26. Using
the generator (divisor) polynomial x3 + x + 1 for CRC, what frame
will be transmitted for the data M = x7 + x5 + x3
+ 1?
Solution
M(x)=10101001
C(x)=1011
r=3
T(x) = 10101001 111
Note:
Cyclic Redundancy Check:
A
major goal in designing error detection algorithms is to maximize the
probability of detecting errors using only a small number of redundant bits.
Add k bits of redundant data to an n-bit
message
Represent n-bit message as n-1 degree
polynomial
Algorithm for computing the checksum
1. shift left r bits (append r zero bits to
low order end of the frame), i.e., M(x)xr
2. divide the bit string corresponding to
G(x) into (xr)M(x).
3. subtract (or add) remainder of M(x)xr
/ G(x) from M(x)xr using XOR, call the result T(x). Transmit T(x).
27. Using
the generator (divisor) polynomial x4 + x + 1 for CRC, what frame
will be transmitted for the data M = x7 + x5 + x3
+ 1?
Solution
M(x)=10101001
C(x)=10011
r=4
T(x) = 10101001 1100
28. A
disadvantage of a broadcast subnet is the capacity wasted when multiple hosts
attempt to access the channel at the same time. As a simplistic example,
suppose that time is divided into discrete slots, with each of the n hosts
attempting to use the channel with probability p during each slot. What fractions
of the slots are wasted due to collisions?
Solution
The fraction of the slots are wasted due to
collisions is
1 - the fraction of the slots without
collisions.
Two situations are without collisions:
1) Successful transmission. One station
transmits and the rest of them idle. The probability is n x p x (1 - p)^(n -
1).
2) All stations are idle. The probability is
(1 - p)^n.
Hence, the fraction that are wasted is
1 - n x p x (1 - p)^(n - 1) - (1 - p)^n.
29. What
is the maximum data rate for a noiseless 1-MHz channel if QAM-256 is used?
Solution
Maximum data rate = 2 B log2 V
= 2 x 1 x log2 256
= 2 x 8 = 16 Mbps
30. If
a binary signal is sent over a 1-MHz channel whose signal-to-noise ratio is 20
dB, what is the maximum achievable data rate?
Solution
Signal-to-Noise ratio (S/N) = 20 dB
=>
10 log10(S/N)=20
=> log10 S/N = 2,
=> S/N =102 = 100
Now, from Shannon's theorem we know,
Max. data rate = B log2(1 + S/N)
bits/sec
= 1 x log2 101 = 6.658 Mbps or
about 7 Mbps
Note:
The Shannon-Hartley theorem states
that the channel capacity is given by
C = B log2(1 + S/N)
where C is the capacity in bits per second, B
is the bandwidth of the channel in Hertz, and S/N is the signal-to-noise ratio.
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