Flashcards
T324
Cards in this deck (24)
- 1A radio wave has a period of 250 µs. Calculate the frequency in MHz. Express your answer to one significant figures. Frequency = MHzPeriod, T = 250 × 10−6 s (or T = 0.00025 s)
Frequency is the reciprocal of time period.
Frequency = 1/T
= 1/(250 × 10−6)
= 4000 Hz
= 4000 / 1000000 MHz
= 0.004 MHz (to 1 sig fig) - 2Calculate the wavelength of a radio wave of frequency 875 MHz assuming that its speed is 3 × 108 m/s. Express your answer in millimetres to two significant figures. Wavelength = mmSpeed = Wavelength × Frequency
Wavelength = Speed / Frequency
Speed = 3 × 108 m/s
Frequency 875 MHz = 875 × 106 Hz
Wavelength = 3 × 108 /
875 × 106
= 0.343 × 1000 mm
= 340 mm (to 2 sig fig) - 3An antenna receives 0.6 µW of power at a distance of 10 km from a transmitter. Assuming isotropic conditions and an inverse square power law, calculate the received power at the antenna if it is moved to 2.5 km from the transmitter. Express your answer in µW to one significant figure. Power = µWAs distance doubles the power reduces by a factor of
1/2^2 = 1/4.
Hence, as distance halves the power increases by a factor of 4.
As distance increases by a factor of 8 the power reduces by a factor of
1/8^2 = 1/64.
Here the new distance is one eighth of the original so the power increases
by a factor of 64 to:
64 × 0.6 µW
= 40 µW (to 1 sig fig) - 4Calculate the power loss (in dB) through a material of thickness 50 mm and attenuation coefficient of 50 dB/m. Express your answer to one significant figure. Power loss = dBPower loss = thickness (in metres) × attenuation coefficient
Thickness = 50 mm = 50 / 1000 m
= 0.05 m
Power loss = 0.05 × 50
= 3 dB (to 1 sig fig) - 5An electronic system comprises three different sections with gains of: 10 dB, -14 dB and -15 dB. Calculate the overall gain in dB. Overall gain = dBAll three dB values are described as gains so, no matter what the sign of the dB
values, you simply add all three together to give and overall gain of:
( 10 ) + ( -14 ) + ( -15 ) =
-19 dB - 6An antenna receives 10 µW of power at a distance of 1 km from a transmitter. Assuming 1/d^4 relationship, calculate the power received if the distance is increased to 9 km. Express your answer in µW to two significant figures. Power received = µWThe distance has increased by a factor of 9 with an inverse power 4
relationship.
Hence, the received power falls by a factor of:
1/9^4
= 1 / 6561 to 10 × 1 / 6561
= 0.0015 µW (to 2 sig fig) - 7It is proposed to use a half-wavelength dipole of physical size 9 m to produce a short wave radio. Calculate the corresponding frequency of the radio station in MHz. Take the speed of light to be 3 × 10^8 m/s. Express your answer to two significant figures. Frequency = MHzThe physical size of the dipole represents half the wavelength of the radio waves.
The wavelength is therefore:
2 × 9 = 18 m.
The corresponding frequency
= radio wavespeed / wavelength
= 3 × 108 / 18 Hz
= 1.67 × 107 Hz
= 1.67 × 107 / 1000000 MHz
= 17 MHz (to 2 sig fig) - 8Calculate the number of bits that can be represented by each symbol for a modulation system with 32 symbols. Number of bits =Either recognise that 2^5 = 32
and therefore 5 bits are required, or us the approximation formula given on page 69 of the Block
whereby:
Number of bits = log to base 10(32)/0.301
= 5
In the latter case the answer has to be a whole number so you obviously round off as appropriate. - 9In a version of MSK one symbol uses a segment of a 5.622 MHz sinusoid, the other symbol using a segment of a 5.635 MHz sinusoid. Calculate the signalling rate of the signal in kbaud. Signalling rate = kbaudThe frequency spacing is equal to:
5.635 - 5.622 = 0.013 MHz
This is numerically equal to half the signalling rate.
Therefore, the signalling rate = 2 × 0.013
= 0.026 Mbaud
= 0.026 × 1000
= 26 kbaud - 10The spectral efficiency for a bandwidth of 65 kHz is 19 bit/s/Hz. Calculate the data rate in Mbit/s. Express your answer to two significant figures. Data rate = Mbit/sSpectral efficiency = Data rate / Bandwidth
Data rate = Spectral efficiency × Bandwidth
= 19 × 65000 = 1.235 × 10^6 bit/s
= 1.235 × 10^6 / 1000000 Mbit/s
= 1.2 Mbit/s ( to 2 sig fig)
Note that the bandwidth has been multiplied by 1000 to convert kHz to Hz before calculating
the data rate because the units of spectral efficiency involve Hz not kHz. - 11Calculate the width of a continuous frequency band containing 250 µW of power if the power density is 0.65 mW/MHz. Express your answer to two significant figures. Bandwidth = MHzPower density = Power / Bandwidth
Bandwidth = Power / Power density
= 0.25 / 0.65 MHz
= 0.38 MHz (to 2 sig fig)
Note that the power has been divided by 1000 to convert µW to mW before calculating
the frequency because the units of power density involve mW not µW. - 12Calculate the noise power in a signal of power of 0.56 mW if the signal-to-noise ratio is 6. Express your answer in µW to two significant figures. Noise power = µWSignal-to-noise ratio = Signal power / Noise power
Noise power = Signal power / Signal-to-noise ratio
= 0.56 mW / 6
= 0.09333 mW
= 0.09333 × 1000 µW
= 93 µW (to 2 sig fig) - 13The theoretical maximum data rate for a bandwidth of 6.0 MHz is 23.4 Mbit/s. Calculate the approximate value of the signal-to-noise ratio. Signal-to-noise ratio =Shannon's formula says C = W log to base 2(1 + S/N)
Maximum data rate, C = 23.4 Mbit/s
Bandwidth, W = 6.0 MHz
log to base 2(1 + S/N) = C/W
= 23.4 / 6.0
= 3.9
Using Figure 58 on page 110 of Block 1 shown above, this corresponds to:
S/N = 14 approximately - 14In an OFDM system, the data rate per subcarrier is 100 bit/s. If the data is evenly distributed across all of 182 subcarriers, calculate the approximate, overall data rate in Mbit/s to three significant figures. Overall data rate = Mbit/sOverall data rate = 100 × 182 = 18200 bit/s
= 18200 / 1000000 Mbit/s
= 0.0182 Mbit/s
= 0.0182 Mbit/s (to 3 sig fig) - 15Calculate the total time (in ns) it takes for a signal to be transmitted over this link and be reflected back to the source, assuming no time delay at reflection. Assume the speed of propagation is 3 × 10^8 m/s. Express your answer to three significant figures. Total time = nsTime taken to travel 155 m
= Distance / Speed
= 155 / 3 × 10^8
= 5.1667 × 10^−7 s
Total time is twice this value to take into account the reflected path.
