Question about bandwith of signal

I want to know: if I want to transmit a signal at the speed S bps.I use ordinary ASK technique. So, Is bandwith of signal equal to 2*S?

with a signal of bandwidth B you can possibily acheive

data rate = 2 B log2 (M)

M is 1 if u use 2ASK.

so in ur case the bandwidth should be just S/2

---- but you have to check the channel capacity also if your channel is noisy then i may happen that ur channel capacity is below S then u will need more bandwidth to acheive this data rate.

The difference between two stages of the system has to be clear

1. sampling a physical(analog) signal, eg. voice, to a digital form: in this stage, if the the analoge signal bandwidth is B the sampling rate has to be 2B min.(Nyquist rate) after that each sample is represented using a specific number of bits (say k), the bit rate will be (2Bk bps), (as an example, encoding the voice 4 kHz* 2(Nyquist rate)*8kbit/sample=64kbps). this stage absolutly has no relation with the transmission or the transmission technique, eg.ASK, FSK, PSK,......, .the sampling may be for storage not for transmission.

2. the second stage is the transmission, the digital date stored need to be transmitted(2Bk bps in 1, S bps in your question), it has to be represented in a from suitable for the transmission channel(eg. ASK, FSK, PSK,......), the bandwidth depends on the technique, the bandwidth for each is available in text books.