Transcript EE302: Lesson 2 Gain and decibels

EC310
16-week
Review
Rules of Engagement
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Teams selected by instructor
Host will read the entire questions. Only after, a team may “buzz” by
raise of hand
A team must answer the question within 5 seconds after buzzing in
(must have answer at hand)
If the answer is incorrect, the team will lose its turn and another
team may buzz in. No score will be deducted. No negative scores.
Maximum score is 100. Once reached, that team will stand down for
others to participate. Teams will earn all points scored at the end of
game.
When selecting a question, Teams must only select questions of
different value, unless there are no others, but may be from different
categories.
All team members will participate and will answer questions
Only one round - No Daily Doubles, Double Jeopardy or Final
Jeopardy … and no partial credits!
Jeopardy!
BGP / BGP
Routing
Comm Sys.
EM Spectrum,
Signals,
Modulation
A/D
Conversion
Digital
Modulation
Power Gain,
SNR,
Antennas
Propagation
/ Electronic
Warfare
10
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40 pts
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40
60 pts
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60
BGP/BGP Routing 10 pts
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Briefly describe each of the following
Autonomous Systems Categories:
(a) Stub AS
(b) Multihomed AS (c) Transit AS
Has only one connection to another AS
Has more than one connection to other ASes, but
doesn’t allow data to pass through it
Connects to more than one AS and allows traffic to
pass through it
Comm Sys., EM Spectrum,
Signals, Modulation 10 pts
Name the four basic components of a
communications system.
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Transmitter
Receiver
Channel
Noise
A/D Conversion 10 pts
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What is the most important advantage of digital
communications?
Relative Noise Immunity – a digital system only
has to decide between a finite set, e.g. binary 0 or
binary 1 for a binary digital receiver, which is
easier even amid of a significant noise.
Digital Modulation 10 pts

What digital modulation scheme does each
constellation represent?
8-PSK
16-PSK
8-QAM
QPSK
BPSK
Gain, SNR, Antennas 10 pts
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Fill in the missing information for each antenna
type.
Antenna
Type
Azimuth
beam pattern
Vertical
beam pattern
Dipole
(vertically oriented)
omnidirectional
null
λ
2
Monopole
(vertically oriented)
omnidirectional
null
λ
4
Depends on gain
(it varies)
Depends on gain
(it varies)
Yagi
Physical size
(of the driven element)
λ
2
Propagation/EW 10 pts

For VLF-HF communications, list and describe
the basic modes of propagation.
Ground wave – EM waves that travel close to the
surface of the earth
Sky wave – EM waves that travel up into the
atmosphere and then bend back to earth
Space wave – EM waves that travel in a straight line
(direct line-of-sight or LOS)
BGP/BGP Routing 20 pts

Describe the steps followed in BGP routing
when selecting a route.
1) a BGP router first attempts to find all paths from the
router to a given destination
2) it then judges these paths against the policies of the AS
administrator
3) it then selects a “good enough” path to the destination
that satisfies the policy constraints
Comm Sys., EM Spectrum,
Signals, Modulation 20 pts

What is the wavelength of a radio station whose
broadcast frequency is 106.9 MHz?
𝑐
3 × 108 𝑚 𝑠
λ= =
= 𝟐. 𝟖𝟏𝒎
6
𝑓
106.9 × 10 𝐻𝑧
A/D Conversion 20 pts
What is quantization error and how can it be
mitigated?
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Quantization error is the difference between the
original analog signal and the reconstructed
(stairstep) signal after A/D and D/A. Quantization
error can be reduced by increasing the number of
bits N for each sample. This will make the
quantization intervals smaller, reducing the difference
between the analog sample values and the
quantization levels.
Digital Modulation 20 pts
(a) Consider the frequency spectrum for an FSK signal below. Express
the Bandwidth in its simplified form in terms of ∆𝑓 and 𝑅𝑏 .
𝐵𝑊 = 2 ∆𝑓 + 𝑅𝑏
(b) Using the equation obtained in part (a), consider the following FFT
for an FSK signal. Given ∆𝑓 = 200 𝑘𝐻𝑧, what bit rate can this channel
support?
𝐵𝑊 = 940 − 460 = 480 𝑘𝐻𝑧
∆𝑓
𝐵𝑊
480 𝑘𝐻𝑧
𝑅𝑏 =
− ∆𝑓 =
− 200 𝑘𝐻𝑧
2
2
= 240 − 200 = 𝟒𝟎 𝒌𝒃𝒑𝒔
Gain, SNR, Antennas 20 pts

