TM111 TMA03
L5490096
TM111 TMA03
Question 1
(a)
685 Hz = 685 cycles per second
1 (second) / 685 = 0.00145985401
0.00145985401 x 10³ = 1.45985401 x 10¯³
Period of the sinewave = 1.46 ms (3.s.f.)
1.45985401 (period of sinewave) / 3 = 0.486618
.486618 = 2.05500002
Frequency of the sine wave = 2.06 kHz (3.s.f.)
Excellent.
Well explained calculations.
(b)
237.63.198.43
Denary 128 64 32 16 8 4 2 1
237- 109- 45- 13-8=5 5-4=1 1-1=0
128=109 64=45 32=13
Binary 1 1 1 0 1 1 0 1
Correct
Denary 32 16 8 4 2 1
63-32=31 31-16=15 15-8=7 7-4=3 3-2=1 1-1=0
Binary 1 1 1 1 1 1
Correct
Denary 128 64 32 16 8 4 2 1
198- 70-64=6 6-4=2 2-2=0
128=70
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1
, TM111 TMA03
L5490096
Binary 1 1 0 0 0 1 1 0
Correct
Denary 32 16 8 4 2 1
43-32=11 11-8=3 3-2=1 1-1=0
Binary 1 0 1 0 1 1
Correct
Binary equivalent of 237.63.198.43 = 1110110 111111 11000110 101011
On the first byte the final 1 has not been copied over.
The other values are numerically correct but since this is an IP address it must have 32 bits.
The answer should therefore be:
11101101.00111111.11000110.00101011
Well explained calculations.
(c)
0.35W = 350,000µW (0.35x10¯³W)
Distance/km Signal Power/W
0 0.35
30 0.35 x 10¯¹
60 0.35 x 10¯²
90 0.35 x 10¯³
The signal can be transmitted up to 90km before the power is attenuated to 350µW.
Excellent, a correct table and deduction.
A high-quality coaxial cable operating at 200Mbps drops power at roughly a factor of 10 every 1km,
which means that this LDF5-50A cable would not be expected transmit a similar amount of power as
the optical fibre. Below is a table to show how far the LDF5-50A cable would be able to transmit
data before the power is attenuated to 350µW.
Distance/km Signal Power/W
0 0.35
1 0.35 x 10¯¹
2 0.35 x 10¯²
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2
L5490096
TM111 TMA03
Question 1
(a)
685 Hz = 685 cycles per second
1 (second) / 685 = 0.00145985401
0.00145985401 x 10³ = 1.45985401 x 10¯³
Period of the sinewave = 1.46 ms (3.s.f.)
1.45985401 (period of sinewave) / 3 = 0.486618
.486618 = 2.05500002
Frequency of the sine wave = 2.06 kHz (3.s.f.)
Excellent.
Well explained calculations.
(b)
237.63.198.43
Denary 128 64 32 16 8 4 2 1
237- 109- 45- 13-8=5 5-4=1 1-1=0
128=109 64=45 32=13
Binary 1 1 1 0 1 1 0 1
Correct
Denary 32 16 8 4 2 1
63-32=31 31-16=15 15-8=7 7-4=3 3-2=1 1-1=0
Binary 1 1 1 1 1 1
Correct
Denary 128 64 32 16 8 4 2 1
198- 70-64=6 6-4=2 2-2=0
128=70
Page |
1
, TM111 TMA03
L5490096
Binary 1 1 0 0 0 1 1 0
Correct
Denary 32 16 8 4 2 1
43-32=11 11-8=3 3-2=1 1-1=0
Binary 1 0 1 0 1 1
Correct
Binary equivalent of 237.63.198.43 = 1110110 111111 11000110 101011
On the first byte the final 1 has not been copied over.
The other values are numerically correct but since this is an IP address it must have 32 bits.
The answer should therefore be:
11101101.00111111.11000110.00101011
Well explained calculations.
(c)
0.35W = 350,000µW (0.35x10¯³W)
Distance/km Signal Power/W
0 0.35
30 0.35 x 10¯¹
60 0.35 x 10¯²
90 0.35 x 10¯³
The signal can be transmitted up to 90km before the power is attenuated to 350µW.
Excellent, a correct table and deduction.
A high-quality coaxial cable operating at 200Mbps drops power at roughly a factor of 10 every 1km,
which means that this LDF5-50A cable would not be expected transmit a similar amount of power as
the optical fibre. Below is a table to show how far the LDF5-50A cable would be able to transmit
data before the power is attenuated to 350µW.
Distance/km Signal Power/W
0 0.35
1 0.35 x 10¯¹
2 0.35 x 10¯²
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2
