EC310 16-week Review - USNA 16...Consider the diagram and formula below. What is the longest line-of-sight communication range between a transmitter on a DDG with an ... Comm Sys.,

EC310

16-week

Review

Rules of Engagement

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 pts 10 pts 10 pts 10 pts 10 pts 10 pts








Briefly describe each of the following

Autonomous Systems Categories:

(a) Stub AS

(b) Multihomed AS

(c) Transit AS

BGP/BGP Routing 10 pts

- 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

Name the four basic components of a

communications system.

Comm Sys., EM Spectrum,

Signals, Modulation 10 pts

Transmitter

Receiver

Channel

Noise

What is the most important advantage of digital

communications?

A/D Conversion 10 pts

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.

What digital modulation scheme does each

constellation represent?

Digital Modulation 10 pts

BPSKQPSK

8-PSK 16-PSK

8-QAM

Antenna

Type

Azimuth

beam pattern

Vertical

beam pattern

Physical size

Dipole

(vertically oriented)

Monopole

(vertically oriented)

Yagi (of the driven element)

Fill in the missing information for each antenna

type.

Gain, SNR, Antennas 10 pts

omnidirectional

omnidirectional

Depends on gain

(it varies)

null

null

Depends on gain

(it varies)

λ

2

λ

4

λ

2

For VLF-HF communications, list and describe

the basic modes of propagation.

Propagation/EW 10 pts

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)

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

What is the wavelength of a radio station whose

broadcast frequency is 106.9 MHz?

λ =𝑐

𝑓=

3 × 108 𝑚 𝑠106.9 × 106𝐻𝑧

= 𝟐. 𝟖𝟏𝒎



What is quantization error and how can it be mitigated?


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.

(a) Consider the frequency spectrum for an FSK signal below. Express

the Bandwidth in its simplified form in terms of ∆𝑓 and 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 = 2 ∆𝑓 + 𝑅𝑏

𝐵𝑊 = 940 − 460 = 480 𝑘𝐻𝑧

𝑅𝑏 =𝐵𝑊

2− ∆𝑓 =

480 𝑘𝐻𝑧

2− 200 𝑘𝐻𝑧

= 240 − 200 = 𝟒𝟎 𝒌𝒃𝒑𝒔

∆𝑓


Given the following radiation pattern, where

each ring represents a 2 dB change in power,

What is the beamwidth?

What is the sidelobe level?

What is the front-to-back ratio?

~ 30º

𝑆𝐿𝐿𝑑𝐵 = 0 𝑑𝐵 − −11 𝑑𝐵 = 𝟏𝟏 𝒅𝑩

𝐹𝐵𝑅𝑑𝐵 = 0 𝑑𝐵 − −11 𝑑𝐵 = 𝟏𝟏 𝒅𝑩


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?

𝑑𝑇𝑜𝑡𝑎𝑙 = 2 ∙ 130 𝑓𝑡 + 2 ∙ 12𝑓𝑡 = 𝟐𝟏𝒎𝒊

𝑑 = 2ℎ


What AS path would an IP packet

from 12.12.12.1 take to reach

17.17.200.2?


from 13.13.13.3 take reach

17.17.200.2?


from 20.20.20.1 take to reach

17.17.200.2?

40 – 2003 – 2005

40 – 2003 – 2005

60 – 40 – 2003 – 2005

Consider the network diagram and BGP route

announcement from Router 3 below, assuming no local

preferences are set.

What is the time-domain expression of the

following modulated signal?

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



sin 𝐴 sin 𝐵 =1

2cos 𝐴 − 𝐵 −

1

2cos 𝐴 + 𝐵

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

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°)?

If the phase shifts are received at 20 µs intervals,

what is the baud?

What is the bit rate?


Ans: 000111001011

𝑅𝑏 = 𝑅𝑆 × 𝑁

= 50,000𝑠𝑦𝑚

𝑠𝑒𝑐∙ 3

𝑏𝑖𝑡

𝑠𝑦𝑚

= 𝟏𝟓𝟎, 𝟎𝟎𝟎 𝒃𝒑𝒔

Phase shift = Symbol

𝑅𝑠 =1

𝑇𝑆=

1

20 𝜇𝑠𝑒𝑐= 𝟓𝟎, 𝟎𝟎𝟎 𝒔𝒚𝒎

𝒔𝒆𝒄


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.

A transmitter feeds a dipole antenna (gain G = 1.64)

with 150 Watts of power. Calculate EIRP.

