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Q1. What is the probability that there will be two errors in a code 16 binary digits long if the
probability of a single error is 10-5 and errors occur independently of each other?
As given; r = 2 : n = 16 : p = 10-5 = 0.000010
SAQ 1.12 : page 10
P (r)= nCr pr (1-p)n-r = 16C2 (10-5)2 (1-10-5)16-2 = 120 (10-10) (0.99999)14 = 1.1998 × 10-8
16
C2 = 16i / 2i (16-2)i = 120

What is the probability that there will be more one errors in a code
5 binary digits long if the probability of a single error is 0.1 and
errors occur independently of each other?
P(>1)= 1- ( P(0) + P(1) )
P(0) =( 5! / 0! (5-0) ! ) * (0.1)^0 (1-0.01)^5 = 0.590
P(1) = ( 5! / 1! (5-1) ! ) * (0.1)^1 (1-0.01)^4 = 0.328
P(>1)= 1- ( P(0) + P(1) )
P(>1)= 1- (0.590 + 0.328 ) = 0.082

What is the probability that there will be more than three errors in a
code 10 binary digits long if the probability of a single error is 0.01
and errors occur independently of each other
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Q2. In table below you can depict the average daily loading of a wireless Ethernet LAN recorded
over a period of 312 days.
loading
<= 10%
> 10% and
> 20% and
> 30% and
> 40% and
> 50%
<= 20%
<= 30%
<= 40%
<= 50%
Number of days
90
84
37
63
17
21
The LAN ‘s performance degrades significantly at loading above 40%. SAQ 1.2 : page 7
What is the probability that the LAN loading will exceed this value on a given day?
The number of days on which the LAN loading exceed 40% can be found by totaling the figure in
the last two rows of the table 21+17= 38 days.
The total numbers of days on which the loading measured is 312 , so the probability that the LAN
loading above 40% is, 38 / 312 = 0.121
Q3. A system is found to have a MTBF of 400 hours and a MTTR of 6 hours. What is the
availability of the system? What is the effect on the availability of doubling the MTBF?
The availability A is given by;
SAQ 1.12 : page 10
-A = MTBF / MTBF + MTTR = 400 / 400 + 6 = 0.9852
-If the MTBF is doubled to 800 (400×2) hours, the availability is 800/806= 0.9925
So that these procedure increase the availability to 0.9925
A system is found to have a MTBF of 500 hours and a MTTR of 5 hours. What is the availability
of the system? What is the effect on the availability of triple the MTBF?
A = MTBF / MTBF + MTTR = 500 / 500 + 5 = 0.990
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-If the MTBF is triple to 1500 (500×3) hours, the availability is 1500/1505= 0.996
So that these procedure increase the availability to 0.996
Q4. Draw using a graph the double-sided spectra representation of the following sinusoid
5 cos(1t+3).
Given:



A= 5, ω = 1, = 3
by use the Acos (ω t+ )= [(A/2)exp[j(ω t + )]] + [(A/2)exp[j(- ω t then 5Cos(1t+3) = [(2.5/2)exp[j(1t+3)]] + [(2.5/2)exp[j(-1t - 3)]]
= [5exp [j (1t+3)]] + [5exp [j (-1t - 3)]]

