College of Computer and Information Sciences
Department of Computer Engineering
First Semester 1437-1438H
HW # 2 (Solution)
CEN342: Introduction to Data Transmission
Section: 22156
Due Date:
Wednesday 11/1/1438H, 12/10/2016G (8:00am)
Name: ……………………………………………………
Student Code:
Student ID: …………………………………………….
Score:
/ 40
Course Learning Outcome #2:
Learn the fundamentals of data transmission principles: time and frequency representation of
signals, relation between data rate and channel bandwidth, and transmission impairments.
1. (1 Point) The maximum amplitude of a simple sine wave is approximately 5 volts and the
frequency is 10 MHz. If the phase of the signal is π/2, express the mathematical equation of
the sine wave.
x(t) = A sin(2ft + )
 x(t) = 5 sin(210106t + π/2)
2. (2 Points) Given the following signal:
s(t) = 20 cos(2.100t) cos(2.300t)
Answer the following questions:
(a) (½ Point) Use the following trigonometric identity and simplify s(t).
1
1
cos A. cos B  cos( A  B)  cos( A  B)
2
2
s(t) = 10 cos(2.200t) + 10cos(2.400t)
(b) (½ Point) Find the spectrum of the s(t) signal.
Spectrum: 200 Hz and 400 Hz
(c) (½ Point) Find the Bandwidth (B) of the s(t) signal.
Bandwidth = fH - fL = 400 – 200 = 200 Hz
(d) (½ Point) Find the average power of the s(t) signal.
P  C02 
1 j 2
 Cn
2 n 1
, where C0: Amplitude of the DC component;
Cn: Amplitude of the harmonic n.
P = 02 + 0.5[102 + 102] = 100 Watts
1
3. (a) (1 Point) A signal has a bandwidth of 700 KHz. The lowest frequency is 2 MHz. What is the
highest frequency?
B = fH – fL
 fH = fL + B
 fH = 2000 KHz + 700 KHz = 2700 KHz = 2.7 MHz
(b) (1 Point) A signal has a bandwidth of 700 KHz. The highest frequency is 2 MHz. What is the
lowest frequency?
B = fH – fL
 fL = fH - B
 fL = 2000 KHz -700 KHz = 1300 KHz = 1.3 MHz
4. (1 Point) In a typical communication system, the transmitted signal s(t) is attenuated by a
factor  and then corrupted by the noise signal n(t).
Write an equation for the received signal r(t) in terms of s(t), , and n(t).
r(t) =  . s(t) + n(t)
5. (4 Points) Consider the following square wave signal:

sin( 2k (1000 )t )
s(t )  
k
k odd k 1
Assume that s(t) is passed through a channel whose bandwidth is 4 KHz. The attenuation
factor of this channel is constant for all frequencies and is equal to  = 0.5.
(a) (1 Point) Write the expression for the transmitted signal in terms of all its frequency
components.
s(t) = sin(2(1000)t) + 1/3 sin(2(3000)t) + 1/5 sin(2(5000)t) + 1/7 sin(2(7000)t) +
1/9 sin(2(9000)t) + …
(b) (1 Point) Write the expression for the received signal r(t).

