King Saud University
College of Computer and Information Sciences
Department of Information Technology
CAP240
Second semester 1429/1430
Sheet 5
Multiple-Choice Questions
1. ASK, PSK, FSK, and QAM are examples of ________ modulation.
b. Digital-to-analog
2. In QAM, both phase and ____of carrier frequency are varied
a. Amplitude
3. Which of the following is most affected by noise?
b. ASK
4. If the baud rate is 400 for a 4-PSK signal, the bit rate is ________ bps.
c. 800
5. If the bit rate for an ASK signal is 1200 bps, the baud rate is ________
d. 1200 (Baud rate = Bitrate  in ASK)
6. If the bit rate for an FSK signal is 1200 bps, the baud rate is _______
d. 1200 (Baud rate = Bitrate  in FSK)
7. If the bit rate for a QAM signal is 3000 bps and a signal unit is represented by a tribit, what
is the baud rate?
c. 1000
8. If the baud rate for a QAM signal is 3000 and a signal unit is represented by a tribit, what is
the bit rate?
d. 9000
9. Tthe baud rate for a QAM signal is 1800 and the bit rate is 9000, how many bits are there
per signal unit?
c. 5
10. In 16-QAM, there are 16 ________
a. Combinations of phase and amplitude
11.Which modulation technique involves tribits, eight different phase shifts ,and amplitude?
b. 8-PSK
12. A modulated signal is formed by
b. Changing the carrier wave by the modulating signal ( review the book pp.117, and study
the differences between the carrier signal and modulating signal, in the book “ In analog
transmission, the sending device produces a high-frequency signal that acts as a basis for
information signal. This base signal is called the carrier signal or carrier frequency. The
receiving device is tuned to the frequency of the carrier signal that it expects from the sender.
Digital information then modulates the carrier signal by modifying one or more of its
characteristics (amplitude, frequency, or phase). This kind of modification is called modulation
or shift keying, and the information signal is called the modulating signal.”
13. When an ASK signal is decomposed, the result is _______
c.An infinite number of sine waves (review the book pp. 118-119 you will find that when the
ASK signal decomposed, we get a spectrum of many simple frequencies……although there is
only one carrier frequency, the process of modulation produces a complex signal that is a
combination of many simple signals, each with different frequency.
14. For a telephone line, the bandwidth for voice is usually _______ the bandwidth for data.
c. Greater than
15. For a given bit rate, the minimum bandwidth for ASK is _______the minimum bandwidth
for FSK.
b. Less than ( the minimum of ASK-BW when d=0 so, Bw= Baud rate, and as the FSK-Bw
= Baud rate +( fc1-fc0) and the last term of the equation is a nonzero value, then the ASK is
less than FSK by (fc1-fc0).
16. As the bit rate of an FSK signal increases, the bandwidth _______
b. Increases
17. For FSK, as the difference between the two carrier frequencies increases, the bandwidth
______
b. Increases
18. In trellis coding the number of data bits is ________ the number of transmitted bits.
b. Less than
19. What is the object of trellis coding?
d. To reduce the error rate
20. The bit rate always equals the baud rate in which type of signal?
a. FSK
21. A modulator converts a(n) _________ signal to a(n) ________ signal.
a. Digital; analog
22. A 56K modern can download at a rate of ______Kbps and upload at a rate of
______ Kbps
c. 56.6; 33.6
Exercises
23. Calculate the baud rate for the given bit rate and type of modulation:
a. 2000 bps, FSK
Baud rate equal bit rate ( each baud carries one bit)
baud rate = 2000 baud/s
b. 4000 bps, ASK
Baud rate equal bit rate ( each baud carries one bit)
baud rate = 4000 baud/s
c. 6000 bps, 2-PSK
baud rate= bit rate /no of bit/baud
= bit rate / log2 (no of phases)
=6000/1 =6000 baud/s
d. 6000 bps, 4-PSK
baud rate= bit rate /no of bit/baud
= bit rate / log2 (no of phases)
=6000/2 =3000 baud/s
e. 6000 bps, 8-PSK
baud rate= bit rate /no of bit/baud
= bit rate / log2 (no of phases)
=6000/3 =2000 baud/s
f. 4000bps, 4-QAM
baud rate= bit rate /no of bit/baud
= bit rate / log2 (no of phases)
=4000/2 =2000 baud/s
g. 6000 bps, 16-QAM
baud rate= bit rate /no of bit/baud
= bit rate / log2 (no of variations)
=6000/4 =1500 baud/s
h. 36,000bps, 64-QAM
baud rate= bit rate /no of bit/baud
= bit rate / log2 (no of variations)
