According to Nyquist Signaling Theorem:
Rmax = 2Wc pulse/second
where R is the maximum transmission rate in pulse/second
Wc is the channel bandwidth
In your case you have R in bit/s, so the theorem becomes
Rmax = 2Wc *m bit/second
where m is the number bits in each pulse, i.e. bit/pulse
So the solution to your problem is
Wc=9600/(2*4)=1200 Hz
Regards,
no of sources: 5 bandwidth required for each source= 400 Hz no of guard times= 5 bandwidth of each guard time = 200 Hz minimum bandwidth = 5 *400 + 5*200 Hz
ssb modulation scheme
answer in www.ent.mrt.ac.lk/~ekulasek/cni/cni4-eck.ppt last slide
43600Hz (43.6KHz)
Starboard side
C = 9600 = 2B*3 = 2B * 3 W = 1600 Hz
The transmission bandwidth refers to the range of frequencies that are being transmitted from one point to another. The channel bandwidth on the other hand refers to the frequencies of a given channel.
to the right or your starboard side
I'm afraid you will need to specify on what exactly are you referring to. We could be talking about a radio channel bandwidth, a computer data transfer channel bandwidth or even a radio communications channel bandwidth. It is difficult to tell with the narrow question you posed.
Low pass channel or medium with the bandwidth that starts from zero.Band pass channel has the bandwidth that does not start from zero.
The coherence bandwidth of a wireless channel is the range of frequencies that are allowed to pass through the channel without distortion.
Starboard side