Given the following radiation pattern, where
each ring represents a 2 dB change in power,
 What
is the beamwidth? ~ 30º
 What is the sidelobe level? 𝑆𝐿𝐿𝑑𝐵 = 0 𝑑𝐵 − −11 𝑑𝐵 = 𝟏𝟏 𝒅𝑩
 What is the front-to-back ratio?
𝐹𝐵𝑅𝑑𝐵 = 0 𝑑𝐵 − −11 𝑑𝐵 = 𝟏𝟏 𝒅𝑩
Propagation/EW 20 pts
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Consider the diagram and formula below. What
is the longest line-of-sight communication range
between a transmitter on a DDG with an
antenna 130 feet high and a receiver on a small
craft with an antenna 12 feet high?
𝑑 = 2ℎ
𝑑 𝑇𝑜𝑡𝑎𝑙 =
2 ∙ 130 𝑓𝑡 + 2 ∙ 12𝑓𝑡 = 𝟐𝟏 𝒎𝒊
BGP/BGP Routing 40 pts

Consider the network diagram and BGP route
announcement from Router 3 below, assuming no local
preferences are set.
 What AS path would an IP packet
from 12.12.12.1 take to reach
17.17.200.2?
40 – 2003 – 2005
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What AS path would an IP packet
from 13.13.13.3 take reach
17.17.200.2?
40 – 2003 – 2005
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What AS path would an IP packet
from 20.20.20.1 take to reach
17.17.200.2?
60 – 40 – 2003 – 2005
Comm Sys., EM Spectrum,
Signals, Modulation 40 pts

What is the time-domain expression of the
following modulated signal?
sin 𝐴 sin 𝐵 =
1
1
cos 𝐴 − 𝐵 − cos 𝐴 + 𝐵
2
2
𝑣𝐴𝑀 𝑡 = 10 sin 2𝜋 ∙ 300,000𝑡 + 5 cos 2𝜋 ∙ 299,000𝑡 − 5 cos 2𝜋 ∙ 301,000𝑡 + 3 cos 2𝜋 ∙ 298,000𝑡 − 3 cos 2𝜋 ∙ 302,000𝑡
A/D Conversion 40 pts
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Consider the frequencies for a bass guitar
below. What minimum sampling frequency must
be exceeded in order to accurately reconstruct
the signal and avoid aliasing?
𝑓𝑠 > 𝑓𝑁 = 2𝑓𝑚𝑎𝑥
𝑓𝑚𝑎𝑥 = 440 𝐻𝑧
𝑓𝑁 = 2 × 440 𝐻𝑧 = 880 𝐻𝑧
𝑓𝑠 > 𝟖𝟖𝟎 𝑯𝒛
Source: http://www.calvin.edu/academic/engineering/2013-14-team10/background.html
Digital Modulation 40 pts

Consider the code provided for an 8-PSK
system. What would be the bit stream for the
phase shifts received as (0°) (225°) (45°) (90°)?
Ans: 000111001011
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If the phase shifts are received at 20 µs intervals,
what is the baud?
Phase shift = Symbol
1
1
𝒔𝒚𝒎
𝑅𝑠 = =
= 𝟓𝟎, 𝟎𝟎𝟎
𝒔𝒆𝒄
𝑇𝑆 20 𝜇𝑠𝑒𝑐

What is the bit rate?
𝑅𝑏 = 𝑅𝑆 × 𝑁
= 50,000
𝑠𝑦𝑚
𝑠𝑒𝑐
∙ 3
= 𝟏𝟓𝟎, 𝟎𝟎𝟎 𝒃𝒑𝒔
𝑏𝑖𝑡
𝑠𝑦𝑚
Gain, SNR, Antennas 40 pts

The signal power at the input to the receiver is 6.8 nW
and the noise power at the input to the receiver is 2.4
nW. Find SNR.
𝑆𝑁𝑅 =