r(t) = 0.5[sin(2(1000)t) + 1/3 sin(2(3000)t) ]
r(t) = 0.5sin(2(1000)t) + 1/6 sin(2(3000)t)
(c) (1 Point) Compute the average power in r(t) without noise.
P  C02 
1 j 2
 Cn
2 n 1
, where C0: Amplitude of the DC component;
Cn: Amplitude of the harmonic n.
P = S = 02 + 0.5[0.52 + (1/6)2] = 0.139 Watt
(d) (1 Point) Compute the output signal to noise ratio (SNR) in dB of this channel assuming
the noise power density is N0 = 0.1  10-3 W/Hz.
B = 4000 Hz
N = B  N0 = 4000 * 0.1  10-3 = 0.4 Watts
S/N = 0.139 / 0.4 = 0.3475
SNR = 10 log10(S/N) = 10 log10(0.3475) = -4.59 dB
2
8. (4 Points) A transmission channel between two DTEs is made up of three sections. The first
introduces an attenuation of 10 dB and the second introduces an amplification of 20 dB. If
the overall attenuation of the channel is 5 dB, determine the mean output power of all
sections (P2 , P3, and P4) given that the mean power input to the first section is 600 mW.
5 dB Attenuation
P1 = 600 mW
10 dB
Attenuation
P2 = ?
20 dB
Amplification
The third section introduces an attenuation of
P2 = P3 = P4 =?
First Section:
Attenuation = 10log10 P1/P2
 10 = 10log10 600/ P 2
 1 = log10 600/ P 2
 101 = 600/ P 2
 P 2 = 600 / 10 = 60 mW
Second Section:
Amplification = 10log10 P3/P2
 20 = 10log10 P 3 / 60
 2 = log10 P 3 / 60
 102 = P 3 / 60
 P 3 = 60 × 100 = 6000 mW
Third Section:
Attenuation = 10log10 P3/P4
 15 = 10log10 6000 / P 4
 1.5 = log10 6000 / P 4
 101.5 = 6000 / P 4
 31.62 = 6000 / P 4
 P 4 = 6000 / 31.62 = 189.74 mW
OR,
Attenuation = 10log10 P1/P4
 5 = 10log10 600 / P 4
 0.5 = log10 600 / P 4
 100.5 = 600 / P 4
 3.162 = 600 / P 4
 P 4 = 600 / 3.162 = 189.74 mW
6
15
P4 = ?
P3 = ?
?
dB
9. (2 Points) A digital signaling system is required to operate at 4800 bps.
(a) (1 Point) If a signal element encodes a 4-bit word, what is the minimum required bandwidth
of a channel?

R = 2Bm
B = 4800 / (2 × 4)  B = 4800 / 8
 B = 600 Hz
(b) (1 Point) Repeat part (a) for the case of 8-bit words.

R = 2Bm
B = 4800 / (2 × 8)  B = 4800 / 16
 B = 300 Hz
10. (2 Points) A modem to be used with the PSTN uses 512 levels per signaling element. Assuming
a 3100 Hz noiseless channel, deduce:
(a) (1 Point) The bit rate.
(b) (1 Point) The bandwidth efficiency of the modulation scheme.
(a)

R= 2B log2M
R = 2 × 3100 × log2 512  R = 2 × 3100 × 9 = 55800bps
(b)



Bandwidth Efficiency = R / B
Bandwidth Efficiency = 2B log2M / B
Bandwidth Efficiency = 2m
Bandwidth Efficiency = 2 × 9= 18 bps / Hz
11. (3 Points) Deduce the maximum bit rates associated with the following transmission channels:
(a) (1 Point) A network with a bandwidth of 3100 Hz and a signal to noise ratio of 20 dB.
(b) (1 Point) A network with a bandwidth of 3100 Hz and a signal to noise ratio of 15 dB.
(c) (1 Point) Which is better choice (a) or choice (b)? Why?
(a)
SNR = 10log10S/N
 20 = 10log10S/N  2 = log10S/N  S/N = 102 = 100
C = B log2(1+S/N)
 C = 3100 × log2(1+100)  C = 3100 × ln101 / ln2
 C = 20640 bps
(b)
SNR = 10log10S/N
 15 = 10log10S/N  1.5 = log10S/N  S/N = 101.5 = 31.62
C = B log2(1+S/N)
 C = 3100 × log2(1+31.62)  C = 3100 × ln32.62 / ln2
 C = 15585 bps
(c) Choice (a) is better than choice (b). High SNR means a high power signal relative to
the noise level, resulting in a good-quality signal
7
12. (2 Points) Answer the following questions related to a TV signal:
(a) (1 Point) Assume that the picture is to be transmitted over a channel with 10 MHz bandwidth
and 20-dB signal-to-noise ratio. Find the capacity of the channel (C).
(b) (0 Point) Suppose that a digitized TV picture is to be transmitted from a source that uses a
matrix of 800 x 600 pixels, where each pixel can take on one of 256 intensity values. Assume
that 25 pictures are sent per second. Find the bit rate (R) .
(c) (1 Point) Can we transmit the TV signal described in (b) over the channel described in (a)?
Why?
8
0
0
1
1
(a)
6
0
0
SNR = 10log10S/N
 20 = 10log10S/N  2 = log10S/N  S/N = 102= 100




