Data Com

DATA COMMUNICATION & NETWROKING
COMP 310
ASSIGNMENT #2
ST.GEORGE'S UNIVERSITY
LEOPOLDO PIRELA
DUE DATE: SEPTEMBER 21ST, 2013

What is the phase shift for the following?
a. A sine wave with the maximum amplitude at time zero
b. A sine wave with maximum amplitude after 1/4 cycle
c. A sine wave with zero amplitude after 3/4 cycle and increasing
Solution:
a. There is no phase shift at time zero
b.

(¼) 360 = 90
90 x 2π / 360 = 1.57 rad

c.

(¾) 360 = 270
270 x 2π / 360 = 4. 7 rad

A periodic composite signal with a bandwidth of 2000 Hz is composed of two sine waves. The first one has a frequency of 100 Hz with a maximum amplitude of 20 V; the second one has a maximum amplitude of 5 V. Draw the bandwidth.
Solution:
bandwidth = Fh - Fl
Bandwidth = 2000
Lowest = 100
Highest = 2100
Fh - Fl = 2100 - 100
Bandwidth = 2000

Which signal has a wider bandwidth, a sine wave with a frequency of 100 Hz or a sine wave with a frequency of 200 Hz
Solution:
Bandwidth only exist in composite waves when there are two or more waves. None of these waves are composite waves. Therefore, the is no bandwidth to analyses.

What is the bit rate for each of the following signals?
a. A signal in which 1 bit lasts 0.001 s
b. A signal in which 1 bit lasts 2 ms
c. A signal in which 10 bits last 20 μs
Formula: bitrate = Number of bits in 1s
Solution:
a. 0.001s = 1 millisecond
1000 milliseconds = 1 Second bitrate = Number of bits in 1s
1000 x 1 bit = 1000 bps
b. 0.002 = 2ms → 1 bit per ms
1000 ms = 1s bitrate = Number of bits in 1s
1000ms /2s x 1 bit = 500 bps
c. 1,000,000 = 1s → 10 bits bitrate = Number of bits in 1s
1,000,000 x 20 x 10 bits= 500,000 bps

A non-periodic composite signal contains frequencies from 10 to 30 KHz. The peak amplitude is
10 V for the lowest and the highest signals and is 30V for the 20-KHz signal. Assuming that the amplitudes change gradually from the minimum to the maximum, draw the frequency spectrum
Solution:
Amplitude : 10 V Lowest / 30 v for when 2KHz
Bandwidth = Fh - Fl
30 KHZ - 10 KHz
Bandwidth = 20 KHz

A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is the maximum data rate supported by this line?
Solution:
SNR → signal to noise ratio
Formula : C=B log2 (1+SNR)
=4000 KHz x log2 (1 +1000)
= 4000 KHz x log 2 (1001)
= 4000 KHz x Log2 1001
= 4000 KHz x 9.967226
= 39,868.904 Kbps

We have a channel with 4 KHz bandwidth. If we want to send data at 100 Kbps, what is the minimum SNRdB? What is SNR?
Solution:
Formulas :
● SNRdb = (C x 3) / B
● SNR = 10 (snr log / 10 )
C=100 Kbps
B = 4 KHz
SNRdb = (100Kbps x 3 ) /4KHz
=300 kbps / 4KHz
SNRdb= 75
SNR = 10(SNRdb /10)
SNR= 10(75/10)
SNR=10 ^ 7.5
SNR = 31,622,776

Draw the graph of the Manchester scheme using each of the following data streams, assuming that the last signa11evel has been positive.
a. 00000000
b. 11111111
c. 01010101
d. 00110011

Draw the graph of the 2B1Q scheme using each of the following datastreams, assuming that the last signa11evel has been positive.
a. 0000000000000000
b. 1111111111111111
c. 0101010101010101
d. 0011001100110011

Calculate the baud rate for the given bit rate and type of modulation.
a. 2000 bps, FSK
b. 4000 bps, ASK
c. 6000 bps, QPSK
d. 36,000 bps, 64-QAM
Solution:
a. 2000 bps, FSK → r=log2 2 = 1
S=(1/r) x N
S=(1/1) x 2000
S= 2000 baud
b. 4000 bps, ASK → r=log2 2 =1
S=(1/r) x N
S=(1/1) x 4000
S= 4000 baud
c. 6000 bps, QPSK → r=log2 4 = 2
S=(1/r) x N
S=(1/2) x 6000
S=3000 bauds
d. 36,000 bps, 64-QAM → r=log 2 64 = 6
S=(1/r) x N
S=(1/6) x 36,000
S=6000 bauds
Calculate the bit rate for the given baud rate and type of modulation.
a. 1000 baud, FSK
b. 1000 baud, ASK
c. 1000 baud, BPSK
d. 1000 baud, 16-QAM
Solution:
Formula: N= r x s
a.1000 baud, FSK → r = log2 2 = 1
N=rxs
N= (1) x (1000)
N=1000 bauds
b. 1000 baud, ASK → r = log 2 2 =1

N=rxs
N = (1) x (1000)
N= 1000 bauds
c.1000 baud, BPSK → r = log 2 2 =1
N=rxs
N =(1) x (1000)
N= 1000 bauds
d. d. 1000 baud, 16-QAM → r= log2 16 = 4
N=rxs
N = (4) x (1000)
N= 4000 bauds
What is the required bandwidth for the following cases if we need to send 4000 bps? Let d = 1.
a. ASK
b. FSK with 2∆ f =4 KHz
c. QPSK
d. 16-QAM
Solution:
a. BW = 4000 bps , baud rate, bit rate and Bandwidth are equal
b. FSK with 2∆ f =4 KHz
BW = (1 + d ) x 2∆f 2
BW = (1+1 )x 4000 + 4000
BW = (2) x 4000 +4000
BW = 8000 + 4000
BW = 12,000 Hz
c. QPSK br = ½ bitrate br = 4000/ 2 = 2000
BW = 2000 Hz
d.16-QAM
log2 16 = 4 bits (each signal carries 4 bits) baudrate= 4000/4 = 1000hz