the answer i accepted is wrong?, may you please add your answer with explanations -Thanks. \$\endgroup\$
    – HappyFeet
    Commented Jun 12, 2015 at 21:52
  • \$\begingroup\$ @TomCarpenter so the fact that there are 2 amplitude levels doesnt affect anything?, why is this? -Thanks. \$\endgroup\$
    – HappyFeet
    Commented Jun 12, 2015 at 21:53

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