NET352 : Principles of Data
communications
Homework 1
Networks and communications
Department
Spring 2014
Assignment Problem(s)
Q1: Select whether the following statements are True or False :
1
The basic building block of any communications facility is the Modem.
F 0.5
2
LAN usually owned by the same organization that owns
T 0.5
3
Signaling propagates along a medium.
T 0.5
4
The common Analog data is the text
F 0.5
5
Odd signals are symmetric about origin
T 0.5
6
In Half-duplex transmission both stations transmit, but only one at a time
T 0.5
7
A general sine wave can be represented by the frequency and the
phase.
8
any signal is made up of components at similar frequencies
F 0.5
9
Absolute bandwidth is the width of spectrum
T 0.5
10 Intermodulation noise produced from the electrons
F 0.5
F 0.5
Q2: find the value of the following unit step functions, then draw their graphs :
1- 𝑢−2 (𝑡). 0.5 for value , 0.5 for graph
0 𝑖𝑓 𝑡 < −2
𝑢−2 (𝑡) = {
1 𝑖𝑓 𝑡 ≥ −2
2- 2𝑢4 (𝑡). 0.5 for value , 0.5 for graph
2𝑢4 (𝑡) = {
0 𝑖𝑓 𝑡 < 4
2 𝑖𝑓 𝑡 ≥ 4
Q3: find the value of the following unit impulse functions, then draw their graphs :
1- 𝜕(𝑡 − (−3)) 0.5 for value , 0.5 for graph
𝜕 (𝑡 − (−3)) = {
+∞ 𝑖𝑓 𝑡 = −3
0 𝑖𝑓 𝑡 ≠ −3
2- ∑2𝑘=−3 𝜕(𝑡 − 𝑘)1 for graph
Q4: find the output value of the LTI system using the convolution sum, when these
values are given:
𝑥[𝑛] = 𝑢[𝑛]
,
1
ℎ[𝑛] = ( )𝑛 𝑢[𝑛]
3
Answer :
Convolution sum : 𝑦[𝑛] = ∑+∞
−∞ 𝑥[𝑛] ℎ[𝑛 − 𝑘]0.25
By using x and h values : 𝑦[𝑛] = ∑+∞
−∞ 𝑢[𝑘]
𝑦[𝑛] =
1𝑛
3
Then : 𝑢[𝑘] 𝑢[𝑛 − 𝑘] = 0
𝑦[𝑛] = 0
0. 5
Second : when n >= 0
Then : 𝑢[𝑘] 𝑢[𝑛 − 𝑘] = 1
1𝑛
𝑦[𝑛] =
𝑦[𝑛] =
3
1𝑛
2
1−𝑘
∑+∞
−∞
3
3
0.25
𝑘
∑+∞
−∞ 3
1𝑛 1−3𝑛+1
0.25
1−3
𝑦[𝑛] = 3−𝑛
1−3𝑛+1
0.25
−2
1
𝑦[𝑛] = 3−𝑛 (−2 −
3−𝑛
3𝑛+1
−2
3−𝑛+𝑛+1
𝑦[𝑛] = ( −2 −
−2
3−𝑛
3
𝑦[𝑛] = ( −2 −
)0.25
)0.25
)0.25
−2
3−𝑛 −3
𝑦[𝑛] = (
−2
3−𝑛 −3
) 𝑢[𝑛]
−2
𝑦[𝑛] = (
3
∑+∞
−∞ 𝑢[𝑘]
First : when n < 0
𝑦[𝑛] =
1𝑛−𝑘
)0.25
0 𝑖𝑓 𝑛 < 0
= { 3−𝑛−3
1
( −2 ) 𝑖𝑓 𝑛 ≥ 0
𝑢[𝑛 − 𝑘]0.25
1−𝑘
3
for law
[𝑛 − 𝑘]0.25
Q5: If a 5-levels signal sent over a 6 KHz bandwidth channel where signal to noise
ratio (SNR) is 15 dB. What is the maximum achievable data rate?(use both Nyquist
and Shannon formula).
0.25 for laws
𝑺𝒉𝒂𝒏𝒏𝒐𝒏 𝑭𝒐𝒓𝒎𝒖𝒍𝒂
𝑆𝑁𝑅𝑑𝐵 = 10 𝑙𝑜𝑔10 (SNR)
15 = 10 𝑙𝑜𝑔10 (SNR) 0.25
SNR =
15⁄
10 10 =
31.6 0.25
C = B 𝑙𝑜𝑔2 (1+SNR) 0.25
C = 6000 𝑙𝑜𝑔2 (1+31.5) 0.25
C = 30 Kbps 0.25
𝑵𝒚𝒒𝒖𝒊𝒔𝒕 𝑭𝒐𝒓𝒎𝒖𝒍𝒂
C = 2 B 𝑙𝑜𝑔2 M
C = 2 * 6000 * 𝑙𝑜𝑔2 5 0.25
C = 27.6 Kbps 0.25