Solutions of Home Work 4

 

5)      What signal-to-noise ratio is needed to put a T1 carrier on a 50-kHz line?

 

Solution:

Data rate on a T1 line is 1.544 Mbits/sec.

 

Hlog2 (1 + S/N) = 1.544 * 106 with H = 50,000.

Therefore, 50000*log(1 + S/N) = 1.544 * 106

 

S/N = 230 1, which is about 93 db.

 

9)      Is the Nyquist theorem true for optical fiber or only for copper wire?

 

Solution:

The Nyquist theorem is a property of mathematics and has nothing to do with technology. It says that if you have a function whose Fourier spectrum does not contain any sines or cosines above f, then by sampling the function at a frequency of 2f you capture all the information there is. Thus, the Nyquist theorem is true for all media.

 

1)      An upper-layer packet is split into 10 frames, each of which has an 80 percent chance of arriving undamaged. If no error control is done by the data link protocol, how many times must the message be sent on average to get the entire thing through?

 

Solution:

Since each frame has a chance of 0.8 of getting through, the chance of the whole message getting through is (0.8)10, which is about 0.107. Call this value p. The expected number of transmissions for an entire message is then

E = ∑ ip(1-p)i-1 = p∑ i(1-p)i-1

i=1 i=1

 

To reduce this, use the well known formula for the sum of an infinite geometric series,

S = ∑ ai = 1

i=1 ------

1 a

Differentiate both sides with respect to a to get,

S' = ∑ iai-1 = 1

i=1 -------

(1-a)2

 

Now use a = 1-p to get E = 1/p. Thus, it takes an average of 1/0.107, or about 9.3 transmissions.

 

 

Variation of problem 1)

Now do error detection on each frame separately. If a frame contains
an error, it must be retransmitted (until it has arrived intact).
Once every frame has arrived intact once, the message has arrived.
How many frames (in average) are transmitted before the message has 
arrived intact?

 

 

Solution:

 

Probability that each frame arrives intact: 0.8

 

Average for a single frame to arrive intact is: 1/0.8 = 1.25.

 

For 10 frames to arrive intact, the average is 1.25 * 10 = 12.5.

 

Hence, the number of frames need to be transmitted before the message arrives intact is 12.5.