=36000/6 =6000 baud/s
24. Calculate the baud rate for the given bit rate and bit combination
a. 2000 bps, dibit
dibit -> 2 bit/baud
baud rate= bit rate /no of bit/baud
=2000 / 2
=1000 baud/s
b. 6000 bps, tribit
tribit -> 3 bit/baud
baud rate= bit rate /no of bit/baud
=6000 / 3
=2000 baud/s
c. 6000 bps, quadbit
quadbit -> 4 bit/baud
baud rate= bit rate /no of bit/baud
=6000 / 4
=1500 baud/s
d. 6000 bps, bit
baud rate= 6000 baud/s
25. Calculate the bit rate for the given baud rate and type of modulation.
a. l000 baud , FSK
bit rate = baud rate =1000
b. l000 baud , ASK
bit rate = baud rate =1000
c. 1000 baud, 8-PSK
bit rate= Nbaud x no of bit/baud
= Nbaud x log2 (no of phases)
= 1000 x 3 = 3000 bps
d. l000 baud, 16-QAM
bit rate= Nbaud x no of bit/baud
= Nbaud x log2 (no of variations)
= 1000 x 4 = 4000 bps
26.The data points of a constellation are at (4, 0) and (6, 0). Draw the constellation.
Show the amplitude and phase for each point. Is the modulation ASK, P5K, or
QAM? How many bits per baud can one send with this constellation?
2 amplitudes (4,6) --> ASK -->1 bit/baud
27. Repeat Exercise 28 if the data points are at (4, 5) and (8, 10).
2 amplitudes
--> ASK -->1 bit/baud
28.Repeat Exercise 28 if the data points are at (4, 0) and (-4, 0).
1 amplitude 4 and 2 phases (0 , 180) -->2 PSK -->1 bit/baud
29. Repeat Exercise 28 if the data points are at (4, 4) and (-4, 4).
1 amplitude 4 and 2 phases (45 , 135) -->2 PSK -->1 bit/baud
30.Repeat Exercise 28 if the data points are at (4, 0), (4, 4), (-4, 0), and (-4, -4).
2 amplitudes (4,
) and 4 phases (0 , 180,45,225) -->8 QAM -->3 bit/baud
31. What is the required bandwidth for the following cases if we need to send
4000bps? Let d=1
1. ASK
ASK Bw = (1+d) Nbaud
= 2 x 4000
[baud rate = bit rate]
= 8000 Hz
2. FSK with difference between carriers frequencies = 4KHZ
FSK BW = fC1- fC0+Nbaud
=4000 + 4000 [baud rate = bit rate]
=8000
3- 4-PSK ( as mentioned in your slides (slide #27 chapter 5), as shown below)
baud rate = bit rate/ log2(number of phases)
= 4000/2
=2000
OR
Baud rate = N
Bit rate = 2N  N= 4000/2 = 2000
Baud rate = 2000
PSK Bw = (1+d) Nbaud
(as assumed d=1)( bandwidth = baud rate in the minimum case d=0)
= 2 * 2000 =4000 Hz
4. 16-QAM
baud rate = bit rate/ log2(number of phases)
= 4000/4
=1000 baud/s
PSK Bw = (1+d) Nbaud
= 2 * 1000 =2000 Hz
32.Can a constellation have 18 points? Why or why not?
No, 18 is not a power of 2.
33.Can you define a general rule for the number of points in a constellation?
Number of points = 2n where n is the number of bits represented by a point .
34. How many bits per baud can we send in which of the following cases if the signal
constellation has one of the following number of points?
1. 2
no of bit/baud=log2 (no of points)
= 1 baud/s
2. 4
= 2 baud/s
3. 8
= 3 baud /s
4. 16
= 4 baud /s
5. 1024
= 10 baud /s
35. Does the constellation in Figure 1 represent ASK, FSK, P5K, or QAM?
ASK
36. Does the constellation in Figure 2 represent ASK, FSK, PSK, or QAM?
8-PSK
37. Does the constellation in Figure 3 represent ASK, FSK, P5K, or QAM?
8-QAM
38. Does the constellation in Figure 4 represent ASK, P5K, PSK, or QAM?
8-QAM