When giving responses in a quiz or exam or homework, it is important
to give some explanation, so it is easy to follow your train of thought.
That way, it also is easier for the grader to give you credit in case
your method is good but you have an error in the arithmetic.
The model solutions below overdo the explanation.
1. OSI: Open Systems Interconnection.
ISO: International Standards Organization.
2. Suppose a 1000 Byte packet is sent over a 20 km link, using
56 Kbit/sec modems (that indeed operate at 56 Kb/s).
Added in class: the speed of the electomagnetic waves in this case
is (2/3)*c.
A. The propagation delay is 20/((2/3)*c) = 20/((2/3)*300,000) =
(20/2)*(3/300,000) = 10/100,000 = 1/10,000 sec = .0001 sec = .1 msec.
B. The packet size is 8000 bits. The serialization delay therefore is
8000/(56,000) = 8/56 = 1/7 = .14286 sec = 142.86 msec.
The serialization delay is the time from the first bit starting to exit
the source until the last bit finishing exit from the sourse.
The propagation delay (or latency) is the time from the last bit exiting
the source until the last bit entering the destination.
C. The total delay is the sum of serialization delay and propagation delay,
i.e. it is 142.96 msec.
D. By the time the first bit starts arriving at the destination (that is
.1 msec after it started leaving the source), .0001*56,000 = 5.6 more
bits have left the source.
From that point on (until the end of the packet), bits are entering
the destination as fast as they are leaving the source: always 5.6
bits ``underway'' (or ``in transit'').
The fraction of the packet ``in transit'' is 5.6/8000 = .0007 .
Interesting to note: 5.6 bits together are 20 km long. One bit is
20/5.6 = 3.571 km long, that is 2.22 miles long.
3. On some co-axial cable we can use the frequencies from 90 MHz to
100 MHz. The signal to noise ratio is 30 dB.
(Unfortunately: dB was missing on the exam. My fault. tjo).
What is the theoretically highest possible bitrate we can achieve
on this channel?
Expected answer:
The signal to noise ratio is 30 dB, i.e. 10*log(S/N) = 30
(log base 10). Hence log(S/N) = 3, S/N = 1000.
The frequency span is (100 - 90) = 10 MHz. H = 10,000,000
The Shannon rate is H*log(1 + S/N) . (log base 2).
That is 10,000,000*log(1 + 1000) = 10,000,000*log(1001) =
10,000,000*(ln 1001)/(ln 2) = 10,000,000 * 6.909/.6931 =
99.672 Mbit/sec.
Also OK, because I goofed, and dropped ``dB'':
10,000,000*log(1 + 30) = 10,000,000*log( 31) (log base 2)
= 49.542 Mbit/sec.
4. Suppose you have to build a Hi-Fi digital sound transmission system
for dogs. Suppose you know dogs can hear frequencies up to 20 KHz.
Suppose you have a transmission channel that can carry up to 640,000
bits/sec.
What sampling rate do you use?
The Nyquist sampling rate is 40,000 times per second.
If you sample at a higher rate, you improve performance at frequencies
the dog can not hear: waste.
If you sample at a lower rate, you loose some frequencies dogs can hear:
loss of quality.
So you sample at the Nyquist rate: 40,000 times per second.
You have available 640,000 bits/sec. At 40,000 samples/sec this is
640,000/40,000 = 16 bits per sample. I.e. 2 bytes per sample.
16 bits per sample gives you 2^(16) = 65536 levels.
(Waste of bandwidth, but given the assumptions made this is the correct
answer).
If a student chooses to sample at a slightly higher rate and further does
OK, he/she gets credit AS LONG AS he/she explains why:
(``I add a little bit of margin over the 20,000 Hz cutoff frequency'').