1. The three groups of wireless technologies are:
a. wired wireless, mobile wireless, IR wireless
b. wired wireless, mobile wireless, wired IR
c. fixed wireless, mobile wireless, IR wireless
d. fixed wireless, mobile wireless, wired wireless
2. Determine the channel bandwidth required for a wideband FSK system.
a. 5 to 10 kHz
b. 5 to 15 kHz
c. 20 to 25 kHz
d. not enough information to answer
Use the following equation to answer questions 3 – 5.
3. Find vo(t) for a phase shift keying system given i = 1, M = 2
a. Vsinωct
b. -Vsinωct +
c. 2Vsinωct
d. not enough information to solve.
4. Find vo(t) for a phase shift keying system given i = 3, M = 8
a. Vsinωct
b. -Vsinωct
c. Vsin(ωct + π/2)
d. not enough information to solve.
5. Find vo(t) for a phase shift keying system given i = 2, M = 4
a. Vsinωct
b. -Vsinωct
c. Vsin(ωct + π/2)
d. not enough information to solve.
6. The output signal of a M-ary system is called a
a. array
b. constellation
c. star
d. none of the above
7. The output data of a digital communication link is shown in Figure 10-1. The
modulation technique being used is:
a. BPSK
b. QPSK
c. 8PSK
d. 8QAM
8. The vector representation of the constellation
shown in Figure 10-2 is an example of :
a. BPSK
b. QPSK
c. 8PSK
d. 16QAM
9. The constellation shown in figure 10-3 is an
example of:
a. BPSK
b. QPSK
c. 8PSK
d. 16QAM
10. Which of the eye patterns shown in Figure 10-4 shown a normal system.
11. Which eye pattern shown in Figure 10-4 shows timing jitter?
a. a
b. b
c. c
d. d
12. The terms sin2A can be written as
a. ½ [1 + cos 2A]
b. ½ [1 x cos 2A]
c. ½ [1 – cos 2A]
d. ½ [1 – cos A]
13. The output of a QPSK phase detector is 0.5A cos (0 + Φd) – 0.5A cos (2ωct).
Identify the high frequency component for this output.
a. 0.5 is the high frequency component
b. Φd is the high frequency componrnt
c. 2ωct is the high frequency component
d. none of the above
14. How is the high frequency component removed for the output of a QPSK
phase detector with a value of 0.5A cos (0 + Φd) – 0.5A( cos 2ωct).
a. low-pass filter
b. band-stop filter
c. high-pass filter
d. notch filter
15. Determine the voltage vpd for the data vector point defined by (0,0) for a
QPSK system with a unit value of 1.
a. 0.707V, 0.707V
b. 0.707V, 0.35V
c. 0.35V, 0.707V
d. 0.35V, 0.35V
16. What is the purpose of a loop back when used in digital modulation
systems?
a. the loop back is compared with the original data to indicate system
performance
b. the loop back is compared with the original data to indicate system data
rate
c. the loop back is never compared with the original data
d. the loop back is only compared with the original data to indicate a system
noise figure
17. This type of code has an output that appears to be noise-like.
a. pseudo noise code
b. pseudo code
c. pseudo spectrum
d. spread code
18. A PN code of length 2n – 1 is said to be of
a. repetitive
b. minimal length
c. maximal length
d. lengthy
19. Determine the sequence length of a properly connected PN sequence
generator containing 5 shift registers.
a. 32
b. 69
c. 36
d. 31
20. Determine the sequence length of a properly connected PN sequence
generator containing 31 shift registers.
a. 2,147,483,648
b 2,147,483,649
c. 2,147,483,647
d. 2,147,483,646
21. The RF spectrum shown in Figure 10-5 is an
example of
a. direct sequence spread spectrum
b. orthogonal frequency division multiplexing
c. frequency division spread spectrum
d. frequency hopping spread spectrum
22. DSSS uses a pseudorandom sequence of pulses shorter than the message bit.
These are called
a. fractals
b. parce bits
c. fractures
d. chips
23. The term for the situation when two spread spectrum transmitters
momentarily transmit at the same time is called
a. lost data
b. design error
c. a hit
d. a data burst
24. The signature sequence in DSSS is
a. eliminates the need for QPSK transmission
b. the pseudo random digital sequence used to spread the signal
c. aids with the reception of weak signals
d. limits the data rate
25. A general rule is that the chip rate is what relative to the data modulation
rate?
a. approximately 5 to 7 times the modulation rate
b. must be less that 12% of the modulation rate
c. is independent of the modulation rate
d. must be greater than 70.7% of the modulation rate
26. Determine the spreading of a DSSS signal given the following parameters
Modulation bit rate: 56 kbps Chip rate: 280 kbps
a. spreading = 0.2
b. spreading = 1.2
c. spreading = 1.0
d. spreading = 5.0
27. Compare the two RF spectrums shown.
What has changed for from Figure 10-6 A to
B?
a. the QPSK signal has been spread
b. the BPSK signal has been spread
c. the 16QAM signal has been spread
d. the 8PSK signal has been spread
28. Another name for OFDM is
a. multitone modulation
b. discrete modulation
c. DSSS
d. QPSK
29. If two signals are sent over the same media without interference the two
signals are said to be
a. symmetrical
b. vertical
c. orthogonal
d. parallel
30. A cyclic prefix in OFDM means
a. the very beginning of a symbol is copied thus leaving no gap
b. the very end of a symbol is copied to the beginning of the data stream
thus leaving a gap
c. the very end of a symbol is copied to the beginning of the data stream
thus leaving a defined gap
d. the very end of a symbol is copied to the beginning of the data stream
thus leaving no gap
31. FSK is a form of:
a. AM
b. FM
c. PM
d. none of the above
32. The LM1871 and LM1872 integrated circuits:
a. can be used to create a radio-telemetry system that operates at a
frequency of approximately 49 MHz.
b. use PWM as a pulse modulation technique.
c. can handle more than one channel of digital data.
d. all of the above.
33. Which of the following is an advantage of transmitting in a digital format?
a. less sensitive to noise
b. less crosstalk
c. lower distortion levels
d. all of the above
34. The term for an analog digital FM signal that shares the same bandwidth is
a. spread spectrum FM
b. SSB FM
c. hybrid FM
d. PCM AM
35. The digital data rate for AM digital is
a. 36 kbps
b. 56 kbps
c. 1.544 Mbps
d. 96 kbps
36. The digital data rate for FM digital is
a. 36 kbps
b. 96 kbps
c. 1.544 Mbps
d. 56 kbps