QUESTION 11 B8ZS is a technique that specifically replaces a bytes (8 bits) cont
ID: 3873600 • Letter: Q
Question
QUESTION 11
B8ZS is a technique that specifically replaces a bytes (8 bits) containing all logical 0's with an unique voltage pattern before signal transmission. Select the correct B8ZS sequence that replaces the byte "00000000", if the last voltage before the byte was a +v.
a. 000+-0-+
b. 000-+0+-
c. 0++--00+
d. 00+-00-+
10 points
QUESTION 12
Which method is used to ensure receive clock synchronization by mapping four bit words into five bit representations?
a. B8ZS
b. AMI
c. 4B/5B
d. DSI
10 points
QUESTION 13
Select the description that best describes "synchronous transmission"
a. The use of start/stop bits for each byte of information
b. Master clock sources are used over the entire network
c. Timing is recovered from the transitions in the received data stream
d. Devices run at nominally the same rate with defined parameters of tolerance for variations
10 points
QUESTION 14
Select the correct statement(s) regarding "Error Control".
a. With "Error Detection", additional bits are added to the data word to help identify bit errors experienced during transmission
b. With "Error Correction", redundant bits are added to the data word to help identify and correct bit errors experienced during transmission
c. Error control provides probabilities of experiencing errors, vice an absolute certainty of experiencing errors
d. All of the above are correct
10 points
QUESTION 15
Select the correct statement(s) regarding error detection.
a. A codeword represents information bits
b. A dataword is comprised of both the codes work plus check bits
c. The dataword goes through a specific error detection algorithm in order to compute error detection bits that are then appended to the dataword for transmission
d. The codeword goes through a specific error detection algorithm in order to compute error detection bits that are then appended to the codeword for transmission
10 points
QUESTION 16
Which multiplexing technique can be used with analog basebands?
a. FDMA
b. TDMA
c. CDMA
d. Both FDMA and TDMA
10 points
QUESTION 17
-30dBW = _________dBm
a. -30dBm
b. 0dBm
c. +30dBm
d. -60dBm
10 points
QUESTION 18
-85 dBm = _________mW
a. 3.16E-9 mW
b. 3.16E-12 mW
c. -115 mW
d. -55 mW
10 points
QUESTION 19
600mW = _________dBW
a. 60 dBW
b. 0.6 dBW
c. -2.2 dBW
d. 27.8 dBW
10 points
QUESTION 20
45 dBW = _________mW
a. 3.16E7 mW
b. 3.16E4 mW
c. 75 mW
d. 45 mW
10 points
QUESTION 21
Given the following information, what is the power received (Rx) at the receiver? Tx=50dBm, FSL=-50dB, given the equation: Rx(dBm) = Tx(dBm) + FSL(dBs)
a. Rx = 100 dBW
b. Rx = 0 dBm
c. Rx = -100dBm
d. Rx = 30 dBW
10 points
QUESTION 22
The Signal-to-Noise Ratio (SNR or S/N) can be represented in decibel or non-decibel form. Select the correct decibel format where the use of brackets (i.e., [ ]) signifies dB values.
a. [N]/[S]
b. [S]/[N]
c. [S] - [N]
d. [S] + [N]
10 points
QUESTION 23
What is the noise temperature for a receiver operating at a temperature of T=305oK and frequency bandwidth of the receiver is B=4kHz? (N=kTB)
a. 1.68E-17 watts or -167.7dBW
b. 1.68E-14 watts or -167.7dBW
c. 1.22E+14 watts or +139dBW
d. 4.07E+21 watts or +204dBW
10 points
QUESTION 24
Determine the SNR (dBs) for a received signal power, Pr=-10dBm, and a receiver noise, N=-203dBm?
a. Pr(dBm)*N(dBm)= 2.03E3 dBm
b. Pr(dBm)/N(dBm)= 49.26 dBm
c. Pr(dBm)+N(dBm)= -213 dBm
d. Pr(dBm)-N(dBm)= 193 dBm
10 points
QUESTION 25
The Shannon-Hartley equation give us the theoretical maximum data rate capacity achievable. In practice, this theoretical rate is rarely, if ever, achieved.
True
False
10 points
QUESTION 26
What does the Shannon-Hartley equation tell us?
a. As the frequency bandwidth increases, so does data rate capacity
b. Increasing signal power will increase data rate capacity
c. Decreasing noise power will increase data rate capacity
d. All of the above
10 points
QUESTION 27
Propagation delay only considers the time it takes for a signal to travel through a medium to its destination.
True
False
10 points
QUESTION 28
For UTP, the greater the copper wire diameter (i.e., smaller AWG), the less the signal attenuation experienced, and the further the signal can travel in distance.
True
False
10 points
QUESTION 29
In contrast to conductive guided medium, smaller diameter fiber optic cores means greater attenuation and decreased transmission distances when compared to larger fiber optic cores.
True
False
10 points
QUESTION 30
Guided UTP is configured as a balanced circuit.
True
False
10 points
QUESTION 31
Delay skew and crosstalk are only problems encountered when using coaxial cables for communications.
True
False
10 points
QUESTION 32
A BER of 10E-6 tells you that there is a probability of having one bit error for every 1,000,000 bits sent.
True
False
10 points
QUESTION 33
Signals on coaxial cables are unbalanced. When compared to UTP, coaxial medium has higher bandwidth and data rate carrying capability compared to UTP.
True
False
10 points
QUESTION 34
Select the correct statement(s) regarding fiber optic medium.
a. TIR is the principle that makes light communications travel efficiently through a fiber optic cable
b. MMF fiber can be step-indexed or graded-index
c. SMF has a greater bandwidth and throughput capability over MMF
d. ITU describes transmission windows associated with signal wavelength
e. All of the above are correct
10 points
QUESTION 35
Both SMF optical cores are either step-index or graded-index. As such, SMF is susceptible to modal dispersion.
True
False
a. 000+-0-+
b. 000-+0+-
c. 0++--00+
d. 00+-00-+
Explanation / Answer
Q11. (a) - 000+-0-+
Q12. (c). 4B/5B
Q.16 (a) FDMA
Q17 (b). 0dBm as 0 dBW = 30dBm
Q.18 (a). 3.16E-9 mW
Q.19 (d) 27.78 dBm
Q20. (a). 3.16E7 mW
Q21. (b). Rx = 50 - 50 = 0 dBm
Q22. (c). [S] - [N]
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