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joja [24]
3 years ago
6

4 Error-Correcting Polynomials (a) Alice has a length 8 message to Bob. There are 2 communication channels available. When n pac

kets are fed through channel A, the channel will only deliver 5 packets (picked at random). Similarity, channel B will only deliver 5 packets (picked at random), but it will also corrupt (change the value) of one of the delivered packets. All channels can only work if at least 10 packets are sent through it. Using the 2 channels, how can Alice send the message to Bob? (b) Alice wishes to send a message to Bob as the coefficients of a degree 2 polynomial P. For a message [mi,m2,m3], she creates polynomial P- mix^ m2x +m3 and sends 5 packets: (0,P(0)), (1,P(1)), (2,P(2)), (3, P(3)), (4, P(4). However, Eve interferes and changes one of the values of a packet before it reaches Bob. If Bob receives and knows Alice's encoding scheme and that Eve changed one of the packets, can he still figure out what the original message was? If so find it as well as the x-value of the packet that Eve changed, if not, explain why he can not. (Work in mod 1 l.) (c) Alice decides that putting the message as the coefficients of a polynomial is too inefficient for long messages because the degree of the polvnomial grows quite large. Instead, she
Engineering
1 answer:
yKpoI14uk [10]3 years ago
7 0

(m_{1}, m_{2}, m_{3}) = (3, -6, 3)

<u>Explanation:</u>

Given data,

p=m_{1} x^{2}+m_{2} x+m_{3}

5 packets are delivered by channel, which was randomly taken.

0(P(0)), 1(P(1)), 2(P(2)), 3(P(3)), 4(P(4))

Bob receives are

(0,3) (1,0) (2,3) (3,0) (4,3)

P(0) = 3

P(1) = 0

P(2) = 3

P(3) = 0

P(4) = 3

The given equation is

p=m_{1} x^{2}+m_{2} x+m_{3}

Solution:

p(0)=m_{1}(0)+m_{2}(0)+m_{3} = 3

m_{3} = 3

p(1)=m_{1}+m_{2}+m_{3}=0

m_{1} + m_{2} = - 3

m_{2} = -3 -m_{1}

p(2)=4 m_{1}+2 m_{2}+m_{3} = 3

4 m_{1}+2 m_{2} = 0

2 m_{1}+m_{2}=0

P(3)=9 m_{1}+3 m_{2}+m_{3}  = 0

3 m_{1}+m_{2}+3=0

P(4)=16 m_{1}+4 m_{2}+m_{3} = 3

4 m_{1}+m_{2} = 0

From P(1) and P(2)

m_{2} = -3 -m_{1}

2 m_{1}+m_{2}=0

2 m_{1}-3-m_{1}=0

-3 + m_{1}  = 0

m_{1} = 3

m_{2} = -6

(m_{1}, m_{2}, m_{3}) = (3, -6, 3)

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Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

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Using the given formula:

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\frac{A}{F} will give the tranfer function

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\frac{A}{F} = \frac{1}{\sqrt{(4K - 120\ ^{2})}}

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2 years ago
A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron cast
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What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit
jeka94

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Explanation:

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hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

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v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s

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4 0
2 years ago
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

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Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0
2 years ago
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