Problems: Physical layer
1. Assume binary modulation and that a data transfer is disturbed by a burst of noise during
2 milliseconds. How many bits may have been affected if the bit rate is
(a) 1500 bps
(b) 12 kbps
(c) 96 kbps
2. Assume that we want to transmit a bit sequence consisting of 10 zeros. Code the sequence
with the following line coding schemes:
(a) Manchester
(b) Differential Manchester
3. The wave forms shown below are Manchester coded bit sequences. Decode them!
(a)
(b)
(c)
4. Repeat Exercise 3. using Differential Manchester coding.
5. The propagation velocity of both twisted pair copper and fibre optics can be approxi-
mated to 2/3 of the speed of light in vacuum. For each of the systems below give the data
rate, signal rate and physical length on the medium for the signal and bit, respectively.
(a) A telegraph operator writing 2 letters per second. Assume a letter represents 5 bits.
(b) In ADLS, in average, a signal is constructed from 400 sub-signals carrying up tp 15
bits each. The total signalling rate is about 4kHz.
(c) Fibre to the home (FttH) gives the best connectivity for the home access of today.
The standard technology is based on GPON (Giga bit Passive Optical Network) and
uses on-off keying (binary ASK).
6. Assume even parity. Then find the parity bit for the each of the following data units:
1
(a) 1001011
(b) 0001100
(c) 1000000
(d) 1110111
7. A receiver receives the bit pattern 01101011. If the system is using even parity, is the
pattern in error?
8. Find the binary equivalent of x8 + x3 + x + 1.
9. Find the polynomial equivalent of 100001110001.
10. Given the 10-bit sequence 1010011110 and a divisor of 1011, find the CRC. Check your
answer.
11. Assume a receiver receives a data unit of 10110011, with a remainder of 111, generated
with a divisor of 1001. Is there an error in the data unit?
12. A receiver in a communication system received the sequence r(x) = F F 01D16 . The
sender used the polynomial g(x) = 3316 . Where there any errors?
13. Suppose that we want to send the message M (x) = 3AB16 . Use g(x) = 1B16 . Find the
CRC
14. Suppose a 4-bit CRC with generator polynomial g(x) = x4 + x3 + 1 has been used. Which,
if any, of the following three messages will be accepted by the receiver?
(a) 11010111
(b) 10101101101
(c) 10001110111
15. Find the checksum of the following bit sequence. Assume a 16-bit segment size.
10010011100100111001100001001101
16. Assume that a receiver receives the following bit sequences. An 8-bit check sum is used.
Have the bit sequences been received correctly?
(a) 10010011 10011011 11011001
(b) 00110011 10110111 00010101
(c) 01110000 00111000 01010111
Solutions: Physical layer
2
1. If the data rate is X bps then each bit lasts 1/X seconds.
(a) 3 bits
(b) 24 bits
(c) 192 bits
2. (a)
(b)
3. The sequence is:
(a) 0 1 0 0 1 1 0 0 0 1
(b) 1 1 0 0 1 1 0 1 1 0
(c) 1000011100
4. The sequence is:
(a) 1 1 1 0 1 0 1 0 0 1
(b) 1 0 1 0 1 0 1 1 0 1
(c) 1100010010
5. The propagation speed is v = 23 c ≈ 2 · 108 m/s.
(a) Rb = 10bps, Rs = 2signal/s (Hz), Lb = Rvb ≈ 20 000km/b (about half way arround
the earth), Ls = Rvs ≈ 100 000km/letter (about 2.5 times arround the earth).
(b) There are maximum 400·15 = 6000bits/signal. With Rs = 4kHz we get Rb = 24kbps.
Then, Lb = 8.3m/b and Ls = 50km/signal.
(c) Since it is binary signalling Rb = Rs = 1Gbps and Lb = Ls = 2dm.
Today there are available equipment (optical equipment for the core network) transmit-
ting at Rb = 60Tbps, which corresponds to Lb = 3.3µm/b. But this is not based on binary
signalling.
6. There should be an even number of ones in the message. Therefore, the parity bits are as
follows:
(a) 0
(b) 0
(c) 1
3
(d) 0
7. The system is using even parity, and there are five positions equal to 1. Thus the pattern
is in error.
8. x8 + x3 + x + 1 ↔ 100001011.
9. 100001110001 ↔ x11 + x6 + x5 + x4 + 1
10. Sequence after adding n = 3 extra zeros: 1010011110000. Divisor (of length n + 1): 1011.
The polynomial division is
x12 + x10 + x7 + x6 + x5 + x4 1
3
= x9 + x6 + x2 + x + 1 + 3 .
x +x+1 x +x+1
The remainder polynomial is 1, so CRC=001.
11. Sequence after adding remainder: 10110011111. Divisor: 1001. The polynomial division
is
x10 + x8 + x7 + x4 + x3 + x2 + x + 1
= x7 + x5 + x2 + x + 1 .
x3 + 1
The remainder polynomial is 0, so there is no error.
12. M=FF01D16 =111111110000000111012 . P=3316 =001100112 . The polynomial division is
x19 + x18 + x17 + x16 + x15 + x14 + x13 + x12 + x4 + x3 + x2 + 1
=
x5 + x4 + x + 1
x4 + x3 + x2 + 1
= x14 + x12 + 5 .
x + x4 + x + 1
The remainder polynomial is x4 + x3 + x2 + 1 6= 0, so there is an error.
13. M=3AB16 =0011101010112 . P=1B16 =000110112 . We have that n + 1 = 5, so the data has to
be extended by n = 4 zeros. The polynomial division is
x13 + x12 + x11 + x9 + x7 + x5 + x4 x
4 3
= x9 + x7 + x4 + x3 + x2 + x + 4 3
.
x +x +x+1 x +x +x+1
The remainder polynomial is x, so CRC=0010.
14. (a)
x7 + x6 + x4 + x2 + x + 1 3 x2 + x
= x + 1 +
x4 + x3 + 1 x4 + x3 + 1
remainder 6= 0, so no acceptance.
(b)
x10 + x8 + x6 + x5 + x3 + x2 + 1 6 5 x3 + x2 + 1
= x + x +
x4 + x3 + 1 x4 + x3 + 1
remainder 6= 0, so no acceptance.
4
(c)
x10 + x6 + x5 + x4 + x2 + x + 1
= x6 + x5 + x4 + x3 + x2 + x + 1
x4 + x3 + 1
remainder = 0, so acceptance.
15. The checksum is 1101010000011110.
16. The message is correct if the complement of the sum is 0.
(a) No
(b) Yes
(c) Yes