𝑃𝑆 6.8 𝑛𝑊
=
= 𝟐. 𝟖𝟑
𝑃𝑁 2.4 𝑛𝑊
A transmitter feeds a dipole antenna (gain G = 1.64)
with 150 Watts of power. Calculate EIRP.
𝐸𝐼𝑅𝑃 = 𝑃𝑡 𝐺𝑡 = 150 𝑊 1.64 = 𝟐𝟒𝟔 𝑾
Propagation/EW 40 pts
During an operation, you discover your communications are being
jammed by an opposing force. Intelligence indicates the jammer is at 9
km from you and can generate an EIRP of 20 Watts. You cannot destroy
the jammer at this time, so you decide to move closer to your source to
mitigate your received J/S. The source is generating an EIRP of 10 dB.
How close do you have to be from the source to achieve a received J/S of
-3 dB?
𝐽
= 𝐸𝐼𝑅𝑃𝐽,𝑑𝐵 − 𝐸𝐼𝑅𝑃𝑆,𝑑𝐵 + 20 log 𝑑𝑆 − 20 log 𝑑𝐽
𝑆 𝑑𝐵
𝐽
𝑑𝐽 = 9000 𝑚 , 𝐸𝐼𝑅𝑃𝐽,𝑑𝐵 = 10 log 20 𝑊 = 13 𝑑𝐵 , 𝐸𝐼𝑅𝑃𝑆,𝑑𝐵 = 10 𝑑𝐵 , 𝑆
= −3 𝑑𝐵

𝑑𝐵
20 log 𝑑𝑆 =
𝐽
𝑆 𝑑𝐵
− 𝐸𝐼𝑅𝑃𝐽,𝑑𝐵 + 𝐸𝐼𝑅𝑃𝑆,𝑑𝐵 + 20 log 𝑑𝐽
= −3 𝑑𝐵 − 13 𝑑𝐵 + 10 𝑑𝐵 + 20 log 9000 𝑚 = 73 𝑑𝐵
𝑑𝑆 = 10
73 𝑑𝐵
20
= 𝟒, 𝟒𝟔𝟔 𝒎
BGP/BGP Routing 60 pts
Name and describe (include negative and positive consequences)
one technical solution that an AS network operator can use to
combat prefix hijacking an MITM attack on BGP networks?

Filtering – Best current practices for AS network operators dictate the use of filters at AS borders
to reject suspicious route announcements or alter malicious route attributes. Filters are manually
established based on the routing policies of an organization. Filtering has both a business cost
and computational cost associated with it.
Internet Routing Registries – These are repositories of the IP prefixes, ASNs, routing policy,
network topology, and human points of contact for those ASes which choose to register their
information. While this method may be effective, the downside is that these registries are only
effective if the registry data is secure, complete, and accurate, which is currently not guaranteed.
Resource Public Key Infrastructure (RPKI) – Similar to the IRRs, RPKI is a repository of Internet
routing information. The key difference is that it uses the X.509 certificate system to provide
cryptographic assurance of (1) the association between an ASN and the IP prefixes it has been
allocated, and (2) the association between an ASN and the IP prefixes it is authorized to originate.

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

There is nothing in RPKI which validates the route attributes, including the AS path, associated with a BGP
route announcement from an AS.
Nor does it provide certainty that the AS which has registered their information used the correct ASN or set of
prefixes.
Nor does it provide network topology information or human points of contact as with IRRs.
Lastly, it does not mandate that network operators use this information when constructing their filters.
How RPKI is applied is entirely dependent on what AS network operators choose to do with the information
available.
Comm Sys., EM Spectrum,
Signals, Modulation 60 pts

Consider the figure below, and calculate the following:
𝑉
17.5𝑉 − 2.5𝑉 2
𝑉𝑚𝑎𝑥 − 𝑉𝑚𝑖𝑛
= 𝟎. 𝟕𝟓 =
(a) 𝑚 = 𝑉𝑚𝑐 =
10𝑉
𝑉𝑚𝑎𝑥 + 𝑉𝑚𝑖𝑛
1
= 𝟓, 𝟎𝟎𝟎 𝑯𝒛 = 5 𝑘𝐻𝑧
(b) 𝑓𝑐 = 0.2𝑚𝑠𝑒𝑐
1
= 𝟒𝟑𝟒. 𝟖 𝑯𝒛
(c) 𝑓𝑚 = 2.3𝑚𝑠𝑒𝑐
A/D Conversion 60 pts
Consider the following analog waveform. It is to be sampled at 2 MHz
and quantized with a 2-bit quantizer (input voltage range is -2.0 V to
+2.0 V). Using digital bits sequentially from low to high, what is the
series of transmitted bits representing the waveform?