This has double-side spectral of amplitude (5/2) = 2.5 , phase + 3 at +1
)]]
Phase radians
Amplitude V
43
6
6 2.5
-8
8
-1
-8
-1
=
18
1
=
Angular frequency radians/sec
1
=
Angular frequency
radians/sec
-3
Notes: if the equation by sin we must first convert it to cos by adding -bi/2
Exp: 5sin(2t)
The answer is:
5cos(2t-bi/2)
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Amplitude V
Phase radians
2.5
-2
Bi/2
2
-2
2
-Bi/2
Final note here may be give us the graph and ask from us the equation
Q5. The angular frequency of a signal is 16 rad-1. What are its frequency in Hz, and its period in
second?
The expressions relating angular frequency w, in rad s-1, and frequency f in Hz, are
W=2π f and f = w/2 π
-In this case f = 16 / 2 π = 2. 55 Hz,
-the period T = 1/ f = 1/2.55 = .39s
 If a source has source symbols which each consist of one binary coded decimal
symbols 0-9,(they give a arabic letters) and each symbol is equally probable. What
is the entropy of the source. What is the efficiency assuming fix length of the
symbols?
p =1/28= .04
H=I
I = - log2(.04)=4.6
H=4.6
E= H/L
4.6/5 = .92
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Q6. If a source has source symbols which each consist of one binary coded decimal symbols 0-9
and each symbol is equally probable. What is the entropy of the source? What is the efficiency
assuming fix length of the symbols?
where symbols 0-9 and each symbol is equally probable so Pi = 1/10 = 0.1 so, By using
entropy rule to find H;
H = {0.1 log2 (0.1) + 0.1 log2 +.....etc} = 3.3219
H = 3.3219 bits
Fristly must find average code work lenth by :
Ii = length of the code word of 9 (1001) binary coded decimal it’s necessary to transmits
4 digits binary word so,
Ii = 4
Pi = 1/10 =0.1
by using formula of length, L= 10*4*(1/10) = 4 so;
E = 3.3219 / 4 = 0.830475
 If a source has source symbols which each consist of one binary coded symbols
(ABCDEFGH) and each symbol is equally probable. What is the entropy of the
source? What is the efficiency assuming fix length of the symbols?
p =1/8= .125
H=I
I = - log2(.125)=3
H=3
E= H/L
3/3 = 1
Q7. If 30 components are tested and, of the 30, 7 fail on day 5, 7 fail on day 6 and 8
fail on day 6 what is the average life of the component?
t1=5 , t2=6 ,t3=6 ........ N0=30 , N1=7 , N2=7,N3=8
ti = (t1 N1 )+ (t2 N2)+ (t3 N3 )/ N0 == ( 5×7) + (6×7)+ (6×8) / 30 = 125 days
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If 30 components are tested and, of the 30, 12 fail on day 6, and 6 fail on day 8, what is the
average life of the component?
-The two failure time of 6 days and 8 days must be weighted in proportion to the number of
components which failed for each so the average lifetime is;
t1 = 6 , t2= 8 N0 = 30 , N1 = 12, N2 = 6
ti = (t1 N1 )+ (t2 N2) / N0 = ( 6×12) + (8×6) / 30 = 4 days
Q8. A communication system uses two exchanges, each with availability A=0.884, linked by
four parallel trunk routes, each with an availability of 0.55. If the selected route fails, the
exchanges automatically switch to another route. What is the availability of this system?
A system = 1- (1-Acomponent) n = 1- (1- 0.55)4 = 0.958
This is in series with 2 exchanges, each with availability 0.884, so overall system availability is:
0.884 ×0.884 × 0.958 = 0.75
Q9. Calculate the 7-bits word by using Hamming code, the Original bits is 1101.
By using party check scheme for 7- digit hamming code:-
Position
Digit number
Parity check 1
Parity check 2
Parity check 3
1-3-5-7
2-3-6-7
4-5-6-7
1
1
2
0
3
1
1
1
4
-
0
5
1
1
1
6
0
0
0
7
1
1
1
1
Original
message
Even Pass
Even Pass
Even Pass
So that the 1010101 which mean the message received by no error occurred
Q10. A channel can be modelled as having a bandwidth of 7.5 kHz, and can be operated at up to
65% of its theoretical capacity. The signal-to-noise power ratio is 4000(36dB).what the value
closest to the maximum bit rate of the channel?
If doesn’t give us
W= 7.5 kHz, S/N =36 dB, theoretical capacity = 65%
By using the decibel approximate to Shannon’s equation C =W [(S/N) dB ÷3)]
C = 7.5 × (36÷3) = 90 kBit/s = 90 000 Bit/s
C = 90 000 Bit/s
Also, using the formula of capacity to find rate
Theoretical capacity = 100% × (transmission rate / C) ....x
65% = 100% × (transmission rate / 9000)  / 100%
0.65 = (transmission rate / 90 000) ×90 000%
the 36 db what is
the equation we
must use it?
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0.65 × 90 000 = transmission rate
Transmission rate =58500 bits x
‫اسئلة من فرع البحرين تحتاج الى شرح‬
‫وفي سؤال عن ال‬hamming distance??
 Hamming distance The number of digits by which any two binary sequences
differ.
‫عدد ارقام التي فيه أي سلسلتان ثنائيه تختلفان‬
43) A network contains 4 routers. Each has a probability of 0.01 of being in operation at any
given time.

Error correction and detection.
What is the minimum number of error recovery bits required to recover from an n bit error?
Explain why this number of bits is sufficient
()‫هذا السؤال جانا تقريبا ًمشابه ولكن فيه ارقام‬