ECE 271 INTRODUCTION TO TELECOMMUNICATION NETWORKS
ANSWERS TO HOMEWORK2
Homework Question 6
Comparing the carriers of microwave and optical communication systems:
a. Which one has higher frequency ?
b. Which one has shorter wavelength ?
c. Write the frequency range of HF.
d. Write the wavelength range of UHF.
Answer to Homework Question 6
a. Optical communication systems have higher frequency
b. Optical communication systems have shorter wavelength
c. 3 – 30 MHz.
d. 300 MHz – 3 GHz corresponds to wavelength range of
300000000 m/sec / 300000000 (1/sec) = 1 m
300000000 m/sec / 3000000000 (1/sec) = 0.1 m = 10 cm
Homework Question 7
a. How many total voice channels are transmitted in a 2.048 Mb/s system ?
b. How many total video channels can be transmitted in a 139.264 Mb/s system ?
Answer to Homework Question 7
a. 2.048 Mb/s / 64 Kb/s = 2048000b/s / 64000 b/s = 32 channels total channels
32 – 1 signalling channel – 1 synchronization channel = 30 voice channels.
b. One video channel in 1.b 96 Mbps or in 1.c 120 Mbps. In either case only one video channel can be transmitted in a 139.264 Mb/s system ?
Homework Question 8
a. For the analog signal v = 3.6 sin 6280 t, find the minimum sampling rate needed.
b. What happens if the sampling rate is 250 Hz ?
c. What happens if the sampling rate is 5 KHz ?
Answer to Homework Question 8
a. For the analog signal v = 3.6 sin 6280 t, find the minimum sampling rate needed = 2KHz.
b. If the sampling rate is 250 Hz, then the signal can not be recovered correctly at the receiver ?
c. If the sampling rate is 5 KHz, the signal can be recovered correctly, however there is no need to sample over the sampling rate since it will bring unnecessary complication and cost to electronics.
Homework Question 9
An analog signal has time variation f(t) = 3 + 0.2 cos (8000π t)  0.3 sin (4000π t) .
a. Minimum how many samples should be taken to satisfy Nyquist requirement?
b. 256 levels is used to represent one sample. How many bits are required to transmit one sample value?
c. What is the mimimum transmission rate of this signal?
d. Would you allocate a 256 kbps channel to transmit this signal? Why?
e. Would you allocate a 32 kbps channel to transmit this signal? Why?
Answer to Homework Question 9
An analog signal has time variation f(t) = 3 + 0.2 cos (8000π t)  0.3 sin (4000π t) .
a. Minimum how many samples should be taken to satisfy Nyquist requirement?
Maximum frequency = 4 KHz, i.e. sampling frequency = 8 KHz
The minimum number of samples per second needed to satisfy Nyquist requirement = 8000
b. 256 levels is used to represent one sample. How many bits are required to transmit one sample value?
256 levels mean that each sample is represented by 8 bits since 256 = 2 ^{8}
The number of bits needed to transmit one sample = 8 bits
c. What is the mimimum transmission rate of this signal?
The mimimum rate in Kbps at which this signal is transmitted
= 8000 samples / sec x 8 bits / sample = 64 Kbps
d. Would you allocate a 256 kbps channel to transmit this signal? Why?
No because more than 64 kbps rate will be unnecessary in the recovery of the original signal.
e. Would you allocate a 32 kbps channel to transmit this signal? Why?
No because with a rate less than 64 kbps, the original signal will only be recovered with loss of information which is not desired.
Homework Question 10
a. For a carrier of sin (2000 π t), the Amplitude Shift Keying (ASK) Modulated signal is given below. Plot the digital information signal x(t).
b. If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented
by sin (4000 π t), plot the Frequency Shift Keying (FSK) Modulated signal for the digital information signal x(t) found in part a.
c. If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented
by cos (2000 π t), plot the Phase Shift Keying (PSK) Modulated signal for the digital information signal x(t) found in part a.
d. If the carrier in part a becomes sin (4000 π t), replot the Amplitude Shift Keying (ASK) Modulated signal given in part a.
e. Find the rate of the digital information signal x(t) found in part a.
Answer to Homework Question 10
a. For a carrier of sin (2000 π t), the Amplitude Shift Keying (ASK) Modulated signal is given below. Plot the digital information signal x(t).

