c) 15 d) -20 e) -10
COMP4336/9336 Mobile data networking W1 Quiz on PHY Fundamentals I
A telephone line is known to have a loss of 15 dB. The input signal power is measured at 1
Watt, and the output signal noise level is measured at 1 dBm. What is the signal to noise ratio
1 Watt = 30dBm
Received signal = 30 – 15 = 15dBm
Noise = 1dBm
SNR = signal power(dBm) – noise(dBm) = 15-1 = 14 dB
a) b) c) d) e)
Nyquist formula is about noise-free channel capacity.
What signal to noise ratio (in dB) is required to achieve 20 Mbps through a 10 MHz channel?
c) 5.77 d) 9.89 e) 10
What is the bandwidth of a noiseless channel supporting a data rate of 240 Mbps while using
24 MHz 10 MHz
Max. data rate = 240×106 = 2xBxlog2(64) = 12xB
B= 240/12 MHz = 20 MHz
© Mahbub Hassan 2017
Shannon’s formula is about noisy channel. 20 Mbps = 10 MHz ́ log2 (1+S/N)
2 = log2 (1+S/N)
In dB: S/N = 10log10 (3) = 4.77 dB
A base station allocates one frequency for the downlink communications, while a separate frequency is allocated for the uplink. Which of the following would represent this allocation?
c) Either TDD or FDD
d) Neither TDD nor FDD
e) Half-duplex communication
As per the definition, FDD means two different frequencies (channels) are allocated for the downlink/uplink, which enables full-duplex communications.
If a wireless signal has a wavelength greater than 40cm, it is likely to represent which of the following mobile networking technologies?
b) Bluetooth
d) Cellular
e) None of these
WiFi 802.11af targets 700MHz, which has a wavelength of 42.8cm (Table 2, page 4, in Chapter 2 of the textbook “Wireless and Mobile Networking”)
© Mahbub Hassan 2017
What is the channel coherence time for a 2.4 GHz WiFi link connecting a car, travelling at
100 km/hr, to a stationary base station?
程序代写 CS代考 加微信: cstutorcs
e) 2.15 ms
v = 100 km/hr = 1000/36 m/s
Doppler spread = 2vf/c = (2x1000x2.4×109)/(36x3x108) = 444.44 Hz Coherence Time = 1/Doppler Spread = 1/444.44 = 2.25 ms
To transmit 2-bit symbols, a transmitter uses the following 5-bit codewords (5-bit codewords are eventually transmitted instead of 2-bit symbols):
Data Codeword
00 –> 00000
01 –> 00111
10 –> 11001
11 –> 11110
What errors this coding scheme can detect?
a) 1-bit errors only
b) 1-bitand2-biterrors
c) 2-bit errors only
d) 3-bit errors only
e) 2-bit and 3-bit errors
HD(1-2): 3 HD(1-3): 3 HD(1-4): 4 HD(2-3): 4 HD(2-4): 3 HD(3-4): 3
Thus, the minimum Hamming distance is 3, which can detect up to 2-bit errors. Q8.
If a mobile error coding system uses a minimum Hamming distance of 4, which of the
following statements is correct?
© Mahbub Hassan 2017
Code Help, Add WeChat: cstutorcs
b) c) d) e)
Any single bit error will still be closer to the original codeword compared to any other codewords, so it can be corrected. Note that, to correct double bit errors, we need a minimum Hamming distance of 2×2+1 = 4.
All single bit errors can be detected
All double bit errors can be corrected
All 4-bit errors can be detected
All triple bit errors can be corrected
Bit errors can be detected, but they CANNOT be corrected
If a mobile technology wants to allow multiple co-located devices to use the same frequency
at the same time, which of the following multiple access techniques would be most
appropriate for them?
d) Either TDMA or CDMA
e) Either TDMA or FDMA
Only CDMA allows co-located devices to use the same frequency at the same time without interfering with each other. This is done by forcing the devices to use a unique code for coding all their data transmissions.
a) 100 thousand bits
b) 1 million bits
c) 10 million bits
d) 100millionbits e) 1 billion bits
10,000 x 10,000 = 100,000,000
To achieve high security, a secret service agent is using a direct-sequence spread spectrum
with a spreading factor of 10,000 for all its transmissions. To transmit a message comprising
of 10,000 bits, the transmitter will have to transmit
© Mahbub Hassan 2017
Code Help