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I have attached the assignment.

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Answered Same Day Mar 18, 2020

Solution

Saurabh answered on Mar 27 2020
148 Votes
Q1.mclc;
clear all;
close all;
P1 = 0:0.0001:0.01;
P = (6.*(P1) - 0.006)./0.3;
M = (1.6.*(P1) + 0.004)./3;
plot(P1,P);
hold on
plot(P1,M);
xlabel('P_1');
ylabel('P_2');
xlim([0 0.01]);
ylim([0 0.01]);
Q2.mclc;
clear all;
close all;
syms x y z
eqn1 = x - 0.05*y - 0.05*z == 0.001;
eqn2 = -0.533*x + y - 0.533*z == 0.00133;
eqn3 = -0.333*x - 0.333*y + z == 0.005;
[A,B] = equationsToMatrix([eqn1, eqn2, eqn3], [x, y, z])
X = linsolve(A,B)
Q3.mclc;
clear all;
close all;
syms x y
eqn1 = 6*x - 7.5*y == 0.6;
eqn2 = 3.5*x + 4*y == -0.7;
[A,B] = equationsToMatrix([eqn1, eqn2], [x, y])
X = linsolve(A,B)
P1 = 0:0.001:1;
P = (6.*(P1) - 0.6)./7.5;
M = (3.5.*(P1) + 0.7)./4;
plot(P1,P);
hold on
plot(P1,M);
xlabel('P_1');
ylabel('P_2');
xlim([0 1]);
ylim([0 1]);
Solution.docxSOLUTION 1:
(i)
The signal to noise ratio is given by,
Given that:
Substituting the values given,
Therefore, we get the following two equations:
Plots are below:
The blue colour plot is for 1st equation.
(ii)
Solving the equations simultaneously,
Substituting the value of from equation 2 to 1,
Therefore, we now substitute the obtained value in expression for to get the other value.
The intersection point in the...
SOLUTION.PDF

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