C = B log2(1+S/N)
C = 10×106 × log2(1+100)
C = 10×106 × ln101/ ln2
C = 10×106 × 6.66
C = 50276848 bps = 66.6 Mbps
(b)
R = 600 × 800 × 8 × 25 = 96000000 bps = 96 Mbps
(c)
R > C We cannot transmit the TV signal over the channel.
13. (3 Points) Answer the following questions:
(a) (1 Point) Consider a channel with 1 MHz bandwidth and 20-dB signal-to-noise ratio. Find the
capacity of the channel (C).
(b) (1 Point) Data is to be transmitted over the above channel using a transmission scheme with
4 levels per signaling element. If the bandwidth of the transmitted signal is 1 MHz, Find
the bit rate (R) .
(c) (1 Point) Can we transmit the the signal described in (b) over the channel described in (a)?
Why?
(a)
SNR = 10log10S/N
 20 = 10log10S/N  2 = log10S/N  S/N = 100
C = B log2(1+S/N)
 C = 1×106 × log2(1+100)
 C = 1×106 × ln101 / ln2
 C = 6658211 bps
(b)
(c)
R = 2 B log2M
 R = 2 × 1×106 × log2 4  R = 2 × 1×106 × 2 = 4,000,000 bps
R < C  We can transmit the signal over the channel.
8
14. (3 Points) Consider a channel with a 1 MHz capacity and an S/N of 63.
(a) (1 Point) What is the upper limit to the data rate that the channel can carry?
(b) (2 Points) The result of part (a) is the upper limit. However, as a practical matter, better
error performance will be achieved at a lower data rate. Assume we choose a data rate
of 2/3 the maximum theoretical limit. How many signal levels are needed to achieve this
data rate?
(a)
C = B log2(1+S/N)
 C = 1×106 × log2(64)
 C = 1×106 × 6
 C = 6 Mbps
(b)
R = 2/3 * 6 Mbps = 4 Mbps



R = 2 B log2M
4 Mbps = 2 × 1×106 × m
m =2
M=4 Levels
15. (3 Points) For binary phase-shift keying (PSK), a bit error rate of 10-5 (one bit error out of every
100,000) is considered. If the effective noise temperature is 290°K (room temperature) and
the data rate is 9600 bps, what received signal level is required?




(Eb/N0)dB = SdBW - 10 log10 R + 228.6 dBW - 10 log10 T
9.5 = S - 10 log10 9600 + 228.6 dBW - 10 log10 290
9.5 = S – 39.82 + 228.6 dBW – 24.62
S = 9.5 + 39.82 - 228.6 dBW + 24.62
S = -154.66 dBW
9
16. (0 Point) Complete the following table:
17. (0 Point) The square wave, with T = 1ms, is passed through a lowpass filter that passes
frequencies up to 8 kHz with no attenuation.
a. Find the power in the output waveform.
b. Assuming that at the filter input there is a thermal noise voltage with N0 = 0.1  Watt/Hz,
find the output signal to noise ratio in dB.
(a)
1
1
1


x(t )  sin(1000 t )  sin( 3000 t )  sin( 5000 t )  sin( 7000 t ) 
3
5
7


1 j 2
P  C   Cn
2 n 1
2
0
1 1 1
1 
Power  0  1   
 0.586 watt
2  9 25 49 
(b)
N  B  No
N  8 103  0.110 6  0.0008Watt
S/N = 0.586 / 0.0008 = 732.5
(SNR)db = 28.65
10