CPS 214 Written Homework 2 Solutions    	    	Fall, 1996

1) There is no maximum data rate on a noiseless channel; one can
always increase the number of encodings per baud to arbitrarily
increase the data rate.


2) According to Nyquist, one must sample an analog source of bandwidth
H 2*H times per second in order to be able to reconstruct  the signal
exactly. Therefore, 2H = 44.1 KHz, or H = 22.05kHz. Thus, the
frequency response on a CD runs from 0-22.05 kHz.

3) Using Nyquist's Theorem,  H = 6MHz, V=6
   Maximum Data Rate = 2Hlog2(V) = 2HLog2(6) = 31Mbps
   Of course, this only applies if the S/N ratio is acceptably high.

4) First, we must determine the signal to noise ratio (expressed as a
ratio). Recall that S/N in db = 10Log10(signal power/noise power).
Or, signal/noise (power ratio) = 10**[(S/Ndb)/10] = 10**4 = 10000.

Max data rate = H*log2(1+S/N) = (68-4)KHz*log2(10001)
    	    	    	      = 64*log2(10001) bits/sec 
			      = 850kbps
			      >> 256kbps, so the answer is yes.

Tanenbaum 2.6) A passive star diffuses a signal out to all the nodes
on the network effectively "splitting" the signal to multiple
destinations.  A repeater regenerates the bits.

Tanenbaum 2.9) Nyquist's Theorem still holds. Although one tends to
think of the bandwidth of fiber is limitless, it is not. But because
the data transmission characteristics of fiber are so much superior to
copper, one tends to forget that it too has limits.

Tanenbaum 2.17) There are four symbols (i.e., 2-bits) carried in a
baud, so the bit rate is 2*1200 = 2400 bps

Tanenbaum 2.18) The modem uses amplitude modulation; there are
encodings in which the phase doesn't change, but the amplitude does.

Tanenbaum 2.22) Telephone channels are approximately 4kHz
wide. According to Nyquist's Theorem, the maximum sampling rate to
recover the signal is 2H samples. 1/8000 samples/sec is equivalent to
1 sample every 125usec.

Tanenbaum 2.33) In a circuit-switched network, each hop adds only 1
bit delay, which we can ignore. Thus, the time to send an x-bit
message is:

	s + x/b + dk

	that is, the circuit setup time, the time to transmit the
message plus the propagation delay.

In the case of the packet-switched network, the delay is:

	x/b + kd +(k-1)(p/b)
    
The first term is the time to send the bits, the second term is the
propagation delay, while the third term is the delay each packet
encounters at a switch. A switch cannot begin transmitting a packet on
an output link until it has completely received it.

Tanenbaum 2.47) 840/7 = 120

Tanenbaum 2.50) With 6 necklaces, each necklace contains 66/6 = 11
satellites. Since they each orbit once every 90 minutes, the
time it takes for one satellite to move away and the next one to
arrive is 90/11 minutes.