X(t)






















1






















0













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t (msec)


































b. If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented
by sin (4000 π t), plot the Frequency Shift Keying (FSK) Modulated signal for the digital information signal x(t) found in part a.
c. If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented
by cos (2000 π t), plot the Phase Shift Keying (PSK) Modulated signal for the digital information signal x(t) found in part a.
d. If the carrier in part a becomes sin (4000 π t), replot the Amplitude Shift Keying (ASK) Modulated signal given in part a.
e. Find the rate of the digital information signal x(t) found in part a.
Rate of x(t) found in part a = 1 bit / msec = 1 bit / 10 ^{3} sec = 10 ^{3} bits / sec = 1 kbps.
Homework Question 11
The digital signal X(t) given below.
X(t)






















1






















0













1

2

3

4

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8

t (msec)



































If the carrier is sin (2000 π t), plot Amplitude Shift Keying (ASK) Modulated signal.
ASK






















1






















0













1

2

3

4

5

6

7

8

t (msec)

1


































If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented
by sin (4000 π t), plot Frequency Shift Keying (FSK) Modulated signal.
FSK






















1






















0













1

2

3

4

5

6

7

8

t (msec)

 1


































If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented
by cos (2000 π t), plot Phase Shift Keying (PSK) Modulated signal.
PSK






















1






















0













1

2

3

4

5

6

7

8

t (msec)

 1
































 Answer to Homework Question 11
The digital signal X(t) given below.
X(t)






















1






















0













1

2

3

4

5

6

7

8

t (msec)



































If the carrier is sin (2000 π t), plot Amplitude Shift Keying (ASK) Modulated signal.

If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented
by sin (4000 π t), plot Frequency Shift Keying (FSK) Modulated signal.

If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented
by cos (2000 π t), plot Phase Shift Keying (PSK) Modulated signal.
Homework Question 12
Eight bits of information is sent in the following modulated signal where time axis is in microseconds:
a. Write the type of modulation used. Why?
b. Find the carrier frequency.
c. Find the rate of the information signal.
d. Plot the information signal if “1” is represented by no signal, and “0” is represented by 0.5 mV and no carrier.
e. Is this information signal convenient to carry 1 digital voice channel ? Why?
Is this information signal convenient to carry 1 digital video channel ? Why?
Answer to Homework Question 12
Eight bits of information is sent in the following modulated signal where time axis is in microseconds:
a. Write the type of modulation used. Why?
Solution: ASK (Amplitude Shift Keying) because digits “1” and “0” are differentiated with different amplitudes.
b. Find the carrier frequency.
Solution: For one bit, duration is 1 sec. and the number of cycles=2
Thus,the carrier frequency = 2 cycle in 1 sec., i.e., 2 x 10 ^{6} cycles/sec = 2 MHz.
c. Find the rate of the information signal.
Solution: One bit has duration of 1 sec.
Thus, the rate of the signal = 1 x 10 ^{6} bits/sec = 1 Mbps.
d. Plot the information signal if “1” is represented by no signal, and “0” is represented by 0.5 mV and no carrier.
Solution:
e. Is this information signal convenient to carry 1 digital voice channel ? Why?
Is this information signal convenient to carry 1 digital video channel ? Why?
Solution: This information signal is convenient to carry 1 digital voice channel because the information signal has a rate of 2 Mbps and 1 digital voice channel needs only 64 kbps.
However, this information signal is not convenient to carry 1 digital video channel because the information signal has a rate of 2 Mbps and 1 digital video channel needs (6 x 10 ^{6 }x 2) samples / sec x 8 bits / sample = 96 Mbps. 