𝑆𝑁𝑅 =𝑃𝑆𝑃𝑁

=6.8 𝑛𝑊

2.4 𝑛𝑊= 𝟐. 𝟖𝟑

𝐸𝐼𝑅𝑃 = 𝑃𝑡𝐺𝑡 = 150 𝑊 1.64 = 𝟐𝟒𝟔𝑾


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

𝑆 𝑑𝐵= 𝐸𝐼𝑅𝑃𝐽,𝑑𝐵 − 𝐸𝐼𝑅𝑃𝑆,𝑑𝐵 + 20 log 𝑑𝑆 − 20 log 𝑑𝐽

𝑑𝐽 = 9000 𝑚 , 𝐸𝐼𝑅𝑃𝐽,𝑑𝐵 = 10 log 20 𝑊 =13 𝑑𝐵 , 𝐸𝐼𝑅𝑃𝑆,𝑑𝐵 = 10 𝑑𝐵 , 𝐽

𝑆 𝑑𝐵= −3 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 = 1073 𝑑𝐵

20 = 𝟒, 𝟒𝟔𝟔𝒎

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.

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.

Consider the figure below, and calculate the following:

(a) 𝑚

(b) 𝑓𝑐(c) 𝑓𝑚

=𝑉𝑚𝑉𝑐

= 17.5𝑉 − 2.5𝑉 2

10𝑉= 𝟎. 𝟕𝟓 =

𝑉𝑚𝑎𝑥 − 𝑉𝑚𝑖𝑛

𝑉𝑚𝑎𝑥 + 𝑉𝑚𝑖𝑛

=1

0.2𝑚𝑠𝑒𝑐= 𝟓, 𝟎𝟎𝟎 𝑯𝒛 = 5 𝑘𝐻𝑧

=1

2.3𝑚𝑠𝑒𝑐= 𝟒𝟑𝟒. 𝟖 𝑯𝒛



time (microsec)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

am

pli

tud

e (

V)

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

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?


Ans: 1111100101101001011010

𝑇𝑠 =1

𝑓𝑠=

1

2 × 106 𝐻𝑧

= 0.5𝜇𝑠𝑒𝑐

𝑁 = 2

𝑞 =𝑣𝑚𝑎𝑥 − 𝑣𝑚𝑖𝑛

2𝑁

=2− −2

22=

4 𝑉

4= 1 𝑉

11

10

01

00

11

11

10

01

01

10

10

01

01

1010

(Underlined bits are borderline)

Consider a system using 16-PSK modulation.

(a) How many bits per symbol are used (N)?

(b) If the baud is 15,000 symbols/second, what is the bit rate 𝑅𝑏 ?

(c) How much bandwidth does this system need?

(d) What would be the major drawback in using a higher M-ary

modulation like 32-PSK?


𝑁 = log2 𝑆 = log2 16 = 𝟒 𝒃𝒊𝒕𝒔 𝒔𝒚𝒎

𝑅𝑏 = 𝑅𝑆 × 𝑁 =15,000 𝑠𝑦𝑚

𝑠𝑒𝑐

4 𝑏𝑖𝑡𝑠

𝑠𝑦𝑚= 𝟔𝟎, 𝟎𝟎𝟎 𝒃𝒑𝒔 = 60 𝑘𝑏𝑝𝑠

𝐵𝑊 =2𝑅𝑏𝑁

=2 × 60,000 𝑏𝑝𝑠

4 𝑏𝑖𝑡𝑠 𝑠𝑦𝑚= 𝟑𝟎, 𝟎𝟎𝟎 𝑯𝒛 = 30 𝑘𝐻𝑧

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.


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 = 10 log16 × 10−3 𝑊

1𝑊= −𝟏𝟖 𝒅𝑩𝑾

Consider the Friis Free Space equation 𝑃𝑟 =𝑃𝑡𝐺𝑡𝐺𝑟λ

2

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 × 10−3𝑊 , 𝐺𝑡 = 102

10 = 1.58 , 𝐺𝑟 = 108

10 = 6.31 ,

λ800 =𝑐

𝑓=

3×108 𝑚 𝑠

800×106 𝐻𝑧= 0.375 𝑚 , λ1900 =

𝑐

𝑓=

3×108 𝑚 𝑠

1900×106 𝐻𝑧= 0.158 𝑚

𝑃𝑟,(800𝑀𝐻𝑧@8𝑘𝑚) =500×10−3 𝑊 1.58 6.31 0.375 𝑚 2

4𝜋∙8×103 𝑚 2 = 𝟔. 𝟗𝟒 × 𝟏𝟎−𝟏𝟏 𝑾

𝑃𝑟,(1900𝑀𝐻𝑧@4𝑘𝑚) =500×10−3 𝑊 1.58 6.31 0.158 𝑚 2

4𝜋∙4×103 𝑚 2 = 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.

Documents

EC310 16-week Review - USNA 16...Consider the diagram and formula below. What is the longest line-of-sight communication range between a transmitter on a DDG with an ... Comm Sys.,