Hence, total time = 2 × 5.1667 × 10^−7 s
= 10.333 × 10^−7 s
= 10.333 × 10^−7
× 10^9 ns
= 1030 ns (to 3 sig fig) - 16Perform a bit by bit XOR operation on the following two binary numbers: 10101010101100010011 XOR =You can use the truth table on page 210 of Block 1 or just remember that in this operation
"like states produce a 0 output" and "unlike states produce a 1 output".
Hence:
10101010101100010011
0110111001 - 17In a WEP encryption system it takes 9 hours for an initialisation vector value to be repeated. If the initialisation vector is 22 bits long and is incremented by 1 for each frame transmitted by an access point, calculate the operating data rate in Mbit/s assuming the frames are, on average, 2000 bytes long. Express your answer to three significant figures. Data rate = Mbit/sWith 22 bits, the number of different initialisation vector values is:
2^22 = 4194304
This is the number of frames that are transmitted before a value is repeated.
Repetition time
= 9 hours = 9 × 60 × 60 s
= 32400 s
Frame transmission rate = number of vector values / repetition time
= 4194304 / 32400
= 129.45 frames/s
Frame size = 2000 × 8 bits
Data rate
= frame transmission rate × frame size
= 129.45 × 2000 × 8 bits
= 2071300 bits
= 2.07 Mbit/s (to 3 sig fig) - 18A wireless sensor device employs a battery of lifetime 27 days when used over a distance of 6 m. Assuming the transmitter power is proportional to 1/d^3, estimate how long the battery will last if the distance of operation is reduced to 2.5 m and the same power is received. Give your answer to the nearest day. Battery life = daysThe 1/d^3 law shows that as distance increases the power decreases. Hence as
distance decreases the power increases.
The distance has decreased by a factor of:
2.5 / 6
= 1 / 2.4
The power, therefore, at a new distance of 2.5 m will increase by a factor
of:
1 / (1 / 2.4^3)
= 13.824
The power received is required to stay the same (not increase) so the battery will last
a lot longer in this new situation by a factor of 13.824.
Hence the battery life will be increased to:
27 × 13.824 days
373.248 days
= 373 days - 19A 0.7 GHz bandwidth is centred on a frequency of 7 GHz Is this situation considered to be ultra wideband? Tick the box if the situation is considered to be ultra wideband. Leave it blank if it is not. Ultra wideband?The fractional bandwidth = Bandwidth / Central frequecy
= 0.7 / 7
= 1 / 10
This is significantly less than 1/4. Hence, the rule of thumb
(page 102 in the Block) says that this situation is NOT
considered to be ultra wideband. - 20A battery can supply a current of 75 µA for 150 days. Calculate its capacity in mAh. Battery capacity = mAhBattery capacity (or charge) = Current × Time
Before performing the multiplication it is necessary to have each value in the required
units.
Current = 75 µA = 75 / 1000 = 0.075 mA
Time = 150 × 24 = 3600 hours
Hence, capacity = 0.075 × 3600
= 270 mAh - 21One source estimated the population of a country in 2007 to be 10.5 million people of which appromimately 3.1 million possessed mobile telephones. Calculate the percentage of the total poplulation in 2007 that possessed mobile phones. Express your answer to three significant figures. Percentage = %To find the percentage we divide 3.1 million by 10.5 million and
multiply by 100, so the required value is:
(3.1 / 10.5) × 100
= 29.5 % - 22In one part of a country the number of people using mobile telephones increased from 250000 to 325000, in one year. Calculate the percentage increase in the number of people using mobile phones. Express your answer to two significant figures. Percentage increase = %To find the percentage increase we subtract the initial figure from the final figure to find the difference, i.e.:
325000 - 250000 = 75000
Then divide this difference by the initial figure and multiply by 100 to give:
(75000 / 250000) × 100
= 30 % (to 2 sig fig) - 23At one time is was estimated that out of a population of 16400000 in a country only about 100000 were users of the internet. Calculate the number of internet users per 100 persons in the population. Express your answer to two significant figures. Internet users/100 persons =To find the required number we divide the number of users by the population
(to give the number of internet users per person) and then multiply by 100 to give:
(100000 / 16400000) × 100
= 0.61 (to 2 sig fig) - 24A country is said to have approximately one landline for every 26 people. If this country has a total of 217400 landlines, calculate the population of the country. Express your answer to two significant figures. Population =Population = 217400 × 26
= 5652400
= 5700000 (to 2 sig fig)
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