2
1
1
𝑇𝑠 =
=
𝑓𝑠 2 × 106 𝐻𝑧
11
11 1.5
11
1
= 0.5𝜇𝑠𝑒𝑐
amplitude (V)
10 0.5
10
10
0
01
10
𝑁=2
10
10
01
01
01-0.5
01
𝑞=
-1
=
00-1.5
-2
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
𝑣𝑚𝑎𝑥 − 𝑣𝑚𝑖𝑛
2𝑁
2− −2
22
=
4𝑉
4
5
time (microsec)
Ans: 1111100101101001011010
(Underlined bits are borderline)
= 1𝑉
Digital Modulation 60 pts

(a)
Consider a system using 16-PSK modulation.
How many bits per symbol are used (N)?
𝑁 = log 2 𝑆 = log 2 16 = 𝟒 𝒃𝒊𝒕𝒔 𝒔𝒚𝒎
(b)
If the baud is 15,000 symbols/second, what is the bit rate 𝑅𝑏 ?
𝑅𝑏 = 𝑅𝑆 × 𝑁 =
(c)
4 𝑏𝑖𝑡𝑠
= 𝟔𝟎, 𝟎𝟎𝟎 𝒃𝒑𝒔 = 60 𝑘𝑏𝑝𝑠
𝑠𝑦𝑚
How much bandwidth does this system need?
𝐵𝑊 =
(d)
15,000 𝑠𝑦𝑚
𝑠𝑒𝑐
2𝑅𝑏 2 × 60,000 𝑏𝑝𝑠
=
= 𝟑𝟎, 𝟎𝟎𝟎 𝑯𝒛 = 30 𝑘𝐻𝑧
𝑁
4 𝑏𝑖𝑡𝑠 𝑠𝑦𝑚
What would be the major drawback in using a higher M-ary
modulation like 32-PSK?
As more symbols are added, their phases get closer making the
system more susceptible to noise as correctly differentiating
between symbols would be more difficult.
Gain, SNR, Antennas 60 pts

For the diagram below, Pin = 2 mW, AP1 = 10, AP2,dB =
-4 dB, and AP3,dB = 3 dB. Calculate Pout,dBW
Note: There are different ways to approach this problem.
𝐴𝑃2 = 10
−4 𝑑𝐵
10
= 0.4 and 𝐴𝑃3 = 2
𝑃𝑜𝑢𝑡 = 𝑃𝑖𝑛 ∙ 𝐴𝑃1 ∙ 𝐴𝑃2 ∙ 𝐴𝑃3 = 2 𝑚𝑊 10 0.4 2 = 16 𝑚𝑊
𝑃𝑜𝑢𝑡,𝑑𝐵𝑊
16 × 10−3 𝑊
= 10 log
= −𝟏𝟖 𝒅𝑩𝑾
1𝑊
Propagation/EW 60 pts

Consider the Friis Free Space equation 𝑃𝑟 =
𝑃𝑡 𝐺𝑡 𝐺𝑟 λ2
.
4𝜋𝑑 2
Your cell phone transmits at a
power level of 500 mW, and has an antenna gain of 2 dB. The cell tower has an
antenna gain of 8 dB. Your cell phone is dual-band, so it can transmit at either 800
MHz or 1900 MHz, depending on your location. What combination of frequency and
distance from the tower would yield the highest power level at the tower, 800 MHz at
a distance of 8 km or 1900 MHz at a distance of 4 km? note: Justify your answer by
comparing power levels.
𝑃𝑡 = 500 × 10
λ800 =
𝑐
𝑓
=
−3
𝑊 , 𝐺𝑡 = 10
3×108 𝑚
800×106
𝑠
𝐻𝑧
𝑃𝑟,(800𝑀𝐻𝑧@8𝑘𝑚) =
𝑃𝑟,(1900𝑀𝐻𝑧@4𝑘𝑚) =
2
10
8
10
3×108 𝑚
= 1.58 , 𝐺𝑟 = 10
= 0.375 𝑚 , λ1900 =
𝑐
𝑓
=
1900×106
500×10−3 𝑊 1.58 6.31 0.375 𝑚 2
4𝜋∙8×103 𝑚 2
𝑠
𝐻𝑧
= 0.158 𝑚
= 𝟔. 𝟗𝟒 × 𝟏𝟎−𝟏𝟏 𝑾
500×10−3 𝑊 1.58 6.31 0.158 𝑚 2
4𝜋∙4×103 𝑚 2
= 6.31 ,
= 4.93 × 10−11 𝑊
Hence, a cell phone transmitting at 800 MHz would yield the highest received power level at
the tower even at double the distance of 8 km.